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Standard XII

Physics

Relative Motion: Rain Example

A firecracker...

Question

A firecracker is thrown with velocity of $30 m s^{- 1}$ in a direction which makes an angle $75^{o}$ withe the vertical axis. At some point on its trajectory, the firecracker split into two identical pieces in such a way that one piece falls 27 m far from the shooting point. Assuming that all trajectories are contained in the same plane, how far will the other piece fall from the shooting point?(Take $g = 10 m s^{- 2}$ and neglect air resistance)

63 m or 144 m

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28 m or 72 m

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72 m or 99 m

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64 m or 117 m

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Solution

The correct option is D 64 m or 117 m
$R = \frac{u^{2} s i n 2 θ}{g} = \frac{30^{2} s i n 150^{o}}{10} = 90 \times \frac{1}{2} = 45 m$
COM of firecracker at 45 m from projection point
$X_{C m} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$
$45 = \frac{\frac{m}{2} (27) + \frac{m}{2} r_{2}}{m}$
$90 = 27 + r_{2}$
$r_{2} = 90 - 27 = 64 m$
If $r_{2} = - 27 m$ then
$90 = - 27 + r_{2}$
$r_{2} = 117 m$ .

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The correct option is D 64 m or 117 mR=u2sin2θg=302sin150o10=90×12=45mCOM of firecracker at 45 m from projection pointXCm=m1r1+m2r2m1+m245=m2(27)+m2r2m90=27+r2r2=90−27=64mIf r2=−27m then90=−27+r2r2=117m.