1. Unit Six: Linear Momentum and Collisions
GE253 Physics
Unit Six: Linear Momentum and Collisions
John Elberfeld

2. Schedule Unit 1 – Measurements and Problem Solving Unit 2 – Kinematics
Unit 3 – Motion in Two Dimensions
Unit 4 – Force and Motion
Unit 5 – Work and Energy
Unit 6 – Linear Momentum and Collisions
Unit 7 – Solids and Fluids
Unit 8 – Temperature and Kinetic Theory
Unit 9 – Sound
Unit 10 – Reflection and Refraction of Light
Unit 11 – Final

3. Chapter 6 Objectives
Compute linear momentum and the components of momentum.
Relate impulse and momentum, and kinetic energy and momentum.
Explain the condition for the conservation of linear momentum and apply it to physical situations.
Describe the conditions on kinetic energy and momentum in elastic and inelastic collisions.
Explain the concept of the center of mass and compute its location for simple systems, and describe how the center of mass and center of gravity are related.
Apply the conservation of momentum in the explanation of jet propulsion and the operation of rockets.

4. Reading Assignment
Read and study College Physics, by Wilson and Buffa, Chapter 6, pages 179 to 210

5. Written Assignments
PREPARE FOR A UNIT EXAM II on all material covered in the course
Do all the assignments on the handouts as partial preparation
Review all your quizzes

6. Introduction We know F = m a, Newton’s Second Law
Now we will manipulate that equation to learn more about impulse and momentum
Masses with large momentums are hard to stop

7. Introduction
Newton’s First Law describes momentum by explaining that an object at rest remains at rest, and an object in motion continues in motion at a constant speed in a straight line unless acted upon by an outside force
The formula, p = mv, give momentum (p) a value

8. Linear Momentum
The momentum of an object is the product of its mass and velocity. It is a vector quantity because velocity is a vector quantity.
The unit of momentum is kilogram times meter divided by second :
p = m v (kg m/s)
Both velocity and momentum are vectors and both use signs to indicate directions

9. Practice
A 100-kg football player runs with a velocity of 4.0 m/s straight down the field. A 1.0-kg artillery shell leaves the barrel of a gun with a muzzle velocity of 500 m/s.
Which has a greater momentum (magnitude), the football player or the shell?

10. Calculations P = mv P = 100kg 4 m/s = 400 kg m/s (player) P = mv
P = 1 kg 500 m/s = 500 kg m/s (shell)
The artillery shell has more momentum, but not that much more

11. Conservation of Momentum
Momentum Before = Momentum After
The total momentum of a system of objects (or group under consideration) is constant - ALWAYS
For example, the momentum of two cars before they collide is equal to the momentum of the two cars are the collision

12. Total Momentum
The total momentum of a system (P) is the sum of the momentum of all its particles (p).
You can write this as:
P = P1 + P2 + P3 + …
Remember to use vector addition and not just the sum of the magnitudes

13. Vector Addition
Add horizontal and vertical components

14. When Angles are Involved
Use vector addition

15. Impulse and Momentum
You can express Newton’s second law in terms of momentum as:
F = m a = m (v-v0)/ Δt = m Δv / Δt
F Δt = m Δv
F Δt = Impulse
m Δv = change in momentum
Impulse = change in momentum
Remember to use vector addition

16. Momentum in a System
Part (a) shows a particle hitting a wall and bouncing straight back.
The change in momentum of the ball is twice the initial momentum and points away from the wall.

17. Force and Momentum
A projectile is a good example of changing momentum.
In the figure shown below, the momentum of the projectile changes constantly in both direction and magnitude.
Force of gravity is constantly changing the momentum

18. Impulse Thus impulse is defined as Fave Δt and: Fave Δt = m Δv = ΔP
Impulse = change in momentum
The unit of impulse is Newton times second, which is the same as that of momentum.
Impulse is used for collisions and other phenomena where a force is applied for a short period of time.

