1. Momentum and Collisions

2. Momentum Momentum is Mass  Velocity p = mv
Momentum has dimensions of: mass  length / time
The SI units are kilogram-meters per second (kgm/s)
Note that momentum is a vector
It has direction as well as magnitude

3. Think-Pair-Share!
A 40-g mouse is moving in the same direction as a 4000-kg elephant. The elephant is walking at 2 m/s.
Is it possible for the mouse to have the same momentum as the elephant?
What would be the speed of the mouse?

4. Impulse-Momentum Theorem
Remember Newton’s second law:
F = ma
Rewrite a as v/t (average acceleration):
F = mv/t
F t = mv = mvf  mvi = p
F t = p is called the Impulse-Momentum Theorem
F t is called the impulse of the force F on an object

5. In words, the Impulse-Momentum Theorem is
impulse = change in momentum
Note that the force in F t is an average force
Example:
A 135-g baseball is pitched to a batter at 33 m/s and is in contact with the bat for s. If the baseball leaves the bat in the opposite direction at 40 m/s, what is the average force on the ball?
p = m(vf  vi) = (.135 kg)(-40m/s  33m/s)
= kgm/s
F = p/ t = (-9.86 kgm/s)/(.020 s) = -490 N

6. Conservation of Momentum
Consider two billiard balls colliding each with mass m:
The total momentum of the two balls before and after the collision is the same
Before: 𝑝 𝑔𝑟𝑒𝑒𝑛 =𝑚 𝑣 𝑖
𝑝 𝑏𝑙𝑢𝑒 =𝑚 0 =0
𝑝 𝑖 =𝑚 𝑣 𝑖 +0=𝑚 𝑣 𝑖
After: 𝑝 𝑔𝑟𝑒𝑒𝑛 =𝑚 0 =0
𝑝 𝑏𝑙𝑢𝑒 =𝑚 𝑣 𝑓
𝑝 𝑓 =0+𝑚 𝑣 𝑓 =𝑚 𝑣 𝑓 vi vf

7. Conservation of Momentum Equation
If two objects collide with different masses (m1 and m2) and different velocities (v1i and v2i) and momentum is always conserved
m1v1i + m2v2i = m1v1f + m2v2f
v2i
v1i m1 m2
v2f
v1f
𝒑 𝒊 = 𝒑 𝒇

8. v1f = v1i  (m2/m1)v2f = 0.50 m/s  [5.6/21.5](1.0 m/s) = 0.24 m/s
Example:
A big marble (21.5 g), traveling 0.50 m/s hits a stationary little marble (5.6 g). After the collision, the little marble has a velocity of 1.0 m/s. What is the final velocity of the big marble?
m1 = 21.5 g
m2 = 5.6 g
v1i = 0.50 m/s v1f = ?
v2i = 0 v2f = 1.0 m/s
m1v1i + 0 = m1v1f + m2v2f
v1f = v1i  (m2/m1)v2f
= 0.50 m/s  [5.6/21.5](1.0 m/s) = 0.24 m/s

9. Two barges full of salted toad guts have a collision
Two barges full of salted toad guts have a collision. The red barge has a mass of kg and is traveling Northwest at 0.25 m/s. The blue barge has a mass of kg and is traveling Southeast at 0.1 m/s. After the collision the blue barge has a velocity of 0.32 m/s to the Northwest. What is the final velocity of the red barge?
V=0.03 m/s

10. Is Conservation of Momentum Always True?
If we have a system of objects that is isolated, then momentum is conserved
What is an isolated system?
It has zero net external force acting on it
Balls hitting each other are internal forces v2 v1

11. Collisions We will study two types of collisions:
inelastic collisions
elastic collisions
An inelastic collision is when two objects collide and energy is lost
An elastic collision is when two objects bounce off each other and KE is conserved

12. Inelastic Collisions
When two objects collide and stick together they become one object
m1v1i + m2v2i = (m1 + m2)vf
Kinetic energy is always lost in inelastic collisions (turned into heat, sound, etc.)
V1i
V2i
before vf
after

13. Example:
A 95.0-kg fullback moving south with a speed of m/s has a perfectly inelastic collision with a kg opponent running north at 3.0 m/s.
Calculate the velocity of the players just after the tackle.
= [(95)(-5.0)+(90)(3.0)]/(95+90)
= -1.1 m/s
Calculate the decrease in total KE.
KEi = ½ m1v1i2 + ½ m2v2i2 = 1593 J
KEf = ½ (m1 + m2)vf2 = 112 J
KE = 112 J  1593 J = J = J
vf = 
m1v1i + m2v2i
m1 + m2

14. Elastic Collisions
Two objects collide and bounce off of each other in such a way that no KE is lost
m1v1i + m2v2i = m1v1f + m2v2f
½m1v1i2 + ½m2v2i2 = ½m1v1f2 + ½m2v2f2
V1,i
V2,i
before
v2,f
after
v1,f

15. Example:
A 0.14-kg baseball travelling at 21 m/s strikes a 0.40-kg stationary soccer ball. The baseball rebounds back the way it came at 9.0 m/s. Assume the collision is elastic.
What is the final velocity of the soccer ball?
Show that kinetic energy is conserved.
V2i = 0
V1i
before
V2f
V1f
after