Question

A $5kg$ stationary bomb is exploded in three parts having mass $1:1:3$ respectively. Parts having same mass move in perpendicular directions with velocity $39m{s}^{-1}$, then the velocity of bigger part will be

• A. $10\sqrt{2}m{s}^{-1}$
• B. $\frac{10}{\sqrt{2}}m{s}^{-1}$
• C. $13\sqrt{2}m{s}^{-1}$
• D. $\frac{15}{\sqrt{2}}m{s}^{-1}$

Answer 1

Answer: (C)

As $m_1:m_2:m_3=1:1:3$
and momentum is conserved,
$\therefore \sqrt{P_1^2+P_2^2+P_3^2}=3v_3$
$\sqrt{1\times 39^2+1\times 39^2}=3v_3$
$39\sqrt{2}=3v_3$
$v_3=\frac{39\sqrt{2}}{3}=13\sqrt{2}m{s}^{-1}$
$u=0,$
$s=l,$
$g\sin \varphi ,$
$u=?$
From $v^2-u^2=2as$
$v^2-0=2g\sin \varphi \times l$
For the rough portion $CO$
$u=v=\sqrt{2g\sin \varphi .l}$
From $v=0,a=g(\sin \varphi =\mu \cos \varphi )$
$s=l$
$v^2-u^2=2as$
$0-2gl\sin \varphi =2g(\sin \varphi -\mu \cos \varphi )l$
$-\sin \varphi =\sin \varphi -\mu \cos \varphi$
$\mu \cos \varphi =2\sin \varphi$
$\mu =2\tan \varphi$