Question

In the given figure, the position-time graph of a particle of mass $0.1\text{kg}$ is shown. The impulse at $\text{t}=2\text{sec}$ is:

• A. $0.2{\text{kg m sec}}^{-1}$
• B. $-0.2{\text{kg m sec}}^{-1}$
• C. $0.1{\text{kg m sec}}^{-1}$
• D. $-0.4{\text{kg m sec}}^{-1}$

Answer 1

Answer: (B)

$\text{Step 1: Initial velocity}$

Velocity is given by the slope of the displacement-time graph.

From $t=0tot=2s$

Slope =$\frac{\text{Displacement}}{\text{time}}=\frac{4-0}{2-0}=2$

$\therefore u=2m/s$

$\text{Step 2: Final velocity}$

At $t=2s$, Slope of $x-t$ graph is zero

$\therefore v=0m/s$

$\text{Step 3: Impulse}$

Impulse is defined as change in momentum.

$I=m(v-u)$

$=0.1kg\times (0-2m/s)$

$=-0.2kgm{s}^{-1}$

Hence, Option(B) is correct.