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A ball whose mass is 0.2kg hits the 铿俹or with a speed of 8 m/s and rebounds upward with a speed of 7m/s. The ball was in contact with the 铿俹or for0.5ms0.510-3s.

(a) What was the average magnitude of the force exerted on the ball by the 铿俹or? (b) Calculate the magnitude of the gravitational force that the Earth exerts on the ball.

(c) In a collision, for a brief time there are forces between the colliding objects that are much greater than external forces. Compare the magnitudes of the forces found in parts (a) and (b).

Short Answer

Expert verified

(a). The average force is Favg=6000N.

(b). The gravitational force is 2 N.

(c). The force exerted by the floor is 3000 times greater.

Step by step solution

01

Given parameters

Mass of the ball ism=0.2kg.

Speed isvi,y=8m/s.

Upward speed isvf,y=7m/s.

Contact time with the floor is,F=0.5ms(0.510-3s).

Acceleration due to gravity is g=9.8m/s2.

02

The formula of change in momentum

The change in momentum is given by:

P=Pf-PiP=mvf,y-mvi,yP=m(vf,y-vi,y)

The average force exerted on the ball can be calculated by dividing the change in momentum by the time interval.

Favg=PF

03

Find the change in momentum

(a)

Substitute the values to the first formula to find for the change in momentum.

P=mvf,y-vi,y=(0.2kg)(-7msec-8msec)=3kg.m/s

04

Find the average force

Use the second formula and substitute the values.

Favg=PFFavg=3kgm/s510-3sFavg=6000N

Therefore, the average force is Favg=6000N.

05

Find the gravitational force

(b)

The magnitude of the gravitational force that the Earth exerts on the ball.

Ball's mass:m=0.2kg

Gravity strength on earth:g=9.8m/s2

The gravitational forceFgrav=mg

Substitute the values I the above formula.

Fgrav=0.29.8Fgrav=2N

Therefore, the gravitational force is 2 N.

06

Finding gravitational force

(c)

From part (a) and (b):

FfloorFgrav=60022=3000

So, the average magnitude of the force exerted on the ball by the floor is 3000 times the magnitude of the force exerted on the ball by the Earth.

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