A bullet weighing 10 g and moving at 300 ms $^{- 1}$ strikes a 5 kg block of ice and drops dead. The ice block is sitting on frictionless level surface. The speed of the block, after the collision is :

60 ms

^{- 1}

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3 ms

^{- 1}

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6 ms

^{- 1}

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0.6 ms

^{- 1}

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Solution

The correct option is B 0.6 ms $^{- 1}$
Here, a bullet weighing 10 g and moving at 300 $m s^{- 1}$ strikes a 5 kg block of ice and drops dead hence
$m_{1} u_{1} + m_{2} u_{2} = m_{2} v$
$10 \times 10^{- 3} (300) + (5) (0) = 5 v$
$3000 \times 10^{- 3} = 5 v$
$v = \frac{3}{5} = 0.6 m s^{- 1}$

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A bullet weighing 10 g and moving at 300 ms−1 strikes a 5 kg block of ice and drops dead. The ice block is sitting on frictionless level surface. The speed of the block, after the collision is :

The correct option is B 0.6 ms−1Here, a bullet weighing 10 g and moving at 300 ms−1 strikes a 5 kg block of ice and drops dead hencem1u1+m2u2=m2v10×10−3(300)+(5)(0)=5v3000×10−3=5vv=35=0.6ms−1

A bullet weighing 10 g and moving at 300 ms $^{- 1}$ strikes a 5 kg block of ice and drops dead. The ice block is sitting on frictionless level surface. The speed of the block, after the collision is :

The correct option is B 0.6 ms $^{- 1}$
Here, a bullet weighing 10 g and moving at 300 $m s^{- 1}$ strikes a 5 kg block of ice and drops dead hence
$m_{1} u_{1} + m_{2} u_{2} = m_{2} v$
$10 \times 10^{- 3} (300) + (5) (0) = 5 v$
$3000 \times 10^{- 3} = 5 v$
$v = \frac{3}{5} = 0.6 m s^{- 1}$