19. Design for High Impulse

20. Practice
A shuffleboard player pushes a 0.25-kg puck, which is initially at rest, so that a constant horizontal force of 6.0 N acts on it through a distance of 0.50 m. (Neglect friction
(a) What is the kinetic energy and the speed of the puck when the force is removed?
(b) How long does the force act?
(c) What is the change in momentum?
(d) What is the impulse?
(Notice this builds on previous chapters!)

21. Calculations Conservation of energy and velocity W = F d = mv2/2
W = 6.0 N .5 m = .25kg v2/2
KE = 3.0 Nm = 3.0 J
v2 = 3.0 Nm·2 / .25kg
v = 4.90 m/s

22. Find the Time F = ma 6.0N = .25kg a a = 24m/s2 v = v0 + a Δt
4.90 m/s = m/s2 Δt
Δt = .204 s

23. Momentum and Impulse v = 4.90 m/s Momentum = P = m v
m v = .25 kg 4.90 m/s = 1.23 kg m/s
Impulse = F Δt
F Δt = 6 N .204 s
Impulse = 1.23 Ns
Impulse = Change in Momentum

24. Practice
A golfer drives a 0.10-kg ball from an elevated tee, giving it an initial horizontal speed of 40 m/s (about 90 mi/h).
The club and the ball are in contact for 1.0 ms (millisecond).
What is the average force exerted by the club on the ball during this time?

25. Calculation FΔt = m Δv F 1 x 10-3s = .10 kg 40m/s
F = 4,000 N in the direction of the ball’s motion

26. Practice
The golfer in the previous example drives the ball with the same average force, but “follows through” on the swing so as to increase the contact time to 1.5 ms.
What effect will this change have on the initial horizontal speed of the drive?

27. Time Affects Results

28. Calculations FΔt = m Δv 4000N 1.5 x10-3s = .10 kg v v = 60 m/s
Applying the same force for a longer period of time creates a greater change in momentum
The ball has a higher initial velocity (60 m/s –not 40m/s) and travels farther

29. Practice
A tennis ball with a mass of .1 kg moving at 25 m/s is smashed back with a velocity of 35 m/s?
What is the change in velocity?
What is the impulse?

30. Sketch a diagram and label
Before
After
.1 kg
25 m/s
.1 kg
35 m/s

31. Solve Pay close attention to signs Set motion to the right as positive
m1 = .1 kg
v1i = +25 m/s
v1f = -35 m/s
m1 v1f - m1v1i = (.1kg)(-35m/s)-(.1kg)(25m/s)
Impulse = change in momentum =
-6.0 kg m/s (to the left)
Impulse = -6.0 N s

32. Find the force F Δt = m (vf – vi) Impulse = -6.0 N s
If the applied force happens over 0.1 seconds, what is the average force?
F Δt = Ns = F (0.1s)
F = -6.0 N s/0.1s = N
A force of 60.0 N to the left was applied for .1 s to make the tennis ball go back with a velocity of 35 m/s.

33. Conservation of Linear Momentum
Momentum BEFORE = Momentum AFTER
Using Impulse = Change in Momentum:
F Δt = m (vf – vi)
For every action, there is an equal and opposite reaction, so the force on one object in a collision is equal but opposite to the force on the other object, and times are equal
In a collision, the positive increase in momentum is balanced by a negative decrease in momentum, so the total momentum is not changed

34. Practice
Two masses, m1 = 1.0 kg and m2 = 2.0 kg, are held on either side of a lightly compressed spring.
They are joined together by a light string.
When you burn the string (negligible external force), the masses move apart on the frictionless surface, with m1 having a velocity of 1.8 m/s to the left.
What is the velocity of m2?
This apparatus illustrates how you can have a system

35. Calculations First, decided motion to the right is positive:
Since there are no outside force affecting material within the system, the total change in momentum = 0
Momentum before the collision = momentum after the collision
m01v01 + m02v02 = m1v1 + m2v2
0 + 0 = 1.0kg(-1.8m/s) + 2 kg v2
1.8 kg m/s = 2 kg v2
v2 = .9 m/s to the right because it is positive

