A Bit About Vectors

IN THIS CHAPTER

Summary: Understand the difference between scalars and vectors, how to draw vectors, how to break down vectors into components, and how to add vectors.

Key Ideas

Scalars are quantities that have a magnitude but no direction—for example, temperature; in contrast, vectors have both magnitude and direction—for example, velocity.

Vectors are drawn as arrows; the length of the arrow corresponds to the magnitude of the vector, and the direction of the arrow represents the direction of the vector.

Any vector can be broken down into its x - and y -components; breaking a vector into its components will make many problems simpler.

Relevant Equations

Note: this assumes that θ is measured from the horizontal. These equations are not on the equation sheet, but should be memorized.

Scalars and vectors are easy. So we’ll make this quick.

Scalars

Scalars are numbers that have a magnitude but no direction.

For example, temperature is a scalar. On a cold winter day, you might say that it is “4 degrees” outside. The units you used were “degrees.” But the temperature was not oriented in a particular way; it did not have a direction.

Another scalar quantity is speed. While traveling on a highway, your car’s speedometer may read “70 miles per hour.” It does not matter whether you are traveling north or south, if you are going forward or in reverse: your speed is 70 miles per hour.

Vector Basics

Vectors, by comparison, have both magnitude and direction.

An example of a vector is velocity. Velocity, unlike speed, always has a direction. So, let’s say you are traveling on the highway again at a speed of 70 miles per hour. First, define what direction is positive—we’ll call north the positive direction. So, if you are going north, your velocity is +70 miles per hour. The magnitude of your velocity is “70 miles per hour,” and the direction is “north.”

If you turn around and travel south, your velocity is −70 miles per hour. The magnitude (the speed) is still the same, but the sign is reversed because you are traveling in the negative direction. The direction of your velocity is “south.”

IMPORTANT: If the answer to a free-response question is a vector quantity, be sure to state both the magnitude and direction. However, don’t use a negative sign if you can help it! Rather than “−70 miles per hour,” state the true meaning of the negative sign: “70 miles per hour, south.”

Graphic Representation of Vectors

Vectors are drawn as arrows. The length of the arrow corresponds to the magnitude of the vector—the longer the arrow, the greater the magnitude of the vector. The direction in which the arrow points represents the direction of the vector. Figure 9.1 shows a few examples:

Figure 9.1 Examples of vectors.

Vector A has a magnitude of 3 meters. Its direction is “60 degrees above the positive x -axis.” Vector B also has a magnitude of 3 meters. Its direction is “â degrees above the negative x -axis.” Vector C has a magnitude of 1.5 meters. Its direction is “in the negative y -direction” or “90 degrees below the x -axis.”

Vector Components

Any vector can be broken into its x - and y -components. Here’s what we mean:

Place your finger at the tail of the vector in Figure 9.2 (that’s the end of the vector that does not have a on it). Let’s say that you want to get your finger to the head of the vector without moving diagonally. You would have to move your finger three units to the right and four units up. Therefore, the magnitude of left–right component (x -component) of the vector is “3 units” and the magnitude of up–down (y -component) of the vector is “4 units.”

Figure 9.2 Breaking vectors into x - and y -components.

If your languages of choice are Greek and math, then you may prefer this explanation:

You may want to check to see that these formulas work by plugging in the values from our last example.

Adding Vectors

Let’s take two vectors, Q and Z , as shown in Figure 9.3a .

Figure 9.3a Two vectors.

Now, in Figure 9.3b , we place them on a coordinate plane. We will move them around so that they line up head-to-tail.

Figure 9.3b Vectors on a coordinate plane.

If you place your finger at the origin and follow the arrows, you will end up at the head of vector Z . The vector sum of Q and Z is the vector that starts at the origin and ends at the head of vector Z . This is shown in Figure 9.3c .

Figure 9.3c Adding vectors.

Physicists call the vector sum the “resultant vector.” Usually, we prefer to call it “the resultant” or, as in our diagram, “ R .”

Vector Components, Revisited

Breaking a vector into its components will make many problems simpler. Here’s an example:

To add the vectors in Figure 9.4a , all you have to do is add their x - and y -components. The sum of the x -components is 3 + (−2) = 1 units. The sum of the y -components is 1 + 2 = 3 units. The resultant vector, therefore, has an x -component of +1 units and a y -component of +3 units. See Figure 9.4b .

Figure 9.4a Adding vectors.

Figure 9.4b Final sum of vectors.

Some Final Hints

1. Make sure your calculator is set to DEGREES, not radians.

2. Always use units. Always. We mean it. Always.

Practice Problems

1 . A canoe is paddled due north across a lake at 2.0 m/s relative to still water. The current in the lake flows toward the east; its speed is 0.5 m/s. Which of the following vectors best represents the velocity of the canoe relative to shore?

(A) 2.5 m/s

(B) 2.1 m/s

(C) 2.5 m/s

(D) 1.9 m/s

(E) 1.9 m/s

2 . Force vector A has magnitude 27.0 N and is direction 74° from the vertical, as shown above. Which of the following are the horizontal and vertical components of vector A ?

3 . Which of the following is a scalar quantity?

(A) electric force

(B) gravitational force

(C) weight

(D) mass

(E) friction

Solutions to Practice Problems

1 . B —To solve, add the northward 2.0 m/s velocity vector to the eastward 0.5 m/s vector. These vectors are at right angles to one another, so the magnitude of the resultant is given by the Pythagorean theorem. You don’t have a calculator on the multiple-choice section, though, so you’ll have to be clever. There’s only one answer that makes sense! The hypotenuse of a right triangle has to be bigger than either leg, but less than the algebraic sum of the legs. Only B, 2.1 m/s, meets this criterion.

2 . A —Again, with no calculator, you cannot just plug numbers in (though if you could, careful: the horizontal component of A is 27.0 N cos 16° because 16° is the angle from the horizontal). Answers B and E are wrong because the vertical component is bigger than the horizontal component, which doesn’t make any sense based on the diagram. Choice C is wrong because the horizontal component is bigger than the magnitude of the vector itself—ridiculous! Same problem with choice D, where the horizontal component is equal to the magnitude of the vector. So the answer must be A.

3 . D —A scalar has no direction. All forces have direction, including weight (which is the force of gravity). Mass is just a measure of how much stuff is contained in an object, and thus has no direction.

Vectors

1 . Five physics students are asked to add the vectors P and Q to find the resultant vector R . The answers from each student are in the following responses. Which student DID NOT add the vectors correctly?

  (A)

  (B)

  (C)

  (D)

  (E)

2 . A pilot flies 180 m/s at an angle 30° west of due south as shown. How quickly is the pilot traveling west?

  (A) 90 m/s

  (B) 104 m/s

  (C) 156 m/s

  (D) 208 m/s

  (E) 360 m/s

  Questions 3 and 4 refer to the following information: A fisherman wishes to pilot his boat directly north across a river to a destination marked by an X on the figure. The river has a westward current of 3 m/s, and the boat can travel 5 m/s in still water. The fisherman points his boat at a heading angle of θ into the current as shown and begins his journey.

3 . What heading angle θ should the fisherman use to travel directly north across the river?

  (A) 30°

  (B) 31°

  (C) 37°

  (D) 53°

  (E) 59°

4 . What is the boat’s resultant velocity while traveling due north across the river?

  (A) 2.0 m/s

  (B) 3.0 m/s

  (C) 3.8 m/s

  (D) 4.0 m/s

  (E) 5.8 m/s

Answers

1 . E —Vectors should be added head to tail. Note: Student B simply showed the correct resultant vector R . Student C chose to break vector P into its x and y components and add them to the vector Q .

2 . A —The westward component of the velocity is given by the following equation:

3 . C —The velocities of the boat and the current must be added to produce a resultant that points directly north as shown. The angle θ is found by using

4 . D —The velocities of the boat and current are added as shown in the figure, which forms a right triangle. The resultant velocity can be found using the Pythagorean theorem or by realizing that this is a 3-4-5 triangle.

Free-Body Diagrams and Equilibrium

IN THIS CHAPTER

Summary: Free-body diagrams can help you see forces as vectors, and we’ll review torque as well as a variety of forces: normal force, tension, friction, forces operating on inclined planes, and static and kinetic friction.

Key Ideas

A free-body diagram is a picture that represents an object, along with the forces acting on that object.

When the net force on an object equals zero, that object is in equilibrium.

The normal force is not always equal to the weight of an object.

Tension is a force applied by a rope or string.

Friction is only found when there is contact between two surfaces.

When an object is on an incline, use tilted axes, one parallel to the incline, one perpendicular.

Torque occurs when a force is applied to an object, and that force can cause the object to rotate.

Relevant Equations

On an inclined plane, the weight vector can be broken into components:

The force of friction is given by

F_f = μF_N

Physics, at its essence, is all about simplification. The universe is a complicated place, and if you want to make sense of it—which is what physicists try to do—you need to reduce it to some simplified representation: for example, with free-body diagrams.

We will refer regularly to forces. A force refers to a push or a pull applied to an object. Something can experience many different forces simultaneously—for example, you can push a block forward while friction pulls it backward, but the net force is the vector sum of all of the individual forces acting on the block.

What Is a Free-Body Diagram?

A free-body diagram is a picture that represents one or more objects, along with the forces acting on those objects. The objects are almost always drawn as rectangles or circles, just for the sake of simplicity, and the forces are always shown as vectors. Figure 10.1 shows a few examples.

Figure 10.1 Two examples of free-body diagrams. As you see, there is no need to be artistic on the AP exam.

Free-body diagrams are important because they help us to see forces as vectors. And if you can add vectors, you can analyze a free-body diagram. (If you can’t add vectors, you didn’t read Chapter 9 carefully enough.)

Let’s look at the two examples in Figure 10.1 . In the first, a force is directed down. This force, which is the force of gravity, was labeled in the diagram as “weight.” The force of gravity on the hippo (that is, the hippo’s weight) pulls downward. In the second example, a force is directed to the right. The pineapple is being pulled by a rope to the right.

You’ll often see weight abbreviated as mg . Just be careful that the mass you use is in kilograms.

For the rest of this chapter, we focus on objects in equilibrium.

Equilibrium

When the net force on an object equals zero, that object is in equilibrium. At equilibrium, an object is either at rest or moving with a constant velocity , but it is not accelerating.

You’ve heard of Newton’s first law, of course: an object maintains its velocity unless acted upon by a net force. Well, an object in equilibrium is obeying Newton’s first law.

How to Solve Equilibrium Problems

We have a tried-and-true method. Follow it every time you see an equilibrium situation.

1. Draw a proper free-body diagram.

2. Resolve force vectors into x - and y -components, if necessary.

3. Write an expression for the vector sum of the left–right forces. Then write an expression for the vector sum of the up–down forces. Set each of these expressions equal to zero.

4. Solve the resulting algebraic equations.

A Brief Interlude: UNITS!

Before we lose ourselves in the excitement of free-body diagrams, we need to pay tribute to the unit of force: the newton. One N (as newtons are abbreviated) equals one kg·m/s² . We discuss why 1 newton equals 1 kg·m/s² in a future chapter. For now, let it suffice that any force can be measured in newtons.

A Really Simple Equilibrium Problem

For those of you who prefer to splash your toes in the metaphorical swimming pool of physics before getting all the way in, this section is for you. Look at this situation:

The solution to this problem is painfully obvious, but we’ll go through the steps just to be thorough.

Step 1 : Draw a proper free-body diagram.

We can skip Step 2 because these vectors already line up with each other, so they do not need to be resolved into components.

Step 3 : Write expressions for the vector sums.

This problem only involves left–right forces, so we only need one expression. Because we have an equilibrium situation, the net force is ZERO:

Step 4 : Solve.

F ₂ = 30 N

Very good. Now, let’s see how closely you were paying attention. Here’s the same problem, with a slightly different twist.

Think for a moment. Does the second astronaut have to apply more, less, or the same force as compared to the previous problem?

The second astronaut applies exactly the same force as in the previous problem! An object moving with constant velocity is in equilibrium, just as if the object were still. This is a central concept in Newtonian mechanics.

Normal Force

Let’s return to Earth for a moment.

In Figure 10.2 , a box is sitting on a table. The force of gravity pulls downward (as with the hippo, we’ve labeled this force “weight”). We know from experience that boxes sitting on tables do not accelerate downward; they remain where they are. Some force must oppose the downward pull of gravity.

Figure 10.2 Normal force.

This force is called the normal force, ¹ and it is abbreviated F_N . In fact, whenever you push on a hard surface, that surface pushes back on you—it exerts a normal force. So, when you stand on the floor, the floor pushes up on you with the same amount of force with which gravity pulls you down, and, as a result, you don’t fall through the floor.

The normal force is not always equal to the weight of an object! Think about this before we get to the practice problems.

Tension

Tension is a force applied by a rope or string. Here are two of our favorite tension problems. The first is super easy, but a good introduction to tension; the second is more involved.

Step 1 : Free-body diagram.

Step 2 : Vector components.

Hey! These vectors already line up. On to Step 3 .

Step 3 : Equations.

Remember, weight is equal to mass times the gravitational field, or mg .

Step 4 : Solve.

T = 50 N

Step 1 : Free-body diagram.

Step 2 : Vector components.

Step 3 : Equations.

Let’s start with the x -direction.

And from vector analysis we know that

so,

Similarly, if we look at the y -direction,

and

so,

Step 4 : Solve.

We can solve Equation 1 and Equation 2 simultaneously and find T ₁ and T ₂ . We’ll let you do this on your own, ² but in case you want to check your answers, T ₁ = 37 N and T ₂ = 45 N. (These are reasonable answers, as the tension in each rope is the same power of 10 as the 50 N weight of the box.)

Steps 1 , 2 , and 3 are the important steps. Step 4 only involves math. “ONLY math?!?” you ask, incredulous. “That’s the toughest part!”

