## AP Physics C Exam

# Part IV

# Content Review for the AP Physics C Exam

# Chapter 7

# Linear Momentum

## INTRODUCTION

When Newton first expressed his Second Law, he didn’t write **F**_{net} = *m***a**. Instead, he expressed the law in the words, *The alteration of motion is … proportional to the … force impressed.…* By “motion,” he meant the product of mass and velocity, a vector quantity known as **linear momentum** which is denoted by **p**:

So Newton’s original formulation of the Second Law read **p** ∝ **F**, or, equivalently, **F** ∝ **p**. But a large force that acts for a short period of time can produce the same change in linear momentum as a small force acting for a greater period of time. Knowing this, if we take the average force that acts over the time interval *t*, we can turn the proportion above into an equation:

This equation becomes **F** = *m***a**, since **p**/*t* = (*m***v**)/*t* = *m* (**v**/*t*) = *m***a** (assuming that *m* remains constant). If we take the limit as *t* → 0, then the equation above takes the form:

**Example 1** A golfer strikes a golf ball of mass 0.05 kg, and the time of impact between the golf club and the ball is 1 ms. If the ball acquires a velocity of magnitude 70 m/s, calculate the average force exerted on the ball.

**Solution.** Using Newton’s Second Law, we find

## IMPULSE

The product of force and the time during which it acts is known as **impulse**; it’s a vector quantity that’s denoted by **J:**

**J** =

In terms of impulse, Newton’s Second Law can be written in yet another form:

**J** = **p**

Sometimes this is referred to as the **impulse–momentum theorem**, but it’s just another way of writing Newton’s Second Law. If **F** varies with time over the interval during which it acts, then the impulse delivered by the force **F** = **F**(*t*) from time *t* = *t*_{1} to *t* = *t*_{2} is given by the following definite integral:

On the equation sheet for the free-response section, this information will be represented as follows:

If a graph of force-versus-time is given, then the impulse of force **F** as it acts from *t _{1}* to

*t*is equal to the area bounded by the graph of

_{2}**F**, the

*t*-axis, and the vertical lines associated with

*t*and

_{1}*t*as shown in the following graph.

_{2}**Example 2** A football team’s kicker punts the ball (mass = 0.4 kg) and gives it a launch speed of 30 m/s. Find the impulse delivered to the football by the kicker’s foot and the average force exerted by the kicker on the ball, given that the impact time is 8 ms.

**Solution.** Impulse is equal to change in linear momentum, so

*J* = *p* = *p*_{f} – *p*_{i} = *p*_{f} = *mv* = (0.4 kg)(30 m/s) = 12 kg·m/s

Using the equation = *J* / *t*, we find that the average force exerted by the kicker is

= *J* / *t* = (12 kg · m / s) / (8 × 10^{−3} s) = 1500 N [≈ 340 lb]

**Example 3** An 80 kg stuntman jumps out of a window that’s 45 m above the ground.

(a) How fast is he falling when he reaches ground level?

(b) He lands on a large, air-filled target, coming to rest in 1.5 s. What average force does he feel while coming to rest?

(c) What if he had instead landed on the ground (impact time = 10 ms)?

**Solution.**

(a) His gravitational potential energy turns into kinetic energy: *mgh*= *mv*^{2}, so

(You could also have answered this question using Big Five #5.)

(b) Using = **p** / *t*, we find that

= = = = =

(c) In this case,

= = = = =

This force is equivalent to about 27 tons(!), more than enough to break bones and cause fatal brain damage. Notice how crucial impact time is: Increasing the slowing-down time reduces the acceleration and the force, ideally enough to prevent injury. This is the purpose of air bags in cars, for instance.

**Example 4** A small block of mass *m* = 0.07 kg, initially at rest, is struck by an impulsive force **F** of duration 10 ms whose strength varies with time according to the following graph:

What is the resulting speed of the block?

