Q 11-04-043NEETJEE MainMedium
A $2$ kg body at rest is acted on by a force that rises uniformly from $0$ to $100$ N in $0.1$ s and then falls uniformly back to zero in the next $0.1$ s. The speed of the body afterwards is
Answer: (D) $5$ m/s
Impulse = area under the force-time graph (a triangle):
$$J = \frac{1}{2} \times 0.2 \times 100 = 10\ \text{N s}$$
$J = m\,\Delta v$, so $v = \dfrac{10}{2} = 5$ m/s.
Solution by Sreeraj P, M.Sc Physics