Question

A stationary body of mass $3kg$ explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocity of $3\hat{i}m{s}^{-1}$ and the other with a velocity of $4\hat{j}m{s}^{-1}$ . If the explosion occurs in ${10}^{-4}s,$ the average force acting on the third piece in newton is

• A. $(3\hat{i}+4\hat{j})\times {10}^{-4}$
• B. $(3\hat{i}-4\hat{j})\times {10}^{-4}$
• C. $(3\hat{i}+4\hat{j})\times {10}^4$
• D. $-(3\hat{i}+4\hat{j})\times {10}^4$
• E. $(4\hat{i}-3\hat{j})\times {10}^4$

Answer 1

Answer: (D)

By law of conservation of linear momentum
$m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+m_3{\overrightarrow{v}}_3=0$
Here: $m_1=m_2=m_3=1kg,$
${\overrightarrow{v}}_1=3\hat{i},{\overrightarrow{v}}_2=4\hat{j}$
$\therefore$ $3\hat{i}+4\hat{j}+{\overrightarrow{v}}_3=0$
The average force acting on the third piece is
$F=\frac{m{\overrightarrow{v}}_3}{t}$
$=\frac{1\times (-3\hat{i}+4\hat{j})}{{10}^{-4}}N$
$=-(3\hat{i}+4\hat{j})\times {10}^{4}N$