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 velocity of $3\ \widehat{\dot{\mathbf{i}}}\ {ms}^{- 1}$ and the other with a velocity of $4\ \widehat{\dot{\mathbf{j}}}\ {ms}^{- 1}$. If the explosion occurs in $10^{- 4}\ s$, the force acting on the third piece in newtons is
A) $(3\ \widehat{\dot{\mathbf{i}}} + 4\ \widehat{\dot{\mathbf{j}}}) \times 10^{- 4}$
B) $(3\ \widehat{\dot{\mathbf{i}}} - 4\ \widehat{\dot{\mathbf{j}}}) \times 10^{- 4}$
C) $(3\ \widehat{\dot{\mathbf{i}}} + 4\ \widehat{\dot{\mathbf{j}}}) \times 10^{4}$
D) $- (3\ \widehat{\dot{\mathbf{i}}} + 4\ \widehat{\dot{\mathbf{j}}}) \times 10^{4}$
Explanation
(d) By law of conservation of linear momentum.
$m_{1}\ \mathbf{v}_{1} + m_{2}\ \mathbf{v}_{2} + m_{3}\mathbf{v}_{3} = 0$
$Here\ \ {\ m}_{1} = m_{2} = m_{3} = 1\ kg,$
$\mathbf{v}_{1} = 3\ \widehat{\dot{\mathbf{i}}},\mathbf{v}_{2} = 4\ \widehat{\dot{\mathbf{j}}}$
$\therefore\ \ 3\ \widehat{\dot{\mathbf{i}}} + 4\ \widehat{\dot{\mathbf{j}}} + \mathbf{u}_{3} = 0$
The average force acting on the third piece is
$F = \frac{m\ \mathbf{v}_{3}}{t}$
$= \frac{1 \times - \left( 3\ \widehat{\dot{\mathbf{i}}} + 4\ \widehat{\dot{\mathbf{j}}} \right)}{10^{4}}\ N$
$= - \left( 3\ \widehat{\dot{\mathbf{i}}} + 4\ \widehat{\dot{\mathbf{j}}} \right) \times 10^{4}\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.