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QJEEAdv18Ph7700: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph7700 Solution: A solid horizontal surface is covered with a thin layer of oil. …

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4m=0.4 kgkg is at rest on this surface. An impulse of 1.01.0 NsNs 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/τ,v\left( t \right) = {v_0}{e^{ - t/\tau }}, where v0{v_0} is a constant and τ=4s.\tau = 4s. The displacement of the block, in metres, at t=τt = \tau is ______________ Take e−1=0.37.{e^{ - 1}} = 0.37.

Step-by-Step Verified Solution

The block was initially at rest and its velocity just after the application of impulse is v(0)=v0e−0/τ=v0v(0) = {v_0}{e^{ - 0/\tau }} = {v_0} . The applied impulse is equal to the change in linear momentum of the block i.e., J = mv0, which gives

v0 = J/m = 1/0.4 = 2.5 m/s.

JEE Advanced 2018 Paper 2 Offline Physics - Laws of Motion Question 15 English Explanation

The velocity of the particle is given as

v(t)=v0e−t/τv(t) = {v_0}{e^{ - t/\tau }} .

Integrate to get the displacement

x(t)=∫0tv0e−t/τdt=v0τ(1−e−t/τ)x(t) = \int_0^t {{v_0}{e^{ - t/\tau }}dt = {v_0}\tau (} 1 - {e^{ - t/\tau }}) .

Substitute t = τ\tau = 4 s and v0 = 2.5 m/s to get x( τ\tau ) = (2.5) (4) (1 −- e −- 1) = 6.3 m.

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