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

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3.0 m/s

1.5 m/s

2.0 m/s

2.5 m/s

answer is C.

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Detailed Solution

Momentum of one piece $= \frac{M}{4} \times 3$
Momentum of the other piece $= \frac{M}{4} \times 4$
Resultant momentum $= \sqrt{\frac{9 M^{2}}{16} + M^{2}} = \frac{5 M}{4}$
The third piece should also have the same momentum. Let its velocity be v, then
$\frac{5 M}{4} = \frac{M}{2} \times v or v = \frac{5}{2} = 2 .5 m / \sec$

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