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 1.0Ns is applied to the block at time t=0 so that it starts moving along the x -axis with a velocity v( t ) = v_0e^ - t/\tau , where v_0 is a constant and \tau = 4s. The displacement of the block, in metres, at t = \tau is \underline\hspace2cm Take e^ - 1 = 0.37.

Laws of Motion PYQ — JEE Advanced 2018 · Shift 2 · Q41 with Solution

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Correct answer: 6.30

The impulse given to the block is $J = 1.0\,\text{N s}$

Impulse equals change in momentum: $J = m(v_0-0) = mv_0$

Given: $m=0.4\,\text{kg}$

So, $v_0 = \frac{J}{m} = \frac{1.0}{0.4} = 2.5\,\text{m/s}$

The velocity is $v(t)=v_0 e^{-t/\tau}$ with $\tau=4\,\text{s}$

Hence, $v(t)=2.5 e^{-t/4}$

Displacement is $x = \int_0^{\tau} v(t)\,dt$

So, $x = \int_0^{4} 2.5 e^{-t/4}\,dt$

Take constant outside: $x = 2.5 \int_0^{4} e^{-t/4}\,dt$

Now, $\int e^{-t/4}\,dt = -4e^{-t/4}$

Therefore, $x = 2.5\left[-4e^{-t/4}\right]_0^4$

$x = 2.5\left(-4e^{-1}+4\right)$

$x = 10(1-e^{-1})$

Using $e^{-1}=0.37$

$x = 10(1-0.37)=10(0.63)=6.3\,\text{m}$

The displacement at $t=\tau$ is $\boxed{6.3\,\text{m}}$