Question

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m=0.4kg$ is at rest on this surface. An impulse of $1.0N$ s is applied to the block at time to $t=0$ so that it starts moving along the x-axis with a velocity $v\left(t\right)=v_0{e}^{\frac{-t}{\tau }}$, where $v_0$ is a constant and $\tau =4s$. The displacement of the block, in metres, at $t=\tau$ is .
[Take $e^{-1}=0.37$]

Answer 1

Given, mass = $0.4kg$
where, impulse= change in liner momentum
So, $\frac{1}{0.4}=v_0-v{v}_0=2.5m/s$

As velocity is decreasing continuously so the area under the graph is given by

so, $d=v_0{\int }_0^{\tau }{e}^{-t/\tau }=\tau v_0[1-e^{-t/\tau }]$
at $t=\tau \Rightarrow \tau v_0[1-\frac{1}{e}]$
where, $v_0=2.5,\tau =4sec$ and $[1-\frac{1}{e}]=0.63$ [by mathematical calculation]
so, $\Rightarrow 4\times 2.5\times 0.636.3m$