Question

A ball of mass 400gm. dropped from a height 5 m. A boy on the ground hits the ball vertically up with a force of 100 newton so that it attains a vertical height 20m. Find the time for which the ball will remain in contact with bat. g=10m/s^2

Answer 1

the initial momentum of the ball will be

pi = mvi

here,

m = mass of the ball = 400g = 0.4kg

vi = initial velocity = [2ghi]1/2

{hi = 5m}

=  [2 x 10 x 5]1/2 = 10 m/s

so,

pi = 0.4 x 10

thus,

pi = 4 kg.m/s

and

the initial momentum of the ball will be

pf = mvf

here,

vf = final velocity = [2ghf]1/2

{hf = 20m}

=  [2 x 10 x 20]1/2 = 20 m/s

so,

pf = 0.4 x 20

thus,

pf = 8 kg.m/s

now,

we know from Newton's Second Law

force = rate of change of momentum

F = dp/dt = [pf - pi] / dt

or

{F = 100 N}

dt =  [pf - pi] / F

= [8 - (-4)] / 100

{negative sign because pi is directed downwards; convention}

thus, contact time of ball with bat will be

dt = 0.12 s