Question

a) Define impulse. Give its SI unit.
(b) A foot ball of mass 200g moving towards a player with a velocity 20m/s is kicked by him. If the ball after being kicked, moves with velocity 20m/s at an angle of $90{}^\circ $with initial direction in the same horizontal plane, find the impulse of the force applied by the player.

Answer 1

Hint: As a first step, one could read the question well and hence note down the important points and values given. Then you could recall the expression for impulse and also its definition for the first part. Then you could substitute accordingly for the second part and hence find the answer.

Formula used:
$I=\dfrac{\Delta p}{t}$

Complete step by step solution:
The given question has subparts and for the first part we are supposed to define impulse.
(a) We could define impulse as the integral of force F over the time interval of t for which the given force acts. Impulse is a vector quantity. The SI unit of impulse is given by newton seconds (Ns).
(b) Here, we are given a ball of mass,
$m=200g=0.2kg$
The velocity with which this ball moves towards a player is given by,
$v=20m/s$
This ball after being kicked attains a velocity of 20m/s perpendicular to the initial direction. We are supposed to find the impulse applied by the player on the ball.
Initial momentum,
$\left| p \right|=mv=0.2\times 20=4kgm/s$
Final momentum,
$\left| p' \right|=0.2\times 20=4kgm/s$
Now the impulse would be given by,
$I=\dfrac{\sqrt{{{p}^{2}}+p{{'}^{2}}-2pp'\cos 90{}^\circ }}{t}$
$\Rightarrow I=\dfrac{\sqrt{{{4}^{2}}+{{4}^{2}}+0}}{1}$
$\therefore I=4\sqrt{2}Ns$
Therefore, we found the impulse to be $4\sqrt{2}Ns$.

Note: We could say that an object experiences a certain impulse in a collision. We could simply find the change in momentum for a particular collision and then divide the value by the time taken for the collision and this will give us the impulse on the body for that collision.