## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 12

### Collisions: Impulse and Momentum

**IN THIS CHAPTER**

**Summary:** Whenever you see a collision, the techniques of impulse and momentum are likely to be useful in describing or predicting the result of the collision. In particular, momentum is conserved in all collisions—this means that the total momentum of all objects is the same before and after the collision. When an object (or a system of objects) experiences a net force, the impulse momentum theorem Δ*p* = *F* · Δ*t* can be used for predictions and calculations.

**Definitions**

A moving object”s **momentum** is its mass times its velocity. Momentum is in the direction of motion.

**Impulse** is defined as a force multiplied by the time during which that force acts. The net impulse on an object is equal to the change in that object”s momentum.

A **system** is made up of several objects that can be treated as a single thing. It”s important to define the system you are considering before you treat a set of objects as a system.

While total momentum is conserved in all collisions, kinetic energy is conserved only in an **elastic collision** .

Momentum is a useful quantity to calculate because it is often **conserved** ; that is, the total amount of momentum available in most situations cannot change. Whenever you see a collision, the techniques of impulse and momentum are most likely to be useful. Try impulse and momentum first, before trying to use force or energy approaches.

**The Impulse-Momentum Theorem**

Here is the impulse-momentum theorem:

**Example 1:** A teacher whose weight is 900 N jumps vertically from rest while standing on a platform scale. The scale reading as a function of time is shown in the preceding figure.

A force versus time graph is essentially an invitation to calculate impulse. Since impulse is defined as *F* · Δ*t* , from a force versus time graph, impulse is the area under the graph.

** Strategy: **When you need to take the area of an experimental graph, approximate as best you can with rectangles and triangles.

This graph for Example 1 is tricky—impulse calculations should use the *net* force on an object. The scale reading on the vertical axis of the graph is not the net force; the net force is the scale reading in excess of the person”s 900-N weight.

To estimate the impulse given to the jumper in this example, draw a horizontal line at the 900-N mark as a zero point for calculating the net force:

Regions 1 and 2 have somewhat close to the same area, one above and one above the zero net force line; so these areas cancel out. Region 3 looks somewhat like a rectangle, with base something like 0.25 s and height (1,800 N − 900 N) = 900 N. The impulse is then 900 N times 0.25 s, or something like 220 N·s.

**Exam Tip from an AP Physics Veteran**

You must be comfortable with this kind of rough approximation. Sure, the impulse could well be more like 228 N·s, or 210 N·s. Who cares? On a free-response item it will be the reasoning behind your calculation that earns credit, much more so than the answer itself. On a multiple-choice item, the choices might be far separated, like (A) 200 N·s; (B) 2,000 N·s; (C) 20,000 N·s; (D) 200,000 N·s. Isn”t the choice obvious?

Impulse by itself doesn”t say much. The more interesting question about this jumping teacher in Example 1 is the speed with which he leaves the scale. Since impulse is the change in an object”s momentum, you know the teacher changed his momentum by 220 N·s. Since he started from rest, his momentum right after leaving the scale is also 220 N·s. Finally, momentum is mass times speed. The teacher”s mass is 90 kg,^{ }^{1}^{ }so plug in to the equation *p* = *mv* : (220 N·s) = (90 kg)*v* . His speed is 2.4 m/s, or thereabouts.

**Conservation of Momentum**

**Example 2:** Cart A, of mass 0.5 kg, moves to the right at a speed of 60 cm/s. Cart B, of mass 1.0 kg, is at rest. The carts collide.

**FACT:** In any system in which the only forces acting are between objects in that system, momentum is conserved. This effectively means that momentum is conserved in *all* collisions.

Define the system for momentum conservation in Example 2—just the two carts. They apply a force to each other in the collision, but that”s it, so momentum is conserved.^{ }^{2}

A common task in a problem with a collision involves calculating the speed of one or both objects after the collision. Even when a collision-between-two-objects question is qualitative or conceptual in nature, it”s often a good idea to try calculating speeds after a collision.

To do this, define a positive direction and then make a chart indicating the mass *m* and speed *v* of each cart before and after the collision. I use a “prime” mark (′) to indicate when we”re dealing with values after a collision rather than before. Indicate the direction of motion with a plus or minus sign on the velocities.

(Note that it”s okay to use centimeters per second [cm/s] rather than meters per second [m/s], as long as you are consistent throughout.)

*m* _{A} = 0.5 kg

*m* _{B} = 1.0 kg

*v* _{A} = + 60 cm/s

*v* _{B} = 0

*v* _{A} ′ = ?