**Solution.** The impulse delivered to the block is equal to the area under the *F* vs. *t* graph. The region is a trapezoid, so its area, (base_{1} + base_{2}) × height, can be calculated as follows:

*J* = (10 + 4) × 20 = 0.14 N · s

Now, by the impulse–momentum theorem,

## CONSERVATION OF LINEAR MOMENTUM

Newton’s Third Law says that when one object exerts a force on a second object, the second object exerts an equal but opposite force on the first. Since Newton’s Second Law says that the impulse delivered to an object is equal to the resulting change in its linear momentum, **J** = **p**, the two interacting objects experience equal but opposite momentum changes (assuming that there are no external forces), which implies that the total linear momentum of the system remains constant. In fact, given any number of interacting objects, each pair that comes in contact will undergo equal but opposite momentum changes, so the result described for two interacting objects will actually hold for any number of objects, given that the only forces they feel are from each other. This means that, in an isolated system, *the total linear momentum will remain constant*. This is the **Law of Conservation of Linear Momentum**. In equation form, for two objects colliding, we have

**Example 5** An astronaut is floating in space near her shuttle when she realizes that the cord that’s supposed to attach her to the ship has become disconnected. Her total mass (body + suit + equipment) is 89 kg. She reaches into her pocket, finds a 1 kg metal tool, and throws it out into space with a velocity of 9 m/s directly away from the ship. If the ship is 10 m away, how long will it take her to reach it?

**Solution.** Here, the astronaut + tool are the system. Because of Conservation of Linear Momentum,

Using *distance = average speed* × *time*, we find

## COLLISIONS

Conservation of Linear Momentum is routinely used to analyze **collisions**. The objects whose collision we will analyze form the *system*, and although the objects exert forces on each other during the impact, these forces are only *internal* (they occur within the system). The system’s total linear momentum is conserved if there is no net external force on the system.

Collisions are classified into two major categories: (1) **elastic** and (2) **inelastic**. A collision is said to be *elastic* if kinetic energy is conserved. Ordinary macroscopic collisions are never truly elastic, because there is always a change in energy due to energy transferred as heat, deformation of the objects, and the sound of the impact. However, if the objects do not deform very much (for example, two billiard balls or a hard glass marble bouncing off a steel plate), then the loss of initial kinetic energy is small enough to be ignored, and the collision can be treated as virtually elastic. *Inelastic* collisions, then, are ones in which the total kinetic energy is different after the collision. An extreme example of inelasticism is **completely** (or **perfectly** or **totally) inelastic**. In this case, the objects stick together after the collision and move as one afterward. In all cases of isolated collisions (elastic or not), Conservation of Linear Momentum states that

total **p**_{before collision} = total **p**_{after colision}

**Example 6** Two balls roll toward each other. The red ball has a mass of 0.5 kg and a speed of 4 m/s just before impact. The green ball has a mass of 0.3 kg and a speed of 2 m/s. After the head-on collision, the red ball continues forward with a speed of 1.7 m/s. Find the speed of the green ball after the collision. Was the collision elastic?

**Solution.** First remember that momentum is a vector quantity, so the direction of the velocity is crucial. Since the balls roll toward each other, one ball has a positive velocity while the other has a negative velocity. Let’s call the red ball’s velocity before the collision positive; then **v**_{red} = +4 m/s, and **v**_{green} = –2 m/s. Using a prime to denote *after the collision*, Conservation of Linear Momentum gives us the following:

Notice that the green ball’s velocity was reversed as a result of the collision; this typically happens when a lighter object collides with a heavier object. To see whether the collision was elastic, we need to compare the total kinetic energies before and after the collision. In this case, however, an explicit calculation is not needed since both objects experienced a decrease in speed as a result of the collision. Kinetic energy was lost, so the collision was inelastic; this is usually the case with macroscopic collisions. Most of the lost energy was transferred as heat; the two objects are both slightly warmer as a result of the collision.

**Example 7** Two balls roll toward each other. The red ball has a mass of 0.5 kg and a speed of 4 m/s just before impact. The green ball has a mass of 0.3 kg and a speed of 2 m/s. If the collision is completely inelastic, determine the velocity of the composite object after the collision.

**Solution.** If the collision is completely inelastic, then, by definition, the masses stick together after impact, moving with a velocity, **v’**. Applying Conservation of Linear Momentum, we find

**Example 8** An object of mass *m*_{1} is moving with velocity *v*_{1} toward a target object of mass *m*_{2} which is stationary (*v*_{2} = 0). The objects collide head-on, and the collision is elastic. Show that the relative velocity before the collision, *v*_{2} – *v*_{1}, has the same magnitude as *v*′_{2} − *v*′_{1}, the relative velocity after the collision.

