Physics · Newton's Laws of Motion

JEE Advanced 2018 — Paper 2 — Question 6

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m=0.4 \mathrm{~kg}$ is at rest on this surface. An impulse of 1.0 Ns is applied to the block at time $t=0$ so that it starts moving along the x -axis with a velocity $\mathrm{v}(\mathrm{t})=\mathrm{v}_{0} \mathrm{e}^{-\mathrm{t} / \tau}$ , where $\mathrm{v}_{0}$ is a constant and $\tau=4 \mathrm{~s}$ . The displacement of the block, in meters, at $\mathrm{t}=\tau$ is $\_\_\_\_$ Take $\mathrm{e}^{-1}=0.37$ .

Answer: 6.3

Numerical answer — enter this value.

Step-by-step solution

Using impulse momentum theorem

$\mathrm{J}=\mathrm{mv}_{0}$

$\mathrm{v}_{0}=\frac{\mathrm{J}}{\mathrm{m}}=\frac{1}{0.4}=2.5 \mathrm{~m} / \mathrm{s}$

$\mathrm{v}=\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{t}}{\tau}}$

$\int d x=v_{0} \int_{0}^{\tau} e^{\frac{-t}{\tau}} d t$

$\Delta x=v_{0} \tau\left[1-e^{\frac{-t}{\tau}}\right]$

at $\mathrm{t}=\tau \mathrm{sec}$

$\Delta \mathrm{x}=6.30 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Physics
Chapter: Newton's Laws of Motion
Topic: Application of NLM and Impulse