Question

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

The correct option is B $-0.2{\text{kg m sec}}^{-1}$

From the $x-t$ graph, velocity of particle is given by:
$v=\text{slope of x-t curve}=\frac{dx}{dt}$
$\Rightarrow$From $t=0$ to $t=2\text{s}$, slope of $x-t$ curve is constant, so velocity is,
$\therefore v=\text{slope of x-t curve}=\frac{4-0}{2-0}=2\text{m/s}$
Just after $t=2\text{s},x=\text{constant}v=\frac{dx}{dt}=0\text{m/s}$
$\Rightarrow$At $t=2\text{s}$, particle is undergoing a transition in velocity from $v_i=2\text{m/s}\text{to}{v}_f=0$
From impulse momentum-theorem on particle:
$J=\Delta P=m{v}_f-m{v}_i...\left(i\right)$
$\therefore J=0.1(0-2)=-0.2\text{kg m}{\text{sec}}^{-1}$