WBJEE · Physics · Center of Mass Momentum and Collision
A bullet of mass \(4.2 \times 10^{-2} \mathrm{kg},\) moving at a speed of \(300 \mathrm{ms}^{-1}\), gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be
- A \(45 \mathrm{cal}\)
- B \(405 \mathrm{cal}\)
- C \(450 \mathrm{cal}\)
- D \(1701 \mathrm{cal}\)
Answer & Solution
Correct Answer
(B) \(405 \mathrm{cal}\)
Step-by-step Solution
Detailed explanation
Let mass of bullet = \(m\) Mass of block \(=\mathrm{M}\) Velocity of bullet \(=v=300 \mathrm{m} / \mathrm{s}\) Velocity of combined system \(M+m=V\) Here, fhom momentum conservation \(\frac{M+m}{m} V=v\)…
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