Answer:
The speed of the receiver is 0.074 m/s.
Explanation:
Given that,
Mass of the receiver, m = 96 kg
Mass of the ball, m' = 0.42 m/s
Initially the receiver is at rest, u = 0
Initial speed of the ball, u' = 17 m/s
We need to find the speed of the receiver after the collision. It is a case of conservation of momentum. Let V is the speed of the receiver.
[tex]mu+m'u'=(m+m')V[/tex]
[tex]m'u'=(m+m')V[/tex]
[tex]V=\dfrac{m'u'}{m+m'}[/tex]
[tex]V=\dfrac{0.42\times 17}{96+0.42}[/tex]
V = 0.074 m/s
So, the speed of the receiver is 0.074 m/s. Hence, this is the required solution.
