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JEE Advanced 2024 · Paper 1, Physics question 4

A block of mass 5 kg5\,\mathrm{kg} moves along the xx-direction subject to the force F=(−20x+10) NF = (-20x + 10)\,\mathrm{N}, with the value of xx in metre. At time t=0 st = 0\,\mathrm{s}, it is at rest at position x=1 mx = 1\,\mathrm{m}. The position and momentum of the block at t=(π/4) st = (\pi/4)\,\mathrm{s} are

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Answer and worked solution

Official answer: C

F=−20(x−0.5)F = -20(x - 0.5): SHM about x=0.5 mx = 0.5\,\mathrm{m} with ω=20/5=2 rad/s\omega = \sqrt{20/5} = 2\,\mathrm{rad/s}.

Starting from rest at x=1x = 1: x=0.5+0.5cos⁡2tx = 0.5 + 0.5\cos 2t and v=−sin⁡2tv = -\sin 2t.

At t=π4t = \frac{\pi}{4}: x=0.5 mx = 0.5\,\mathrm{m} and v=−1 m/sv = -1\,\mathrm{m/s}, so p=mv=−5 kg m/sp = mv = -5\,\mathrm{kg\,m/s}.

Topics:Spring-mass systemsSimple harmonic motion
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