1. Momentum, Impulse, and Collisions
Physics 7C lecture 08
Momentum, Impulse, and Collisions
Tuesday October 22, 8:00 AM – 9:20 AM
Engineering Hall 1200

2. Goals for Chapter 8
To learn the meaning of the momentum of a particle and how an impulse causes it to change
To learn how to use the conservation of momentum
To learn how to solve problems involving collisions
To learn the definition of the center of mass of a system and what determines how it moves
To analyze situations, such as rocket propulsion, in which the mass of a moving body changes

3. Introduction
In many situations, such as a bullet hitting a carrot, we cannot use Newton’s second law to solve problems because we know very little about the complicated forces involved.
In this chapter, we shall introduce momentum and impulse, and the conservation of momentum, to solve such problems.

4. Momentum and Newton’s second law
The momentum of a particle is the product of its mass and its velocity:
Newton’s second law can be written in terms of momentum as

5. Momentum and Newton’s second law
The momentum p = mV is a vector, along the same direction of the velocity.
Note that kinetic energy K = ½ m v2 is a scalar.

6. Impulse and momentum
The impulse of a force is the product of the force and the time interval during which it acts.
On a graph of Fx versus time, the impulse is equal to the area under the curve, as shown in Figure 8.3 to the right.
Impulse-momentum theorem: The change in momentum of a particle during a time interval is equal to the impulse of the net force acting on the particle during that interval.

7. Impulse and momentum
Do we have something similar to the work-energy theorem?
Impulse-momentum theorem: The change in momentum of a particle during a time interval is equal to the impulse of the net force acting on the particle during that interval.
dP = F dt = dJ
proof: m dv/dt = F, thus d (mv) = F dt, or dP = dJ

8. Compare momentum and kinetic energy
Changes in momentum depend on the time over which the net force acts, but changes in kinetic energy depend on the distance over which the net force acts. (See Figure to the right.)
Think of examples where K remains constant while P changes.

9. A ball hits a wall
A ball rebounds from a wall.
Calculate the change of momentum and kinetic energy.
Find out the impulse and work done by the normal force from the wall.

10. A ball hits a wall
Calculate the change of momentum and kinetic energy.
P2-P1 = m v2x – m v1x = 0.4 (20) – 0.4 (-30) = 20 Ns
impulse J = P2 – P1=20 Ns during the time of impact

11. Calculate the change of momentum and kinetic energy.
A ball hits a wall
Calculate the change of momentum and kinetic energy.
K2-K1 = ½ m v2x2 – ½ m v1x2 = ½ 0.4 (20)2 – ½ 0.4 (-30)2 = -100 J
Work by the normal force is W = K2 – K1= – 100 J
What is the distance associated with this work?

12. Q8.1
A ball (mass 0.40 kg) is initially moving to the left at 30 m/s. After hitting the wall, the ball is moving to the right at 20 m/s. What is the impulse of the net force on the ball during its collision with the wall?
A. 20 kg • m/s to the right
B. 20 kg • m/s to the left
C. 4.0 kg • m/s to the right
D. 4.0 kg • m/s to the left
E. none of the above
Answer: A

13. A8.1
A ball (mass 0.40 kg) is initially moving to the left at 30 m/s. After hitting the wall, the ball is moving to the right at 20 m/s. What is the impulse of the net force on the ball during its collision with the wall?
A. 20 kg • m/s to the right
B. 20 kg • m/s to the left
C. 4.0 kg • m/s to the right
D. 4.0 kg • m/s to the left
E. none of the above

14. Kicking a soccer ball A kick changes the direction of a soccer ball.
What is the direction of the impulse?

15. An isolated system
The total momentum of a system of particles is the vector sum of the momenta of the individual particles.
No external forces act on the isolated system consisting of the two astronauts shown below, so the total momentum of this system is conserved.

16. Conservation of momentum
External forces (the normal force and gravity) act on the skaters shown in Figure at the right, but their vector sum is zero. Therefore the total momentum of the skaters is conserved.
Conservation of momentum: If the vector sum of the external forces on a system is zero, the total momentum of the system is constant.

