An object of mass 0.005kg moving with a velocity of 200 m/s gets embedded in a hanging wooden block of mass 0.5 kg. The velocity acquired by the block is

190 m/s

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19 m/s

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1.9 m/s

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0.19 m/s

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Solution

The correct option is C 1.9 m/s
Given : $m_{1} = 0.5$ kg $u_{1} = 0$ m/s $m_{2} = 0.005$ kg $u_{2} = 200$ m/s
Let the velocity acquired by the block after the collision be $V$ .
Using conservation of linear momentum before and after the collision : $P_{i} = P_{f}$
$∴$ $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$
OR $0.5 \times 0 + 0.005 \times 200 = (0.5 + 0.005) V$ $⟹ V \approx 1.9$ $m / s$

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An object of mass 0.005kg moving with a velocity of 200 m/s gets embedded in a hanging wooden block of mass 0.5 kg. The velocity acquired by the block is

The correct option is C 1.9 m/sGiven : m1=0.5 kg u1=0 m/s m2=0.005 kg u2=200 m/s Let the velocity acquired by the block after the collision be V.Using conservation of linear momentum before and after the collision : Pi=Pf∴ m1u1+m2u2=(m1+m2)VOR 0.5×0+0.005×200=(0.5+0.005)V ⟹V≈1.9 m/s