*v* _{B} ′ = ?

Then write the equation for conservation of momentum.

Uh, where do you get such an equation? I know it”s not on the equation sheet on the old or new AP physics exams.

The relevant equation comes from the definition of “conservation,” meaning an unchanging quantity. The total change in momentum for the system of the two carts must be zero. Any momentum lost by Cart A is gained by Cart B. Set zero equal to Cart A”s change in momentum, plus Cart B”s change in momentum:

Then, knowing that *p* = *mv* , plug in what you know. I”m going to leave off the units to make the mathematics clearer; since the table above has values and units, it”s clear what units are intended.

Is this solvable? Not yet, because it”s only one equation with two variables. The information about this collision is incomplete. The collision in Example 2 could thus have all sorts of results.

One possibility is that the carts stick together. In that case, the carts share the same speed: in the notation above, *v* _{A} ′ = *v* _{B} ′. That makes the calculation solvable; replace the *v* s with a single variable to get *v* = 20 cm/s.

Perhaps the problem statement continues to tell us the speed of one of the carts after the collision. Then the problem is solvable: plug in the value given, and solve for the other *v* .

The only tricky part here would be, say, if Cart A rebounded after the collision. Then *v* _{A} ′ would take a negative value. But the solution would be approached the same way.

**When Is the Momentum of a System Not Conserved?**

The simple answer goes back to the definition of momentum conservation: The momentum of a system is *not* conserved when a force is exerted by an object that”s not in the system.

**Example 3:** Two identical balls are dropped from the same height above the ground, such that they are traveling 50 cm/s just before they hit the ground. Ball A rebounds with speed 50 cm/s; Ball B rebounds with a speed of 10 cm/s. Each is in contact with the ground for the same amount of time.

Define the system here. If the system is just Ball A, say, then is the momentum of Ball A conserved? Of course not! The problem says that Ball A rebounds, which means it changed its direction and thus its momentum.

**Mistake:** It”s tempting to say that since Ball A didn”t change its mass, and since its speed was 50 cm/s before *and* after the collision, that Ball A didn”t change its momentum. This is not correct; momentum has direction. An object that changes direction loses all its momentum and then gains some more. If Ball A had mass 2 kg, then it lost 1 N·s of momentum in stopping, and then gained another 1 N·s of momentum in order to rebound—for a total change in momentum of 2 N·s.

*Wait* . You said in the preceding *Fact* that momentum is conserved in all collisions. What happened?

Well, yes, momentum *is* conserved in all collisions, if you define the system to include the two (or three) objects that are colliding. In Example 3, Ball A is effectively colliding with the entire Earth. If we consider the system of the Earth and Ball A, then momentum is, in fact, conserved. The change in Ball A”s momentum is equal to the change in the Earth”s momentum. Since the Earth is so mind-bogglingly massive, its speed won”t change in any measureable amount.

**Mistake:** The total momentum for a system of objects is always the same. So in a single collision, the total momentum cannot change. In a problem like Example 3, though, Balls A and B are involved in two separate collisions. Therefore, they can”t be part of the same system! Don”t use “conservation of momentum” as a reason for anything about Balls A and B to be equal, when Balls A and B are involved in separate collisions.

The point is that momentum conservation is not an effective approach to consider when a ball collides with the entire Earth.^{ }^{3}^{ }Instead, use the impulse-momentum theorem to find out what you can.

The easy question is as follows: Which ball changes its momentum by a greater amount? That”d be Ball A. Both balls lost the same amount of momentum in coming to a brief rest, then rebounded; since Ball A rebounded faster, and since the balls have the same mass, Ball A changed its momentum by a greater amount.^{ }^{4}

The harder question is this: Which ball exerted a larger force on the ground during its collision? We know that momentum change equals force times time.^{ }^{5}^{ }With the same time of collision, the bigger force is exerted by the ball with the greater momentum change—that”s Ball A.

Similar reasoning can explain why airbags make a car safer. You^{ }^{6}^{ }lose all your momentum in a crash regardless of how you come to rest. Airbags extend the time of the collision between you and the car. In the equation Δ*p* = *F* · Δ*t* with the same Δ*p* , a bigger Δ*t* gives a smaller *F* , so the force you experience is less in an airbag collision.

