WBJEE · Physics · Rotational Motion
A uniform thin rod of length \(L\), mass \(m\) is lying on a smooth horizontal table. A horizontal impulse \(P\) is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is
- A \(\frac{\mathrm{P}^{2}}{\mathrm{M}}\)
- B \(\frac{7 \mathrm{P}^{2}}{8 \mathrm{M}}\)
- C \(\frac{13 \mathrm{P}^{2}}{2 \mathrm{M}}\)
- D \(\frac{2 P^{2}}{M}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 P^{2}}{M}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}=(\mathrm{mv}-0)\) angular impulse \(\mathrm{J}=\mathrm{P} \times \frac{\mathrm{L}}{2}=\mathrm{L}_{\mathrm{f}}-0\) \(\mathrm{LE}_{\mathrm{f}}=\frac{\mathrm{PL}}{2}\)…
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