A bullet of mass $10 g$ moving horizontally with a velocity of $500 m / s$ strikes a wooden block of mass $2 k g$ which is suspended by a light inextensible string of length $5 m .$ As a result, the centre of gravity of the block is found to rise a vertical distance of $10 c m .$ The speed of the bullet after it emerges out horizontally from the block will be

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$120 \frac{m}{s}$

$220 \frac{m}{s}$

$200 \frac{m}{s}$

$160 \frac{m}{s}$

answer is A.

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Detailed Solution

Applying momentum conservation

$\frac{10}{1000} \times 500 + 0 = 2 \times v_{1} + \frac{10}{1000} \times v_{2} \Rightarrow 5 = 2 v_{1} + 0.01 v_{2} . . . . . . . (1)$

Applying work energy theorem for block

$W = ∆ K E \Rightarrow 2 \times 10 \times 0.1 = \frac{1}{2} \times 2 \times v_{1}^{2} \Rightarrow v_{1} = \sqrt{2} = 1.4 \frac{m}{s}$

Putting the value of $v_{1}$ in equation (1)

$5 = 2 \times 1.4 + 0.01 v_{2} \Rightarrow v_{2} = 220 \frac{m}{s}$

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