Q. In the figure given below, the position –time graph of a particle of mass $0.1\ kg$ is shown. The impulse at $t = 2\ sec$ is
A) $0.2\ kg\ m\ \sec^{- 1}$
B) $- 0.2\ kg\ m\ \sec^{- 1}$
C) $0.1\ kg\ m\ \sec^{- 1}$
D) $- 0.4\ kg\ m\ \sec^{- 1}$
Explanation
(b) Velocity between $t = 0$, and $t = 2\ sec$
$\Rightarrow v_{i} = \frac{dx}{dt} = \frac{4}{2} = 2\ m/s$
Velocity at $t = 2\ \sec,\ v_{f} = 0$
Impulse = Change in momentum $= m(v_{f} - v_{i})$
$= 0.1(0 - 2) = - 0.2\ kg\ m\ \sec^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.