Question

A child bounces a 52 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 14 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.

Answer 1

0.982 N

Step-by-step explanation:

mass of ball, m = 52 g = 0.052 kg

initial velocity, u = - 20 m/s (downward)

final velocity, v = 14 m/s (upward)

time of contact, t = 1.8 s

According to Newton's second law, the rate of change of momentum of the body is equal to the forced exerted on that body.

initial momentum , pi = mass x initial velocity = 0.052 x (-20) = - 1.04 kg m/s

final momentum, pf = mass x final velocity = 0.052 x 14 = 0.728 kg m/s

Change in momentum = final momentum - initial momentum

= 0.728 - (- 1.04)= 1.768 kg m/s

So, force = change in momentum / time

Force = 1.768 / 1.8 = 0.982 N

Answer 2

Force on superball will be
`=9.8222* 10^(-4)N`

Step-by-step explanation:

We have given mass of superball m = 52 gram = 0.052 kg

Velocity change from 20 m/sec downward to 14 m/sec upward

Let downward velocity is positive then upward velocity is negative

So downward velocity is + 20 m/sec and upward velocity is -14 m/sec

Time is given as 1800 sec

We know that acceleration is rate of change of velocity

So
`a=(20-(-14))/(1800)=0.0188m/sec^2`

According to newton second law

Force = ma = 0.052×0.0188
`=9.8222* 10^(-4)N`