Q. A body of mass $M$ at rest explodes into three pieces, two of which of mass $M/4$ each are thrown off in perpendicular directions with velocities of $3\ m/s\ $ and $4\ m/s$ respectively. The third piece will be thrown off with a velocity of
A) $1.5\ m/s$
B) $2.0\ m/s$
C) $2.5\ m/s$
D) $3.0\ m/s$
Explanation
(c) Momentum of one piece $= \frac{M}{4} \times 3$
Momentum of the other piece $= \frac{M}{4} \times 4$
$\therefore$ Resultant momentum $= \sqrt{\frac{9M^{2}}{16} + M^{2}} = \frac{5M}{4}$
The third piece should also have the same momentum
Let its velocity be $v$, then
$\frac{5M}{4} = \frac{M}{2} \times v\ \Rightarrow v = \frac{5}{2} = 2.5\ m/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.