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  4. Two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 m/s. After collision the two balls come to rest when the velocity of B is

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Q. Two solid rubber balls $A$ and $B$ having masses $200\, g$ and $400 \,g$ respectively are moving in opposite directions with velocity of A equal to $0.3\, m/s$. After collision the two balls come to rest when the velocity of $B$ is

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Solution:

According to the law of conservation of linear momentum
$m_{A} u_{A}+m_{B}u_{B}=m_{A}V_{A}+m_{B}V_{B}$
$\therefore (200\,g)(0.3 m/s)+(400\,g)(u_{b})$
$=(200\,g)(0)+(400\,g)(0)$
$\therefore u_{B}=-\frac{\left(200\,g\right)\left(0.3 m/ s\right)}{\left(400 \,g \right)}$
$=-0.15 m/s$