Question

An explosion blows a stationary rock into three parts. Two parts of masses $1 \,kg$ and $2 \,kg$ moves at right angles to one another with velocities $12\, ms ^{-1}$ and $8\, ms ^{-1}$, respectively. If the velocity of third part is $4 \,ms ^{-1}$, the mass of the rock is

• A. 8 kg
• B. 5 kg
• C. 17 kg
• D. 3 kg

Answer 1

Answer: (A)

Given, $m_{1}=1 kg , m_{2}=2 \,kg$,
$v_{1}=12 ms ^{-1}, v_{2}=8 ms ^{-1}$ and $v_{3}=4 ms ^{-1}$
Since, in the explosion of stationary rock, the momentum is conserved,
so $ p _{i} = p _{f} $
$0 = p _{f}= p _{1}+ p _{2}+ p _{3} $
where, $p _{1} =m_{1} v _{1}, p _{2}=m_{2} v _{2} $ and $ p _{3}=m_{3} v _{3} $
$p _{3} =-\left( p _{1}+ p _{2}\right)$
$p _{3}=\sqrt{ p _{1}^{2}+ p _{2}^{2}+2 p _{1} \cdot p _{2} \cos \theta} $
$p _{3}=\sqrt{12^{2}+16^{2}+2 \times 12 \times 16 \cos 90^{\circ}}$
$P_3=20$
$m_{3} v_{3}=m_{3} \times 4=20 $
$\Rightarrow \,m_{3}=5\, kg$
Hence, the mass of the rock is,
$m=m_{1}+m_{2}+m_{3} $
$m=1+2+5=8\, kg$