Question

A ball of mass 0.5 kg collides with a wall at a speed of 15.0 ms−1 and bounces back with

a speed of 12.5 ms−1

. If the average force exerted by the ball is 1100 N calculate the

impulse and the time for which the collision lasted.

Answer 1

mass(v-u)=0.5(12.5-(-15)
0.5*27.5
137.5 this is change in momentum in other words this is also called as Impulse
So impulse=Force* time
137.5=1100*time
time =137.5/1100sec
#above we take -15 because motion will be in opposite direction after collision from wall.
if any query please ask if need any correction please specify

Answer 2

Answer: The impulse is 13.75 kgm/s and time is 0.0125 sec.

Explanation:

Given that,

Mass of the ball m = 0.5 kg

Speed $v_{1}=15.0 m/s$

$v_{2}=12.5 m/s$

We know that,

The impulse is the change of momentum.

$I = \Delta P$

$I = m\Delta v$

$I = m(v_{2}-v_{1})$

$I = 0.5\times(-12.5-15.0)$

$I = -13.75\ kgm/s$

Negative sign represents the force on the wall is opposite to force on the ball.

Now, the  impulse is

$I = F\Delta t$

$t = \dfrac{I}{F}$

$t = \dfrac{13.75}{1100}$

$t = 0.0125\ sec$

Hence, The impulse is 13.75 kgm/s and time is 0.0125 sec.