**Elastic/Inelastic Collisions**

In elastic collisions, the total kinetic energy of both objects combined is the same before and after the collision. A typical AP problem might pose a standard collision problem and then ask, “Is the collision elastic?” To figure that out, add up the kinetic energies (½*mv* ^{2} ) of both objects before the collision, add up the kinetic energies of both objects after the collision, and compare. If these kinetic energies are essentially the same, the collision is elastic. If the final kinetic energy is less than the initial kinetic energy, the collision was *not* elastic—kinetic energy was converted, generally to work done by nonconservative forces exerted by one colliding object on the other.^{ }^{7}

**Example 4:** Two carts of equal mass move toward each other with identical speeds of 30 cm/s. After colliding, the carts bounce off each other, each regaining 30 cm/s of speed, but now moving in the opposite direction.

**Mistake:** Never start a collision problem writing anything about kinetic energy. Always start with conservation of momentum. Only move on to kinetic energy conservation if you have to, that is, if you don”t have enough information to solve with just momentum conservation, *and* if the problem is explicit in saying that the collision is elastic.

Is momentum conserved in this collision? Yes, and you don”t have to do any calculations to show it. In a collision, momentum is always conserved because the only forces acting on the carts are exerted by the carts themselves.

Is kinetic energy conserved in this collision? You”ve got to do the calculation to check.

But the carts bounced off each other. Doesn”t that automatically mean the collision is elastic?

No. When carts stick together, the collision cannot be elastic. But when carts bounce off each other, the collision might be elastic, or might not be.

Start with the kinetic energy before the collision. In this case, make up a mass for each cart: they”re identical, so call them 1 kg each. Each cart has speed 0.30 m/s, so the kinetic energy of each cart before the collision is ½(1 kg)(0.30 m/s)^{2} = 0.045 J. The combined kinetic energy before collision is thus 0.090 J.

*Whoa there* —One cart was moving right, the other left; that means the kinetic energies subtract, giving zero total kinetic energy. Right?

**Wrong.** Kinetic energy is a scalar, which means it has no direction. Kinetic energy can never take on a negative value. Always add the kinetic energies of each object in a system to get the total kinetic energy. After the collision, the calculation is the same: total kinetic energy is still 0.090 J. So the collision is, in fact, elastic.

**2-d Collisions**

**Example 5:** Maggie, of mass 50 kg, glides to the right on a frictionless frozen pond with a speed of 2.5 m/s. She collides with a 20-kg penguin. After the collision, the directions of the penguin”s and Maggie”s motion is shown in the following figure.

** Strategy: **When objects move in both an

*x-*and a

*y*-direction after a collision, analyze the collision with momentum conservation

*separately*in each direction.

You will not likely be asked to do quantitative analysis of a two-dimensional collision, but you do need to understand conceptually how momentum conservation works here. Be able to explain how you would carry out the analysis of momentum conservation in each direction and be able to answer simple qualitative questions.

For example, who has a greater magnitude of momentum in the *y* -direction after collision? Before the collision, there was no momentum in the *y-* direction. After the collision, the total *y* -momentum must also be zero. Since both the penguin and Maggie are moving in the *y* -direction, their momentums must be equal and opposite so as to subtract to zero. The answer is neither—both the penguin and Maggie have the same amount of *y* -momentum.

What about the *y-* component of their velocities? We”ve already established that they have the same *y-* momentum, which is equal to mass times *y* -velocity. Since Maggie has the bigger mass, she must have the smaller *y* -component of velocity.

Is momentum conserved in the *x* -direction? Of course it is. The total momentum before collision is all due to Maggie”s movement: (50 kg)(2.5 m/s) = 125 N·s, all in the *x* -direction. After collision, the total *x* -momentum is also 125 N·s. The *x* -component of the penguin”s momentum after collision is just his momentum *mv* times the cosine of 60°; Maggie”s *x* -momentum is her momentum times the cosine of 30°. In this problem, the only way to get values for these components is to do some complicated algebra, which is beyond the scope of AP Physics 1. But you should be able to explain everything about this collision in words, as discussed here.

**Motion of the Center of Mass**

**FACT:** The center of mass of a system of objects obeys Newton”s second law.

Two common examples illustrate this fact:

**Example:** 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.

**Example:** 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.

** Practice Problems**

** 1 .** A 2-kg coconut falls from the top of a tall tree, 30 m above a person”s head. The coconut strikes and comes to rest on the person”s head. Justify all answers thoroughly.

(a) Calculate the magnitude momentum of the coconut just before it hits the person in the head.

(b) Calculate the magnitude and direction of the impulse experienced by the coconut in colliding with the person”s head.

(c) The person”s head experienced a force of 10,000 N in the collision. How long was the coconut in contact with the person”s head?

