Question

A ball of mass 50 gram moving with velocity ofo50 meter per second strike a wall and reflected by a velocity of 18 metre per second. If the type of contact between wall and wall is 1/20 second then find the change in momentum of the ball and also find the impulse imparted to the ball?

Answer 1

Change in momentum
= final momentum- initial momentum

final momentum= m×v
initial momentum
= - m×u
(negative sign implies that direction is opposite)

gives,

change in momentum
= mv - (-mu)
=mv+mu
=m(u+v)

given,
m=50g
=0.05kg
u=50 m/s
v= 18 m/s

So,

change in momentum
=0.05×(50+18)
=0.05×68
=3.4 kg.m/s

impulse is same as change in momentum

Impulse= Change in momentum

So impulse= 3.4 kg.m/s