Question

n balls each of mass m impinge elastically in each second on a surface with velocity u. the average force experienced by the surface with be?

Answer 1

Given :
mass of each ball = m kg
velocity= u m/s
In elastic collision :
Change in momentum= mu-(-mu)
=mu+mu
=2mu
For N balls change in momentum:
Δp=N*2mu

Rate of change of momentum=Δp/t=N2mu/1 sec
=N2mu

According to Netwon's Second Law
F=Δp/t

F=N2mu

Answer 2

Answer:

Your answer is in the attachment given above

I hope it is useful

Have a great day