Well, maybe for some people. Getting the actual correct answer does depend on your algebra skills. But, and this is important, this is AP Physics, NOT AP Algebra . The graders of the AP exam will assign most of the credit just for setting up the problem correctly! If you’re stuck on the algebra, skip it! Come up with a reasonable answer for the tensions, and move on!

We’re not kidding. Look at Chapter 7 , which discusses approaches to the free-response section, for more about the relative importance of algebra.

Friction

Friction is only found when there is contact between two surfaces.

For example, let’s say you slide a book at a constant speed across a table. The book is in contact with the table, and, assuming your table isn’t frictionless, the table will exert a friction force on the book opposite its direction of motion. Figure 10.3 shows a free-body diagram of that situation.

Figure 10.3 Free-body diagram of a book sliding on a table.

We know that because the book represented in Figure 10.3 is not being shoved through the table or flying off it, F_N must equal the book’s weight. And because the book moves at constant velocity, the force you exert by pushing the book, F _push , equals the force of friction, F_f . Remember, being in equilibrium does not necessarily mean that the book is at rest. It could be moving at a constant velocity .

How do we find the magnitude of F_f ?

Mu (μ ) is the coefficient of friction. This is a dimensionless number (that is, it doesn’t have any units) that describes how big the force of friction is between two objects. It is found experimentally because it differs for every combination of materials (for example, if a wood block slides on a glass surface), but it will usually be given in AP problems that involve friction.

And if μ isn’t given, it is easy enough to solve for—just rearrange the equation for μ algebraically:

Remember, when solving for F_f , do not assume that F_N equals the weight of the object in question. Here’s a problem where this reminder comes in handy:

The free-body diagram looks like this:

Now, in the vertical direction, there are three forces acting: F_N acts up; weight and the vertical component of P act down.

Notice that when we set up the equilibrium equation in the vertical direction, F_N - (mg + P_y ) = 0, we find that F_N is greater than mg .

Let’s finish solving this problem together. We’ve already drawn the vertical forces acting on the buffer, so we just need to add the horizontal forces to get a complete free-body diagram with the forces broken up into their components (Steps 1 and 2 ):

Step 3 calls for us to write equations for the vertical and horizontal directions. We already found the equilibrium equation for the vertical forces,

F_N - (mg + P_y ) = 0,

and it’s easy enough to find the equation for the horizontal forces,

F_f - P_x = 0.

To solve this system of equations (Step 4 ), we can reduce the number of variables with a few substitutions. For example, we can rewrite the equation for the horizontal forces as

μ ·F_N - P ·cos 37° = 0.

Furthermore, we can use the vertical equation to substitute for F_N ,

F_N = mg + P ·sin 37°.

Plugging this expression for F_N into the rewritten equation for the horizontal forces, and then replacing the variables m, g , and μ with their numerical values, we can solve for P . The answer is P = 93 N.

Static and Kinetic Friction

You may have learned that the coefficient of friction takes two forms: static and kinetic friction. Use the coefficient of static friction if something is stationary, and the coefficient of kinetic friction if the object is moving. The equation for the force of friction is essentially the same in either case: F_f = μF_N .

The only strange part about static friction is that the coefficient of static friction is a maximum value. Think about this for a moment … if a book just sits on a table, it doesn’t need any friction to stay in place. But that book won’t slide if you apply a very small horizontal pushing force to it, so static friction can act on the book. To find the maximum coefficient of static friction, find out how much horizontal pushing force will just barely cause the book to move; then use F_f = μF_N .

Inclined Planes

These could be the most popular physics problems around. You’ve probably seen way too many of these already in your physics class, so we’ll just give you a few tips on approaching them.

In Figure 10.4 we have a block of mass m resting on a plane elevated an angle θ above the horizontal. The plane is not frictionless. We’ve drawn a free-body diagram of the forces acting on the block in Figure 10.5a .

Figure 10.4 Generic inclined-plane situation.

Figure 10.5a Forces acting on the block in Figure 10.4 .

F_f is directed parallel to the surface of the plane, and F_N is, by definition, directed perpendicular to the plane. It would be a pain to break these two forces into x - and y -components, so instead we will break the “weight” vector into components that “line up” with F_f and F_N , as shown in Figure 10.5b .

Figure 10.5b Forces acting on the block in Figure 10.4 , with the weight vector resolved into components that line up with the friction force and the normal force.

Memorize this

This rule always works, as long as the angle of the plane is measured from the horizontal.

Even Physics C Students Must Use Free-Body Diagrams

It must be emphasized that even Physics C students must go through the four-step problem-solving process described in this chapter. Frequently, Physics C students try to take shortcuts, thinking that equilibrium problems are easy, only to miss something important. If free-body diagrams are good enough for professional physicists to use, they are good enough for you.

Torque

Torque occurs when a force is applied to an object, and that force can cause the object to rotate.

In other words, the torque exerted on an object equals the force exerted on that object (F ) multiplied by the distance between where the force is applied and the fulcrum (d ) as long as the force acts perpendicular to the object.

Figure 10.6 shows what we mean:

Figure 10.6 The torque applied to this bar equals Fd .

The unit of torque is the newton-meter.

Here’s an example.

Step 1 : Free-body diagram.

We’ll use point A as the fulcrum to start with. Why? In a static equilibrium situation, since the bridge isn’t actually rotating, any point on the bridge could serve as a fulcrum. But we have two unknown forces here, the forces of the supports A and B . We choose the location of one of these supports as the fulcrum, because now that support provides zero torque—the distance from the fulcrum becomes zero! Now all we have to do is solve for the force of support B .

The diagram below isn’t a true “free-body diagram,” because it includes both distance and forces, but it is useful for a torque problem. Bob’s weight acts downward right where he stands.

The bridge’s weight is taken into account with a force vector acting at the bridge’s center of mass; that is, 40 m to the right of pillar A . This is a generally valid approach—replace the weight of an extended object with a single weight vector acting at the center of mass.

Step 2 : Vector components.

We don’t have to worry about vector components here. (We would have if the forces had not acted perpendicular to the bridge.)

Step 3 : Equations.

Torque_net = counterclockwise - clockwise = 0
(F_B )(80 m) - [(100 kg·10 N/kg)(20 m) + (10,000 N)(40 m)] = 0

Step 4 : Solve. F_B = 5300 N

This is reasonable because pillar B is supporting less than half of the 11,000 N weight of the bridge and Bob. Because Bob is closer to pillar A , and otherwise the bridge is symmetric, A should bear the majority of the weight.

The Physics C exam will often expect you to find the torque provided by a force that acts at an angle. For example, consider a force F acting on a bar at an angle θ , applied a distance x from a pivot. How much torque does this force provide? See Figure 10.7 .

Figure 10.7 Force F acting on a bar at an angle θ.

To solve, break the force vector into horizontal and vertical components, as shown in Figure 10.8 .

Figure 10.8 Break the force vector into horizontal and vertical components.

The vertical component of F applies a torque of (F sin θ )x . The horizontal component of F does not apply any torque, because it could not cause the bar to rotate. So, the total torque provided by F is (F sin θ )x .

Lever Arm

The “lever arm” for a force is the closest distance from the fulcrum to the line of that force. Then, the torque provided by a force is the force times the lever arm.

Consider Figure 10.9 , which represents the same situation as Figure 10.7 . Instead of breaking F into components, continue the line of the force. The torque is F times the lever arm shown in the diagram. By trigonometry, you can see that the lever arm is equal to x sin θ . No matter how you look at it, the torque provided by F is still (F sin θ )x .

Figure 10.9 Force F acting on a bar at an angle θ

Practice Problems

1 . A 50-g mass is hung by string as shown in the picture above. The left-hand string is horizontal; the angled string measures 30° to the horizontal. What is the tension in the angled string?

2 . A 6000-kg bus sits on a 30° incline. A crane attempts to lift the bus off of the plane. The crane pulls perpendicular to the plane, as shown in the diagram. How much force must the crane apply so that the bus is suspended just above the surface? [cos 30° = 0.87, sin 30° = 0.50]

(A) 52,000 N

(B) 30,000 N

(C) 6000 N

(D) 5200 N

(E) 300 N

3 . Give two examples of a situation in which the normal force on an object is less than the object’s weight. Then give an example of a situation in which there is NO normal force on an object.

4 . A 150-N box sits motionless on an inclined plane, as shown above. What is the angle of the incline?

5 . A 50-g meterstick is to be suspended by a single string. A 100-g ball hangs from the left-hand edge of the meterstick. Where should the string be attached so that the meterstick hangs in equilibrium?

(A) at the left-hand edge

(B) 40 cm from left-hand edge

(C) 30 cm from right-hand edge

(D) 17 cm from left-hand edge

(E) at the midpoint of the meterstick

Solutions to Practice Problems

1 .

  Call the tension in the angled rope T ₂ . In the y-direction, we have T _{2, y} = T ₂ (sin 30 °) acting up, and mg acting down. Set “up” forces equal to “down” forces and solve for tension: T ₂ = mg /(sin 30°). Don’t forget to use the mass in KILOgrams, i.e., 0.050 kg. The tension thus is (0.050 kg)(10 N/kg)/(0.5) = 1.0 N. This is reasonable because the tension is about the same order of magnitude as the weight of the mass.

2 .

  A —Because the force of the crane, F _c , acts perpendicular to the plane, the parallel-to-the-plane direction is irrelevant. So all we need to do is set F _c equal to mg (cos 30°) = (6000 kg)(10 N/kg)(.87) and plug in. F _c = 52,000 N. This is a reasonable answer because it is less than—but on the same order of magnitude as—the weight of the bus.

3 . When a block rests on an inclined plane, the normal force on the block is less than the block’s weight, as discussed in the answer to #2. Another example in which the normal force is less than an object’s weight occurs when you pull a toy wagon.

  In any situation where an object does not rest on a surface (for example, when something floats in space), there is no normal force.

4 . This free-body diagram should be very familiar to you by now.

  The box is in equilibrium, so F_f must equal mg (sin θ ), and F_N must equal mg (cos θ ).

μ ·F_N = μ ·mg (cos θ ) = mg (sin θ ).

  Plugging in the values given in the problem we find that μ = 17°. This answer seems reasonable because we’d expect the incline to be fairly shallow.

5 . D —This is a torque problem, and the fulcrum is wherever the meterstick is attached to the string. We know that the meterstick’s center of mass is at the 50-cm mark, so we can draw the following picture.

Because the stick is in equilibrium, the clockwise torques equal the counterclockwise torques: (1 N)(x ) = (0.5 N)(50 - x ). So x = something in the neighborhood of 25/1.5 ∼ 17 cm. This answer is less than 50 cm, and is closer to the edge with the heavy mass, so it makes sense.

Rapid Review

• A free-body diagram is a simplified representation of an object and the forces acting on it.

• When the net force on an object is zero, it is in equilibrium. This means that it is either at rest or that it is moving at a constant velocity.

• To solve an equilibrium problem, draw a good free-body diagram, resolve all forces into x - and y -components, and then set the vector sum of the x -components equal to zero and the vector sum of the y -components equal to zero.

• The units of force are newtons, where 1 N = 1 kg·m/s² .

• Torque equals the force exerted on an object multiplied by the distance between where that force is applied and the fulcrum (the point about which an object can rotate). When an object is in equilibrium, the counterclockwise torques equal the clockwise torques.

• A “normal force” means the force of a solid surface pushing perpendicular to that surface. The normal force is NOT always equal to an object’s weight.

¹ When physicists say “normal,” they mean “perpendicular.” The word “normal” in its conventional meaning simply does not apply to physicists.

² Try solving the x -axis equation for T ₁ , then plug that into the y -axis equation:

    Plug in the value of cos 45°, os 30°, sin 30°, sin 45° … and now it’s easy to solve for T₂ = 45 N.

Free Body Diagrams and Equilibrium

1 . A car is parked on a hill. Which of the following figures correctly shows the forces acting on the car?

  (A)

  (B)

  (C)

  (D)

  (E)

  Questions 2 and 3 refer to the following information: A student pulls a 20 kg dolly to the right at a constant rate of 1 m/s with an 80 kg force angled 60° above the horizontal ground as shown in the figure. The coefficient of friction between the dolly and the ground is 0.1.

2 . The normal force from the ground on the dolly is closest to

  (A) 100 N

  (B) 130 N

  (C) 160 N

  (D) 170 N

  (E) 200 N

3 . Which of the following equations correctly models the relationship between the forces in the horizontal direction?

  (A)

  (B)

  (C)

  (D)

  (E)

4 . An eight-spoke wheel with a radius of R is free to rotate on a horizontal axis with masses hung from the end of three of the spokes, as shown in the figure. The spokes are evenly spaced, and the wheel is in equilibrium. Two masses are known and have a mass of m . What must the third mass be in order to maintain equilibrium?

  (A) m

  (B) 2m

  (C) m (sin 45°)

  (D)

  (E) m =0 because the two existing masses already create equilibrium.

Answers

1 . E —Gravity (mg ) always points downward, and normal force (F_N ) is always perpendicular to the contact surface. Finally, you need friction to keep the car stationary in equilibrium. Friction is always parallel to the contact surface and opposite to the direction the object is trying to slide.

2 . B —

3 . A —The forces in the horizontal direction must cancel each other out because the object is moving at a constant velocity; thus, it is in dynamics equilibrium. The horizontal forces are from the student to the right 80 cos 60° and from friction μF_N . to the left, where F_N = mg - 80 sin 60°. Equating the forces in the positive and negative directions we get

  80 cos 60° = μF_N

4 . D —The center mass does not produce a torque since it is hung directly below the pivot point/fulcrum. The right mass produces a clockwise torque of Rmg . The unknown mass must produce a counterclockwise torque to counter this clockwise torque. Torque is calculated as rF sinθ. Thus,

Kinematics

IN THIS CHAPTER

Summary: As soon as an object’s velocity changes, you need to analyze the problem using kinematics, which deals with aspects of motion separate from considerations of mass and force.

