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A body of mass M at rest explodes into three pieces, in the ratio of masses 1:1:2. Two smaller pieces fly off perpendicular to each other with velocity of $30 m s^{- 1}$ and $40 m s^{- 1}$ respectively. The velocity of the third piece will be _____m/s

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answer is 25.

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Detailed Solution

$\bar{P_{1}} + \bar{P_{2}} + \bar{P_{3}} = 0,$

${\bar{P}}_{3} = - ({\bar{P}}_{1} + {\bar{P}}_{2})$

$As {\bar{P}}_{1} & {\bar{P}}_{2} are perpendicular$

$so P_{3} = \sqrt{P_{1}^{2} + P_{2}^{2}}$

$⇒ P_{3} = \frac{M}{4} \sqrt{30^{2} + 40^{2}} = 50 \times \frac{M}{4}$

Also, $P_{3} = \frac{2 M}{4} \times v$

Comparing, we get $v = 25 m / s$

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