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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

  1. A \(\frac{\mathrm{P}^{2}}{\mathrm{M}}\)
  2. B \(\frac{7 \mathrm{P}^{2}}{8 \mathrm{M}}\)
  3. C \(\frac{13 \mathrm{P}^{2}}{2 \mathrm{M}}\)
  4. D \(\frac{2 P^{2}}{M}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{2 P^{2}}{M}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{P}=(\mathrm{mv}-0)\) angular impulse \(\mathrm{J}=\mathrm{P} \times \frac{\mathrm{L}}{2}=\mathrm{L}_{\mathrm{f}}-0\) \(\mathrm{LE}_{\mathrm{f}}=\frac{\mathrm{PL}}{2}\)…