Advanced Placement

PHYSICS B Linear Momentum & Impulse

Student

2013-2014

Linear Momentum & Impulse

What I Absolutely Have to Know to Survive the AP* Exam

Momentum (p) is a vector. It has the same direction as the velocity of the object. Momentum can be thought of as inertia in motion. It is the product of mass times velocity. To have momentum an object must be moving. Impulse (J) equals the change in momentum of an object and is a vector that has the same direction as the net Force. It is derived from Newton’s second law. For a collision, the area under a force versus time graph yields the impulse applied during the collision. There are two types of collisions, elastic and inelastic. In both types of collisions momentum is conserved. In elastic collisions, kinetic energy is conserved as well.

Key Formulas and Relationships

p = mv Momentum Units: kg ⋅ m / s Impulse J = FΔt Units: N ⋅ s Note: kg ⋅ m / s and N ⋅ s are equivalent units Impulse – momentum theorem J = Δp

FΔt = mv f − mv i Conservation of linear momentum p a + p b = p 'a + p 'b

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Linear Momentum & Impulse

Important Concepts

• •

Momentum (p) is a vector quantity. It has the same direction as the velocity. Momentum can be thought of as inertia in motion. To have momentum an object must be moving. Impulse (J) equals the change in momentum of an object and is a vector that has the same direction as the net Force. It is derived from Newton’s second law. F = ma

F = m( v f − v i )

Δt FΔt = m( v f − v i ) FΔt = mv f − mv i FΔt = Δp J = FΔt is used on the AP exam for impulse. The impulse is the applied force multiplied by the time over which it acts. Impulse is equal to the change in the momentum of a system. Hence, if we apply a large force for a short time we can generate a momentum change of the same magnitude as having a small force for a long time.

F

t

•

=

F

t

The area under a Force vs. Time curve = the change in momentum of the object

Force (N)

Time (s) •

The impulse—momentum theorem says that Impulse (J) is the product of the average force acting on an object and the time interval during which the force acts. This impulse produces a change in an object’s momentum J = Ft = Δp and can be used to derive the law of conservation of linear momentum. ®

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Linear Momentum & Impulse

Derivation: Imagine two objects traveling in opposite directions that collide. According to Newton’s third law they exert equal and opposite forces upon each other F1 = −F2 Since the forces act upon each other for the same time, F1t = −F2t Since Ft = Δp we can write Δp1 = −Δp2

p1' − p1 = −(p'2 − p2 ) therefore

m1 v1' − m1 v1 = −m2 v '2 + m2 v 2 this gives us the law of conservation of momentum

m1 v1 + m2 v 2 = m1 v1' + m2 v '2 •

•

When no external forces act on a system, the total momentum remains the same. o Example 1: Two objects collide. The momentum of object one before the collision plus the momentum of object two before the collision equals the momentum of objects one and two after the collision. m1v1 + m2v2 = m1 v1' + m2 v2' o Example 2: A gun is fired and recoils. The sum of the momentum of the gun and bullet before it is fired (which is 0) equals the sum of the momentum of the gun and bullet after it is fired (also 0). The gun kicks and has a negative velocity and therefore a negative momentum. The bullet has a positive velocity and momentum, and when added together, the resulting momentum is zero. 0 + 0 = − p a + pb = 0 When objects collide, there are two types of collisions that can occur-- elastic and inelastic. o Elastic: In this type both kinetic energy and momentum are conserved. If an equation is classified as elastic, both the conservation of linear momentum and conservation of KE equations can be used p a + p b = p 'a + p 'b

KEa + KEb = KEa' + KEb' When two objects collide and bounce off each other, there are two possible energy interactions. Kinetic energy can be conserved or not conserved. If Kinetic energy is conserved, then we have an elastic collision. Momentum will always be conserved. o Inelastic: In these collisions only momentum is conserved When two objects collide and bounce off each other, some of the kinetic energy is ®

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Linear Momentum & Impulse

transformed into other forms of energy. Real world collisions are inelastic. On the AP exam, you may be asked whether kinetic energy is conserved and, if it isn’t, then what happened to that energy or how much kinetic energy is transferred elsewhere. When two objects collide and they stick together after they collide, then this is called a perfectly inelastic collision. The key thing to remember here is that after the collision both objects stick together and have the same velocity. Only momentum is conserved in this type of collision. A third example of an inelastic collision on the AP Exam is the two-body explosion. The idea of an explosion is that you have two bodies that are at rest, i.e. they have no momentum. After the explosion takes place, the bodies end up moving away from each other. Because momentum is conserved, their final momentum, when added together, must still equal zero (the momentum before the explosion). Remember momentum is a vector.