Key Ideas

Kinematics problems involve five variables: initial velocity, final velocity, displacement, acceleration, and time interval.

Use the three kinematics equations whenever acceleration is constant.

Average speed is the total distance in a given time divided by the time it takes you to travel that distance.

Velocity is just like speed, except it’s a vector.

Acceleration is the change in velocity divided by a time interval.

Displacement is the vector equivalent of distance.

The key rule of projectile motion is that an object’s motion in one dimension does not affect its motion in any other dimension.

Relevant Equations

The constant-acceleration kinematics equations, which we refer to as the “star” equations:

The equilibrium problems we saw in the last chapter all had something in common: there was no acceleration. Sure, an object can move at a constant velocity and still be in equilibrium, but as soon as an object’s velocity changes, you need a new set of tricks to analyze the situation. This is where kinematics comes in.

Velocity, Acceleration, and Displacement

We’ll start with a few definitions.

In this definition, Δx means “displacement” and Δt means “time interval.” Average speed is the total displacement you travel in a straight line in a given time divided by the time it takes you to travel that distance. This is different from “instantaneous speed,” which is your speed at any given moment. WARNING: The formula you learned in seventh grade, “speed = distance/time” is ONLY valid for an average speed, or when something is moving with constant speed. If an object speeds up or slows down, and you want to know its speed at some specific moment, don’t use this equation! ¹

Questions on the AP exam tend to focus on velocity more than speed, because velocity says more about an object’s motion. (Remember, velocity has both magnitude and direction.)

Acceleration occurs when an object changes velocity.

The symbol Δ means “change in.” So Δv = v_f − v ₀ , where v _f means “final velocity” and v ₀ means “initial velocity” and is pronounced “v-naught.” Similarly, Δt is the time interval during which this change in velocity occurred.

Just as velocity is the vector equivalent of speed, displacement is the vector equivalent of distance—it has both magnitude and direction.

So, let’s say that you head out your front door and walk 20 m south. If we define north to be the positive direction, then your displacement was “−20 m.” If we had defined south to be the positive direction, your displacement would have been “+20 m.” Regardless of which direction was positive, the distance you traveled was just “20 m.” (Or consider this: If you walk 20 m north, followed by 5 m back south, your displacement is 15 m, north. Your displacement is not 25 m.)

Constant-Acceleration Kinematics Equations

Putting all of these definitions together, we can come up with some important lists. First, we have our five variables:

Using just these five variables, we can write the three most important kinematics equations. An important note: The following equations are valid ONLY when acceleration is constant. We repeat: ONLY WHEN ACCELERATION IS CONSTANT . Which is most of the time. ²

We call these equations the “star equations.” You don’t need to call them the “star equations,” but just be aware that we’ll refer to the first equation as “^* ,” the second as “^** ,” and the third as “^*** ” throughout this chapter.

These are the only equations you really need to memorize for kinematics problems.

Constant-Acceleration Kinematics Problem-Solving

Step 1 : Write out all five variables in a table. Fill in the known values, and put a “?” next to the unknown values.

Step 2 : Count how many known values you have. If you have three or more, move on to Step 3 . If you don’t, find another way to solve the problem (or to get another known variable).

Step 3 : Choose the “star equation” that contains all three of your known variables. Plug in the known values, and solve.

Step 4 : Glory in your mastery of physics. Feel proud. Put correct units on your answer .

Be sure that you have committed these steps to memory. Now, let’s put them into action.

Before we solve this problem—or any problem, for that matter—we should think about the information it provides. The problem states that acceleration is constant, so that means we can use our kinematics equations. Also, it asks us to find two values, a time and a distance. Based on the information in the problem, we know that the time needed for the car to reach 120 m/s is greater than 6 s because it took 6 s for the car just to reach 60 m/s. Moreover, we can estimate that the car will travel several hundred meters in total, because the car’s average velocity must be less than 120 m/s, and it travels for several seconds.

So now let’s solve the problem. We’ll use our four-step method.

Step 1 : Table of variables.

The car begins at rest, so v ₀ = 0 m/s. The final velocity of the car is 120 m/s. We’re solving for time and displacement, so those two variables are unknown. And, at least for right now, we don’t know what the acceleration is.

Step 2 : Count variables.

We only have two values in our chart, but we need three values in order to use our kinematics equations. Fortunately, there’s enough information in the problem for us to solve for the car’s acceleration.

Acceleration is defined as a change in velocity divided by the time interval during which that change occurred. The problem states that in the first 6 s, the velocity went from 0 m/s to 60 m/s.

We now have values for three of our variables, so we can move to Step 3 .

Step 3 : Use “star equations” to solve.

All three of our known values can be plugged into *, which will allow us to solve for t .

Now that we know t , we can use either ** or *** to solve for displacement. Let’s use **:

Step 4 : Units.

Always remember units! And make sure that your units are sensible—if you find that an object travels a distance of 8 m/s, you’ve done something screwy. In our case, the answers we found have sensible units. Also, our answers seem reasonable based on the initial estimates we made: It makes sense that the car should travel a bit more than 6 s, and it makes sense that it should go several hundred meters (about half a mile) in that time.

Freefall

Problems that involve something being thrown off a cliff ³ are great, because vertical acceleration in these problems equals g in just about every case.

Falling-object problems should be solved using the method we outlined above. However, you have to be really careful about choosing a positive direction and sticking to it. That is, figure out before you solve the problem whether you want “up” to be positive (in which case a equals −10 m/s² ) or “down” to be positive (where a would therefore equal +10 m/s² ).

Here’s a practice problem:

Begin by defining the positive direction. We will call “up” positive. Then use the four-step method to solve the problem.

Step 1 : Table of variables.

Remember that displacement is a vector quantity. Even though the melon goes up before coming back down, the displacement is simply equal to the height at which the melon ends its journey (0 m) minus its initial height (30 m). Another way to think about displacement: In total, the melon ended up 30 m BELOW where it started. Because down is the negative direction, the displacement is −30 m.

Step 2 : Count variables.

Three! We can solve the problem.

Step 3 : Solve.

The rest of this problem is just algebra. Yes, you have to do it right, but setting up the problem correctly and coming up with an answer that’s reasonable is more important than getting the exact right answer. Really! If this part of an AP free-response problem is worth 5 points, you might earn 4 of those points just for setting up the equation and plugging in values correctly, even if your final answer is wrong.

But which equation do you use? We have enough information to use ** (x − x ₀ = v ₀ t + ¹ /₂ at ² ) to solve for t . Note that using ** means that we’ll have to solve a quadratic equation; you can do this with the help of the quadratic formula. ⁴ Or, if you have a graphing calculator, you can use it to solve. But almost always there’s a way to avoid the quadratic.

What If Acceleration Isn’t Constant?

A typical Physics C kinematics question asks you to use calculus to find position, velocity, or acceleration functions. Then you can solve a motion problem even if acceleration is not constant. The way to remember what to do is, first and foremost, to understand graphical kinematics as discussed in the section below. Then, we know that the slope of a graph is related to the derivative of a function; the area under a graph is related to the integral of a function. Therefore:

• To find velocity from a position function, take the derivative with respect to time:

• To find acceleration from a velocity function, also take the time derivative:

• To find position from a velocity function, take the integral with respect to time:

• To find velocity from an acceleration function, take the time integral:

Most of the time, even on the Physics C exam, you’ll be able just to use the star equations to solve a kinematics problem. Reserve your use of calculus for those problems that explicitly include an unusual function for position, velocity, or acceleration.

Projectile Motion

Things don’t always move in a straight line. When an object moves in two dimensions, we look at vector components.

The super-duper-important general rule is this: An object’s motion in one dimension does not affect its motion in any other dimension .

The most common kind of two-dimensional motion you will encounter is projectile motion. The typical form of projectile-motion problems is the following:

“A ball is shot at a velocity v from a cannon pointed at an angle è above the horizontal …”

No matter what the problem looks like, remember these rules:

• The vertical component of velocity, v_y , equals v (sin θ ).

• The horizontal component of velocity, v_x , equals v (cos θ ) when θ is measured relative to the horizontal.

• Horizontal velocity is constant.

• Vertical acceleration is g , directed downward.

Here’s a problem that combines all of these rules:

We begin by defining “up” to be positive and writing our tables of variables, one for horizontal motion and one for vertical motion.

Note that because horizontal velocity is constant, on the horizontal table, v _f = v ₀ , and a = 0. Also, because the ball lands at essentially the same height it was launched from, Δx = 0 on the vertical table. You should notice, too, that we rounded values in the tables to two significant figures (for example, we said that v ₀ in the vertical table equals 13 m/s, instead of 12.5 m/s). We can do this because the problem is stated using only two significant figures for all values, so rounding to two digits is acceptable, and it makes doing the math easier for us.

We know that t is the same in both tables—the ball stops moving horizontally at the same time that it stops moving vertically (when it hits the ground). We have enough information in the vertical table to solve for t by using equation **.

Using this value for t , we can solve for x − x ₀ in the horizontal direction, again using **.

The cannonball traveled 57 m, about half the length of a football field.

You may have learned in your physics class that the range of a projectile (which is what we just solved for) is

If you feel up to it, you can plug into this equation and show that you get the same answer we just got. There’s no need to memorize the range equation, but it’s good to know the conceptual consequences of it: the range of a projectile on level earth depends only on the initial speed and angle, and the maximum range is when the angle is 45°.

A Final Word About Kinematics Charts

The more you practice kinematics problems using our table method, the better you’ll get at it, and the quicker you’ll be able to solve these problems. Speed is important on the AP exam, and you can only gain speed through practice. So use this method on all your homework problems, and when you feel comfortable with it, you might want to use it on quizzes and tests. The other benefit to the table method, besides speed, is consistency; it forces you to set up every kinematics problem the same way, every time. This is a time-tested strategy for success on the AP exam.

Motion Graphs

You may see some graphs that relate to kinematics on the AP test. They often look like those in Figure 11.1 . We call these graphs by the names of their axes: For example, the top graph in Figure 11.1 is a “position–time graph” and the second one is a “velocity–time graph.”

Figure 11.1 Typical motion graphs. (As an excercise, you may want to describe the motion these represent; answers are at the end of this section.)

Here are some rules to live by:

• The slope of a position–time graph at any point is the velocity of the object at that point in time.

• The slope of a velocity–time graph at any point is the acceleration of the object at that point in time.

• The area under a velocity–time graph between two times is the displacement of the object during that time interval.

It’s sometimes confusing what is meant by the area “under” a graph. In the velocity–time graphs below, the velocity takes on both positive and negative values. To find the object’s displacement, we first find the area above the t -axis; this is positive displacement. Then we subtract the area below the t -axis, which represents negative displacement. The correct area to measure is shown graphically in Figure 11.2a . Whatever you do, don’t find the area as shown in Figure 11.2b ! When we say “area,” we measure that area to the t -axis only.

Figure 11.2a : Do this.

Figure 11.2b : Don’t do this.

Problems involving graphical analysis can be tricky because they require you to think abstractly about an object’s motion. For practice, let’s consider one of the most common velocity–time graphs you’ll see:

Whenever you have to describe motion in words, do so in everyday language, not physics-speak. Don’t say the word “it”; instead, give the object some specificity. Never say “positive” or “negative”; + and − merely represent directions, so name these directions. ⁵

In this case, let’s consider a ball going up (positive) and down (negative). Here’s how we’d answer the question:

  “At first the ball is moving upward pretty fast, but the ball is slowing down while going upward. (I know this because the speed is getting closer to zero in the first part of the graph.) The ball stops for an instant (because the v–t graph crosses the horizontal axis); then the ball begins to speed up again, but this time moving downward.”

Now, no numerical values were given in the graph. But would you care to hazard a guess as to the likely slope of the graph’s line if values were given? ⁶

Figure 11.1 Graphs

The position–time graph has a changing slope, so the speed of the object is changing. The object starts moving one way, then stops briefly (where the graph reaches its minimum, the slope, and thus the speed is zero). The object then speeds up in the other direction. How did I know the object’s velocity changed direction? The position was at first approaching the origin, but then was getting farther away from the origin.

The second graph is a velocity–time graph, and it must be analyzed differently. The object starts from rest, but speeds up; the second part of the motion is just like the example shown before, in which the object slows to a brief stop, turns around, and speeds up.

Air Resistance and the First-Order Differential Equation

The force of air resistance is usually negligible in kinematics problems. You probably don’t believe me, though. After all, unless you’re on the moon, or unless your teacher is using a vacuum chamber for demonstrations, air is all around us. And if I dropped a piece of paper simultaneously and from the same height as a lead weight, the weight would hit the ground first—and by a huge margin. Certainly. Obviously.

Why don’t you try it? But be sure you crinkle up the paper first and drop it from about waist height. Notice how the lead weight hits the ground WAY before …

Oops.

The weight and the paper hit the ground essentially at the same time. If the weight did hit first, the difference wasn’t anything you could reliably time or even be sure enough to gamble on.

Conclusion: As long as we’re not throwing objects out of our car on the freeway, air resistance is not important in kinematics.

The most common questions asked:

1. Find the terminal speed. Terminal speed means that, after a long time, the object’s speed becomes constant. To find that terminal speed, do an equilibrium problem: Free body, set up forces equal down forces, and left forces equal right forces. If something’s falling straight down with no other forces, usually you’ll get bv = mg . Then solve for v . That’s the terminal speed.

2. Sketch a graph of the speed of the object as a function of time. Perhaps the problem will say the object was dropped from rest, or give an initial velocity. Well, you can plot that point at time t = 0. Then you can find the terminal velocity using the method above—the terminal speed is the constant velocity after a long time. Plot a horizontal line for the terminal velocity near the right-hand side of the t -axis.

