A bullet of mass 0.1 kg moving horizontal with speed 400 ms^-1hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is ….. (given, g = 10m/s²)

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0.25

0.50

0.90

0.65

answer is C.

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Detailed Solution

By using conservation of momentum,
$\begin{array}{l} mv = (m + M) v^{'} \Rightarrow (0.1) 400 = 4 \times v^{'} \\ \Rightarrow 40 = v^{'} (4) \Rightarrow v^{'} = 10 m / \sec \end{array}$
After hitting, Bullet and blocks is combinedly moving on frictional surface. So by using work energy theorem,
$\begin{array}{l} w = Δk \\ \Rightarrow μ (m + M) gx = \frac{1}{2} (M + m) v^{' 2} \\ \Rightarrow μ (4) 10 \times 20 = \frac{1}{2} \times 4 \times (10)^{2} \\ \Rightarrow μ = \frac{1}{4} = 0.25 \end{array}$