A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m =0.4 k

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4-1.Newton's Laws of Motion

hard

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m =0.4 kg$ 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 $v ( t )= v _0^{ e ^{-1 / \tau}}$, where $v_0$ is a constant and $\tau=4 s$. The displacement of the block, in metres, at $t=\tau$ is. . . . . . . Take $e ^{-1}=0.37$

A

$6.2$

B

$6.3$

C

$6.4$

D

$6.5$

(IIT-2018)

Solution

$v = v _0 e ^{-t / \tau}$

$v =\frac{ dS }{ dt }= v _0 e ^{- t / t }$

$\int_0^{ s } dS = v _0 \int_0^\tau e ^{- t / t } dt$

$\Delta S = v _0\left[-\tau e ^{- t / \tau }\right]_0^{ t }$

$\Delta S = v _0 \tau\left[1-\frac{1}{ e }\right]$

$\text { Impulse }=\Delta P = m \Delta v$

$\Delta v =\frac{1}{0.4}=2.5$

$\Delta s =4 \times 2.5[1-0.37]=10 \times 0.63=6.3 m / s$

Standard 11

Physics

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