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

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Solution

Q.
On a frictionless surface, 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 [2015]

1 $\frac{3}{\sqrt{2}} v$

2 $\frac{\sqrt{3}}{2} v$

3 $\frac{2 \sqrt{2}}{3} v$

4 $\frac{3}{4} v$

Ans.
(3)

The situation is shown in the figure.

Let $v'$ be speed of second block after the collision.

As the collision is elastic, so kinetic energy is conserved.

According to conservation of kinetic energy,

$\frac{1}{2} M v^{2} + 0 = \frac{1}{2} M {(\frac{v}{3})}^{2} + \frac{1}{2} M v'^{2}$

$v^{2} = \frac{v^{2}}{9} + v'^{2} or v'^{2} = v^{2} - \frac{v^{2}}{9} = \frac{9 v^{2} - v^{2}}{9} = \frac{8}{9} v^{2}$

$v' = \sqrt{\frac{8}{9} v^{2}} = \frac{\sqrt{8}}{3} v = \frac{2 \sqrt{2}}{3} v$

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