JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Laws of Motion

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

m = 0.4 k g

is at rest on this surface. An impulse of

1.0 N s

is applied to the block at time

t = 0

so that it starts moving along the

x

-axis with a velocity

v (t) = v_{0} e^{- t / τ}

, where

v_{0}

is a constant and

τ = 4 s

. The displacement of the block, in metres, at

t = τ

is ____. Take

e^{- 1} = 0.37

Answer: 6.3

Solution

From the impulse-momentum theorem,

J = m v_{0}

, so

v_{0} = \frac{J}{m} = \frac{1.0}{0.4} = 2.5

m/s. The displacement is

x = \int_{0}^{τ} v_{0} e^{- t / τ} d t = v_{0} τ [1 - e^{- 1}] = 2.5 \times 4 \times (1 - 0.37) = 10 \times 0.63 = 6.30

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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