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A bullet of mass 10g 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 centre 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:
A
120 $ms^{-1}$
B
160 $ms^{-1}$
C
100 $ms^{-1}$
D
80 $ms^{-1}$
Explanation
Get block speed from its rise, then conserve momentum.
Detailed Solution
Momentum conservation: $\frac{10}{1000}\times400 + 0 = 2\times v_1 + \frac{10}{1000}\times v_2 \Rightarrow 4 = 2v_1 + 0.01v_2$ ...(1)
Work–energy theorem for the block: $2\times10\times0.1 = \frac{1}{2}\times2\times v_1^2 \Rightarrow v_1 = \sqrt{2} = 1.4$ m/s
Putting $v_1$ in (1): $4 = 2\times1.4 + 0.01v_2 \Rightarrow v_2 = 120$ m/s