Question

A softball of mass 0.220 kg that is moving with a speed of 6.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward with a speed of 4.8 m/s.

Calculate the mass of the target ball when the final velocity of the target ball is 1.7 m/s

Answer 1

Conservation of momentum tells you that
0.22*6.5 = -0.22*4.8 + m v
mv = 2.486 kg m/s (positive direction)

Conservation of kinetic energy tells you that
0.22[(6.5)^2 - (4.8)^2] = (1/2) m v^2
mv^2 = 8.452 kg m^2/s^2

m = (mv)^2/mv^2 = 0.714 kg

Answer 2

Given:
M1 = 0.220kg, V1 = 6.5 m/s.
M2 = ?, V2 = 0.
V3 = -4.8 m/s = velocity of M1 after the collision.
V4 = 1.7m/s = Velocity of M2 after the collision.

M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.22*6.5 + M2*0 = 0.22*(-4.8) + M2*1.7,
1.43 + 0 = -1.06 + M2*1.7,
M2*1.7 = 2.49,
M2 = 1.46kg.