1. Momentum Objectives: Momentum and Newton’s Second Law
Impulse of a Force
Collisions

2. Momentum Momentum = (Mass ) x (Velocity)
Magnitude
Direction
Newton’s Second Law Revisited:
Net Force =
(Time)
(Change in Momentum)
No Net Force
Net Force
Momentum changes due to the Force
Momentum Before = Momentum After

3. Conservation of Momentum
Q1. You fall into a deep pool and begin sinking. Can you grab yourself by the hair and pull yourself upward?
(a) Yes
(b) No
(c) I don’t know
Q2. You are performing an extravehicular activity (EVA) just outside the International Space Station (ISS) . You have some instruments in your pocket, but your safety rope has just broken. What can you do to begin moving toward the ISS?

4. Conservation of Momentum
Q3. A bullet with a mass 60.0 g is fired with an initial velocity of 575 m/s from a gun with a mass 4.50 kg. What is the speed of the recoil of the gun?

5. Conservation of Momentum
Q3. A bullet with a mass 60.0 g is fired with an initial velocity of 575 m/s from a gun with a mass 4.50 kg. What is the speed of the recoil of the gun?
Before collision:
(0.060 kg)(0 m/s) + (4.50 kg)(0 m/s) = 0 kg.m/s

6. Conservation of Momentum
Q3. A bullet with a mass 60.0 g is fired with an initial velocity of 575 m/s from a gun with a mass 4.50 kg. What is the speed of the recoil of the gun?
Before collision:
(0.060 kg)(0 m/s) + (4.50 kg)(0 m/s) = 0 kg.m/s
After collision:
575 m/s
???

7. Conservation of Momentum
Q3. A bullet with a mass 60.0 g is fired with an initial velocity of 575 m/s from a gun with a mass 4.50 kg. What is the speed of the recoil of the gun?
Before collision:
(0.060 kg)(0 m/s) + (4.50 kg)(0 m/s) = 0 kg.m/s
After collision:
575 m/s
???
(0.060)(575 m/s) + (4.50 kg) (velocity) =
= (34.5 kg.m/s) + (4.50 kg) (velocity)

8. Conservation of Momentum
Q3. A bullet with a mass 60.0 g is fired with an initial velocity of 575 m/s from a gun with a mass 4.50 kg. What is the speed of the recoil of the gun?
Before collision:
(0.060 kg)(0 m/s) + (4.50 kg)(0 m/s) = 0 kg.m/s
After collision:
575 m/s
???
(0.060)(575 m/s) + (4.50 kg) (velocity) =
= (34.5 kg.m/s) + (4.50 kg) (velocity)
0 = (4.50 kg) (velocity)
-34.5 = 4.50 (velocity)
v = 7.67 m/s

9. Impulse The force acting during a certain interval of time Impulse
(change in momentum) =
(Force) x
(Time Interval)
Q4. Which way hurts less? Jumping from a tree with your knees stiff or bent?
(a) stiff
(b) bent
(c) It doesn’t matter

10. Impulse The force acting during a certain interval of time Impulse
(change in momentum) =
(Force) x
(Time Interval)
Q5. Which way hurts less? Jumping from a tree with your knees stiff or bent?
(a) stiff
(b) bent
(c) It doesn’t matter
Q6. Can Superman really save Lois Lane by catching her just before she hits the ground?
(a) Yes
(b) No
(c) I do not know

11. Momentum and Impulse 20
P1. Find the momentum of a bullet of 20.0 g traveling at velocity of 185 m/s.
Ans.
P2. What force is required to stop a 1,250-kg car traveling at 95.0 km/h within 4.00 s?
Ans.
P3. A 125-g glider traveling to the right at 2.5 m/s collides head-on with a 220-g glider traveling to the left at 3.1 m/s. What is the total momentum of the system?
Ans.
P4. While standing on ice, you take your kg hat and throw it horizontally at 0.5 m/s. If your mass is 70 kg, what is your recoil velocity?
Ans.

12. Example: Collisions Momentum Before = Momentum After
Q7. A 200-g glider moves at 1.2 m/s to the right and collides with a 100-g stationary glider. In which direction will the 200-g glider go after the collisions?
(a) right
(b) left
(c) It stops

13. Example: Collisions Momentum Before = Momentum After
Q8. A 200-g glider moves at 1.2 m/s to the right and collides with a 100-g that is moving to the left at 0.8 m/s. After the collisions, the two gliders stick together. With what velocity are they moving?

14. Example: Collisions Momentum Before = Momentum After
Q8. A 200-g glider moves at 1.2 m/s to the right and collides with a 100-g that is moving to the left at 0.8 m/s. After the collisions, the two gliders stick together. With what velocity are they moving?
Before collision:
(0.200 kg)(1.2 m/s) + (0.100 kg)(-0.8) = 0.16 kg.m/s
After collision:
(0.200 kg kg)(velocity) = (0.3 kg) (velocity)
0.16 = 0.3 v
v = 0.53 m/s

15. Collisions 20
P1. A 125-g glider moves east at 1.2 m/s and collides with a 90-g stationary glider. After the collision, the 125-g glider continues to move east at 0.7 m/s. What is the velocity of the 90-g glider?
Ans.
P2. A 75-g glider moves east at 1.2 m/s and collides head-on with a 45-g glider moving west at 0.8 m/s. If both collide and stick together, what is their velocity after the collision?
Ans.