   In between the initial velocity and the terminal velocity, just know that the velocity function will look like an exponential function, changing rapidly at first, and changing less rapidly as time goes on. Sketch a curve in between the point and the line you drew. Done.

3. Describe the motion in words, including what’s happening to the acceleration and/or the velocity. Be sure to use a free-body diagram, and to separate the motion if necessary into parts when the object is moving up, and moving down. When the speed is zero, the force of air resistance is zero. This doesn’t mean no acceleration, of course. When the speed is not zero, use a free body to figure out the amount and direction of the net force. Remind yourself of the basics of kinematics—the net force is the direction of acceleration. If the acceleration is in the direction of motion, the object speeds up; if the acceleration is opposite motion, the object slows down.

4. Solve a differential equation to find an expression for the velocity as a function of time. Again, start with a free-body diagram, and write F _net = ma . Now, though, you’ll need to do some calculus: acceleration . Perhaps your Newton’s second law equation might say something like mg – bv = ma .

   Okay, solve for a and write the calculus expression for . This type of equation is called a differential equation, where a derivative of a function is proportional to the function itself. Specifically, since the first derivative is involved, this is called a “first order” differential equation.

   Your calculus class may well have taught you how to solve this equation by a technique known as separation of variables: put all the v terms on one side, the t terms on the other, and integrate. If you know how to do that, great; if not, it’s complex enough not in any way to be worth learning in order to possibly—possibly —earn yourself one or two points. Everyone, though, should be able to recognize and write the answer using the knowledge that the solution to a first-order differential equation will involve an exponential function . You can use facts about the initial velocity and the terminal velocity to write this function without an algorithmic solution.

Imagine that a ball was dropped from rest in the presence of air resistance F _air = bv . Writing the second law gives you the equation shown above. What’s the solution? Well, the initial velocity is zero; the terminal velocity can be found by setting acceleration to zero, meaning . The answer will always be something like v (t ) = Ae ^{– kt} or v (t ) = A (1 − e ^{− kt} ). Start by figuring out which: does the speed start large, and get small? If so, use the first expression. Or, does the speed start small, and get larger? If so, use the second expression. In this case, the speed starts from zero and ends up faster. So we use v (t ) = A (1 − e ^{− kt} ). Generally, the k term is going to be whatever’s multiplying the v in the original equation. In this case, then, k = b/m .

Now, look at the initial and final conditions to find the value of A . At time t = 0, v = 0; that works no matter the value of A , because e ⁰ = 1. But after a long time, we know the terminal velocity is mg /b . And in the equation, e ^−kt goes to zero for large t . Meaning: after a long time, the velocity function equals A . This A must be the terminal velocity!

So our final equation looks like: .

Practice Problems

Multiple Choice:

1 . A firework is shot straight up in the air with an initial speed of 50 m/s. What is the maximum height it reaches?

(A) 12.5 m

(B) 25 m

(C) 125 m

(D) 250 m

(E) 1250 m

2 . On a strange, airless planet, a ball is thrown downward from a height of 17 m. The ball initially travels at 15 m/s. If the ball hits the ground in 1 s, what is this planet’s gravitational acceleration?

(A) 2 m/s²

(B) 4 m/s²

(C) 6 m/s²

(D) 8 m/s²

(E) 10 m/s²

3 . An object moves such that its position is given by the function x (t ) = 3t ² − 4t + 1. The units of t are seconds, and the units of x are meters. After 6 s, how fast and in what direction is this object moving?

(A) 32 m/s in the original direction of motion

(B) 16 m/s in the original direction of motion

(C) 0 m/s

(D) 16 m/s opposite the original direction of motion

(E) 32 m/s opposite the original direction of motion

Free Response:

4 . An airplane attempts to drop a bomb on a target. When the bomb is released, the plane is flying upward at an angle of 30° above the horizontal at a speed of 200 m/s, as shown below. At the point of release, the plane’s altitude is 2.0 km. The bomb hits the target.

(a) Determine the magnitude and direction of the vertical component of the bomb’s velocity at the point of release.

(b) Determine the magnitude and direction of the horizontal component of the bomb’s velocity at the point when the bomb contacts the target.

(c) Determine how much time it takes for the bomb to hit the target after it is released.

(d) At the point of release, what angle below the horizontal does the pilot have to look in order to see the target?

Solutions to Practice Problems

1 . Call “up” the positive direction, and set up a chart. We know that v _f = 0 because, at its maximum height, the firework stops for an instant.

  Solve for Δx using equation ***: . The answer is (C) 125 m, or about skyscraper height.

2 . Call “down” positive, and set up a chart:

  Plug straight into ** (Δx = v ₀ t + ¹ /₂ at ² ) and you have the answer. This is NOT a quadratic, because this time t is a known quantity. The answer is (B) 4 m/s² , less than half of Earth’s gravitational field, but close to Mars’s gravitational field.

3 . First find the velocity function by taking the derivative of the position function: v (t ) = 6t − 4. Now plug in t = 6 to get the velocity after 6 s; you get 32 m/s. Note that this velocity is positive. Was the object originally moving in the positive direction? Plug in t = 0 to the velocity formula to find out … you find the initial velocity to be −4 m/s, so the object was originally moving in the negative direction, and has reversed direction after 6 s. The answer is (E).

4 . (a) Because the angle 30° is measured to the horizontal, the magnitude of the vertical component of the velocity vector is just (200 m/s) (sin 30°), which is 100 m/s. The direction is “up,” because the plane is flying up.

(b) The horizontal velocity of a projectile is constant. Thus, the horizontal velocity when the bomb hits the target is the same as the horizontal velocity at release, or (200 m/s)(cos 30°) = 170 m/s, to the right.

(c) Let’s call “up” the positive direction. We can solve this projectile motion problem by our table method.

  Don’t forget to convert to meters, and be careful about directions in the vertical chart.

  The horizontal chart cannot be solved for time; however, the vertical chart can. Though you could use the quadratic formula or your fancy calculator to solve x − x ₀ = v ₀ t + ¹ /₂ at ² , it’s much easier to start with ***, v _f ² = v ₀ ² + 2a (x − x ₀ ), to find that v_f vertically is −220 m/s (this velocity must have a negative sign because the bomb is moving down when it hits the ground). Then, plug in to *(v _f = v ₀ + at ) to find that the bomb took 32 s to hit the ground.

(d) Start by finding how far the bomb went horizontally. Because horizontal velocity is constant, we can say distance = velocity × time. Plugging in values from the table, distance = (170 m/s)(32 s) = 5400 m. Okay, now look at a triangle:

  By geometry, tan θ = 2000 m/5400 m. The pilot has to look down at an angle of 20°.

Rapid Review

• Average speed is total distance divided by total time. Instantaneous speed is your speed at a particular moment.

• Velocity is the vector equivalent of speed.

• Acceleration is a change in velocity divided by the time during which that change occurred.

• Displacement is the vector equivalent of distance.

• The three “star equations” are valid only when acceleration is constant.

• To solve any constant-acceleration kinematics problem, follow these four steps:

° Write out a table containing all five variables—v ₀ , v _f , x − x ₀ , a, t —and fill in whatever values are known.

° Count variables. If you have three known values, you can solve the problem.

° Use the “star equation” that contains your known variables.

° Check for correct units.

• When an object falls (in the absence of air resistance), it experiences an acceleration of g , about 10 m/s² . It’s particularly important for problems that involve falling objects to define a positive direction before solving the problem.

• An object’s motion in one dimension does not affect its motion in any other dimension.

• Projectile motion problems are usually easier to solve if you break the object’s motion into “horizontal” and “vertical” vector components.

• The slope of a distance–time graph is velocity.

• The slope of a velocity–time graph is acceleration.

• The area under a velocity–time graph is displacement.

¹ Use the “star equations,” which we will address in detail momentarily.

² When can’t you use kinematics, you ask? The most common situations are when a mass is attached to a spring, when a roller coaster travels on a curvy track, or when a charge is moving in a non-uniform electric field produced by other charges. To approach these problems, use conservation of energy, as discussed in Chapter 14 .

³ The writers of the AP exam love to throw things off cliffs.

⁴

⁵ Why shouldn’t I say “positive” and “negative,” you ask? Well, how do these directions to the store sound: “Define north as positive. Start from zero, and go positive 20 constantly; then come back at –20, also constantly.” You’d never say that! But, this is what you’ll sound like unless you use common language.

⁶ −10m/s² , if we’re on Earth.

⁷ Occasionally you might see a different form, F air = bv ² . In this case, you might be asked about the terminal speed, but you will NOT have to solve a differential equation.

Kinematics

1 . A cannonball is shot straight up in the air and then falls back to Earth. The force of air resistance is given by the equation F_air = -bv ² . If air resistance cannot be neglected, the acceleration of the cannonball is

  (A) constant while traveling upward.

  (B) increasing while traveling upward.

  (C) greater while traveling upward than while traveling downward.

  (D) equal to 9.8 m/s² while traveling downward.

  (E) constant while traveling downward.

2 . A shortstop throws a baseball to the first baseman. Which graphs correctly depict the horizontal velocity and horizontal acceleration of the baseball while in flight, assuming air resistance can be ignored?

  (A)

  (B)

  (C)

  (D)

  (E)

3 . You drop a baseball, then after 1.0 s, you drop a second baseball. Assuming air resistance can be neglected, which of the following statements is correct?

  (A) The difference in velocity between the two baseballs decreases as they fall.

  (B) The difference in velocity between the two baseballs remains constant as they fall.

  (C) The difference in velocity between the two baseballs increases as they fall.

  (D) The distance between the baseballs decreases as they fall.

  (E) The distance between the baseballs stays constant as they fall.

4 . An object begins at the origin and has a velocity as shown in the figure. When does the object return to the origin?

  (A) 2 s

  (B) 3 s

  (C) Between 3 and 4 s

  (D) 4 s

  (E) The object never returns to the origin.

Answers

1 . C —On the way up, the acceleration of the cannonball is . On the way down, the acceleration is . Therefore, the acceleration on the way up is larger in magnitude because both gravity and drag are acting in the same direction. The acceleration is not constant, as the velocity changes during the cannonball’s flight. The velocity decreases on the upward journey, so the acceleration will also decrease in magnitude. Finally, the acceleration is only g when the velocity is zero at the very top of the flight.

2 . A —The horizontal velocity is constant, and there is no horizontal acceleration in the absence of air resistance.

3 . B —Both balls start from rest and pick up speed at the same constant rate of g = 9.8 m/s² . The first ball simply starts 1.0 s ahead of the second ball and will remain a constant 9.8 m/s ahead. Since the first ball is always 9.8 m/s faster than the second, the distance between the balls will always increase as they fall.

4 . C —The displacement of the object is given by the area of the graph. The positive area from 0 to 2 s is 3 m. The negative area from 2 to 3 s is 1 m, which is not enough to return to the origin. The negative area from 2 to 4 s is 4 m, which is too large. Therefore, the object will pass through the origin between 3 and 4 s as it travels in the negative direction.

Newton’s Second Law, F_net = ma

IN THIS CHAPTER

Summary: Chapter 10 explained how to deal with objects in equilibrium, that is, with zero acceleration. The same problem-solving process can be used with accelerating objects.

Key Ideas

Only a NET force, not an individual force, can be set equal to ma .

Use a free-body diagram and the four-step problem-solving process when a problem involves forces.

When two masses are connected by a rope, the rope has the same tension throughout. (One rope = one tension.) ¹

Newton’s third law: force pairs must act on different objects.

Relevant Equations

Um, the chapter title says it all …

What this means is that the net force acting on an object is equal to the mass of that object multiplied by the object’s acceleration . And that statement will help you with all sorts of problems.

The Four-Step Problem-Solving Process

If you decide that the best way to approach a problem is to use F _net = ma , then you should solve the problem by following these four steps.

1. Draw a proper free-body diagram.

2. Resolve vectors into their components.

3. For each axis, set up an expression for F _net , and set it equal to ma .

4. Solve your system of equations.

Note the marked similarity of this method to that discussed in the chapter on equilibrium.

Following these steps will get you majority credit on an AP free-response problem even if you do not ultimately get the correct answer. In fact, even if you only get through the first one or two steps, it is likely that you will still get some credit.

Only Net Force Equals ma

THIS IS REALLY IMPORTANT. Only F _net can be set equal to ma . You cannot set any old force equal to ma . For example, let’s say that you have a block of mass m sitting on a table. The force of gravity, mg , acts down on the block. But that does not mean that you can say, “F = mg , so the acceleration of the block is g , or about 10 m/s² .” Because we know that the block isn’t falling! Instead, we know that the table exerts a normal force on the block that is equal in magnitude but opposite in direction to the force exerted by gravity. So the NET force acting on the block is 0. You can say “F _net = 0, so the block is not accelerating.”

A Simple Example

We follow our four-step process. First, draw a proper free-body diagram.

Second, we break the F _push vector into components that line up with the horizontal and vertical axes.

We can now move on to Step 3 , writing equations for the net force in each direction:

Now we can plug in our known values into these equations and solve for the acceleration of the block. First, we solve for the right–left direction:

Next, we solve for the up–down direction. Notice that the block is in equilibrium in this direction—it is neither flying off the table nor being pushed through it—so we know that the net force in this direction must equal 0.

So the acceleration of the block is simply 2.2 m/s² to the right.

F _net on Inclines

This is a really boring problem. But it’s also a really common problem, so it’s worth looking at. ²

Note that no numbers are given, just variables. That’s okay. It just means that our answer should be in variables. Only the given variables—in this case m, θ , and m —and constants such as g can be used in the solution. And you shouldn’t plug in any numbers (such as 10 m/s² for g ), even if you know them.

Step 1 : Free-body diagram.

Step 2 : Break vectors into components.

If you don’t know where we got those vector components, refer back to Chapter 9 .

