A bullet of mass $10 g$ moving horizontally with a velocity of $400 m s^{- 1}$ strikes a wood block of mass $2 k g$ which is suspended by light inextensible string of length $5 m$ . As a result, the centre of gravity of the block 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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$120 m s^{- 1}$

$80 m s^{- 1}$

$100 m s^{- 1}$

$160 m s^{- 1}$

answer is C.

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Detailed Solution

We can apply work energy theorem for the bullet and block before strike and after the bullet is ejected out of the block

$From conservation of linear momentum \frac{10}{1000} \times 400 + 0 = 2 v_{1} + \frac{10}{1000} v_{2} From work energy theorem W_{gr} = ∆ K . E \Rightarrow 2 g \times y_{C . G} = \frac{1}{2} \times 2 \times {v_{1}}^{2}$

solving we get $v_{1} = \sqrt{2} m / \sec and v_{2} = 120 m / \sec$

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