5 Steps to a 5: AP Physics C - Greg Jacobs 2019

Momentum
Review the knowledge you need to score high

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 = FΔt

Location of the center of mass:

Mx_cm = m₁x₁ + m₂x₂ + …

If an object is moving, it has momentum. The formal definition of momentum¹ 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 FΔt 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.

A man jumps off the roof of a building, 3.0 m above the ground. His mass is 70 kg. He estimates (while in free-fall) that if he lands stiff-legged, it will take him 3 ms (milliseconds) to come to rest. However, if he bends his knees upon impact, it will take him 100 ms to come to rest. Which way will he choose to land, and why?

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.

A satellite floating through space collides with a small UFO. Before the collision, the satellite was traveling at 10 m/s to the right, and the UFO was traveling at 5 m/s to the left. If the satellite’s mass is 70 kg, and the UFO’s mass is 50 kg, and assuming that the satellite and the UFO bounce off each other upon impact, what is the satellite’s final velocity if the UFO has a final velocity of 3 m/s to the right?

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 = m₁x₁ + m₂x₂ + …

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.

Maggie has decided to go ice-skating. While cruising along, she trips on a crack in the ice and goes sliding. She slides along the ice at a velocity of 2.5 m/s. In her path is a penguin. Unable to avoid the flightless bird, she collides with it. The penguin is initially at rest and has a mass of 20 kg, and Maggie’s mass is 50 kg. Upon hitting the penguin, Maggie is deflected 30° from her initial path, and the penguin is deflected 60° from Maggie’s initial path. What is Maggie’s velocity, and what is the penguin’s velocity, after the collision?

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 v₀.

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 v₁ = 4 m/s to the right and v₂ = 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 v₀ 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.

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