Step 3 : Write equations for the net force in each direction.

Note that the block is in equilibrium in the direction perpendicular to the plane, so the equation for F _{net, perpendicular} (but not the equation for F _{net, down the plane} ) can be set equal to 0.

Step 4 : Solve.

We can rewrite F _f , because

F_f = μF_N = μ mg (cos θ )

Plugging this expression for F _f into the “F _{net, down the plane} ” equation, we have

It always pays to check the reasonability of the answer. First, the answer doesn’t include any variables that weren’t given. Next, the units work out: g has acceleration units; neither the sine or cosine of an angle nor the coefficient of friction has any units.

Second, compare the answer to something familiar. Note that if the plane were vertical, θ = 90°, so the acceleration would be g —yes, the block would then be in free fall! Also, note that friction tends to make the acceleration smaller, as you might expect.

Constant speed means a = 0. The solution for F_N in the perpendicular direction is the same as before: F _N = mg (cosθ ). But in the down-the-plane direction, no acceleration means that F _f = mg (sinθ ). Because μ = F _f /F _N ,

Canceling terms and remembering that sin/cos = tan, you find that μ = tanθ when acceleration is zero.

You might note that neither this answer nor the previous one includes the mass of the block, so on the same plane, both heavy and light masses move the same way!

F _net for a Pulley

Before we present our next practice problem, a few words about tension and pulleys are in order. Tension in a rope is the same everywhere in the rope, even if the rope changes direction (such as when it goes around a pulley) or if the tension acts in different directions on different objects. ONE ROPE = ONE TENSION. If there are multiple ropes in a problem, each rope will have its own tension. TWO ROPES = TWO TENSIONS. ³

When masses are attached to a pulley, the pulley can only rotate one of two ways. Call one way positive, the other, negative.

We arbitrarily call counterclockwise rotation of the pulley “positive.”

Step 1 : Free-body diagrams.

The tension T is the same for each block—ONE ROPE = ONE TENSION. Also, note that because the blocks are connected, they will have the same acceleration, which we call a .

Step 2 : Components.

The vectors already line up with one another. On to Step 3 .

Step 3 : Equations.

Notice how we have been careful to adhere to our convention of which forces act in the positive and negative directions.

Step 4 : Solve.

Let’s solve for T using the first equation:

T = Mg − Ma .

Plugging this value for T into the second equation, we have

(Mg − Ma ) − mg = ma

Our answer is

Why don’t you work this one out for yourself? We have included our solution on the following page.

Solution to Example Problem

Step 1 : Free-body diagrams.

Step 2 : Components.

Again, our vectors line up nicely, so on to Step 3 .

Step 3 : Equations.

Before we write any equations, we must be careful about signs: we shall call counterclockwise rotation of the pulley “positive.”

For the more massive block, we know that, because it is not flying off the table or tunneling into it, it is in equilibrium in the up–down direction. But it is not in equilibrium in the right–left direction.

For the less massive block, we only have one direction to concern ourselves with: the up–down direction.

F _net = T − mg = ma

We can solve for T from the “F _{net, x} ” equation for the more massive block and plug that value into the “F _net ” equation for the less massive block, giving us

(−Ma ) − mg = ma

We rearrange some terms to get

Now we plug in the known values for M and m to find that

To finish the problem, we plug in this value for a into the “F _{net, x} ” equation for the more massive block.

More Thoughts on F _net = ma

The four example problems in this chapter were all solved using only F _net = ma . Problems you might face in the real world—that is, on the AP test—will not always be so straightforward. Here’s an example: imagine that this last example problem asked you to find the speed of the blocks after 2 seconds had elapsed, assuming that the blocks were released from rest. That’s a kinematics problem, but to solve it, you have to know the acceleration of the blocks. You would first have to use F _net = ma to find the acceleration, and then you could use a kinematics equation to find the final speed. We suggest that you try to solve this problem: it’s good practice.

Also, remember in Chapter 12 when we introduced the unit of force, the newton, and we said that 1 N = 1 kg·m/s² ? Well, now you know why that conversion works: the units of force must be equal to the units of mass multiplied by the units of acceleration.

Newton’s Third Law

We’re sure you’ve been able to quote the third law since birth, or at least since 5th grade: “Forces come in equal and opposite action-reaction pairs,” also known as “For every action there is an equal and opposite reaction.” If I push down on the Earth, the Earth pushes up on me; a football player who makes a tackle experiences the same force that he dishes out.

What’s so hard about that? Well, ask yourself one of the most important conceptual questions in first-year physics: “If all forces cause reaction forces, then how can anything ever accelerate?” Pull a little lab cart horizontally across the table … you pull on the cart, the cart pulls on you, so don’t these forces cancel out, prohibiting acceleration?

Well, obviously, things can move. The trick is, Newton’s third law force pairs must act on different objects , and so can never cancel each other.

Consider the lab cart. The only horizontal force that it experiences is the force of your pull. So, it accelerates toward you. Now, you experience a force from the cart, but you also experience a whole bunch of other forces that keep you in equilibrium; thus, you don’t go flying into the cart.

This Chapter Was Not as Easy as You Thought

Be careful with this chapter. Most Physics C students say, “Oh, come on, this stuff is easy … let’s move on to something challenging.” Okay, you’re right—if you’re at the level you need to be for Physics C, basic Newton’s second law problems need to be easy for you. What you must remember from this and the equilibrium chapter is the absolute necessity of free-body diagrams.

No matter how easy or hard an F _net problem may seem, you must start the problem with a free-body diagram. Points are awarded for the free-body diagram, and that diagram will ensure that you don’t make minor mistakes on the rest of the problem. My own Physics C students frequently mess up on what should be straightforward problems when they try to take shortcuts. If you draw the FBD and follow the four-step problem-solving procedure, it’s hard to go wrong. Even professional physicists use free-body diagrams. You must, too.

Practice Problems

Multiple Choice:

1 . A 2.0-kg cart is given a shove up a long, smooth 30° incline. If the cart is traveling 8.0 m/s after the shove, how much time elapses until the cart returns to its initial position?

(A) 1.6 s

(B) 3.2 s

(C) 4.0 s

(D) 6.0 s

(E) 8.0 s

2 . A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?

(A) The normal force must be equal to the car’s weight.

(B) The normal force must be less than the car’s weight.

(C) The normal force must be greater than the car’s weight.

(D) The normal force must be zero.

(E) The normal force could have any value relative to the car’s weight.

3 . In the diagram above, a 1.0-kg cart and a 2.0-kg cart are connected by a rope. The spring scale reads 10 N. What is the tension in the rope connecting the two carts? Neglect any friction.

(A) 30 N

(B) 10 N

(C) 6.7 N

(D) 5.0 N

(E) 3.3 N

4 . The velocity–time graph above represents the motion of a 5-kg box. The only force applied to this box is a person pushing. Assuming that the box is moving to the right, what is the magnitude and direction of the force applied by the person pushing?

(A) 2.0 N, right

(B) 2.0 N, left

(C) 0.4 N, right

(D) 0.4 N, left

(E) 12.5 N, left

Free Response:

5 . A 2-kg block and a 5-kg block are connected as shown above. The coefficient of friction between the 5-kg block and the flat surface is θ = 0.2.

(A) Calculate the magnitude of the acceleration of the 5-kg block.

(B) Calculate the tension in the rope connecting the two blocks.

6 . Bert, Ernie, and Oscar are discussing the gas mileage of cars. Specifically, they are wondering whether a car gets better mileage on a city street or on a freeway. All agree (correctly) that the gas mileage of a car depends on the force that is produced by the car’s engine—the car gets fewer miles per gallon if the engine must produce more force. Whose explanation is completely correct?

  Bert says: Gas mileage is better on the freeway. In town the car is always speeding up and slowing down because of the traffic lights, so because F _net = ma and acceleration is large, the engine must produce a lot of force. However, on the freeway, the car moves with constant velocity, and acceleration is zero. So the engine produces no force, allowing for better gas mileage.

  Ernie says: Gas mileage is better in town. In town, the speed of the car is slower than the speed on the freeway. Acceleration is velocity divided by time, so the acceleration in town is smaller. Because F _net = ma , then, the force of the engine is smaller in town giving better gas mileage.

  Oscar says: Gas mileage is better on the freeway. The force of the engine only has to be enough to equal the force of air resistance—the engine doesn’t have to accelerate the car because the car maintains a constant speed. Whereas in town, the force of the engine must often be greater than the force of friction and air resistance in order to let the car speed up.

Solutions to Practice Problems

1 . B— “Smooth” usually means, “ignore friction.” So the only force acting along the plane is a component of gravity, mg (sin 30°). The F _net equation becomes mg (sin 30°) − 0 = ma . The mass cancels, leaving the acceleration as 5 m/s² . What’s left is a kinematics problem. Set up a chart, calling the direction down the plane as positive:

  Use ** (Δx = v ₀ t + ¹ /₂ at ² ) to find that the time is 3.2 s.

2 . B— The normal force exerted on an object on an inclined plane equals mg (cos θ ), where θ is the angle of the incline. If θ is greater than 0, then cos θ is less than 1, so the normal force is less than the object’s weight.

3 . E— Consider the forces acting on each block separately. On the 1.0-kg block, only the tension acts, so T = (1.0 kg)a . On the 2.0-kg block, the tension acts left, but the 10 N force acts right, so 10 N − T = (2.0 kg)a . Add these equations together (noting that the tension in the rope is the same in both equations), getting 10 N = (3.0 kg )a ; acceleration is 3.3 m/s² . To finish, T = (1.0 kg)a , so tension is 3.3 N.

4 . B— The acceleration is given by the slope of the v –t graph, which has magnitude 0.4 m/s² . F _net = ma , so 5 kg × 0.4 m/s² = 2.0 N. This force is to the left because acceleration is negative (the slope is negative), and negative was defined as left.

5 . The setup is the same as in the chapter’s example problem, except this time there is a force of friction acting to the left on the 5-kg block. Because this block is in equilibrium vertically, the normal force is equal to the block’s weight, 50 N. The friction force is μF _N , or 10 N.

  Calling the down-and-right direction positive, we can write two equations, one for each block:

(A) To solve for acceleration, just add the two equations together. The tensions cancel. We find the acceleration to be 1.4 m/s² .

(B) Plug back into either equation to find the final answer, that the tension is 17 N. This is more than the 14 N we found for the frictionless situation, and so makes sense. We expect that it will take more force in the rope to overcome friction on the table.

6 . Although Bert is right that acceleration is zero on the freeway, this means that the NET force is zero; the engine still must produce a force to counteract air resistance. This is what Oscar says, so his answer is correct. Ernie’s answer is way off—acceleration is not velocity/time, acceleration is a CHANGE in velocity over time.

Rapid Review

• The net force on an object equals the mass of the object multiplied by the object’s acceleration.

• To solve a problem using F _net = ma , start by drawing a good free-body diagram. Resolve forces into vector components. For each axis, the vector sum of forces along that axis equals ma _i , where a _i is the acceleration of the object along that axis.

• When an object is on an inclined plane, resolve its weight into vector components that point parallel and perpendicular to the plane.

• For problems that involve a massless pulley, remember that if there’s one rope, there’s one tension.

¹ But if the rope is attached across a massive pulley, the tension is different on each side of the pulley. See Chapter 16 .

² If you want to make this problem more interesting, just replace the word “block” with the phrase “maniacal tobogganist” and the word “plane” with the phrase “highway on-ramp.”

³ Except for the physics C corollary, when the pulley is massive—this situation is discussed in Chapter 16 .

Forces

1 . A light string holds three identical masses over a lightweight pulley as shown in the figure. When the masses are released, the acceleration of the masses will be

  (A) 2g

  (B) g

  (C) 2g/3

  (D) g/2

  (E) g/3

2 . While removing paint, a carpenter accelerates a rough sanding block up a wall as shown. The coefficient of friction between the block and the wall is μ. What is the net force on the sanding block?

  (A) F + mg + F_f

  (B) F cos θ - mg - μF sin θ

  (C) F sin θ - mg - μF cos θ

  (D) F sin θ + F_N

  (E) F + mg + F_N

3 . The net force applied to an object as a function of time is shown in the figure. Which of the following position-time and velocity-time graphs are consistent with the force-time graph?

  (A)

  (B)

  (C)

  (D)

  (E)

4 . A force (F ) accelerates three blocks across a rough horizontal surface as shown in the figure. The coefficient of friction (μ) between the blocks and the surface is the same for each block. Which of the following statements is correct?

  (A) The acceleration of all the blocks will be equal to .

  (B) The net force acting on each block will be the same.

  (C) The net force acting on block m is the largest.

  (D) The net force acting on block 2m is the largest.

  (E) The net force acting on block 3m is the largest.

Answers

1 . E —

2 . B —The horizontal forces acting on the block cancel each other out: Fsin θ=F_N . The vertical forces accelerate the block upward:

3 . A —The net force is a constant negative value, and the acceleration must also be constant and negative. Therefore, the velocity-time graph must have a negative slope. This eliminates choice B. Choices C and D start with a velocity of zero, yet their position-time graphs both begin with a positive slope, which is inconsistent. Choice E ends with a zero velocity, yet the position-time graph ends with a negative slope, which is inconsistent. This leaves choice A as the only viable graph that matches the force-time graph.

4 . E —All the blocks move as a unit, with the same acceleration of . Since each block has the same acceleration, the block with the largest mass will require the largest net force acting on it to maintain the same acceleration.

Momentum

IN THIS CHAPTER

Summary: The impulse–momentum relationship can explain how force acts in a collision. Momentum is conserved in all collisions, allowing a prediction of objects’ speeds before and after a collision.

Key Ideas

Impulse can be expressed both as force times a time interval, and as a change in momentum.

The total momentum of a set of objects before a collision is equal to the total momentum of a set of objects after a collision.

