Work Energy and Power - Result Question 48

51. 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

$\frac{v}{3}$. The second block’s speed after the collision is :

[2015 RS]

(a) $\frac{3}{4} v$

(b) $\frac{3}{\sqrt{2}} v$

(c) $\frac{\sqrt{3}}{2} v$

(d) $\frac{2 \sqrt{2}}{3} v$

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Answer:

Correct Answer: 51. (d)

Solution:

  1. (d) Here, $M_1=M_2$ and $u_2=0$

$ u_1=V, \quad V_1=\frac{V}{3} ; \quad V_2=? $

From figure, along $x$-axis,

$M_1 u_1+M_2 u_2=M_1 V_1 \cos \theta+M_2 V_2 \cos \phi \ldots$ (i) Along $y$-axis

$0=M_1 V_1 \sin \theta-M_2 V_s \sin \phi$

By law of conservation of kinetic energy,

$\frac{1}{2} M_1 u_1^{2}+\frac{1}{2} M_2 u_2^{2}=\frac{1}{2} M_1 V_1^{2}+\frac{1}{2} M_2 V_2^{2}$

Putting $M_1=M_2$ and $u_2=0$ in equation (i), (ii) and (iii) we get,

$\theta+\phi=\frac{\pi}{2}=90^{\circ}$

and $u_1^{2}=V_1^{2}+V_2^{2}$

$V^{2}=(\frac{V}{3})^{2}+V_2^{2}[\because u_1=V.$ and $.V_1=\frac{V}{3}]$

or, $V^{2}-(\frac{V}{3})^{2}=V_2^{2}$

$V^{2}-\frac{V^{2}}{9}=V_2^{2}$

or $V_2^{2}=\frac{8}{9} V^{2} \Rightarrow V_2=\frac{2 \sqrt{2}}{3} V$