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

Correct answer: D. 2 2 3 v In elastic collision energy of system remains same. so, ((KE)_ before collision =(KE)_ after collision ) Let speed of second body after collision is (v^ ). aligned & (1)/(2) Mv^2+0=(1)/(2) M( v 3 )^2+(1)/(2) M(v^ )^2 & v^ = 2 2 3 v aligned )

NEET 2015

Exam: NEET
Year: 2015
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

NEET 2015 · Physics · Center of Mass, Momentum and Collision

On a frictionless surfaces, a block of mass M moving at speed v...

On a frictionless surfaces, a block of mass

\text{M}

moving at speed

v

collides elastically with another block of same mass

\text{M}

which is initially at rest. After collision the first block moves at an angle

\theta

to its initial direction and has a speed

\dfrac{v}{3}

. The second block's speed after the collision is:

Options

A $\dfrac{3}{4}v$
B $\dfrac{3}{\sqrt{2}}v$
C $\dfrac{\sqrt{3}}{2}v$
D $\dfrac{2\sqrt{2}}{3}v$

Correct answer

D. $\dfrac{2\sqrt{2}}{3}v$

Step-by-step solution

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} \mathrm{Mv}^2+0=\frac{1}{2} \mathrm{M}\left(\frac{\mathrm{v}}{3}\right)^2+\frac{1}{2} \mathrm{M}\left(\mathrm{v}^{\prime}\right)^2 \\ & \Rightarrow \mathrm{v}^{\prime}=\frac{2 \sqrt{2}}{3} \mathrm{v} \end{aligned}

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