17. Remember that momentum is a vector!
When applying conservation of momentum, remember that momentum is a vector quantity!
Use vector addition to add momenta, as shown in Figure 8.10 at the right.

18. Recoil of a rifle A rifle fires a bullet, causing the rifle to recoil.
What is the impulse from the bullet?

19. Objects colliding along a straight line
Two gliders collide on an air track.
Find out va2x.

20. Objects colliding along a straight line
mA vA + mB vB is a constant before/after collision.
Why is the total momentum converved?

21. Elastic collisions
In an elastic collision, the total kinetic energy of the system is the same after the collision as before.
Figure at the left illustrates an elastic collision between air track gliders.

22. Elastic collisions
What if mA= mB, and vB1=0? Find out vB2

23. Elastic collisions What if mA= mB, and vB1=0? Find out vB2
vB2 = vA1 and vA2 = 0
The two masses exchange their velocity!

24. Inelastic collisions
In an inelastic collision, the total kinetic energy after the collision is less than before the collision.
A collision in which the bodies stick together is called a completely inelastic collision (see Figure at the right).
In any collision in which the external forces can be neglected, the total momentum is conserved.

25. Some inelastic collisions
Cars are intended to have inelastic collisions so the car absorbs as much energy as possible.
A completely inelastic collision

26. Some inelastic collisions
A completely inelastic collision

27. A. the same total momentum and the same total kinetic energy.
Q8.4
Two objects with different masses collide and stick to each other. Compared to before the collision, the system of two objects after the collision has A B
A. the same total momentum and the same total kinetic energy.
B. the same total momentum but less total kinetic energy.
C. less total momentum but the same total kinetic energy.
D. less total momentum and less total kinetic energy.
E. not enough information given to decide
Answer: B

28. A8.4
Two objects with different masses collide and stick to each other. Compared to before the collision, the system of two objects after the collision has A B
A. the same total momentum and the same total kinetic energy.
B. the same total momentum but less total kinetic energy.
C. less total momentum but the same total kinetic energy.
D. less total momentum and less total kinetic energy.
E. not enough information given to decide

29. A. the same total momentum and the same total kinetic energy.
Q8.5
Two objects with different masses collide and bounce off each other. Compared to before the collision, the system of two objects after the collision has A B
A. the same total momentum and the same total kinetic energy.
B. the same total momentum but less total kinetic energy.
C. less total momentum but the same total kinetic energy.
D. less total momentum and less total kinetic energy.
E. not enough information given to decide
Answer: E

30. A8.5
Two objects with different masses collide and bounce off each other. Compared to before the collision, the system of two objects after the collision has A B
A. the same total momentum and the same total kinetic energy.
B. the same total momentum but less total kinetic energy.
C. less total momentum but the same total kinetic energy.
D. less total momentum and less total kinetic energy.
E. not enough information given to decide

31. The ballistic pendulum
Ballistic pendulums are used to measure bullet speeds.
How to deduce bullet speed v1 from height h?

32. Elastic collisions
The behavior of the colliding objects is greatly affected by their relative masses.

33. A. 1/9 the KE of block B. B. 1/3 the KE of block B.
Q8.7
Block A on the left has mass 1.00 kg. Block B on the right has mass 3.00 kg. The blocks are forced together, compressing the spring. Then the system is released from rest on a level, frictionless surface. After the blocks are released, the kinetic energy (KE) of block A is
Answer: C
A. 1/9 the KE of block B. B. 1/3 the KE of block B.
C. 3 times the KE of block B. D. 9 times the KE of block B.
E. the same as the KE of block B.

34. A8.7
Block A on the left has mass 1.00 kg. Block B on the right has mass 3.00 kg. The blocks are forced together, compressing the spring. Then the system is released from rest on a level, frictionless surface. After the blocks are released, the kinetic energy (KE) of block A is
A. 1/9 the KE of block B. B. 1/3 the KE of block B.
C. 3 times the KE of block B. D. 9 times the KE of block B.
E. the same as the KE of block B.

35. An elastic straight-line collision
find out final velocities.

36. An elastic straight-line collision

37. Neutron collisions in a nuclear reactor
find out final velocities.

38. Neutron collisions in a nuclear reactor