Question

A 150 g ball, travelling at 30 m/s strikes the palm of a
player's hand and is stopped in 0.05 second. Find the
force exerted by the ball on the hand.​

Answer 1

let the force be F

t=0.05 sec

u=30 m/s

v=0 m/s

m=150 gm

=0.15 kg

we know that

F*t=pf-pi

pf=final momentum

pi=initial momentum

pf=m*v

=m*0

=0

pi=m*u

=0.15*30

=4.5 kg m/s

plugging the values

F*0.05=0-4.5

F=-90 N

the '-' sign indicates negative acceleration or deceleration

hope this helps