1. Momentum
Copyright Sautter 2003

2. F = m x v / t F = m x a a = v / t F x t = m x v
Newton’s Second Law of Motion
F = m x a
Acceleration = velocity / time
a = v / t
Combining the two equations
F = m x v / t
Rearranging the equation
F x t = m x v
Momentum
Impulse

3. Impulse & Momentum
As seen on the previous slide, momentum and impulse equations are derived from Newton’s Second Law of Motion (F = ma). Impulse is defined as force times time interval during which the force is applied and momentum is mass times the resulting velocity change.
The symbol p is often used to represent momentum, therefore p = mv.
The question maybe however, “why develop a momentum equation when it is merely a restatement of F = ma which we already know ?”
The answer is that no conservation principles can be applied to forces. There is no such thing as “conservation of force”. However, Conservation of Momentum is a fundamental law which can be applied to a vast array of physics problems. Therefore, manipulating Newton’s Second Law into the impulse – momentum format helps us to implement a basic principle of physics !

4. Conservation of Momentum
A collision is a momentum exchange.
In all collisions the total momentum before a
collision equals the total momentum after
the collision. Momentum is merely redistributed
among the colliding objects.
Momentum = Mass x Velocity
p = m x v
Σ Momentum before = Σ Momentum after
collision collision

5. Momentum & Kinetic Energy
Recall the following facts about kinetic energy:
Kinetic energy is the energy of motion. In order to possess kinetic energy an object must be moving.
As the speed (velocity) of an object increases its kinetic energy increases. The kinetic energy content of a body is also related to its mass. Most massive objects at the same speed contain most kinetic energy.
Since the object is in motion, the work content is called kinetic energy and therefore: K.E. = ½ m v2

6. Types of Collisions (Momentum Transfers)
Collisions occur in three basic forms:
(1) Elastic collisions – no kinetic energy is lost during the collision. The sum of the kinetic energies of the objects before collision equals the sum of the kinetic energies after collision.
It is important to realize that kinetic energy is conserved in these types of collisions. Total energy is conserved in all collisions. (The Law of Conservation of Energy requires it!)
The only collisions which are perfectly elastic are those between atoms and molecules. In the macroscopic world some collisions are close to perfectly elastic but when examined closely, they do lose some kinetic energy.

7. Types of Collisions (Momentum Transfers)
(2) Inelastic collisions – the objects stick together after collision and remain as a combined unit. Kinetic energy is not conserved !
(3) Partially elastic collisions – kinetic energy is not conserved but the colliding objects do not remain stuck together after collision. Most collisions are of this type. In these collision some of the energy is converted to heat energy or used to deform the colliding objects. That which remains is retained as kinetic energy.

8. Types of Collisions (Momentum Transfers
Example collisions:
Elastic collisions – pool balls colliding, a golf ball struck with a club, a hammer striking an anvil. Remember, these collisions are not perfectly elastic but they are close!
Inelastic collisions – an arrow shot at a pumpkin and remaining embedded, a bullet shot into a piece of wood and not fully penetrating, two railroad cars colliding and coupling together.
Partially elastic collisions - two cars colliding and then separating, a softball struck by a bat, a tennis ball hit with a racket.

9. Σ K.E. before collision = Σ K.E. after collision
Elastic Collisions
CLICK
HERE
M1U M2U2 = M1V M2V2
Σ K.E. before collision = Σ K.E. after collision
e = 1.0

10. Σ K.E. before collision = Σ K.E. after collision
Inelastic Collisions
CLICK
HERE
M1U1 + M2U2 = (M M2) V
Σ K.E. before collision = Σ K.E. after collision /
e = 0

11. Measuring Collision Elasticity
The elastic nature of a collision can be measured by comparing the K.E. content of the system before and after collision. The closer they are to being equal, the more elastic the collision.
Another way to measure elasticity is to use the Coefficient of Restitution. The symbol for this value is e. If e is equal to 1.0 the collision is perfectly elastic. If e is equal to zero the collision is inelastic. If e lies in between 0 and 1.0 the collision is partially elastic.
The closer e is to 1.0 the more elastic the collision, the closer e is to 0 the more inelastic.

12. Coefficient of Restitution
Ball 1 2
Velocity of ball 1
Before collision u1
Velocity of ball 2
Before collision u2
Ball 1 2
Point of
collision
Velocity of ball 1
after collision V1
Velocity of ball 2
after collision V2
e = v v
u u o1 2 1

13. Momentum – A Vector Quantity
Momentum is a vector quantity (direction counts). In one dimensional collisions motion lies exclusively along the x plane or the y plane.
One Dimensional Collisions - The direction of the momentum vectors are identified by the usual conventions of plus on horizontal plane (x axis) is to the right and minus is to the left. In the vertical plane (y axis) , plus is up and minus is down.
Two Dimensional Collisions – The momentum vectors can be resolved into x – y components and combined using vector addition methods to obtain the momentum sums before and after collision.
In two dimensional collisions the sums of the vertical momentum components must be equal before and after collision just as the sums of horizontal momentum components must be equal before and after collision.