36. Collisions
When you study the collision of two objects, you need to know only their masses and velocities before and/or after the collision.
You may be interested to know that in physics, three-body problems (three objects colliding with each other at once) are impossible to solve to exact precision

37. Elastic Collisions
In elastic collisions, total kinetic energy is conserved.
In other words, the total kinetic energy of all the particles before and after the collision remains the same.
Kf = Ki
Final KE = Initial KE
When a Super Ball hits a hard surface, it bounces off with nearly the same speed as before the collision, therefore (ideally) its kinetic energy is the same after the collision and the collision is elastic.

38. Inelastic Collision
In contrast, an inelastic collision is one in which some (or all) of the total kinetic energy of the particles is lost.
Kf < Ki
Since a soft rubber ball does not bounce up from the ground back to the same height it is dropped from, we know that some of its kinetic energy was lost in the collision, so its collision with the ground is inelastic.
The lost kinetic energy turns into heat

39. Collisions and Impulse
There are two types of collisions
Perfectly elastic collision
When kinetic energy is conserved in a collision, it is called a perfectly elastic collision.
Perfectly Inelastic collisions
If kinetic energy is not conserved, the collision is called inelastic.

40. Elastic Collisions
In an elastic collision between two particles, the conservation of momentum and kinetic energy is given by:
m1v012/2 + m2v022/2 = m1v12/2 + m2v22/2
m1v01 + m2v02 = m1v1 + m2v2
Kinetic energy before the collision = Kinetic energy after the collision
Momentum before the collision = momentum after the collision
When one of these masses is initially at rest, these equations become simpler.

41. Elastic Collisions
When one of the particles is stationary, the equations for conservation of kinetic energy and momentum become:
m1v012/2 = m1v12/2 + m2v22/2
m1v01 = m1v1 + m2v2
By manipulating these equations:
v1 = v01(m1-m2)/(m1+m2)
v2 = v01(2m1)/(m1+m2)

42. Practice
A 0.30-kg object with a speed of 2.0 m/s in the positive x-direction has a head-on elastic collision with a stationary 0.70-kg object located at x = 0 m.
What is the distance separating the objects 2.5 s after the collision?

43. Calculations m1v012/2 = m1v12/2 + m2v22/2 m1v01 = m1v1 + m2v2
These equations lead to:
v1 = v01(m1-m2)/(m1+m2)
v2 = v01(2m1)/(m1+m2)
v1 = 2m/s(.3kg -.7kg )/(.3kg +.7kg) = -.8m/s
v2 = 2m/s(2 x .3kg )/(.3kg +.7kg) = +1.2 m/s
Distance = (1.2m/s)2.5s – (-.8m/s) 2.5s
Distance = 5 m

44. Extreme Elastic Collisions
A small ball bounces off a large ball
A .5 kg ball moves at 8 m/s to the right and collides with a stationary ball with a mass of 9 kg
Obviously, the small ball is going to bounce back off the big ball
The big ball will move very slowly to the right
To conserve kinetic energy, since the bowling ball is barely moving, the small ball must be moving almost as fast away from the big ball as it was just before it hit the ball

45. Calculations m1v012/2 = m1v12/2 + m2v22/2 m1v01 = m1v1 + m2v2
These equations lead to:
v1 = v01(m1-m2)/(m1+m2)
v2 = v01(2m1)/(m1+m2)
v1 = 8m/s(.5kg - 9kg )/(.5kg + 9kg) =-7.15m/s
v2 = 8m/s(2x.5kg )/(. 5kg + 9kg) = +.84 m/s
Velocity of the big ball is small, and the velocity of the small ball is slightly less, but in the opposite direction (just like common sense!)

