Question

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

• A. $0.2\text{kg-m/s}$
• B. $-0.2\text{kg-m/s}$
• C. $0.1\text{kg-m/s}$
• D. $-0.4\text{kg-m/s}$

Answer 1

The correct option is B $-0.2\text{kg-m/s}$

The equation of the line from $t=0$ to $2s$ can be written as
$x=2t$
$v=\frac{dx}{dt}=2\text{m/s}$

After $t=2s$ the position does not vary hence the velocity after two seconds is zero.

Initial momentum $=m{v}_i=0.1\times 2=0.2\text{kg-m/s}$
Final momentum $=m{v}_f=0.1\times 0=0$
Impulse = change in momentum = $0-0.2=-0.2\text{kg-m/s}$