Momentum is a vector, so leftward momentum can “cancel out” rightward momentum.

Relevant Equations

The definition of momentum:

p = mv

The impulse–momentum theorem:

Δp = F Δt

Location of the center of mass:

Mx_cm = m ₁ x ₁ + m ₂ x ₂ + …

If an object is moving, it has momentum. The formal definition of momentum ¹ is that it’s equal to an object’s mass multiplied by that object’s velocity. However, a more intuitive way to think about momentum is that it corresponds to the amount of “oomph” an object has in a collision. Regardless of how you think about momentum, the key is this: the momentum of a system upon which no net external force acts is always conserved.

Momentum and Impulse

The units of momentum are kg·m/s, which is the same as N·s. Momentum is a vector quantity, and it is often abbreviated with a p .

Impulse (designated as I ) is an object’s change in momentum. It is also equal to the force acting on an object multiplied by the time interval over which that force was applied. The above equation is often referred to as the “impulse–momentum theorem.”

The F Δt definition of impulse explains why airbags are used in cars and why hitting someone with a pillow is less dangerous than hitting him or her with a cement block. The key is the Δt term. An example will help illustrate this point.

This is a multistep problem. We start by calculating the man’s velocity the instant before he hits the ground. That’s a kinematics problem, so we start by choosing a positive direction—we’ll choose “down” to be positive—and by writing out our table of variables.

We have three variables with known values, so we can solve for the other two. We don’t care about time, t , so we will just solve for v_f .

Now we can solve for the man’s momentum the instant before he hits the ground.

p = mv = (70)(7.7) = 540 kg·m/s

Once he hits the ground, the man quickly comes to rest. That is, his momentum changes from 540 kg·m/s to 0.

If the man does not bend his knees, then

The negative sign in our answer just means that the force exerted on the man is directed in the negative direction: up.

Now, what if he had bent his knees?

If he bends his knees, he allows for his momentum to change more slowly, and as a result, the ground exerts a lot less force on him than had he landed stiff-legged. More to the point, hundreds of thousands of newtons applied to a person’s legs will cause major damage—this is the equivalent of almost 20 tons sitting on his legs. So we would assume that the man would bend his knees upon landing, reducing the force on his legs by a factor of 30.

Calculus Version of the Impulse–Momentum Theorem

Conceptually, you should think of impulse as change in momentum, also equal to a force multiplied by the time during which that force acts. This is sufficient when the force in question is constant, or when you can easily define an average force during a time interval.

But what about when a force is changing with time? The relationship between force and momentum in the language of calculus is

A common AP question, then, gives momentum of an object as a function of time, and asks you to take the derivative to find the force on the object.

It’s also useful to understand this calculus graphically. Given a graph of momentum vs. time, the slope of the tangent to the graph gives the force at that point in time. Given a graph of force vs. time, the area under that graph is impulse, or change in momentum during that time interval.

Conservation of Momentum

Momentum in an isolated system, where no net external forces act, is always conserved. A rough approximation of a closed system is a billiard table covered with hard tile instead of felt. When the billiard balls collide, they transfer momentum to one another, but the total momentum of all the balls remains constant.

The key to solving conservation of momentum problems is remembering that momentum is a vector .

Let’s begin by drawing a picture.

Momentum is conserved, so we write

The tick marks on the right side of the equation mean “after the collision.” We know the momentum of each space traveler before the collision, and we know the UFO’s final momentum. So we solve for the satellite’s final velocity. (Note that we must define a positive direction; because the UFO is moving to the left, its velocity is plugged in as negative.)

Now, what if the satellite and the UFO had stuck together upon colliding? We can solve for their final velocity easily:

Motion of the Center of Mass

The center of mass of a system of objects obeys Newton’s second law. Two common examples might illustrate the point:

(1) Imagine that an astronaut on a spacewalk throws a rope around a small asteroid, and then pulls the asteroid toward him. Where will the asteroid and the astronaut collide?

   Answer: at the center of mass. Since no forces acted except due to the astronaut and asteroid, the center of mass must have no acceleration. The center of mass started at rest, and stays at rest, all the way until the objects collide.

(2) A toy rocket is in projectile motion, so that it is on track to land 30 m from its launch point. While in the air, the rocket explodes into two identical pieces, one of which lands 35 m from the launch point. Where does the first piece land?

   Answer: 25 m from the launch point. Since the only external force acting on the rocket is gravity, the center of mass must stay in projectile motion, and must land 30 m from the launch point. The two pieces are of equal mass, so if one is 5 m beyond the center of mass’s landing point, the other piece must be 5 m short of that point.

Finding the Center of Mass

Usually the location of the center of mass (cm) is pretty obvious … the formal equation for the cm of several objects is

Mx_cm = m ₁ x ₁ + m ₂ x ₂ + ^…

Multiply the mass of each object by its position, and divide by the total mass M , and voila, you have the position of the center of mass. What this tells you is that the cm of several equal-mass objects is right in between them; if one mass is heavier than the others, the cm is closer to the heavy mass.

Very rarely, you might have to find the center of mass of a continuous body (like a baseball bat) using calculus. The formula is

Do not use this equation unless (a) you have plenty of extra time to spend, and (b) you know exactly what you’re doing. In the highly unlikely event it’s necessary to use this equation to find a center of mass, you will usually be better off just guessing at the answer and moving on to the rest of the problem. (If you want to find out how to do such a problem thoroughly, consult your textbook. This is not something worth reviewing if you don’t know how to do it already.)

Elastic and Inelastic Collisions

This brings us to a couple of definitions.

If you’re unfamiliar with the concept of kinetic energy (KE), take a few minutes to skim Chapter 14 right now.

When the satellite and the UFO bounced off each other, they experienced a perfectly elastic collision. If kinetic energy is lost to heat or anything else during the collision, it is called an inelastic collision.

The extreme case of an inelastic collision is called a perfectly inelastic collision.

The second collision between the satellite and the UFO was a perfectly inelastic collision. BUT, MOMENTUM IS STILL CONSERVED, EVEN IN A PERFECTLY INELASTIC COLLISION!

Two-Dimensional Collisions

The key to solving a two-dimensional collision problem is to remember that momentum is a vector, and as a vector it can be broken into x and y components. Momentum in the x -direction is always conserved, and momentum in the y -direction is always conserved.

We want to analyze the x -component of momentum and the y -component of momentum separately. Let’s begin by defining “right” and “up” to be the positive directions. Now we can look at the x -component.

We can’t do much more with the x -component of momentum, so now let’s look at the y -component.

(Note the negative sign on Maggie’s y -velocity!)

Okay. Now we have two equations and two unknowns. It’ll take some algebra to solve this one, but none of it is too hard. We will assume that you can do the math on your own, but we will gladly provide you with the answer:

The algebra is not particularly important here. Get the conceptual physics down—in a two-dimensional collision, you must treat each direction separately. If you do so, you will receive virtually full credit on an AP problem. If you combine vertical and horizontal momentum into a single conservation equation, you will probably not receive any credit at all.

Practice Problems

Multiple Choice:

First two questions: A ball of mass M is caught by someone wearing a baseball glove. The ball is in contact with the glove for a time t ; the initial velocity of the ball (just before the catcher touches it) is v ₀ .

1 . If the time of the ball’s collision with the glove is doubled, what happens to the force necessary to catch the ball?

(A) It doesn’t change.

(B) It is cut in half.

(C) It is cut to one-fourth of the original force.

(D) It quadruples.

(E) It doubles.

2 . If the time of collision remains t , but the initial velocity is doubled, what happens to the force necessary to catch the ball?

(A) It doesn’t change.

(B) It is cut in half.

(C) It is cut to one-fourth of the original force.

(D) It quadruples.

(E) It doubles.

3 . Two balls, of mass m and 2m , collide and stick together. The combined balls are at rest after the collision. If the ball of mass m was moving 5 m/s to the right before the collision, what was the velocity of the ball of mass 2m before the collision?

(A) 2.5 m/s to the right

(B) 2.5 m/s to the left

(C) 10 m/s to the right

(D) 10 m/s to the left

(E) 1.7 m/s to the left

4 . Two identical balls have initial velocities v ₁ = 4 m/s to the right and v ₂ = 3 m/s to the left, respectively. The balls collide head-on and stick together. What is the velocity of the combined balls after the collision?

(A) m/s to the right

(B) m/s to the right

(C) ½ m/s to the right

(D) m/s to the right

(E) 1 m/s to the right

Free Response:

5 . A 75-kg skier skis down a hill. The skier collides with a 40-kg child who is at rest on the flat surface near the base of the hill, 100 m from the skier’s starting point, as shown above. The skier and the child become entangled. Assume all surfaces are frictionless.

(a) How fast will the skier be moving when he reaches the bottom of the hill? Assume the skier is at rest when he begins his descent.

(b) What will be the speed of the skier and child just after they collide?

(c) If the collision occurs in half a second, how much force will be experienced by each person?

Solutions to Practice Problems

1 . B— Impulse is force times the time interval of collision, and is also equal to an object’s change in momentum. Solving for force, F = Δp /Δt . Because the ball still has the same mass, and still changes from speed v ₀ to speed zero, the ball’s momentum change is the same, regardless of the collision time. The collision time, in the denominator, doubled; so the entire expression for force was cut in half.

2 . E— Still use F = Δp /Δt , but this time it is the numerator that changes. The ball still is brought to rest by the glove, and the mass of the ball is still the same; but the doubled velocity upon reaching the glove doubles the momentum change. Thus, the force doubles.

3 . B— The total momentum after collision is zero. So the total momentum before collision must be zero as well. The mass m moved 5 m/s to the right, giving it a momentum of 5m units; the right-hand mass must have the same momentum to the left. It must be moving half as fast, 2.5 m/s, because its mass it twice as big; then its momentum is (2m )(2.5) = 5m units to the left.

4 . C— Because the balls are identical, just pretend they each have mass 1 kg. Then the momentum conservation tells us that

  (1 kg)(+4 m/s) + (1 kg)(−3 m/s) = (2 kg)(v ′).

  The combined mass, on the right of the equation above, is 2 kg; v ′ represents the speed of the combined mass. Note the negative sign indicating the direction of the second ball’s velocity. Solving, v ′ = +0.5 m/s, or 0.5 m/s to the right.

5 . (a) This part is not a momentum problem, it’s a Newton’s second law and kinematics problem. (Or it’s an energy problem, if you’ve studied energy.) Break up forces on the skier into parallel and perpendicular axes—the net force down the plane is mg (sin 45°). So by Newton’s second law, the acceleration down the plane is g (sin 45°) = 7.1 m/s² . Using kinematics with intitial velocity zero and distance 100 m, the skier is going 38 m/s (!).

(b) Now use momentum conservation. The total momentum before collision is (75 kg)(38 m/s) = 2850 kg·m/s. This must equal the total momentum after collision. The people stick together, with combined mass 115 kg. So after collision, the velocity is 2850 kg·m/s divided by 115 kg, or about 25 m/s.

(c) Change in momentum is force multiplied by time interval … the child goes from zero momentum to (40 kg)(25 m/s) = 1000 kg·m/s of momentum. Divide this change in momentum by 0.5 seconds, and you get 2000 N, or a bit less than a quarter ton of force. Ouch!

Rapid Review

• Momentum equals an object’s mass multiplied by its velocity. However, you can also think of momentum as the amount of “oomph” a mass has in a collision.

• Impulse equals the change in an object’s momentum. It also equals the force exerted on an object multiplied by the time it took to apply that force.

• Momentum is always conserved. When solving conservation of momentum problems, remember that momentum is a vector quantity.

• In an elastic collision, kinetic energy is conserved. When two objects collide and bounce off each other, without losing any energy (to heat, sound, etc.), they have engaged in an elastic collision. In an inelastic collision, kinetic energy is not conserved. The extreme case is a perfectly inelastic collision. When two objects collide and stick together, they have engaged in a perfectly inelastic collision.

¹ This chapter deals only with linear momentum. Angular momentum is covered in Chapter 16 .

Momentum

1 . Two spheres traveling in opposite directions collide head-on as shown. The graph shows the force between the spheres during the collision. The velocity of the 10 kg sphere after the collision is

  (A) 1 m/s to the left.

  (B) 1.75 m/s to the left.

  (C) 1.75 m/s to the right.

  (D) 4 m/s to the left.

  (E) 5 m/s to the right.

2 . A skateboard rolls under a tree at a constant velocity; a cat drops vertically out of a tree and lands on it. The skateboard and cat move off together. Which of the following statements is correct?

  (A) The skateboard speeds up because the cat adds kinetic energy to the system.

  (B) The skateboard speeds up because the cat adds vertical momentum to the system.

  (C) The skateboard continues moving at the same speed as the cat and skateboard exert equal and opposite forces on each other that cancel out during the collision.

  (D) The skateboard slows down due to the cat’s loss of kinetic energy when it lands on the skateboard and stops moving vertically.

  (E) The skateboard slows down due to the added mass of the cat to the horizontally moving skateboard.

3 . The initial velocity (v ₁ ) of a cue ball is to the left. After striking the eight ball, the cue ball has a final velocity (v ₂ ), as shown in the figure. The direction of the impulse from the eight ball on the cue ball is best described by which of the following vectors?

  (A)

  (B)

  (C)

  (D)

  (E)

4 . An air force jet has a Gatling gun mounted in its nose such that it can fire an ordnance round directly forward at a velocity of v ₀ . The total mass of the jet and ordinance is M . The ordnance mass is m . While flying at a velocity of v_j , the jet fires the ordnance directly forward as shown in the figure. What will be the final velocity (v_f ) of the jet after it fires the ordnance?