(A) Much more than 10 seconds

(B) Just a bit more than one second

(C) Just a bit less than one second

(D) Much less than 1/10 second

(d) In a different situation, explain how it could be possible for an identical coconut dropped from the same height to hit the person”s head, but produce *less* than 10,000 N of force.

** 2 .** A car on a freeway collides with a mosquito, which was initially at rest. Justify all answers thoroughly.

(a) Did the total momentum of the car-mosquito system increase, decrease, or remain the same after the collision?

(b) Did the momentum of the mosquito increase, decrease, or remain the same after the collision?

(c) Did the momentum of the car increase, decrease, or remain the same after the collision?

(d) Which changed its speed by more in the collision, the car or the mosquito? (Or did they change speed by the same amount?)

(e) Which changed its momentum by more in the collision, the car or the mosquito? (Or did they change momentum by the same amount?)

(f) Which experienced a greater impulse in the collision, the car or the mosquito? (Or did they experience the same impulse?)

(g) Which experienced a greater magnitude of net force during the collision, the car or the mosquito? (Or did they experience the same net force?)

** 3 .** Car A has a mass of 1,500 kg and travels to the right with a speed of 20 m/s. Car B initially travels to the left with a speed of 10 m/s. After the vehicles collide, they stick together, moving left with a common speed of 5 m/s. Justify all answers thoroughly.

(a) Calculate the mass of Car B.

(b) This collision is not elastic. Explain why not.

(c) Describe specifically a collision between these two cars with the same initial conditions, but which is *not* elastic, and in which the cars bounce off one another.

(d) Is the collision elastic when Car B remains at rest after the collision?

** Solutions to Practice Problems**

** 1 .** (a) Momentum is mass times speed. To find the coconut”s speed, use kinematics with

*v*

_{0}= 0,

*a*= 10 m/s per second, and Δ

*x*= 30 m. The equation

*v*

_{f}

^{2}=

*v*

_{0}

^{2}+ 2

*a*Δ

*x*solved for

*v*

_{f}gives 24 m/s. Multiplying by the 2-kg mass gives a momentum of 48 N·s.

^{ }

^{8}(b) Impulse is the change in momentum and is in the direction of the net force experienced by an object. The coconut”s momentum after colliding with the head is zero—the coconut comes to rest. So its change in momentum, and thus the magnitude of the impulse it experiences, is 48 N·s. The direction of this impulse is upward, because the net force on the coconut must be opposite its speed in order to slow it down.

(c) Impulse is also equal to force times the time interval of collision. Setting 48 N·s equal to (10,000 N) (Δ*t* ), we find the time interval of collision is 48/10,000 of a second—much less than 1/10 second, even without reference to a calculator.

(d) The impulse-momentum theorem says that Δ*p* = *F* Δ*t* . Solving for force, . Here the momentum change has to be the same no matter what—the coconut will be traveling 24 m/s and will come to rest on the person”s head. But if the person is wearing a soft helmet, or if the coconut has a rotten spot on it somewhere, then the time of collision could be larger than before. Since Δ*t* is in the denominator of the force equation, a bigger time interval of collision leads to a smaller force on the coconut (and therefore on the person”s head).

** 2 .** (a) Momentum is conserved when no forces are exerted, except for those on and by objects in the system. Here the only forces are of the car on mosquito and mosquito on car. Therefore, momentum was conserved. That means that the total momentum of the car-mosquito system remains the same.

(b) The mosquito went from rest to moving freeway speeds after it hit the car. The mosquito”s mass didn”t change.^{ }^{9}^{ }Momentum is mass times speed, so the mosquito”s momentum increased.

(c) Since total momentum of the car-mosquito system doesn”t change, and the mosquito gained momentum, the car has to lose that same amount of momentum.

(d) The mosquito”s speed went from, say, zero to 60 miles per hour. While the car must lose the same amount of momentum that the mosquito gained, the car”s mass is so much larger than the mosquito”s that the car”s speed will hardly change. And you knew that, because a car hitting a mosquito on the freeway doesn”t cause the car to stop.

(e) The momentum change is the same for both, because total momentum remains unchanged. Any momentum gained by the mosquito must be lost by the car.

(f) Impulse is the same thing as momentum change, so the same for both.

(g) Newton”s third law says the force of the mosquito on the car is equal to the force of the car on the mosquito. So they”re equal.

