1. Total momentum of system:
Consider systems consisting of more than a single particle and how these multiple particles interact with each other
“Many–body” problems  more complicated, so we need effective ways of describing “systems” of particles
Start with Newton’s 2nd Law:
Quantity is called the momentum of a particle:
For a system of particles: m1 m3
m1v1
m3v3
m2v2 m2
Total momentum of system:

2. Impulse Thus Newton’s 2nd Law can be written as:
This is (essentially) the original way Isaac Newton wrote it!
Assumes constant force producing constant acceleration
The shorter the time period over which a given momentum change occurs, the larger the external force
Product of the net force and the time interval over which it acts is called the impulse of the net force:
Assumes constant force
If net force is not constant, impulse is determined by 1 of 2 ways:
Area under F vs. t curve
Fave  Dt

3. Impulse – Momentum Theorem
So and thus
Same result holds when net force is not constant
Now consider a system consisting of the Sun and the Earth:
Sun and Earth exert equal magnitude (but opposite direction) gravitational forces on each other
No other external forces are acting on them
Apply Newton’s 2nd Law to each mass:
(Impulse – Momentum Theorem)
Sun (m1)
Earth (m2)

4. Conservation of Momentum
Thus
Where = total momentum of the system
This means that the total momentum of the system is conserved (or constant) if the net external force acting on the system is zero
Note that this also follows from Newton’s 3rd Law
Objects exchange momentum through mutual interactions (gravity, collisions, etc.)
Momentum conservation can hold even when mechanical energy isn’t conserved

5. CQ 1: Two 1-kg carts with spring bumpers undergo a collision on a frictionless track as shown in the before and after pictures below. The total momentum of the system is equal to:
0 kg-m/s before the collision and 0 kg-m/s after the collision.
–4 kg-m/s before the collision and 4 kg-m/s after the collision.
–8 kg-m/s before the collision and 8 kg-m/s after the collision.
8 kg-m/s before the collision and 0 kg-m/s after the collision.

6. CQ2: Interactive Conceptual Example: Basketball vs. Flower Pot Drop
How does the change in momentum of the basketball compare with the change in momentum of the flowerpot after each object hits the table?
The change in momentum of the basketball is the same as the change in momentum of the flowerpot.
The change in momentum of the basketball is twice the change in momentum of the flowerpot.
The change in momentum of the basketball is half of the change in momentum of the flowerpot.
The change in momentum of the flowerpot is twice the change in momentum of the basketball.
PHYSLET Exercise 8.5.1, Prentice Hall (2001)

7. Example Problem #6.24
A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2-kg physics textbook horizontally toward the north shore at a speed of 5.0 m/s. How long does it take him to reach the south shore?
Solution (details given in class):
62 s

8. CQ 3: Interactive Example Problem: Saving an Astronaut
How much time elapses from when the astronaut removes her oxygen tank until she arrives back at the ship?
5 s
10 s
15 s
20 s
60 s
(ActivPhysics Online Problem #6.6, Pearson/Addison Wesley)

9. CQ 4: A boy is sliding down a long icy hill on his sled
CQ 4: A boy is sliding down a long icy hill on his sled. In order to decrease his mass and increase his velocity, he drops his heavy winter coat and heavy boots from the sled while he is moving. Will his strategy work?
No, because he loses the potential energy of the objects that he leaves behind.
No, because although his kinetic energy increases, his momentum decreases.
Yes, because although his kinetic energy decreases, his momentum increases.
Yes, because although his momentum decreases, his kinetic energy decreases.