  (A)

  (B)

  (C)

  (D)

  (E)

Answers

1 . A —The area (integral) of the force-time graph equals the impulse, and the impulse equals the spheres’ change in momentum:

  Note that the impulse is directed to the left on the 10 kg sphere, which is the opposite direction of its velocity. This makes the impulse negative:

2 . E —This a perfectly inelastic collision; therefore, kinetic energy is not conserved. Thus, we can ignore answer choices that contain kinetic energy. Momentum is only conserved in the horizontal direction because the ground does not allow the cat-skateboard system to move vertically. Using conservation of horizontal momentum, we can see that the skateboard speed must decrease because the cat adds mass to the system:

  Note: The equal and opposite forces between the cat and the skateboard do not cancel! The force from the cat on the skateboard slows the skateboard down, while the force from the skateboard on the cat speeds the cat up. Both end up with the same final velocity and move off together.

3 . B —Impulse equals the change in momentum. Because the mass of the cue ball remains the same during the collision, the impulse will be in the direction of the change in velocity. J = Δp = Δ(mv = m (Δv ). The change in velocity is best described by choice B.

4 . A —Conservation of momentum gives us the following:

  Note that we need to subtract the mass of the ordnance from the combined mass (M ) after it has been fired from the jet.

Energy Conservation

IN THIS CHAPTER

Summary: While kinematics can be used to predict the speeds of objects with constant acceleration, energy conservation is a more powerful tool that can predict how objects will move even with a changing acceleration.

Key Ideas

Work is related to kinetic energy through the work–energy theorem.

There are many types of potential energy. Two (due to gravity and due to a spring) are discussed in this chapter.

To use conservation of energy, add potential + kinetic energy at two positions in an object’s motion. This sum must be the same everywhere.

A potential energy function can be derived for any conservative force.

Relevant Equations

The definition of work:

W = F ·d _||

The work–energy theorem:

W _net = ΔK

The force of a spring:

F = −kx

Two different types of potential energy:

Power:

Relationship between a conservative force F and the potential energy U it creates:

As with momentum, the energy of an isolated system is always conserved. It might change form—potential energy can be converted to kinetic energy, or kinetic energy can be converted to heat—but it’ll never simply disappear.

Conservation of energy is one of the most important, fundamental concepts in all of physics … translation: it’s going to show up all over the AP exam. So read this chapter carefully.

Kinetic Energy and the Work-Energy Theorem

We’ll start with some definitions.

What this second definition means is that work equals the product of the distance an object travels and the component of the force acting on that object directed parallel to the object’s direction of motion. That sounds more complicated than it really is: an example will help.

A box is pulled along the floor, as shown in Figure 14.1 . It is pulled a distance of 10 m, and the force pulling it has a magnitude of 5 N and is directed 30° above the horizontal. So, the force component that is PARALLEL to the 10 m displacement is (5 N)(cos 30°).

Figure 14.1 Box is pulled along the floor.

One newton·meter is called a joule, abbreviated as 1 J.

• Work is a scalar. So is energy.

• The units of work and of energy are joules.

• Work can be negative … this just means that the force is applied in the direction opposite displacement.

This means that the kinetic energy of an object equals one-half the object’s mass times its speed squared.

The net work done on an object is equal to that object’s change in kinetic energy. Here’s an application:

Here, because the only horizontal force is the force of the brakes, the work done by this force is W _net .

Let’s pause for a minute to think about what this value means. We’ve just calculated the change in kinetic energy of the train car, which is equal to the net work done on the train car. The negative sign simply means that the net force was opposite the train’s displacement.

To find the force:

Potential Energy

Potential energy comes in many forms: there’s gravitational potential energy, spring potential energy, electrical potential energy, and so on. For starters, we’ll concern ourselves with gravitational potential energy.

Gravitational PE is described by the following equation:

U = mgh

In this equation, m is the mass of an object, g is the gravitational field of 10 N/kg on Earth, and h is the height of an object above a certain point (called “the zero of potential”). ¹ That point can be wherever you want it to be, depending on the problem. For example, let’s say a pencil is sitting on a table. If you define the zero of potential to be the table, then the pencil has no gravitational PE. If you define the floor to be the zero of potential, then the pencil has PE equal to mgh , where h is the height of the pencil above the floor. Your choice of the zero of potential in a problem should be made by determining how the problem can most easily be solved.

REMINDER: h in the potential energy equation stands for vertical height above the zero of potential.

Conservation of Energy: Problem-Solving Approach

Solving energy-conservation problems is relatively simple, as long as you approach them methodically. The general approach is this: write out all the terms for the initial energy of the system, and set the sum of those terms equal to the sum of all the terms for the final energy of the system. Let’s practice.

If we were to approach this problem using kinematics equations (which we could), it would take about a page of work to solve. Instead, observe how quickly it can be solved using conservation of energy.

We will define our zero of potential to be the height of the box after it has slid the 5 m down the plane. By defining it this way, the PE term on the right side of the equation will cancel out. Furthermore, because the box starts from rest, its initial KE also equals zero.

The initial height can be found using trigonometry: h _i = (5m) (sin 30°) = 2.5 m.

In general, the principle of energy conservation can be stated mathematically like this:

The term W in this equation stands for work done on an object. For example, if there had been friction between the box and the plane in the previous example, the work done by friction would be the W term. When it comes to the AP exam, you will include this W term only when there is friction (or some other exteral force) involved. When friction is involved, W = F _f d , where F _f is the force of friction on the object, and d is the distance the object travels.

Let’s say that there was friction between the box and the inclined plane.

We start by writing the general equation for energy conservation:

K_i + U_i + W = E_f + U_f

W equals F _f d , where F _f is the force of friction, and d is 5 m. ²

The value for W is negative because friction acts opposite displacement. You may want to draw a free-body diagram to understand how we derived this value for F _N .

Now, plugging in values we have

We rearrange some terms and cancel out m from each side to get

v _f = 5.7 m/s

This answer makes sense—friction on the plane reduces the box’s speed at the bottom.

Springs

Gravitational potential energy isn’t the only kind of PE around. Another frequently encountered form is spring potential energy.

The force exerted by a spring is directly proportional to the amount that the spring is compressed. That is,

In this equation, k is a constant (called the spring constant), and x is the distance that the spring has been compressed or extended from its equilibrium state. The negative sign is simply a reminder that the force of a spring always acts opposite to displacement—an extended spring pulls back toward the equilibrium position, while a compressed spring pushes toward the equilibrium position. We call this type of force a restoring force, and we discuss it more in Chapter 17 on simple harmonic motion. However, we can ignore this sign unless we are doing calculus.

When a spring is either compressed or extended, it stores potential energy. The amount of energy stored is given as follows.

Similarly, the work done by a spring is given by W _spring = ½kx ² . Here’s an example problem.

It’s important to recognize that we CANNOT use kinematics to solve this problem! Because the force of a spring changes as it stretches, the block’s acceleration is not constant. When acceleration isn’t constant, try using energy conservation.

We begin by writing our statement for conservation of energy.

Now we fill in values for each term. PE here is just in the form of spring potential energy, and there’s no friction, so we can ignore the W term. Be sure to plug in all values in meters!

Plugging in values for k and m , we have

Power

Whether you walk up a mountain or whether a car drives you up the mountain, the same amount of work has to be done on you. (You weigh a certain number of newtons, and you have to be lifted up the same distance either way!) But clearly there’s something different about walking up over the course of several hours and driving up over several minutes. That difference is power.

Power is, thus, measured in units of joules/second, also known as watts. A car engine puts out hundreds of horsepower, equivalent to maybe 100 kilowatts; whereas, you’d work hard just to put out a power of a few hundred watts.

Potential Energy vs. Displacement Graphs

A different potential energy function can actually be derived for ANY conservative force. (A conservative force means that energy is conserved when this force acts … examples are gravity, spring, and electromagnetic forces; friction and air resistance are the most common nonconservative forces.) The potential energy U for a force is given by the following integral:

Note that this equation works for the gravitational force F = −mg (where − is the down direction) and the spring force F = −kx ; the potential energy attributable to gravity integrates to mgh , and the spring potential energy becomes ½kx ² .

Chris on a Skateboard

Once a potential energy of an object is found as a function of position, making a U vs. x graph tells a lot about the long-term motion of the object. Consider the potential energy of a spring, ½kx ² . A graph of this function looks like a parabola, as shown in Figure 14.2 .

Figure 14.2 Potential energy vs. displacement graph for a spring.

You can get a general feel for how the mass on a spring moves by imagining that Chris is riding a skateboard on a ramp shaped like the graph. A ramp shaped like this looks like a half-pipe. If he starts from some height above the bottom, Chris will oscillate back and forth, going fastest in the middle, and turning around when he runs out of energy at the right or left end. Although this is not precisely how a mass on a spring moves—the mass only moves back and forth, for example—the long-term properties of Chris’s motion and the motion of the mass on a spring are the same. The mass oscillates back and forth, with its fastest speed in the middle, just like Chris does.

Thinking about Chris on a skateboard works for all U vs. x graphs. Consider a model of the energy between two atoms that looks like the graph in Figure 14.3 .

Figure 14.3 Potential energy vs. displacement graph for two atoms.

If Chris on his skateboard released himself from rest near position x ₁ , he’d just oscillate back and forth, much like in the mass on a spring problem. But if he were to let go near the position labeled x ₂ , he’d have enough energy to keep going to the right as far as he wants; in fact, he’d make it off the page, never coming back. This is what happens to the atoms in molecules, too. If a second atom is placed pretty close to a distance x ₁ from the first atom, it will just oscillate back and forth about that position. However, if the second atom is placed very close to the first atom, it will gain enough energy to escape to a faraway place.

Practice Problems

Multiple Choice:

Questions 1 and 2

A block of weight mg = 100 N slides a distance of 5.0 m down a 30-degree incline, as shown above.

1 . How much work is done on the block by gravity?

(A) 500 J

(B) 430 J

(C) 100 J

(D) 50 J

(E) 250 J

2 . If the block experiences a constant friction force of 10 N, how much work is done by the friction force?

(A) −43 J

(B) −25 J

(C) −500 J

(D) −100 J

(E) −50 J

3 . A mass experiences a potential energy U that varies with distance x as shown in the graph above. The mass is released from position x = 0 with 10 J of kinetic energy. Which of the following describes the long-term motion of the mass?

(A) The mass eventually comes to rest at x = 0.

(B) The mass slows down with constant acceleration, stopping at x = 5 cm.

(C) The mass speeds up with constant acceleration.

(D) The mass oscillates, never getting farther than 5 cm from x = 0.

(E) The mass oscillates, never getting farther than 10 cm from x = 0.

4 . Two identical balls of mass m = 1.0 kg are moving towards each other, as shown above. What is the initial kinetic energy of the system consisting of the two balls?

(A) 0 joules

(B) 1 joules

(C) 12 joules

(D) 18 joules

(E) 36 joules

Free Response:

5 . A 1500-kg car moves north according to the velocity–time graph shown.

(a) Determine the change in the car’s kinetic energy during the first 7 s.

(b) To determine how far the car traveled in these 7 s, the three basic kinematics equations can not be used. Explain why not.

(c) Use the velocity–time graph to estimate the distance the car traveled in 7 s.

(d) What was the net work done on the car in these 7 s?

(e) Determine the average power necessary for the car to perform this motion.

Solutions to Practice Problems

1 . E— The force of gravity is straight down and equal to 100 N. The displacement parallel to this force is the vertical displacement, 2.5 m. Work equals force times parallel displacement, 250 J.

2 . E— The force of friction acts up the plane, and displacement is down the plane, so just multiply force times distance to get 50 J. The negative sign indicates that force is opposite displacement.

3 . D— Think of Chris on a skateboard—on this graph, he will oscillate back and forth about x = 0. Because he starts with a KE of 10 J, he can, at most, have a potential energy of 10 J, which corresponds on the graph to a maximum displacement of 5 cm. (The mass cannot have constant acceleration because constant acceleration only occurs for a constant force; a constant force produces an energy graph that is linear. The mass will not come to rest because we are assuming a conservative force, for which KE can be converted to and from PE freely.)

4 . E— Kinetic energy is a scalar, so even though the balls move in opposite directions, the KEs cannot cancel. Instead, kinetic energy ½(1 kg)(6 m/s)² attributable to different objects adds together algebraically, giving 36 J total.

5 . (a) The car started from rest, or zero KE. The car ended up with ½(1500 kg)(40 m/s)² = 1.2 × 10⁶ J of kinetic energy. So its change in KE is 1.2 × 10⁶ J.

(b) The acceleration is not constant. We know that because the velocity–time graph is not linear.

(c) The distance traveled is found from the area under the graph. It is easiest to approximate by counting boxes, where one of the big boxes is 10 m. There are, give-or-take, 19 boxes underneath the curve, so the car went 190 m.

(d) We cannot use work = force × distance here, because the net force is continually changing (because acceleration is changing). But W_net = ΔKE is always valid. In part (a) the car’s change in KE was found to be 1.2 × 10⁶ J; so the net work done on the car is also 1.2 × 10⁶ J.

(e) Power is work divided by time, or 1.2 × 10⁶ J/7 s = 170 kW. This can be compared to the power of a car, 220 horsepower.

Rapid Review

• Energy is the ability to do work. Both energy and work are scalars.

• The work done on an object (or by an object) is equal to that object’s change in kinetic energy.

• Potential energy is energy of position, and it comes in a variety of forms; for example, there’s gravitational potential energy and spring potential energy.

• The energy of a closed system is conserved. To solve a conservation of energy problem, start by writing K _i + U_i + W = K _f + U_f , where “i ” means “initial,” “f ” means “final,” and W is the work done by friction or an externally applied force. Think about what type of U you’re dealing with; there might even be more than one form of U in a single problem!

• Power is the rate at which work is done, measured in watts. Power is equal to work/time , which is equivalent to force multiplied by velocity.

• If the functional form of a conservative force is known, then the potential energy attributable to that force is given by

When this U is graphed against displacement, the motion of an object can be predicted by imagining “Chris on a skateboard” skating on the graph.

