A bullet of mass 10 g strikes a ballistic pendulum of mass 2 kg. The CM of the pendulum rises a vertical distance of 12 cm. Assuming that the bullet gets embedded in the pendulum calculate its initial speed

425 ms

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256 ms

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321 ms

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308 ms

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Solution

The correct option is D 308 ms
$\begin{matrix} I n t h i s p r o b l e m a l l t h e k i n e t i c e n e r g y o f t h e b u l l e t i s c o n v e r t e d t o p o t e n t i a l e n e r g y f o r t h e w h o l e s y s t e m t h e r e f o r e w e h a v e \frac{1}{2} (m_{b} + m_{d}) V^{2} = (m_{b} + m_{d}) g h U sin g t h e c o n s r v a t i o n o f m o m e n t u m m_{b} v = (m_{b} + m_{d}) V T h e r e f o r e w e h a v e V = \frac{m_{b} + m_{d}}{m_{b}} B y s u b s t i t u t i o n t o f i n d t h e s p e e d o f t h e b u l l e t v = \frac{m_{b} + m_{d}}{m_{b}} \sqrt{2 g h} = \frac{2.01}{0.01} \sqrt{2 (9.8) (0.12)} = 308.25 \sim 308 m / s . \end{matrix}$
Hence, the option $D$ is the correct answer.

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