14. Momentum - A vector Quantity
One Dimensional Collision
Σ p before collision = Σ pafter collision
Two Dimensional Collision
Σ px before collision = Σ px after collision
Σ py before collision = Σ py after collision

15. Momentum and Explosions
In an explosion, momentum is always conserved.
Before an explosion occurs, if the exploding object is at rest, the sum of all momentum is zero. P= mv and if v is zero then momentum (p) must be zero !
After the explosion occurs the sum of the momentum must again be zero. This means that the momentum carried out in one direction by a fragment of the exploded item must be balanced by the momentum of another piece carried off in the opposite direction.
Since momentum is a vector quantity, this must be true in all directions and in all plane surrounding the exploded object.

16. Conservation of Momentum in a Simple "Explosion"
Play movie
Σ p before explosion = Σ pafter collision
Click here
To continue
Since the carts are not in motion before the explosion, the
momentum of the system is zero. Therefore after the
explosion the momentum of the system must be zero.
The carts are moving in opposite directions after the explosion. The
momentum of each of the carts then must be equal but opposite
in order to give a sum of zero after the explosion.
The more massive cart has a lesser velocity while the smaller cart
has a greater velocity so that the momentum of each must be equal !

17. p = momentum before collision
Two Dimensional Collisions Vectors
p = momentum before collision
p’ = momentum after collision 1
p’y1
p’x1
px1 θ1 1 2 θ2
p’x2
p’y2
px = p cos θ
py = p sin θ
py1 before collision = 0
px2 and py2 are both
zero (ball 2 is at rest) 2

18. Momentum Problems A 2400 lb vehicle moving at mph is brought to a stop in 2 seconds. What is the average force acting on the vehicle ?
2400 lbs
Vo = 20 mph
Vi = 0 mph
F x t = m x v
W = mg, m = w/g, m = 2400 lbs/ 32ft/s2 = 75 slugs
Vo = 20 mph = (20 x 5280) / 3600 = 29.4 ft/s, Vf = 0
Δv = vi – vo = 29.4 – 0 = 29.4 ft / s
F x Δt = m x Δv
F (2 sec) = 75 slugs x 29.4 ft/s
F = 1100 lbs

19. An Inelastic Collision
Momentum Problems A 40 kg skater traveling at 20 m/s collides with a 60 kg skater moving in the same direction at 2 m/s. (a) what is their velocity if they hold on to each other? (b) how much kinetic energy is lost?
Vf = ?
-20 m/s
-2 m/s
40 kg
60 kg
M1U1 + M2U2 = (M M2) V
An Inelastic Collision
(a) Mass # 1 = 40 kg, velocity # 1 = -20 m/s (negative means to the left)
Mass # 2 = 60 kg, velocity # 2 = - 2 m/s (negative means to the left)
V = velocity of the combined bodies after collision
M1U1 + M2U2 = (M M2) V
(40 kg x –20 m/s) + (60 kg x – 2 m/s) = (40 kg + 60 kg) V
V = m/s
(b) K.E. = ½ mv2
K.E before collision = ½ (40 kg)(-20 m/s)2 + ½ (60 kg)(-2 m/s)2 = 8120 j
K.E. after collision = ½ (40kg + 60 kg) (-9.2 m/s)2 = 4232 j
ΔK.E. = 8120 j – 4232 j = 3888 j

20. An Inelastic Collision
Momentum Problems A 40 kg skater traveling at 20 m/s collides with a 60 kg skater moving in the opposite direction at 2 m/s. (a) what is their velocity if they hold on to each other? (b) how much kinetic energy is lost?
2 m/s
60 kg
-20 m/s
40 kg
M1U1 + M2U2 = (M M2) V
An Inelastic Collision
(a) Mass # 1 = 40 kg, velocity # 1 = -20 m/s (negative means to the left)
Mass # 2 = 60 kg, velocity # 2 = - 2 m/s (negative means to the left)
V = velocity of the combined bodies after collision
M1U1 + M2U2 = (M M2) V
(40 kg x –20 m/s) + (60 kg x 2 m/s) = (40 kg + 60 kg) V
V = m/s
(b) K.E. = ½ mv2
K.E before collision = ½ (40 kg)(-20 m/s)2 + ½ (60 kg)(-2 m/s)2 = 8120 j
K.E. after collision = ½ (40kg + 60 kg) (-6.8 m/s)2 = 2312 j
ΔK.E. = 8120 j – 2312 j = 5808 j