46. Another Extreme Elastic
A bowling ball (9kg) moves down the gutter (4m/s) and collides with a identical, stationary ball
What happens defies common sense
The moving ball stops completely, and the ball it hits moves off with the same velocity as the incoming ball in this ELASTIC collision

47. Calculations m1v012/2 = m1v12/2 + m2v22/2 m1v01 = m1v1 + m2v2
These equations lead to:
v1 = v01(m1-m2)/(m1+m2)
v2 = v01(2m1)/(m1+m2)
v1 = 4m/s(9kg - 9kg )/(9kg + 9kg) = 0 m/s
v2 = 4m/s(2x9kg )/(9kg + 9kg) = 4 m/s
The first ball stops, and the second moves off with the same velocity as the incoming ball in the ELASTIC collision.

48. Identifying Collisions
Usually collision between bowling balls, pool balls, superballs… are elastic and kinetic energy is conserved
Collisions between cars and trucks or velcro covered balls are inelastic and kinetic energy is NOT conserved
Momentum is ALWAYS conserved for all types of collisions

49. Inelastic Collisions
In inelastic collisions, the total kinetic energy after and before the collisions does not remain the same
Final momentum ALWAYS equals initial momentum
In the first diagram, two velcro-covered balls with equal and opposite momentum collide head-on; as a result of the collision, they stick together and stop.

50. Inelastic Collisions
In the second diagram, one velcro-covered ball is moving and the other is at rest.
As a result of the collision, they stick and move together at a speed less than that of the initially moving ball.

51. Inelastic Collisions
In the previous screens, the first collision is clearly inelastic because the total kinetic energy of the two balls after the collision is zero.
In the second collision, however, you need to calculate total kinetic energy of the two balls to confirm KE was lost.
Since there are no external forces acting on the two balls, momentum is conserved:
m1v01 + m2v02 = m1v1 + m2v2

52. Practice
In the example of inelastic collision on the previous screens, a 1.0-kg ball with a speed of 4.5 m/s strikes a 2.0-kg stationary ball and sticks to it.
(a) What is the speed of the balls after the collision?
(b) What percentage of the initial kinetic energy do the balls have after the collision?
(c) What is their total momentum after the collision?

53. Calculations m1v01 + m2v02 = m1v1 + m2v2
1kg 4.5m/s + 0 = (1kg + 2kg) v
v = 1.5 m/s
KB = m1v012/2 + m2v022/2
KA= m1v12/2 + m2v22/2
KB = 1kg (4.5m/s)2/2 = J
KA = (3kg)(1.5m/s)2/2 = J
KA/KB =3.375 J / J = 33%
P = m1v01 = 1kg 4.5m/s = 4.5 kg m/s – same before and after

54. Results
The end velocity of the joined objects is in the same direction as the object that had the greatest magnitude of momentum
Because Kinetic energy turned to heat, the collision was inelastic

55. Center of Mass
The center of mass of a solid object or a collection of particles is a point at which the entire mass of that solid object or collection of particles is considered to be concentrated.
The location of the center of mass for a system with n particles is defined as:

56. Practice
Three masses, 7.0 kg, 5.0 kg, and 4.0 kg, are located at positions (-4.0, 0), (3.0, 0), and ( 6.0, 0), respectively, in meters from the origin.
Where is the center of mass of this system? -4 3 6 ?

57. Calculations xcm = 7kg(-4m) +5kg3m+4kg6m 7kg + 5kg + 4kg
The center of mass is .688 m to the right of the origin. A single force at that point will hold up all the masses -4 3 6

58. Center of mass
There does not have to be any actual material at the location of the center of mass
Where is the center of mass of a doughnut? Is it inside the material part of the doughnut?
Where is the center of mass for the letter “L”

59. Practice Find the center of mass if 4kg is at .4m and 7kg is a .9m?
4kg .4m
7kg .9m 0m

60. Calculations Xcm = (m1X1+m2x2)/(m1+m2)
Xcm = (4kg .4m+7kg .9m)/(4kg+7kg)
Xcm = .718 m

61. Summary What is momentum? What is the impulse-momentum theorem?
Why is linear momentum conserved?
What is the difference between elastic and inelastic collisions?
What is the center of mass of an object?
How does the conservation of momentum explain jet propulsion?