¹ Note that 10 N/kg is exactly the same as 10 m/s² .

² Note this difference carefully. Although potential energy involves only a vertical height, work done by friction includes the entire distance the box travels.

Energy

1 . A 2 kg mass experiences a potential energy that varies with x as shown in the graph. Which statement correctly describes the behavior of the mass?

  (A) When at a location of 1 m, the mass will experience a force in the negative x direction.

  (B) When at a location of 1 m, the mass will experience no force.

  (C) If released from rest at a location of 3 m, the mass will reach a maximum speed of .

  (D) If released from rest at a location of 3 m, the mass will oscillate about an equilibrium at location of 2 m on the x -axis.

  (E) If released from rest at a location of 5 m, the mass will reach a maximum speed of 20 m/s.

2 . An 80 kg roller-coaster car traveling 10 m/s passes over the crest of a 40 m hill, as shown in the figure. The speed of the car at the bottom of the hill is 24 m/s. The energy lost to friction is most nearly

  (A) 60 J

  (B) 1,000 J

  (C) 5,000 J

  (D) 13,000 J

  (E) 22,500 J

3 . A block of mass (m ) is released from the top of a frictionless incline at a distance of l from the bottom. The block slides down the incline and then across a frictionless horizontal plane. At the far end of the horizontal plane is a fixed spring with force constant k . What will be the maximum compression of the spring?

  (A)

  (B)

  (C)

  (D)

  (E)

4 . The following graphs represent the force (F ) applied to an experimental car of mass m . The engine of the car can supply a maximum force (F _max ). Which force distribution will cause the greatest change in kinetic energy of the car from x = 0 to x _f ?

  (A)

  (B)

  (C)

  (D)

  (E)

Answers

1 . A — . This can be written in differential form as , which shows that the force on the object will be the negative of the slope of the potential energy graph. At 1 m, this indicates that the force will be in the negative x -direction. Note: The peak of the potential energy graph at 2 m is not a stable equilibrium point, and the forces will push the 2 kg mass away from this point.

2 . C —Using conservation of energy and g = 10 m/s² , we get

3 . C —Using conservation of energy,

4 . D —The net work done on the car is the integration (area) of the F-x graph. The graph with the most area is choice D.

Gravitation and Circular Motion

IN THIS CHAPTER

Summary: When an object moves in a circle, it is accelerating toward the center of the circle. Any two massive objects attract each other due to gravity.

Key Ideas

Centripetal (center-seeking) acceleration is equal to .

Circular motion (and gravitation) problems still require a free-body diagram and the four-step problem-solving process.

The gravitational force between two objects is bigger for bigger masses, and smaller for larger separations between the objects.

Kepler’s laws apply to the orbits of planets.

Relevant Equations

Centripetal acceleration:

Gravitational force between any two masses:

Gravitational potential energy a long way from a planet:

It might seem odd that we’re covering gravitation and circular motion in the same chapter. After all, one of these topics relates to the attractive force exerted between two massive objects, and the other one relates to, well, swinging a bucket over your head.

However, as you’re probably aware, the two topics have plenty to do with each other. Planetary orbits, for instance, can be described only if you understand both gravitation and circular motion. And, sure enough, the AP exam frequently features questions about orbits.

So let’s look at these two important topics, starting with circular motion.

Velocity and Acceleration in Circular Motion

Remember how we defined acceleration as an object’s change in velocity in a given time? Well, velocity is a vector, and that means that an object’s velocity can change either in magnitude or in direction (or both). In the past, we have talked about the magnitude of an object’s velocity changing. Now, we discuss what happens when the direction of an object’s velocity changes.

When an object maintains the same speed but turns in a circle, the magnitude of its acceleration is constant and directed toward the center of the circle. This means that the acceleration vector is perpendicular to the velocity vector at any given moment, as shown in Figure 15.1 .

Figure 15.1 Velocity and acceleration of an object traveling in uniform circular motion.

The velocity vector is always directed tangent to the circle, and the acceleration vector is always directed toward the center of the circle. There’s a way to prove that statement mathematically, but it’s complicated, so you’ll just have to trust us. (You can refer to your textbook for the complete derivation.)

Centripetal Acceleration

On to a few definitions.

We know that the net force acting on an object is related to the object’s acceleration by F _net = ma . And we know that the acceleration of an object in circular motion points toward the center of the circle. So we can conclude that the centripetal force acting on an object also points toward the center of the circle.

The formula for centripetal acceleration is

In this equation, v is the object’s velocity, and r is the radius of the circle in which the object is traveling.

Centripetal acceleration is real; centrifugal acceleration is nonsense, unless you’re willing to read a multipage discussion of “non-inertial reference frames” and “fictitious forces.” So for our purposes, there is no such thing as a centrifugal (center-fleeing) acceleration. When an object moves in a circle, the acceleration (and also the net force) must point to the center of the circle .

The main thing to remember when tackling circular motion problems is that a centripetal force is simply whatever force is directed toward the center of the circle in which the object is traveling . So, first label the forces on your free-body diagram, and then find the net force directed toward the center of the circle. That net force is the centripetal force. But NEVER label a free-body diagram with “F_c .”

Mass on a String

Let’s begin by drawing a free-body diagram of the block at the top of the circle and another of the block at the bottom of the circle.

Next, we write Newton’s second law for each diagram. Acceleration is always toward the center of the circle.

The acceleration is centripetal, so we can plug in v ² /r for both accelerations.

At both top and bottom, the speed v and the radius r are the same. So Mv ² /r has to be the same at both the top and bottom, allowing us to set the left side of each equation equal to one another.

With M = 2 kg and T _top = 10 N, we solve to get T _bottom = 50 N.

Car on a Curve

This next problem is a bit easier than the last one, but it’s still good practice.

Before we draw our free-body diagram, we should consider how friction is acting in this case. Imagine this: what would it be like to round a corner quickly while driving on ice? You would slide off the road, right? Another way to put that is to say that without friction, you would be unable to make the turn. Friction provides the centripetal force necessary to round the corner. Bingo! The force of friction must point in toward the center of the curve.

We can now write some equations and solve for μ , the coefficient of friction.

The net force in the horizontal direction is F _f , which can be set equal to mass times (centripetal) acceleration.

We also know that F _f = μ _N . So,

Furthermore, we know that the car is in vertical equilibrium—it is neither flying off the road nor being pushed through it—so F _N = mg .

Solving for μ we have

Note that this coefficient doesn’t depend on mass. Good—if it did, we’d need tires made of different materials depending on how heavy the car is.

Newton’s Law of Gravitation

We now shift our focus to gravity. Gravity is an amazing concept—you and the Earth attract each other just because you both have mass!—but at the level tested on the AP exam, it’s also a pretty easy concept. In fact, there are only a couple equations you need to know. The first is for gravitational force:

This equation describes the gravitational force that one object exerts on another object. m ₁ is the mass of one of the objects, m ₂ is the mass of the other object, r is the distance between the center of mass of each object, and G is called the “Universal Gravitational Constant” and is equal to 6.67 × 10⁻¹¹ (G does have units—they are N·m² /kg² —but most problems won’t require your knowing them). The negative sign indicates that the force is attractive. We can leave it off unless we are doing calculus with the equation.

This is simple plug-and-chug (remember to convert km to m).

Notice that the amount of force that the Earth exerts on the sun is exactly the same as the amount of force the sun exerts on the Earth.

We can combine our knowledge of circular motion and of gravity to solve the following type of problem.

The force of gravity exerted by the sun on the Earth is what keeps the Earth in motion—it is the centripetal force acting on the Earth.

v = 29,000 m/s. (Wow, fast … that converts to about 14 miles every second—much faster than, say, a school bus.)

Along with the equation for gravitational force, you need to know the equation for gravitational potential energy.

Why negative? Objects tend get pushed toward the lowest available potential energy. A long way away from the sun, the r term gets big, so the potential energy gets close to zero. But, since a mass is attracted to the sun by gravity, the potential energy of the mass must get lower and lower as r gets smaller.

We bet you’re thinking something like, “Now hold on a minute! You said a while back that an object’s gravitational potential energy equals mgh . What’s going on?”

Good point. An object’s gravitational PE equals mgh when that object is near the surface of the Earth. But it equals no matter where that object is.

Similarly, the force of gravity acting on an object equals mg (the object’s weight) only when that object is near the surface of the Earth.

The force of gravity on an object, however, always equals regardless of location.

Kepler’s Laws

Johannes Kepler, the late 1500s theorist, developed three laws of planetary motion based on the detailed observations of Tycho Brahe. You need to understand each law and its consequences.

1. Planetary orbits are ellipses, with the sun at one focus . Of course, we can apply this law to a satellite orbiting Earth, in which case the orbit is an ellipse, with Earth at one focus. (We mean the center of the Earth—for the sake of Kepler’s laws, we consider the orbiting bodies to be point particles.) In the simple case of a circular orbit, this law still applies because a circle is just an ellipse with both foci at the center.

2. An orbit sweeps out equal areas in equal times . If you draw a line from a planet to the sun, this line crosses an equal amount of area every minute (or hour, or month, or whatever)—see Figure 15.2 . The consequence here is that when a planet is close to the sun, it must speed up, and when a planet is far from the sun, it must slow down. This applies to the Earth as well. In the northern hemisphere winter, when the Earth is slightly closer to the sun, ¹ the Earth moves faster in its orbit. (You may have noticed that the earliest sunset in wintertime occurs about two weeks before the solstice—this is a direct consequence of Earth’s faster orbit.)

Figure 15.2 Kepler’s second law. The area “swept out” by a planet in its orbit is shaded. In equal time intervals Δt ₁ and Δt ₂ , these swept areas A ₁ and A ₂ are the same.

3. A planet’s orbital period squared is proportional to its orbital radius cubed . In mathematics, we write this as T ² = cR ³ . Okay, how do we define the “radius” of a non-circular orbit? Well, that would be average distance from the sun. And what is this constant c ? It’s a different value for every system of satellites orbiting a single central body. Not worth worrying about, except that you can easily derive it for the solar system by solving the equation above for c and plugging in data from Earth’s orbit: c = 1 year² /AU³ , where an “AU” is the distance from Earth to the sun. If you really need to, you can convert this into more standard units, but we wouldn’t bother with this right now.

Energy of Closed Orbits

When an object of mass m is in orbit around the sun, its potential energy is , where M is the mass of the sun, and r is the distance between the centers of the two masses.

The kinetic energy of the orbiting mass, of course, is K = ½mv ² . The total mechanical energy of the mass in orbit is defined as U + K . When the mass is in a stable orbit, the total mechanical energy must be less than zero. A mass with positive total mechanical energy can escape the “gravitational well” of the sun; a mass with negative total mechanical energy is “bound” to orbit the sun. ²

All of the above applies to the planets orbiting in the solar system. It also applies to moons or satellites orbiting planets, when (obviously) we replace the “sun” by the central planet. A useful calculation using the fact that total mechanical energy of an object in orbit is the potential energy plus the kinetic energy is to find the “escape speed” from the surface of a planet … at r equal to the radius of the planet, set kinetic plus potential energy equal to zero, and solve for v . This is the speed that, if it is attained at the surface of the planet (neglecting air resistance), will cause an object to attain orbit.

Practice Problems

Multiple Choice:

Questions 1 and 2:

Two stars, each of mass M , form a binary system. The stars orbit about a point a distance R from the center of each star, as shown in the diagram above. The stars themselves each have radius r .

1 . What is the force each star exerts on the other?

(A)

(B)

(C)

(D)

(E)

2 . In terms of each star’s tangential speed v , what is the centripetal acceleration of each star?

(A)

(B)

(C)

(D)

(E)

Questions 3 and 4: In the movie Return of the Jedi , the Ewoks throw rocks using a circular-motion device. A rock is attached to a string. An Ewok whirls the rock in a horizontal circle above his head, then lets go, sending the rock careening into the head of an unsuspecting stormtrooper.

3 . What force provides the rock’s centripetal acceleration?

(A) The vertical component of the string’s tension

(B) The horizontal component of the string’s tension

(C) The entire tension of the string

(D) The gravitational force on the rock

(E) The horizontal component of the gravitational force on the rock

4 . The Ewok whirls the rock and releases it from a point above his head and to his right. The rock initially goes straight forward. Which of the following describes the subsequent motion of the rock?

(A) It will continue in a straight line forward, while falling due to gravity.

(B) It will continue forward but curve to the right, while falling due to gravity.

(C) It will continue forward but curve to the left, while falling due to gravity.

(D) It will fall straight down to the ground.

(E) It will curve back toward the Ewok and hit him in the head.

5 . A Space Shuttle orbits Earth 300 km above the surface. Why can’t the Shuttle orbit 10 km above Earth?

(A) The Space Shuttle cannot go fast enough to maintain such an orbit.

(B) Kepler’s laws forbid an orbit so close to the surface of the Earth.

(C) Because r appears in the denominator of Newton’s law of gravitation, the force of gravity is much larger closer to the Earth; this force is too strong to allow such an orbit.

(D) The closer orbit would likely crash into a large mountain such as Everest because of its elliptical nature.

(E) Much of the Shuttle’s kinetic energy would be dissipated as heat in the atmosphere, degrading the orbit.

Free Response:

6 . Consider two points on a rotating turntable: Point A is very close to the center of rotation, while point B is on the outer rim of the turntable. Both points are shown above. A penny could be placed on the turntable at either point A or point B .

(a) In which case would the speed of the penny be greater, if it were placed at point A , or if it were placed at point B ? Explain.

(b) At which point would the penny require the larger centripetal force to remain in place? Justify your answer.

(c) Point B is 0.25 m from the center of rotation. If the coefficient of friction between the penny and the turntable is μ = 0.30, calculate the maximum linear speed the penny can have there and still remain in circular motion.