•

Momentum is conserved in two (and three) dimensions as well. Whenever we analyze a collision in two dimensions the momentum has to be split up into components. The sum of the momenta of all components in the x and y directions before a collision must equal the sum of the momenta after the collision. ∑ px = ∑ px' and ∑ py = ∑ p'y

o The diagram below shows the momentum vector of ball A colliding with ball B which is at rest. Ball A has momentum in the x direction but none in the y while ball B has no momentum at all. Ball A

Ball B

After the collision: Ball A

Ball B

The components of the x and y momentum vectors after the collision must equal the components before the collision. ®

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Linear Momentum & Impulse

= In the x direction

=

+

In the y direction since the momentum before the collision was zero, the y components of the momentum after the collision must be equal to zero. Since they act in opposite directions, their magnitudes must be equal to each other. =

On the AP Exam, collisions in two dimensions are frequent. Momentum still has to be conserved, but the problems can become fairly complex. Note: keep track of what’s going on by breaking things into x and y components: momentum is a vector, not a scalar.

®

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1992 AP Physics B 2. A 30-kilogram child moving at 4.0 meters per second jumps onto a 50-kilogram sled that is initially at rest on a long, frictionless, horizontal sheet of ice. (a) Determine the speed of the child-sled system after the child jumps onto the sled. (b) Determine the kinetic energy of the child-sled system after the child jumps onto the sled.

After coasting at constant speed for a short time, the child jumps off the sled in such a way that she is at rest with respect to the ice. (c) Determine the speed of the sled after the child jumps off it. (d) Determine the kinetic energy of the child-sled system when the child is at rest on the ice. (e) Compare the kinetic energies that were determined in parts (b) and (d). If the energy is greater in (d) than it is in (b), where did the increase come from? If the energy is less in (d) than it is in (b), where did the energy go?  

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1996 AP Physics B

1. (15 points) Two identical objects A and B of mass M move on a one-dimensional, horizontal air track. Object B initially moves to the right with speed vo. Object A initially moves to the right with speed 3vo, so that it collides with object B. Friction is negligible. Express your answers to the following in terms of M and vo. (a) Determine the total momentum of the system of the two objects. (b) A student predicts that the collision will be totally inelastic (the objects stick together on collision). Assuming this is true, determine the following for the two objects immediately after the collision. i.

The speed

ii.

The direction of motion (left or right)

When the experiment is performed, the student is surprised to observe that the objects separate after the collision and that object B subsequently moves to the right with a speed 2.5vo . (c) Determine the following for object A immediately after the collision. i. The speed ii. The direction of motion (left or right) (d) Determine the kinetic energy dissipated in the actual experiment.  

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2001 AP® PHYSICS B FREE-RESPONSE QUESTIONS

2. (15 points) An incident ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of negligible friction shown above. It makes a head-on collision with a target ball B of mass 0.50 kg at rest at the edge of the table. As a result of the collision, the incident ball rebounds, sliding backwards at 0.70 m/s immediately after the collision. (a) Calculate the speed of the 0.50 kg target ball immediately after the collision. The tabletop is 1.20 m above a level, horizontal floor. The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision. (b) Calculate the horizontal displacement d.

In another experiment on the same table, the target ball B is replaced by target ball C of mass 0.10 kg. The incident ball A again slides at 1.4 m/s, as shown above left, but this time makes a glancing collision with the target ball C that is at rest at the edge of the table. The target ball C strikes the floor at point P, which is at a horizontal displacement of 0.15 m from the point of the collision, and at a horizontal angle of 30° from the +x-axis, as shown above right. (c) Calculate the speed u of the target ball C immediately after the collision. (d) Calculate the y-component of incident ball A’s momentum immediately after the collision.

Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

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2002 AP® PHYSICS B FREE-RESPONSE QUESTIONS

2. (15 points) A 3.0 kg object subject to a restoring force F is undergoing simple harmonic motion with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown above. This particular object has a total energy E of 0.4 J. (a) What is the object’s potential energy when its displacement is +4 cm from its equilibrium position? (b) What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning. (c) Determine the object’s kinetic energy when its displacement is –7 cm. (d) What is the object’s speed at x = 0 ?

(e) Suppose the object undergoes this motion because it is the bob of a simple pendulum as shown above. If the object breaks loose from the string at the instant the pendulum reaches its lowest point and hits the ground at point P shown, what is the horizontal distance d that it travels?

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

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