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A bullet of mass $10 g$ moving with a speed of $20 m / s$ hits an ice block of mass $990 g$ kept on a frictionless floor and gets stuck in it. How much ice will melt if $50 %$ of the lost KE goes to ice ? (initial temperature of the ice block and bullet = $0 ° C$ )

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$0.001 g$

$0.002 g$

$0.003 g$

$0.004 g$

answer is C.

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Detailed Solution

$K E =$ $\frac{1}{2} (10 \times 10^{- 3}) {(20)}^{2} = 2 J$

Now $m L =$ $50 %$ of KE

$\Rightarrow m = \frac{1 J}{80 \times 4.2 J / g m} \Rightarrow m = 0.003 g m$

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