Question

A ball of mass 0.220 kg that is moving with a speed of 6.3 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 3.6 m/s. (Assume the positive direction is forward.)
(a) Calculate the velocity of the target ball after the collision.
(b) Calculate the mass of the target ball.

Answer 1

Answer:

(a) Velocity of target is 2.7 m/s.

(b) Mass of target is 0.807 kg.

Explanation:

Mass of ball $m_{1}=0.22\;kg$

initial speed of the ball $v_{1i}=6.3\;m/s$

Final speed of the ball $v_{2f}=-3.6\;m/s$

Part (a) The second ball is initially at rest. So, by the conservation of momentum,

$p_{i}=p_{f}\\m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f}\\0.22\times6.3+m_{2}\times 0=0.22\times(-3.6)+m_2v_{2f}\\m_2v_{2f}=2.178\:\:\:\:\;\;\;\;\;\;\;\;\;\; ...(1)$

Now, the velocity of approach is equal to the velocity of separation,

$v_{1i}-v_{2i}=v_{2f}-v_{1f}\\6.3-0=v_{2f}-(-3.6)\\v_{2f}=2.7\;m/s$

Part (b): From equation (1),

$m_2v_{2f}=2.178\\m_2=\frac{2.178}{2.7}\\m_2=0.806\;kg$