Q. A stationary body of mass $3\ kg$ explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocity of $3\widehat{i}\ ms^{- 1}$ and the other with a velocity of $4\widehat{j}\ ms^{- 1}$. If the explosion occurs in $10^{- 4}\ s$, the average force acting on the third piece in newton is
A) $(3\widehat{i} + 4\widehat{j}) \times 10^{- 4}$
B) $(3\widehat{i} - 4\widehat{j}) \times 10^{- 4}$
C) $(3\widehat{i} + 4\widehat{j}) \times 10^{4}$
D) $- (3\widehat{i} + 4\widehat{j}) \times 10^{4}$
Explanation
(d) According to law of conservation of momentum the third piece has momentum
$= 1 \times - \left( 3i + 4\widehat{j} \right)kg\ ms^{- 1}$
Impulse = Average force $\times$ time
$\Rightarrow Average\ force = \frac{Impulse}{time}$
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$= \frac{Change\ in\ momentum}{time}$
$= \frac{- \left( 3\widehat{i} + 4\widehat{j} \right)kg\ ms^{- 1}}{10^{- 4}s} = - \left( 3\widehat{i} + 4\widehat{j} \right) \times 10^{4}\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.