Question

A softball of mass 0.200 kg that is moving with a speed of 8.2 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.9 m/s.
1. Calculate the velocity of the target ball after the collision.
2. Calculate the mass of the target ball.

Answer 1

Conservation of momentum leads to

.2kg*8.2m/s=.2kg*(-3.9)m/s + Mass*V

Solve for V in therms of all the other.
Then put that into this for V
conservation of energy.
1/2 .2 *8.2^2=1/2 .2*(3.9)^2 + 1/2 MV^2
then solve for mass, after that, go back and solve for V

Answer 2

Given:
M1 = 0.200kg, V1 = 8.2 m/s.
M2 = ?, V2 = 0.
V3 = -3.9 m/s = Velocity of M1 after the collision. +
V4 = ?, = Velocity of V2 after the collision.

Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.2*8.2 + M2*0 = 0.2*(-3.9) + M2*V4,
1.64 + 0 = -0.78 + M2*V4,
Eq1: M2*V4 = 2.42.

Conservation of KE Eq:
V4 = (V2(M1-M2) + 2M1*V1)/(M1+M2).
V4 = (0(M1-M2) + 3.28)/(0.2+M2),
V4 = 3.28/(0.2+M2).
In Eq1, replace V4 with 3.28/(0.2+M2) and solve for M2:
M2*(3.28/(0.2+M2) = 2.42.
Multiply both sides by (0.2+M2)
M2*3.28 = 2.42(0.2+M2),
M2*3.28 = 0.484 + M2*2.42,
M2*0.86 = 0.484,
M2 = 0.563kg.

V4 = 3.28/(0.2+0.563) = 4.30 m/s.

KE before and after the collision:
KEb = 0.5*0.2*8.2^2 = 6.72 J.
KEa = 0.5*0.2*(-3.9^2) + 0.5*0.563*4.3^2 = 6.73 J.