10. Collisions Inelastic collisions What is vf? Conservation of momentum:
Example: cars stick together after a collision
BEFORE collision:
v1i
You (m1)
Me (m2)
v2i = 0
AFTER collision (cars locked together): vf
Energy used to do work (destroy fenders)
What is vf?
Conservation of momentum:

11. Collisions Note that if m1 = m2 then vf = ½ v1i
Note that if m1 = m2 then vf = ½ v1i
In inelastic collisions, the kinetic energy of system after collision < kinetic energy of system before collision (since mechanical energy is typically expended to do work)
K.E. before collision = K1 = ½ m1v1i2
K.E. after collision = K2 = ½ (m1 + m2)vf2 = ½ (m1 + m2)[m1 / (m1 + m2)]v1i2
So K2 / K1 = m1 / (m1 + m2) < 1 so K2 < K1
Elastic collisions
Forces between colliding bodies are conservative
Kinetic energy is conserved (may be temporarily converted to elastic potential energy)
Momentum is conserved

12. Elastic Collisions
Head-on collision where one object is at rest before the collision:
Conservation of kinetic energy gives:
½ m1v1i2 = ½ m1v1f2 + ½ m2v2f2
Conservation of momentum gives:
m1v1i = m1v1f + m2v2f
Combining these 2 equations:
m2, v2i = 0
m1, v1f
m2, v2f
m1, v1i

13. Elastic Collisions Interesting cases: (1) m1 = m2 (2) m1 << m2
v1f = 0 and v2f = v1i
(2) m1 << m2
v1f  – v1i
v2f << v1i
Similar to result of ping-pong ball (m1) striking stationary bowling ball (m2)
(3) m1 >> m2
v1f  v1i
v2f  2v1i
Similar to result of bowling ball (m1) striking stationary ping-pong ball (m2)
If v2i ≠ 0, then in general: v1i – v2i = –(v1f – v2f)

14. CQ 5: A block of mass m1 slides across a frictionless surface with speed v1 and collides with a stationary block of mass m2. The blocks stick together after the collision and move away with speed v2. Which of the following statements is (are) true about the blocks?
I only
II only
I and II only
I, II and III

15. Example Problem #6.73 Solution (details given in class): 4.85 m/s
A tennis ball of mass 57.0 g is held just above a basketball of mass 590 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m, as shown at right.
Find the magnitude of the downward velocity with which the basketball reaches the ground.
Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?
Solution (details given in class):
4.85 m/s
8.41 m

16. 2-D Collisions Interactive
Can also have a “glancing” collision:
Now the final velocities (and hence the momenta) have horizontal and a vertical components
Use the component form of conservation of momentum for the system of both balls:
px,initial = px,final
py,initial = py,final
True for elastic or inelastic collisions q j
m1, v1f
m2, v2f
m2, v2i = 0 x
m1, v1i
2-D Collisions Interactive

17. Example Problem #6.49 BEFORE collision: AFTER collision:
A 2000-kg car moving east at 10.0 m/s collides with a 3000-kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 40.0° north of east and a speed of 5.22 m/s. Find the speed of the 3000-kg car before the collision.
y (N)
BEFORE collision:
AFTER collision:
y (N)
v1 = 10 m/s
vf = 5.22 m/s
vfy
vfx
x (E)
40°
m1 = 2000 kg v2
x (E)
m1 + m2
vfx = vf cos40° = 4.00 m/s
vfy = vf sin40° = 3.36 m/s
m2 = 3000 kg
Solution (details given in class):
5.59 m/s

18. CQ 6: Ball A moving at 12 m/s collides elastically with ball B as shown. If both balls have the same mass, what is the final velocity of ball A? (Note: sin60° = 0.87; cos60° = 0.5)
3 m/s
6 m/s
9 m/s
12 m/s

19. Example Problem #6.41
A 12.0-g bullet is fired horizontally into a 100-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet–block system compresses the spring by a maximum of 80.0 cm, what was the speed of the bullet at impact with the block?
Solution (details given in class):
273 m/s

20. Example Problem #6.56
A bullet of mass m and speed v passes completely through a pendulum bob of mass M as shown above. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the bob will barely swing through a complete vertical circle?
Solution (details given in class):

21. Rocket Propulsion Rocket propulsion can be understood in two ways:
1. Conservation of momentum: Spent fuel (gasses) have momentum directed away from rocket; rocket has momentum in opposite direction
2. Newton’s Third Law: Rocket exerts force on spent fuel gasses, expelling them outward; gasses in response exert force on rocket in opposite direction
Rocket needs nothing to “push against” in order to move, otherwise it would not work in the vacuum of space!