**Solution.** Since the collision is elastic, both total linear momentum and kinetic energy are conserved. Therefore,

Now for some algebra. Cancel the ’s in the second equation and factor to get the following pair of equations:

Next, dividing the second equation by the first gives *v*_{1} + *v*′_{1} = *v*′_{2}, so we can write

Adding Equations (1″) and (2″) gives

Substituting this result into Equation (2″) gives

Now we have calculated the final velocities, *v*_{1} and *v*_{2}. To verify the claim made in the statement of the question, we notice that

so the relative velocity after the collision, *v*′_{2} − *v*′_{1}, is equal (in magnitude) but opposite (in direction) to *v*_{2} – *v*_{1}, the relative velocity before the collision. *This is a general property that characterizes elastic collisions.*

**Example 9** An object of mass *m* moves with velocity **v** toward a stationary object of mass 2*m*. After impact, the objects move off in the directions shown in the following diagram:

Before the collision After the collision

(a) Determine the magnitudes of the velocities after the collision (in terms of *v*).

(b) Is the collision elastic? Explain your answer.

**Solution.**

(a) Conservation of Linear Momentum is a principle that establishes the equality of two vectors: **p**_{total} before the collision and **p**_{total} after the collision. Writing this single vector equation as two equations, one for the *x* component and one for the *y*, we have

*x* component: *mv* = *mv*′_{1} cos 30° + 2 *mv*′_{2} cos 45° (1)

*y* component: 0 = *mv*′_{1} sin 30° − 2 *mv*′_{2} sin 45° (2)

Adding these equations eliminates *v*_{2}, because cos 45° = sin 45°.

*mv* = *mv*′_{1}(cos 30° + sin 30°)

and lets us determine *v*_{1}:

Substituting this result into Equation (2) gives us

(b) The collision is elastic only if kinetic energy is conserved. The total kinetic energy after the collision, *K*’, is calculated as follows:

However, the kinetic energy before the collision is just *K* = *mv*^{2}, so the fact that

tells us that *K*’ is less than *K*, so some kinetic energy is lost; the collision is inelastic.

## CENTER OF MASS

The center of mass is the point where all of the mass of an object can be considered to be concentrated; it’s the dot that represents the object of interest in a free-body diagram.

For a homogeneous body (that is, one for which the density is uniform throughout), the center of mass is where you intuitively expect it to be: at the geometric center. Thus, the center of mass of a uniform sphere or cube or box is at its geometric center.

If we have a collection of discrete particles, the center of mass of the system can be determined mathematically as follows. First consider the case where the particles all lie on a straight line. Call this the *x* axis. Select some point to be the origin (*x* = 0) and determine the positions of each particle on the axis. Multiply each position value by the mass of the particle at that location, and get the sum for all the particles. Divide this sum by the total mass, and the resulting *x* value is the center of mass:

On the equation sheet for the free-response section, this information will be represented as follows:

The system of particles behaves in many respects as if all its mass, *M* = *m*_{1} + *m*_{2} +…+ *m _{n}*, were concentrated at a single location,

*x*

_{cm}.

If the system consists of objects that are not confined to the same straight line, use the equation above to find the *x*-coordinate of their center of mass, and the corresponding equation,

to find the *y*-coordinate of their center of mass (and one more equation to calculate the *z*-coordinate, if they are not confined to a single plane).

From the equation

we can derive

*Mv*_{cm} = *m*_{1}*v*_{1} + *m*_{2}*v*_{2} + … + *m _{n}v_{n}*

So, the total linear momentum of all the particles in the system (*m*_{1}*v*_{1} + *m*_{2}*v*_{2} +…+ *m _{n}v_{n}*) is the same as

*Mv*

_{cm}, the linear momentum of a single particle (whose mass is equal to the system’s total mass) moving with the velocity of the center of mass.

We can also differentiate again and establish the following:

**F**_{net} = *M***a**_{cm}

This says that the net (external) force acting on the system causes the center of mass to accelerate according to Newton’s Second Law. In particular, *if the net external force on the system is zero, then the center of mass will not accelerate*.

**Example 10** Two objects, one of mass *m* and one of mass 2*m*, hang from light threads from the ends of a uniform bar of length 3*L* and mass 3*m*. The masses *m* and 2*m* are at distances *L* and 2*L*, respectively, below the bar. Find the center of mass of this system.

**Solution.** The center of mass of the bar alone is at its midpoint (because it is uniform), so we may treat the total mass of the bar as being concentrated at its midpoint. Constructing a coordinate system with this point as the origin, we now have three objects: one of mass *m* at (–3*L*/2, –*L*), one of mass 2*m* at (3*L*/2, –2*L*), and one of mass 3*m* at (0, 0):

We figure out the *x*- and *y*-coordinates of the center of mass separately:

Therefore, the center of mass is at

(*x*_{cm}, *y*_{cm}) = (*L*/4, –5*L*/6)

relative to the midpoint of the bar.

