A boy of mass 60 kg is standing over a platform of mass 40 kg placed over a smooth horizontal surface. He throws a stone of mass 1 kg with velocity $u = 10 m / s$ at an angle of $45^{0}$ with respect to ground. The displacement of platform (with boy) on the horizontal surface, when the stone land on the ground is $(g = 10 m / s^{2})$

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50 cm

10 cm

5 cm

25 cm

answer is C.

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Detailed Solution

Range of stone $= \frac{u^{2} \sin 2 θ}{g}$

= $\frac{{(10)}^{2} \sin 90^{0}}{10} = 10 m$

Net force on system in horizontal direction is zero. Therefore, centre of mass will

Stationary in horizontal direction, Hence,

$(60 + 40) x = (1) (10)$

Here, x = displacement of boy + platform

$∴$ $x = \frac{10}{100} = 0.1 m = 10 c m$

alternatively :m u cos $θ$ = $(m_{1} + m_{2}) v$

x = $v \times \frac{2 u \sin θ}{g}$

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