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A bullet of mass $50 g$ is shot at the block of a ballistic pendulum whose mass is $2 k g$ . However, the block is too small, and the bullet passes through the block. If the initial speed of the bullet is $425 m / s$ and the block rises a vertical distance of $45 c m$ , how fast is the bullet going when it leaves the block ?

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$390 m / s$

$375 m / s$

$345 m / s$

$305 m / s$

answer is D.

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Detailed Solution

$\frac{1}{2} m v^{2} = m g h v = \sqrt{2 g h} = \sqrt{2 \times 10 \times \frac{45}{100}} = 3 m / s 2 \times 3 + \frac{50}{1000} v = \frac{50}{1000} \times 425 v = 20 (21.25 - 6) = 305 m / s$

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