** 3 .** (a) Before the collision, Car A has a momentum of 30,000 N·s to the right. If we call the mass of Car B “

*M*

_{B},” then Car B has momentum of

*M*

_{B}(10 m/s) to the left. Afterward, the total momentum is (

*M*

_{B}+ 1,500 kg)(5 m/s) to the left. Let”s call right the positive direction. Then the relevant equation for conservation of momentum is 30,000 − 10

*M*

_{B}_{ }= −5(1,500 +

*M*

_{B}_{ }), where I”ve left off the units so the algebra is clearer. Solve for

*M*

_{B}to get 7,500 kg. This makes sense—Car B was initially moving slower, yet after the collision the cars moved off together in the direction Car B was going. Car B must therefore have more momentum than Car A initially and more mass because it was going slower.

(b) “Elastic” means that kinetic energy (= ½*mv* ^{2} ) of all objects combined is the same before and after collision. Before collision, Car A had 300 kJ of kinetic energy, and Car B had 375 kJ, for a total of 675 kJ before the collision.^{ }^{10}^{ }After collision, the kinetic energy of the combined cars is 112 kJ. Kinetic energy was lost in the collision. (Note that it”s legitimate to remember that collisions in which objects stick together can never be elastic.)

(c) Imagine that Car B keeps moving left, but much slower, say, 1 m/s. Momentum is conserved in a collision, regardless of whether the collision is elastic or not. The total momentum of Car A before collision is 30,000 N·s to the right; the total momentum of Car B before collision is 75,000 N·s to the left. This gives a total momentum of 45,000 N·s to the left before collision. If Car B moves 1 m/s after collision, it has 7,500 N·s of momentum to the left, leaving 37,500 N·s to the left for Car A. Dividing by Car A”s 1,500-kg mass, Car A is found to be moving 25 m/s after the collision.

Now check total kinetic energy after collision. Car A has 469 kJ of kinetic energy and Car B has 4 kJ of kinetic energy, for a total of 473 kJ. Before the collision the total kinetic energy was 675 kJ, as calculated in (b). Therefore, kinetic energy is lost and the collision is inelastic. The whole point here is that not all collisions in which cars bounce are elastic.

(d) We need a total of 675 kJ afterward in order to have an elastic collision. Conservation of momentum means that the total momentum after collision is 45,000 N·s to the left. Since Car A is the only moving car, it has all that momentum. Dividing by Car A”s 1,500-kg mass, we find Car A moving 30 m/s. Only Car A has kinetic energy, too; its kinetic energy is ½*mv* ^{2} = 675 kJ, so the collision is elastic.

** Rapid Review**

- In any system in which the only forces acting are between objects in that system, momentum is conserved. This effectively means that momentum is conserved in
*all*collisions. - The center of mass of a system of objects obeys Newton”s second law.
- The impulse-momentum theorem is Δ
*p*=*F*· Δ*t*. - The impulse-momentum theorem is always valid, but it is most useful when objects collide.
- The only time when momentum of a system is
*not*conserved is when a force is exerted by an object that”s not in the system.

__ ^{1}__ This is because his weight is 900 N, and on Earth 1 kg weighs 10 N.

__ ^{2}__ If friction between the track and the carts were significant, then sure, momentum wouldn”t be conserved—the track wasn”t considered part of the system, and it”s applying a force to the carts. But

*even with friction*, if you consider the moments just before and just after the collision, momentum will be essentially conserved. See the following discussion for a time when the momentum of a system is

*not*conserved.

__ ^{3}__ Or equivalently, this is not an effective approach to consider when a car collides with a concrete pillar, or a bird collides with the window of a building, etc.

__ ^{4}__ If you need to make up a mass of 2 kg for each ball and plug in numbers (including a plus and minus sign for the direction of velocity) to calculate the total momentum change for each ball, feel free. That”s not a bad approach if the words are confusing you.

^{5}**Mistake** : The relevant time in the impulse-momentum theorem is always the time of the collision, *not* the time it takes for a ball to fall through the air.

__ ^{6}__ Hopefully not

*you*, personally.

__ ^{7}__ If the final kinetic energy is

*greater*than the initial kinetic energy, something weird happened; like a coiled spring was released during the collision, or a firecracker exploded. You”ll most often see this sort of “superelastic” collision when the objects are initially at rest and then they are blown apart.

__ ^{8}__ If you used units of kg·m/s, that”s fine, too.

__ ^{9}__ … though its mass was likely redistributed around the windshield a bit.

__ ^{10}__ No, the total is not −75 kJ. Kinetic energy is a scalar, meaning it cannot have a direction; and kinetic energy cannot be negative. The total kinetic energy of a system is the sum of all the kinetic energies of the constituent objects, regardless of which way the objects are moving.