Question

N small balls each of mass m impinge elastically in each second on a surface normally with velocity u the force experienced by the surface will be

Answer 1

Answer:

Given :

mass of each ball =m kg

velocity= um/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/1sec

=N2mu

According to Netwon's Second Law

F=Δp/t

F=2Nmu