Question

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

Answer 1

The answer is in the image .

Answer 2

Answer:

$8\ \hat{i}\ Ns$

Explanation:

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

Since acceleration is the rate of change in the velocity, the differentiation of velocity vector with respect to time gives the acceleration vector.

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

From here, we conclude that the particle has a constant acceleration. This means the particle is acted upon by a constant force. Let us find out the constant force.

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

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

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