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:

1.5 m / s

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2.0 m / s

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2.5 m / s

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3.0 m / s

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Solution

The correct option is C $2.5 m / s$
Momentum of A $= \frac{M}{4} \times 3 = \frac{3 M}{4}$
Momentum of B $= \frac{M}{4} \times 4 = M$
Resultant momentum $p_{R} = \sqrt{{(\frac{3 M}{4})}^{2} + M^{2}} = \frac{5}{4} M$
Now C must have a momentum $\frac{5 M}{4}$ along OC
Mass of C: $M - 2 \times \frac{M}{4} = \frac{M}{2}$
$∴ \frac{M}{2} v^{'} = \frac{5 M}{4}$ or $v^{'} = 2.5 m / s$

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