Question

A ball of mass 20g droppedover a large horizontal surface from a height of 5m collides elastically with tje surface from a height of 5m, collides elastically with the surface . if the collision lasts for 10^-2s find the impulse

Answer 1

0.4N-s is the answer

Answer 2

Answer:

The impulse will be $I=0.392\ kg-m/s$.

Explanation:

Given that,

Mass m = 20 g

Height h = 5 m

Time $t =10^{-2}\ s$

The velocity is given by

Using equation of motion

$v^2=u^2+2gh$

$v^2=0+2\times 9.8\times5$

$v= \sqrt{2\times 9.8\times5}$

$v = 9.8\ m/s$

The ball moves with the velocity v and after collision the ball come back with the double velocity 2v in opposite direction.

Therefore, The velocity will be double

$v = 2\times9.8$

$v=19.6\ m/s$

The impulse is the change of momentum.

The impulse is defined as:

$I =m(v-u)$

Here, m = mass

$v$=final velocity

$u$=initial velocity

Put the value of mass, and velocity in equation (I)

$I = 20\times10^{-3}\times 19.6$

$I= 0.392\ kg-m/s$

Hence, The impulse will be $I=0.392\ kg-m/s$.