Question

Velocity of a particle of mass $2kg$ varies with time $t$ according to the equation $\overrightarrow{v}=(2t\hat{i}+4\hat{j})m/s$. Here $t$ is in seconds. Find the impulse imparted to the particle in the time interval from $t=0$ to $t=2s$.

• A. $8\hat{i}N-s$
• B. $10\hat{i}N-s$
• C. $12\hat{i}N-s$
• D. $16\hat{i}N-s$

Answer 1

The correct option is A $8\hat{i}N-s$

velocity vector $=v=(2t\hat{i}+4\hat{j})m/s$

since acceleration is the rate of change in velocity,

$\therefore a=\frac{dv}{dt}=\frac{d(2t\hat{i}+4\hat{j})m/s}{dt}=2\hat{i}m/s^2$

$\therefore$ particle has constant acceleration. This means particle is acts upon by constant force.

$\therefore F=ma=2kg\times \left(2\hat{i}\right)m/s^2=4\hat{i}N$

Since a constant force acts on the particle the impulse imparts to the particle is given by :-

Impulse $=\hat{i}\Delta t=\left(4\hat{i}N\right)\times (2s-0s)=8\hat{i}Ns$

$\therefore$ option $A$ is correct.