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Chapter 1: Q52P (page 41)

A basketball has a mass of 570g. Heading straight downward, in the y-direction, it hits the floor with the speed of 5 m/s and rebounds straight up with nearly the same speed. What was the momentum change ∆p⇀?

Short Answer

Expert verified

Answer

The momentum change will be ( 0,5.7,0 ) kg m/s.

Step by step solution

01

Identification of given data

The given value is below as,

  • Mass of the basketball m= 570 g =0.57 kg.
  • Value of initial and final velocities,vi⇀=0,-5,0m/s,vf⇀=0,5,0m/s
02

Momentum 

Momentum is nothing but the product of the mass of the object and the velocity of the object.

03

Calculation for the momentum change

Initial momentum can be calculated as:

pi⇀=mvi⇀

Initial momentum can be calculated as:

pf⇀=mvf⇀

Change in the momentum can be written as follows:

∆p⇀=pf⇀-pi⇀

Putting the values and we get,

∆p⇀=mvf⇀-mvi⇀=0.57kg×0,5,0m/s-0.57kg×0,-5,0m/s=2×0.57kg×0,5,0m/s=0,5.7,0kg.m/s

Therefore, the change in momentum in the tennis ball is=0,5.7,0kg.m/s .

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Most popular questions from this chapter

Answer the following questions about the factorγ (gamma) in the full relativistic equation for momentum:

(a) Isγ a scalar or a vector quantity?

(b) What is the minimum possible value of γ?

(c) Doesγ reach its minimum value when an object’s speed is high or low?

(d) Is there a maximum possible value for γ?

(e) Doesγ become large when an object’s speed is high or low?

(f) Does the approximationγ≈1 apply when an object’s speed is low or when it is high?

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