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A ball $A$ of mass $m$ moving along positive $x$-direction with kinetic energy $K$ and momentum $p$ undergoes elastic head on collision with a stationary ball $B$ of mass $M$ after collision the ball $A$ moves along negative $x$-direction with kinetic energy $\frac{K}{9}$, final momentum of $B$ is
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The correct answer is:
$\frac{4 p}{3}$
Kinetic energy
$\begin{gathered}K=\frac{1}{2} m u_1^2 \\ u_1=\sqrt{\frac{2 K}{m}} \\ \frac{1}{2} m v_1^2=\frac{K}{9} \\ v_1^2=\frac{2 K}{9 m} \\ \text { Velocity, } v_1=\sqrt{\frac{2 K}{9 m}}\end{gathered}$
Kinetic energy $K=\frac{p^2}{2 m}$
From conservation of the momentum
$\begin{aligned} m u_1 & =-m v_1+p_B \\ m\left(u_1+v_1\right) & =p_B\end{aligned}$
Final momentum of $B$
$\begin{aligned} p_B & =m\left[\sqrt{\frac{2 K}{m}}+\sqrt{\frac{2 K}{9 m}}\right] \\ & =\sqrt{2 m K}+\sqrt{\frac{2 m K}{3}}=p+\frac{p}{3}=\frac{4 p}{3}\end{aligned}$
$\begin{gathered}K=\frac{1}{2} m u_1^2 \\ u_1=\sqrt{\frac{2 K}{m}} \\ \frac{1}{2} m v_1^2=\frac{K}{9} \\ v_1^2=\frac{2 K}{9 m} \\ \text { Velocity, } v_1=\sqrt{\frac{2 K}{9 m}}\end{gathered}$
Kinetic energy $K=\frac{p^2}{2 m}$
From conservation of the momentum
$\begin{aligned} m u_1 & =-m v_1+p_B \\ m\left(u_1+v_1\right) & =p_B\end{aligned}$
Final momentum of $B$
$\begin{aligned} p_B & =m\left[\sqrt{\frac{2 K}{m}}+\sqrt{\frac{2 K}{9 m}}\right] \\ & =\sqrt{2 m K}+\sqrt{\frac{2 m K}{3}}=p+\frac{p}{3}=\frac{4 p}{3}\end{aligned}$
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