**Example 11** A man of mass *m* is standing at one end of a stationary, floating barge of mass 3*m*. He then walks to the other end of the barge, a distance of *L* meters. Ignore any frictional effects between the barge and the water.

(a) How far will the barge move?

(b) If the man walks at an average velocity of *v*, what is the average velocity of the barge?

**Solution.**

(a) Since there are no external forces acting on the man + barge system, the center of mass of the system cannot accelerate. In particular, since the system is originally at rest, the center of mass cannot move. Letting *x* = 0 denote the midpoint of the barge (which is its own center of mass, assuming it is uniform), we figure out the center of mass of the man + barge system:

So, the center of mass is a distance of *L*/8 from the midpoint of the barge, and since the mass is originally at the left end, the center of mass is a distance of *L*/8 to the left of the barge’s midpoint.

When the man reaches the other end of the barge, the center of mass will, by symmetry, be *L*/8 to the *right* of the midpoint of the barge. But, since the position of the center of mass cannot move, this means the barge itself must have moved a distance of

*L*/8 + *L*/8 = 2*L*/8 = *L*/4

to the left.

(b) Let the time it takes the man to walk across the barge be denoted by *t*; then *t* = *L*/*v*. In this amount of time, the barge moves a distance of *L*/4 in the *opposite* direction, so the velocity of the barge is

So far we have dealt with objects that can be considered point masses, or masses with uniform density. Now we will learn how to find the center of mass of objects with non-uniform density.

Take the example of a bar that becomes denser along its length. Here we will deal with the linear density λ as a function of *x*, λ(*x*). Each small segment of the bar, *x*, has a different mass, *m*. We treat each *m* as a point mass and then take the limit as *x* approaches zero. Using the formula for calculating the center of mass of point masses, and replacing *m* with *dm*, we get the integral shown below.

where *M* is the total mass, *x* is the distance to each *dm*, and you can substitute for *dm* in terms of *x*. Linear density is mass per length, so the equation is *λ* = , therefore *M* = ∫*dm* = ∫λ*dx*.

**Example 12** A bar with of length of 30 cm has a linear density λ = 10 + 6*x*, where *x* is in meters and λ is in kg/m. Determine the mass of the bar and the center of mass of this bar.

**Solution.** We can determine the mass of the bar by using the definition of linear density, *λ* = . Therefore

To calculate the center of mass we will use the equation

This answer makes sense because the center of mass of the bar is beyond the midpoint.

## CHAPTER 7 REVIEW QUESTIONS

The answers and explanations can be found in Chapter 16.

### Section I: Multiple Choice

1. An object of mass 2 kg has a linear momentum of magnitude 6 kg·m/s. What is this object’s kinetic energy?

(A) 3 J

(B) 6 J

(C) 9 J

(D) 12 J

(E) 18 J

2. A ball of mass 0.5 kg, initially at rest, acquires a speed of 4 m/s immediately after being kicked by a force of strength 20 N. For how long did this force act on the ball?

(A) 0.01 s

(B) 0.02 s

(C) 0.1 s

(D) 0.2 s

(E) 1 s

3. A box with a mass of 2 kg accelerates in a straight line from 4 m/s to 8 m/s due to the application of a force whose duration is 0.5 s. Find the average strength of this force.

(A) 2 N

(B) 4 N

(C) 8 N

(D) 12 N

(E) 16 N

4. A ball of mass *m* traveling horizontally with velocity **v** strikes a massive vertical wall and rebounds back along its original direction with no change in speed. What is the magnitude of the impulse delivered by the wall to the ball?

(A 0

(B) *mv*

(C) *mv*

(D) 2*mv*

(E) 4*mv*

5. Two objects, one of mass 3 kg and moving with a speed of 2 m/s and the other of mass 5 kg and speed 2 m/s, move toward each other and collide head-on. If the collision is perfectly inelastic, find the speed of the objects after the collision.

(A) 0.25 m/s

(B) 0.5 m/s

(C) 0.75 m/s

(D) 1 m/s

(E) 2 m/s

6. Object 1 moves toward Object 2, whose mass is twice that of Object 1 and which is initially at rest. After their impact, the objects lock together and move with what fraction of Object 1’s initial kinetic energy?

