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Chapter 3: Q25P (page 124)

A steel ball of mass m falls from a height h onto a scale calibrated in newtons. The ball rebounds repeatedly to nearly the same height h. The scale is sluggish in its response to the intermittent hits and displays an average force F avg , such that FavgT=F鈻�t, Z where F鈻�tis the brief impulse that the ball imparts to the scale on every hit, and T is the time between hits.

Question: Calculate this average force in terms of m, h and physical constants. Compare your result with the scale reading if the ball merely rests on the scale. Explain your analysis carefully (but briefly).

Short Answer

Expert verified

The value of average forceFavg=2m2ghT

Step by step solution

01

Identification of given data

  • Mass of steel ball - m
  • Height - h
02

Calculation of the average force

Striking velocity, v=2gh

Rebound velocity v=-2gh

Then the impulse

I=m鈻�VI=m(2gh-(-2gh))I=2m2gh

Force,

F=m鈻�V鈻�t

F=2m2gh鈻�t

As given in the question

localid="1658076769643" FavgT=F鈻�t

FavgT=F鈻�tFavgT=2m2gh鈻�t脳鈻�tFavg2m2ghT

Thus, the value of average force becomes Favg2m2ghT.

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