Question

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

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

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}} =3 v_{3} $
$\sqrt{1 \times 39^{2}+1 \times 39^{2}} =3 v_{3} $
$39 \sqrt{2} =3 v_{3}$
$v_{3} =\frac{39 \sqrt{2}}{3}=13 \sqrt{2}\, ms ^{-1}$
$u=0,$
$ s=l, $
$g \,\sin \phi,$
$ u=?$
From $v^{2}-u^{2}=2 a s$
$v^{2}-0=2 g \sin \phi \times l$
For the rough portion $CO$
$u=v=\sqrt{2 g \sin \phi . l}$
From $ v=0, a =g(\sin \phi=\mu \cos \phi) $
$ s =l $
$ v^{2}-u^{2} =2 \,a s $
$0-2\, g l \sin \phi=2\, g(\sin \phi-\mu \cos \phi) l$
$-\sin \phi =\sin \phi-\mu \cos \phi $
$\mu \cos \phi =2 \sin \phi $
$ \mu =2 \tan \phi $