Question

A $1kg$ stationary bomb is exploded in three parts having mass $1:1:3$ respectively. Part having same mass move in perpendicular direction with velocity $30m/s$, then the velocity of bigger part will be

• A. $10\sqrt{2}m/\sec$
• B. $\frac{10}{\sqrt{2}}m/\sec$
• C. $15\sqrt{2}m/\sec$
• D. $\frac{15}{\sqrt{2}}m/\sec$

Answer 1

Answer: (A)

Apply conservation of linear momentum.
Total momentum before explosion = total momentum after explosion
$0=\frac{m}{5}{v}_1\vec{i}+\frac{m}{5}{v}_2\vec{j}+\frac{3m}{5}\overrightarrow{v_3};$
$\frac{3m}{5}\overrightarrow{v_3}=-\frac{m}{5}[v_1\vec{i}+v_2\vec{j}]$
$\overrightarrow{v_3}=\frac{-v_1}{3}\vec{i}-\frac{v_2}{3}\hat{j}$
$\because v_1=v_2=30m/\sec .$
$\overrightarrow{v_3}=-10\vec{i}-10\vec{j};v_3$
$=10\sqrt{2}m/\sec .$