Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $\theta$ to its initial direction and has a speed $v/3$ . The second block's speed after the collision is:

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

$\frac{3v}{4}$

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

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

Step-by-Step Solution

For an elastic collision, the total kinetic energy of the system is conserved .

Let the speed of the second block after the collision be $v_2$ . Initial Kinetic Energy ( $K_i$ ) = Final Kinetic Energy ( $K_f$ ) $\frac{1}{2}Mv^2 = \frac{1}{2}M\left(\frac{v}{3}\right)^2 + \frac{1}{2}Mv_2^2$

Canceling $\frac{1}{2}M$ from both sides: $v^2 = \frac{v^2}{9} + v_2^2$ $v_2^2 = v^2 - \frac{v^2}{9}$ $v_2^2 = \frac{8v^2}{9}$ $v_2 = \frac{\sqrt{8}v}{3} = \frac{2\sqrt{2}v}{3}$

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