1. Conservation of Momentum in Collisions 2

2. Elastic or Inelastic?
In an inelastic collision, energy is lost and the deformation may be permanent.
An elastic collision loses no energy. The deform-ation on collision is fully restored.

3. Completely Inelastic Collisions
Collisions where two objects stick together and have a common velocity after impact.
Before
After

4. mAvAa + mBvBa = mAvAb + mBvBb
Example 1: A 60-kg football player stands on a frictionless lake of ice. He catches a 2-kg football and then moves at 40 cm/s. What was the initial velocity of the football? A
Given: mA= 2 kg; vABa= 0.4 m/s;
mB= 60 kg; vBb= 0m/s B
mAvAa + mBvBa = mAvAb + mBvBb
Momentum:
(mA + mB)vABa = mAvAb
Inelastic collision:
(2 kg + 60 kg)(0.4 m/s) = (2 kg)vAb
vAb= 12.4 m/s

5. pbefore= p after pbefore= mAvAb +mB vBb pafter= mAvAa +mB vBa
Example 2: A 1-kg ball moving to the right at 1 m/s strikes a 2-kg ball moving left at 3 m/s. The 1-kg moves back to the left with a speed of 3 m/s after the collision. If the two balls do not stick together, what is the velocity of the 2-kg ball after the collision?
pbefore= p after + A B
3 m/s
1 m/s
vBa
1 kg
2 kg
pbefore= mAvAb +mB vBb
pbefore= (1kg)(+1m/s) +(2kg) (-3m/s)
pafter= mAvAa +mB vBa
pafter= (1kg)(-3m/s) +(2kg) vBa
-5=-3+2vBa
vBa = - 1 m/s
B moves to the left after collision

6. Example 3. An 87-kg skater B collides with a 22-kg skater A initially at rest on ice. They move together after the collision at 2.4 m/s. Find the velocity of the skater B before the collision. A B
vBa = ?
vAb = 0
Common speed after colliding: 2.4 m/s.
22 kg
87 kg
vBa= vAa= 2.4 m/s
mAvAb + mBvBb=(mA+mB)vABa
(87 kg)vBb = (87 kg + 22 kg)(2.4 m/s)
(87 kg) vBb=262 kg m/s
vBb = 3.01 m/s

7. Example 4: A 50 g bullet strikes a 1-kg block, passes all the way through, then lodges into the 2 kg block. Afterward, the 1 kg block moves at 1 m/s and the 2 kg block moves at 2 m/s. What was the entrance velocity of the bullet?
1 kg
2 kg
vAb= ?
2 kg
1 kg
1 m/s
2 m/s
Continued…

8. mAvAb + mBvBb + mCvCb = mBvBa + (mA+mC) vACa
50 g
Find entrance velocity of bullet: mA= 0.05 kg; vAb= ?
2 kg
1 kg
1 m/s
2 m/s
Momentum Before =
Momentum After
mAvAb + mBvBb + mCvCb = mBvBa + (mA+mC) vACa
(0.05 kg)vAb =(1 kg)(1 m/s)+(2.05 kg)(2 m/s)
(0.05 kg) vAb =(5.1 kg m/s)
vAb= 102 m/s