Centre of Mass 4 Question 4
5. 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 $v(t)=v _0 e^{-t / \tau}$, where $v _0$ is a constant and $\tau=4 s$. The displacement of the block, in metres, at $t=\tau$ is
(Take, $e^{-1}=0.37$ ).
(2018 Adv.)
Analytical & Descriptive Questions
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Answer:
Correct Answer: 5. $(6.30)$
Solution:
- Linear impulse, $J=m v _0$
$$ \begin{array}{ll} \therefore & v _0=\frac{J}{m}=2.5 m / s \\ \therefore & v=v _0 e^{-t / \tau} \end{array} $$
$$ \frac{d x}{d t}=v _0 e^{-t / \tau} \quad \int _0^{x} d x=v _0 \int _0^{\tau} e^{-t / \tau} d t $$
$\tau$
$$ \begin{aligned} x & =v _0 \frac{e^{-t / \tau}}{-\frac{1}{\tau}} \\ x & =2.5(-4)\left(e^{-1}-e^{0}\right) \\ & =2.5(-4)(0.37-1) \\ x & =6.30 m \end{aligned} $$
