what is the average force exerted by the plate on the ball during the impact?

Question

A 5.00kg steel ball is dropped onto a copper plate from a height of 10.0m.the ball leaves a dent 3.20 mm deep.

Answer 1

First we need to determine V2 at impact

V2^2=V1^2 + 2ad, a=9.81, V1=0 and d=10m
=0+2*9.81*10
=196.2, V2 = 14 m/s

To find average F, we need the acceleration from time of impact to when the ball stops at the bottom of the dent

again we'll use

V2^2=V1^2 + 2ad
a=(V2^2-V1^2)/(2d), V2=0, V1=14, d=0.0032
a=(14^2)/(2*0.0032)
=30625

so F=30625x5 =153125 N

Answer 2

Must be close to 153,300N.

Use the v^2 - u^2 = 2fs type equation - twice - once to get the velocity of the ball on impact, again to get the deceleration over 3.2mm.