21. An Inelastic Collision
Momentum Problems A 1000 kg car travels north at 20 m/s and collides with a 1500 kg car moving west at 12 m/s. They stick together. What is the velocity of the wreckage? 1
20 m/s 2
12 m/s
An Inelastic Collision
M1U1 + M2U2 = (M M2) V
Resolve the momentum of each car into x and y components
P1x = m1v1 cos 00 = 0
P1y = m1v1sin 900 = 1000kg x 20 m/s x 1.0 = 20,000 kg m /s
P2y = m2v2 sin 1800 = 0
P2x = m2v2 cos 1800 = 1500kg x 12 m/s x 1.0 = 18,000 kg m /s
Adding the vectors, Pr = ((0 + 18,000)2 + (20, )2)1/2
Pr= 26,900, Pr = (m1 + m2) v,
26,900 kg m/s = (1000kg kg) v, v = 10.8 m/s
Angle = tan-1 (20,000 / 18,000) = 480 north of west

22. Partially Elastic Collision
Momentum Problems A 1.0 kg ball moving at 5 m/s collides head on with a 2.0 kg ball moving at 4 m/s in the opposite direction. The coefficient of restitution is 0.7. Find the velocity of each ball after collision.
1 kg
+ 5 m/s
e = 0.7
2 kg
- 4 m/s
e = v v
u u 1 2
Partially Elastic Collision
M1U M2U2 = M1V M2V2
m1 = 1 kg, m2 =2 kg, u1 = 5 m/s , u2 = - 4m/s
e = (v2 – v1) / (u1 – u2) , 0.7 = (v2 – v1) / (5 – (-4))
v2 – v1 = 6.3, v2 = v1
M1U M2U2 = M1V M2V2,
(1 kg x 5 m/s) + (2 kg x (-4 m/s) = (1 kg x v1) + (2 kg) x (6.3 + v1)
5 + (-8) = 1v v1 , v1 = (-15.6) /3 = m/s
v2 = v1, v2 = (-5.2) = m/s

23. Momentum Problems A 5 kg gun shoots a 15 gram bullet at 600 m/s
Momentum Problems A 5 kg gun shoots a 15 gram bullet at 600 m/s. What is the recoil velocity of the gun ?
+ 600 m/s
Since both the gun and the bullet are at rest before the bullet is fired, the total momentum of the system is zero. The total momentum after firing must also be zero.
The mass of the bullet is 15 g = kg
Mb ub + mg ug = mb vb + mg vg
(0.015 x 0 ) + (5 x 0) = (0.015 x 600) + 5 vg
5 vg =- 9, vg = - 9 / 5 = m/ s (negative means recoil)

24. Momentum Problems A ball is dropped from 3 meters and bounces up 1 meter. Find the coefficient of restitution.
Recall:
s = (vi2 – vo2) / 2 a
h = (vi2 – vo2) / 2 g
If vo = 0, solve for vi
Vi = (2gh)1/2 3m 1m
Down = -
Up = +
Object 1 is the ball, object 2 is the ground (which does not move !)
e = (v2 – v1) / (u1 – u2), u1 = (2 x 9.8 x 3)1/2 = -7.7 m/s , u2 = 0
v1 = (2 x 9.8 x1)1/2 = m/s, v2 = 0
E = (0 – 4.4) / ( – 0) = 0.57

25. Now it's time for you to try some problems on your own !
The problems are similar to the ones which have been solved
so look back and review the appropriate problem
if you get stuck !

26. A 60 kg skater pushes off a 50 kg skater at 2.0 m/s. What is the
Velocity of the 60 kg skater?
(A) 0.6 m/s (B) 1.7 m/s (C) 2.0 m/s (D)2.4 m/s
Click
here for
answers
A 1500 kg vehicle moving at 60 km/hr rear ends a 4000 kg vehicle moving at
35 km/hr. They remain attached. What is the velocity of the wreckage ?
(A) 9.1 km/hr (B) 42 km/hr (C) 48 km/hr (D) 53 km/hr
An 800 kg car going south at 40 km/hr strikes a 1200 kg truck going west
at 25 km/hr. What is the speed of the wreckage if the stick together ?
(A) 22 km/hr (B) 31 km/hr (C) 33 km/hr (D) 47 km/hr
Find the direction the wreckage moves in the previous problem.
(A) 200 W of S (B) 430 W of S (C) 470 W of S (D) 700 W of S
A 5 kg ball moving a 3 m/s collides with a 3 kg ball moving in the same
Direction at 2.4 m/s. Find the final speeds of both balls if the e value is 0.8.
(A) 2.3 and 2.8 m/s (B) –2.3 and 2.8 m/s (C) – 3 and –2.4 m/s (D) none

27. The End