Question

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

• A. $10\sqrt2\, m/sec$
• B. $\frac{10}{\sqrt2}\,m/sec$
• C. $15\sqrt2\, m/sec$
• D. $\frac{15}{\sqrt2}\,m/sec$

Answer 1

Answer: (A)

$10\sqrt2\, m/sec$

Answer 2

Apply conservation of linear momentum.
Total momentum before explosion
= total momentum after explosion
$0=\frac{m}{5}v_1\widehat i+\frac{m}{5}v_2\widehat j+\frac{3m}{5}\overrightarrow{v_3};$
$\frac{3m}{5}\overrightarrow{v_3}=-\frac{m}{5}[v_1\widehat i+v_2\widehat j]$
$\overrightarrow{v_3}=\frac{-v_1}{3}\widehat i-\frac{v_2}{3}\widehat j$ $\because v_1=v_2=30 m/sec.$
$\overrightarrow{v_3}=-10\widehat i-10\widehat j; v_3=10\sqrt2 m/sec.$