Question

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. $-0.4\, kg \,m \,sec^{-1}$

Answer 1

Answer: (B)

Velocity between $t=0$ and $t=2\,sec$
$\Rightarrow v_{1}=\frac{dx}{dt}$
$=\frac{4}{2}=2\, m/s$
Velocity at $t=2\,sec, v_{f}=0$
Impulse = Change in momentum $=m(v_{f}-v_{t})$
$0.1 (0-2)=-0.2\,kg\,m\, sec^{-1}$