An isolated particle of mass m is moving is horizontal plane (x-y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses $\frac{m}{4}$ and $\frac{3 m}{4}$ . aaaaaan instant later, the smaller fragment is at y = +15 cm. the larger fragment at this instant is at :-

y = -5 cm

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y = +20 cm

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y = +5 cm

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y = -20 cm

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Solution

The correct option is A y = -5 cm
Since the fragments exploded, at that instant the momentum of the explode bodies by law of conservation of energy is $0$ .

In $Y -$ direction,

$m_{1} v_{1} + m_{2} v_{2} = 0$

$\frac{m}{4} v_{1} + \frac{3 m}{4} v_{2} = 0$

$v_{1} = - 3 v_{2}$

Since $y_{1} = 15 c m$ for $v_{1}$ velocity at time $t$

At same instant $t = \frac{y_{1}}{v_{1}} = \frac{y_{2}}{v_{2}}$

$⟹ \frac{15}{- 3 v_{2}} = y_{2} v_{2}$

Larger fragment distance at that instance $y_{2} = - 5 c m$

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