A bullet of mass 10 g moving horizontally with a velocity of 400 ${ms}^{- 1}$ strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the center of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

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${160 ms}^{- 1}$

$120 {ms}^{- 1}$

$80 {ms}^{- 1}$

$100 {ms}^{- 1}$

answer is A.

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Detailed Solution

$\frac{10}{1000} \times 400 + 0 = 2 \times v_{1} + \frac{10}{1000} \times v_{2}$

$\Rightarrow 4 = 2 v_{1} + 0.01 v_{2}$ ---------------(1)

Applying work energy theorem for block
W = $∆$ KE
= 2 x 10 x 0.1 = $\frac{1}{2} \times 2 \times {v^{2}}_{1}$
$\Rightarrow v_{1} = \sqrt{2} = 1.4 m / s$
Putting the value of v₁, in equation (l),
4 = 2 x 1.4 + 0.01 $v_{2}$
$\Rightarrow$ $v_{2} =$ 120 m/s