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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2018 · Paper 2 · Physics · Laws of Motion

A solid horizontal surface is covered with a thin layer of oil. A...

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4kgm = 0.4kg is at rest on this surface. An impulse of 1.0Ns1.0Ns is applied to the block at time t=0t = 0 so that it starts moving along the x-axis with a velocity v(t)=v0e−tτv\left(t\right) = {v}_{0}{e}^{- \dfrac{t}{τ}} , where v0{v}_{0} is a constant and τ=4τ = 4 . The displacement of the block, in meter, at t=τt = τ is __________. Take e−1=0.37{e}^{- 1} = 0.37A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4kg is at rest on

Correct answer

0

Step-by-step solution

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4kg is at rest on
v=v0e−tτv = {v}_{0}{e}^{- \dfrac{t}{τ}}
v0=Jm=2.5m/s{v}_{0} = \dfrac{J}{m} = 2.5m / s
v=v0e−tτv = {v}_{0}{e}^{- \dfrac{t}{τ}}
dxdt=v0e−tτ\dfrac{dx}{dt} = {v}_{0}{e}^{- \dfrac{t}{τ}}
∫0xdx=v0∫0τe−tτdt{∫}_{0}^{x}dx = {v}_{0}{∫}_{0}^{τ}{e}^{- \dfrac{t}{τ}}dt
∫e−xdx=e−x−1∫ {e}^{- x}dx = \dfrac{{e}^{- x}}{- 1}
x=v0[e−tτ−1τ]0τx = {v}_{0}{\left[\dfrac{{e}^{- \dfrac{t}{τ}}}{- \dfrac{1}{τ}}\right]}_{0}^{τ}
x=2.5(−4)(e−1−e0)x = 2.5\left(- 4\right)\left({e}^{- 1} - {e}^{0}\right)
x=25(−4)(0.37−1)x = 25\left(- 4\right)\left(0.37 - 1\right)
x=6.30x = 6.30
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