Question

A ball of mass 0.310 kg that is moving with a speed of 5.7 m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.1 m/s . 1. Calculate the velocity of the target ball after the collision. 2. Calculate the mass of the target ball.

Answer 1

initial momentum
= 0.310 * 5.7 + 0.310 *0 = 1.767
final momentum
= 0.310 * (- 3.1) + m * v
assuming that mass of target ball is m kg
conservation of momentum
0.310 * (- 3.1) + m * v = 1.767
mv =2.728 ---(1)

initial KE = 0.5 * 0.310 * 5.7^2 + 0.5 *0.310 *0^2 =5.035
final KE = 0.5* 0.310* (- 3.1)^2 + 0.5 * m * v^2
collides head-on and elastically, conservation of KE
0.5* 0.310 * (- 3.1)^2 + 0.5 * m * v^2 = 5.035
m v^2 =13.05 ---(2)
(2) / (1)
v = 4.8 m/s answer
velocity of the target ball after the collision
sub into (1)
m = 0.57kg