1. Ch. 9 Linear Momentum

3. Momentum (p)
Momentum is a measure of how hard it is to stop or turn a moving object.

4. What characteristics of an object would make it hard to stop or turn?

5. Calculating Momentum For one particle p = mv
Note that momentum is a vector with the same direction as the velocity!
For a system of multiple particles
P = Spi add up the vectors
The unit of momentum is…
kg-m/s or Ns

6. Example p = mv p = (0.14 kg) (35 m/s) p = 4.9 kg·m/s
A baseball of mass 0.14 kg is moving at 35 m/s. Find its momentum.
p = mv
p = (0.14 kg) (35 m/s)
p = 4.9 kg·m/s

7. p = mv 4.9 kg · m/s = (7.26 kg) v v = 0.67 m/s Momentum
What is the velocity of a 7.26 kg bowling ball that has the same momentum as the baseball? (4.9 kg · m/s)
p = mv
4.9 kg · m/s = (7.26 kg) v
v = 0.67 m/s

8. Change in momentum (DP)
Like any change, change in momentum is calculated by looking at final and initial momentums.
Dp = pf – pi
Dp: change in momentum
pf: final momentum
pi: initial momentum

9. Impulse F = ma F = m Δ v/ Δ t F Δ t = m Δv F Δ t = m (vf - vo)
Let’s review Newton’s Second Law: F = ma
F = ma
F = m Δ v/ Δ t
F Δ t = m Δv
F Δ t = m (vf - vo)
F Δ t = Δ p = pf – pi
The Change in Momentum is called IMPULSE
J = Ft = Δ p = m Δ v
Units: N-s or kg m/s (same as momentum)

11. Area tells you Impulse

12. Example
A kg baseball is pitched at 38 m/s, and is hit straight back at 38 m/s. What Impulse did the bat deliver to the ball?
J = m Δ v
J = (0.144 kg) (vf - vo)
J = (0.144 kg) (-38 m/s - 38 m/s)
J = (0.144 kg) (-76 m/s)
J = kg · m/s

13. Example
If the bat and ball were in contact for 8 ms, what was the average force exerted on the ball by the bat?
J = Ft
-10.9 kg · m/s = F (0.008 s)
F = N

14. Example
Find the average acceleration of the ball during its contact with the bat.
ΣF = ma
-1360 N = (0.144 kg) a
a = N / kg
a = m/s2

15. This force acts on a 1. 2 kg object moving at 120. 0 m/s
This force acts on a 1.2 kg object moving at m/s. The direction of the force is aligned with the velocity. What is the new velocity of the object?

17. Law of Conservation of Momentum
If the resultant external force on a system is zero, then the vector sum of the momentums of the objects will remain constant.
SPbefore = SPafter

18. Sample problem: A 75-kg man sits in the back of a 120-kg canoe that is at rest in a still pond. If the man begins to move forward in the canoe at 0.50 m/s relative to the shore, what happens to the canoe?

19. Conservation of Momentum
For any closed system. (No external net force.)
Initial system momentum = final system momentum
System momentum is found by adding the momentum of each part of the system.
For two objects A and B,
pA1 + pB1 = pA2 + pB2

20. Conservation of Momentum An Example
An Astronaut at rest (mass = 84 kg) fires a thruster on his jetpack, expelling 35 g of hot gas at 875 m/s. How fast does the Astronaut move?
pA + pG = pA’ + pG’
0 + 0 = mAvA’ + mGvG’
0 = (84 kg) vA’ + (0.035 kg) (875 m/s)
0 = (84 kg) vA’ kg · m/s
vA’ = m/s 20

21. Another Example
A 10,000 kg truck moves East at 20 m/s. A 1,500 kg car moves West at 50 m/s. After a head-on collision, they stick together, and move as a single object. What is the velocity of this mess?
pt + pc = ptc
mtvt + mcvc = (mt + mc) vtc
(10,000 kg) (20 m/s) + (1,500 kg) (-50 m/s) = (11,500 kg) vtc
125,000 kg · m/s = (11,500 kg) vtc
vtc = 10.9 m/s – to the East 21

22. External versus internal forces
External forces: forces coming from outside the system of particles whose momentum is being considered.
External forces change the momentum of the system.

23. Internal forces cannot change momentum of the system.
Internal forces: forces arising from interaction of particles within a system.
Internal forces cannot change momentum of the system.

24. An external force in golf
The club head exerts an external impulsive force on the ball and changes its momentum.
The acceleration of the ball is greater because its mass is smaller.

26. Explosions
When an object separates suddenly, as in an explosion, all forces are internal.
Momentum is therefore conserved in an explosion.
There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion.

27. Recoil Guns and cannons “recoil” when fired.
This means the gun or cannon must move backward as it propels the projectile forward.
The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.

28. Sample problem: Suppose a 5
Sample problem: Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?

29. Sample Problem: An exploding object breaks into three fragments. A 2
Sample Problem: An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity?

