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← JEE Advanced 2018 · Paper 2
PhysicsNumerical+3Question 7 of 54

JEE Advanced 2018 · Paper 2, Physics question 7

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4m = 0.4 kg is at rest on this surface. An impulse of 1.0 N s is applied to the block at time t=0t = 0 so that it starts moving along the xx-axis with a velocity v(t)=v0e−t/τv(t) = v_0 e^{-t/\tau}, where v0v_0 is a constant and τ=4\tau = 4 s. The displacement of the block, in metres, at t=τt = \tau is __________. Take e−1=0.37e^{-1} = 0.37.

Answer and worked solution

Official answer: 6.29

Initial speed. The impulse gives mv0=1.0mv_0=1.0, so v0=10.4=2.5 m s−1v_0=\dfrac1{0.4}=2.5\ \mathrm{m\,s^{-1}}.

Displacement.
x(τ)=∫0τv0e−t/τdt=v0τ(1−e−1)=2.5×4×0.63=6.3 m.x(\tau)=\int_0^\tau v_0e^{-t/\tau}dt=v_0\tau\left(1-e^{-1}\right)=2.5\times4\times0.63=6.3\text{ m}.

Answer: 6.30