Question

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

Answer 1

Step 1: Given data

1. The collision is elastic.
2. The mass of the two blocks is M.
3. After the collision, the first block deviated at an angle $\theta$.
4. The speed of the first block before and after the collision is $v$ and $\frac{v}{3}$.
5. The initial velocity of the second block is zero.

Step 2: Diagram

Step 3: Finding the speed of the second block

Let, $v\text{'}$ be the speed of the second block.

As we know that the collision is elastic. And in an elastic collision, the kinetic energy is conserved after and before the collision.

So,

$\frac{1}{2}M{v}^2+0=\frac{1}{2}M{\left(\frac{v}{3}\right)}^2+\frac{1}{2}Mv{\text{'}}^2$ (since kinetic energy is $\frac{1}{2}\times mass\times velocit{y}^2$ ).

$$\begin{gathered} \frac{1}{2}M{v}^2+0=\frac{1}{2}M{\left(\frac{v}{3}\right)}^2+\frac{1}{2}Mv{\text{'}}^2 \\ or\frac{1}{2}M{v}^2=\frac{M}{2}\left(\frac{v^2}{9}+\frac{v{\text{'}}^2}{2}\right) \\ or{v}^2=\frac{v^2}{9}+v{\text{'}}^2 \\ orv{\text{'}}^2=v^2-\frac{v^2}{9}=\frac{8v^2}{9} \\ orv{\text{'}}^2=\frac{8}{9}{v}^2 \\ orv\text{'}=\left(\sqrt{\frac{8}{9}{v}^2}\right)=\frac{\sqrt{8}}{3}v \\ orv\text{'}=\frac{\sqrt{8}}{3}v \end{gathered}$$

Therefore, the speed of the second block is $\frac{\sqrt{8}}{3}v$.