31. Ballistic Pendulum
One way to find the ‘muzzle’ velocity of a bullet involves shooting into a block of wood hanging on a string, and measuring how high the block swings.
Let’s use conservation of momentum and energy to do this. 31

32. Ballistic Pendulum
One way to find the ‘muzzle’ velocity of a bullet involves shooting into a block of wood hanging on a string, and measuring how high the block swings.
Let’s use conservation of momentum and energy to do this. 32

33. Ballistic Pendulum
Suppose a bullet (5g) is fired into a 2.5kg block. As a result, it swings with a height change of 4.7cm.
Find the initial velocity of the bullet. 33

34. Ballistic Pendulum Two parts—Collision and Swinging Pendulum.
Use Conservation of Momentum for the collision part. We don’t know the initial velocity of the bullet or the final velocity of the bullet-block after impact. 34

35. Ballistic Pendulum Two parts—Collision and Swinging Pendulum.
Use Conservation of Energy for the pendulum part. – This will allow us to find the velocity just after impact, which we will use for the conservation of momentum part. 35

36. Collisions
When two moving objects make contact with each other, they undergo a collision.
Conservation of momentum is used to analyze all collisions.
Newton’s Third Law is also useful. It tells us that the force exerted by body A on body B in a collision is equal and opposite to the force exerted on body B by body A.

37. Collisions
During a collision, external forces are ignored. The time frame of the collision is very short. The forces are impulsive forces (high force, short duration).

38. Elastic Collisions
Momentum is conserved when there is no external Net Force.
Momentum involves Mass and Velocity.
Kinetic Energy also involves Mass and Velocity.
A collision in which Kinetic Energy is also conserved is called an Elastic Collision.

39. Inelastic Collisions
Momentum is conserved when there is no external Net Force.
So momentum is conserved in every collision.
Most collisions are not elastic (meaning that kinetic energy is not conserved).
When two objects collide and stick together, this is sometimes called a “sticky collision.” Another name for this is a “perfectly inelastic” collision.
Explosions are the reverse of perfectly inelastic collisions in which kinetic energy is gained!

40. Elastic Collisions Conservation of Momentum:
m1v1 + m2v2 = m1v1’ + m2v2’
Conservation of Kinetic Energy:
½ m1v12 + ½ m2v22 = ½ m1v1’ 2 + ½ m2v2’ 2
Or simply: m1v12 + m2v22 = m1v1’ 2 + m2v2’ 2 40

41. Combining Equations m1(v1 – v1’) = m2(v2’ – v2) eq#1
m1v1 + m2v2 = m1v1’ + m2v2’
m1v1 – m1v1’ = m2v2’ – m2v2
m1(v1 – v1’) = m2(v2’ – v2) eq#1
m1v12 + m2v22 = m1v1’ 2 + m2v2’ 2
m1v12 – m1v1’ 2 = + m2v2’ 2 – m2v22
m1(v12 – v1’2) = m2(v2’ 2 – v22)
m1(v1 – v1’)(v1 + v1’) = m2(v2’ – v2)(v2’ + v2) eq#2 41

42. v1 + v1’ = v2 + v2’ Combining Equations
m1(v1 – v1’) = m2(v2’ – v2) eq#1
m1(v1 – v1’)(v1 + v1’) = m2(v2’ – v2)(v2’ + v2) eq#2
Now, divide Eq#2 by Eq#1:
v1 + v1’ = v2’ + v2
Or, as I like to write it:
v1 + v1’ = v2 + v2’ 42

43. Elastic Collision Problems
Use the two equations:
m1v1 + m2v2 = m1v1’ + m2v2’
v1 + v1’ = v2 + v2’ 43

44. Example
Block A (6kg) is moving to the right at 7 m/s. Block B (3kg) is moving to the left at 1.5 m/s. They collide elastically. After they collide, find the velocity of both blocks. 44

45. m1v1 + m2v2 = m1v1’ + m2v2’ 6 ∙ 7 + 3 ∙ -1.5 = 6 ∙ v1’ + 3 ∙ v2’
Block A (6kg) is moving to the right at 7 m/s. Block B (3kg) is moving to the left at 1.5 m/s. They collide elastically. After they collide, find the velocity of both blocks.
m1v1 + m2v2 = m1v1’ + m2v2’
6 ∙ ∙ -1.5 = 6 ∙ v1’ + 3 ∙ v2’
37.5 = 6 v1’ + 3 v2’
v1 + v1’ = v2 + v2’
7 + v1’ = v2’
v1’ = v2’ 45

46. 37.5 = 6 (-8.5 + v2’) + 3 v2’ 37.5 = -51 + 6 v2’ + 3 v2’ 88.5 = 9 v2’
Block A (6kg) is moving to the right at 7 m/s. Block B (3kg) is moving to the left at 1.5 m/s. They collide elastically. After they collide, find the velocity of both blocks.
37.5 = 6 ( v2’) + 3 v2’
37.5 = v2’ + 3 v2’
88.5 = 9 v2’
9.83 m/s = v2’
(positive = to the right)
v1’ = v2’
v1’ =
v1’ = 1.33 m/s
(positive = to the right) 46

47. Sample Problem: An 80-kg roller skating grandma collides inelastically with a 40-kg kid. What is their velocity after the collision?
How much kinetic energy is lost?

48. Sample Problem A fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner?

49. Conservation of Momentum
Sample Problem: Suppose three equally strong, equally massive astronauts decide to play a game as follows: The first astronaut throws the second astronaut towards the third astronaut and the game begins. Describe the motion of the astronauts as the game proceeds. Assume each toss results from the same-sized "push." How long will the game last?

50. 2D-Collisions Momentum in the x-direction is conserved.
SPx (before) = SPx (after)
Momentum in the y-direction is conserved.
SPy (before) = SPy (after)
Treat x and y coordinates independently.
Ignore x when calculating y
Ignore y when calculating x
Let’s look at a simulation: