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Chapter 9: Q25P (page 249)

A 1.2 kgball drops vertically onto a floor, hitting with a speed of 25 m/sIt rebounds with an initial speed of10m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.020s, what is the magnitude of the average force on the floor from the ball?

Short Answer

Expert verified

a) The impulse acting on the ball during contact, J=42 kg⋅ms.

b) The magnitude of the average force on the floor from the ball, Favg=2.1×103 N.

Step by step solution

01

Understanding the given information

The mass of the ball, m=1.2 kg .

The velocity of the ball before impact,v1=-25m/s.

The velocity of the ball after impact,v2=10m/s .

The time of contact, t=0.020 s.

02

Concept and formula used in the given question

We can use the equation of impulse related to the change in momentum to calculate the impulse on the ball. The magnitude of the average force can be calculated using the equation relating impulse, force, and time and they are given as follows.

J=∆P∆P=m×vv2-v1J=Favg×1

03

(a) Calculation for the impulse which acts on the ball during the contact

The equation of the impulse-momentum theorem is

J=∆P=m×v2-v1=1.2kg×10m/s--25m/s=42kg./s
04

(b) Calculation for the magnitude of the average force on the floor from the ball when the ball is in contact with the floor for 0.020 s ,  

We have,

J=Favg×t

So,

Favg=Jt=42kg.m/s0.020s=2.1×103N

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