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 $2.5 m / s$
By momentum conservation horizontal.
$\frac{2}{4} m u_{2} - \frac{m}{4} \times 4 = 0 u_{2} = 2 m s$
Vertical $\frac{m}{4} \times 3 - d \frac{2 m}{4} u_{1} = 0$
$u_{1} = \frac{3}{2} u = \sqrt{u_{1}^{2} + u_{2}^{2}} = \sqrt{4 + \frac{g}{4}} = \sqrt{\frac{25}{4}} u = 2.5$

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