1. CONSERVATION OF LINEAR MOMENTUM
Newton’s Third Law states what:
For every action there is an equal and opposite reaction.
If two objects collide, what is true of the forces they exert on one another
They are equal and opposite.

2. Consider two objects (1 and 2) that collide.
Newton’s Third Law dictates F12 = F21
So m 1 • v1 = m2 •  v2
t t
Since the time of impact is the same :
m 1 •v1 = m2 • v2

3. This leads us to the law of conservation of linear momentum
m1v1 + m2v2 = m1V1 + m2V2
where : m1 and m2 are the masses of the two objects
v1 and v2 are the initial velocities of the two objects
V1 and V2 are the final velocities of the two objects

4. Example: A 40 kg object is moving to the right at 4 m/s and collides with a 10 kg object moving left at 1 m/s. After the collision, the 40 kg object continues moving right at 2.3 m/s. What is the velocity of the 10 kg object after the collision?
Solution: First draw a set of two diagrams for before and after the collision

5. m1v1 + m2v2 = m1V1 +m2V2 BEFORE AFTER 40 kg 40 kg V1 = 2.3 m/s
m1 = 40 kg
m2 = 10 kg

6. Applying the law of conservation of linear momentum and substituting:
m1v1 + m2v2 = m1V1 + m2V2
(40)(4) + (10)(-1) = (40)(2.3) + (10)V2
so V2 = 5.8 m/s

7. BEFORE AFTER so. . . . . . m1v1 + m2v2 = m1V1 + m2V2 V2 = -13 m/s
A bullet of mass 65 grams is fired from a 4.0 kg gun. If the bullet leaves the gun at 800 m/s, what is the recoil velocity of the gun?
? m/s
800 m/s
BEFORE
AFTER
V1 = 800 m/s
V2 = ? m/s
v1 = 0 m/s
v2 = 0 m/s
m1 = .065 kg
m2 = 4 kg so
V2 = -13 m/s
m1v1 + m2v2 = m1V1 + m2V2
(.065)(0) + (4)(0) = (.065)(800) + (4)V2

8. MOMENTUM TRIX
If the objects stick together upon colliding (perfectly inelastic collision)
V1 = V2
• If both masses are the same, masses cancel