On a frictionless surfaces, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v 3 . The second block's speed after the collision is:

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Aicharya · Accepted Answer

In elastic collision energy of system remains same. so, $(\mathrm{KE})_{\text {before collision }}=(\mathrm{KE})_{\text {after collision }}$
Let speed of second body after collision is $\mathrm{v}^{\prime}$.
$\begin{aligned} & \frac{1}{2} Mv^2+0=\frac{1}{2} M\left(\frac{v}{3}\right)^2+\frac{1}{2} M\left(v^{\prime}\right)^2 \\ & \Rightarrow v^{\prime}=\frac{2 \sqrt{2}}{3} v \end{aligned}$

On a frictionless surfaces, a block of mass M moving at speed v collides elastically with another block of…

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