A steel ball of mass m falls from a height h onto a scale calibrated in Newtons.

Question

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
Favg, such that
FavgT = F t,
where F t is the brief impulse that the ball
imparts to the scale on every hit, and T is
the time between hits. Calculate this average
force in terms of m, h, and other physical
constants.
1. Favg =mg/T
2. Favg =mg/h
3. Favg = mg
4. Favg = mgT
5. Favg = mgh

Explanation / Answer

Basically , the only option which is dimensionally correct is mg . Still , let's try to find it out . B/w any two instants when the ball just leaves the scale (when its velocity is maximum upward) , there is surely no change in momentum . Thus the average force acting must be zero during this period of T+t (I'm using t instead of delta t ) . So, if F is the force the scale exerts on the ball during t , F*t = mg*(T+t) So, F = mg*T/t [ Since t

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