(A) 1/18

(B) 1/9

(C) 1/6

(D) 1/3

(E) None of the above

7. Two objects move toward each other, collide, and separate. If there was no net external force acting on the objects, but some kinetic energy was lost, then

(A) the collision was elastic and total linear momentum was conserved

(B) the collision was elastic and total linear momentum was not conserved

(C) the collision was not elastic and total linear momentum was conserved

(D) the collision was not elastic and total linear momentum was not conserved

(E) None of the above

8. Three thin, uniform rods each of length *L* are arranged in the shape of an inverted U:

The two rods on the arms of the U each have mass *m*; the third rod has mass 2*m*. How far below the midpoint of the horizontal rod is the center of mass of this assembly?

(A) *L*/8

(B) *L*/4

(C) 3*L*/8

(D) *L*/2

(E) 3*L*/4

9. A wooden block of mass *M* is moving at speed *V* in a straight line.

How fast would the bullet of mass *m* need to travel to stop the block (assuming that the bullet became embedded inside)?

(A) *mV*/(*m* + *M*)

(B) *MV*/(*m* + *M*)

(C) *mV*/*M*

(D) *MV*/*m*

(E) (*m* + *M*)*V*/*m*

10. Which of the following best describes a perfectly inelastic collision free of external forces?

(A) Total linear momentum is never conserved.

(B) Total linear momentum is sometimes conserved.

(C) Kinetic energy is never conserved.

(D) Kinetic energy is sometimes conserved.

(E) Kinetic energy is always conserved.

### Section II: Free Response

1. A steel ball of mass *m* is fastened to a light cord of length *L* and released when the cord is horizontal. At the bottom of its path, the ball strikes a hard plastic block of mass *M* = 4*m*, initially at rest on a frictionless surface. The collision is elastic.

(a) Find the tension in the cord when the ball’s height above its lowest position is *L*. Write your answer in terms of *m* and *g*.

(b) Find the speed of the block immediately after the collision.

(c) To what height *h* will the ball rebound after the collision?

2. A *ballistic pendulum* is a device that may be used to measure the muzzle speed of a bullet. It is composed of a wooden block suspended from a horizontal support by cords attached at each end. A bullet is shot into the block, and as a result of the perfectly inelastic impact, the block swings upward. Consider a bullet (mass *m*) with velocity *v* as it enters the block (mass *M*). The length of the cords supporting the block each have length *L*. The maximum height to which the block swings upward after impact is denoted by *y*, and the maximum horizontal displacement is denoted by *x*.

(a) In terms of *m*, *M*, *g*, and *y*, determine the speed *v* of the bullet.

(b) What fraction of the bullet’s original kinetic energy is lost as a result of the collision? What happens to the lost kinetic energy?

(c) If *y* is very small (so that *y*^{2} can be neglected), determine the speed of the bullet in terms of *m*, *M*, *g*, *x*, and *L*.

(d) Once the block begins to swing, does the momentum of the block remain constant? Why or why not?

3. An object of mass *m* moves with velocity **v** toward a stationary object of the same mass. After their impact, the objects move off in the directions shown in the following dagram:

Assume that the collision is elastic.

(a) If *K*_{1} denotes the kinetic energy of Object 1 before the collision, what is the kinetic energy of this object after the collision? Write your answer in terms of *K*_{1} and *θ*_{1}.

(b) What is the kinetic energy of Object 2 after the collision? Write your answer in terms of *K*_{1} and *θ*_{1}.

(c) What is the relationship between *θ*_{1} and *θ*_{2}?

## Summary

### Momentum

- Linear momentum is given by the equation:
**p**=*m***v** - Linear momentum is conserved when no external force acts on a system. This is known as the Law of Conservation of Linear Momentum. It can be written as

total **p**_{before collision} = total **p**_{after collission}

- Elastic collisions conserve
*kinetic energy*(in general, every collision conserves energy but not necessarily kinetic energy). - Inelastic collisions do not conserve
*kinetic energy.* - When the objects stick together, the collision is known as perfectly inelastic.
- Anytime you are given a problem that involves a collision or separation, first consider whether you can use the Law of Conservation of Linear Momentum.

### Impulse

- Impulse is given by the equation:
*J*=*t* - The Impulse-Momentum Theorem states that the impulse on an object is equal to the change in momentum of the object. The equation is:
*J*=*t*=*p*

### Center of Mass

- Usually the motion of an object is describing the motion of the center of mass. When you use Newton’s Second Law,
*F = ma*, the acceleration you calculate is the acceleration of the center of mass. - For point masses,
*r*_{cm}= , where*r*is used for the position of each mass. - For distributed mass (for example, a bar with non-uniform density) the equation is:
*r*_{cm}= ∫*r dm* - You usually use linear density,
*λ**dr*=*dm*, for*dm*and then integrate to solve for the center of mass.