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 $F_{a} v g T = F △_{t}$ , Z where $F △_{t}$ is 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).

Expert verified

The value of average force $F_{a v g} = \frac{2 m \sqrt{2 g h}}{T}$

Step by step solution

Striking velocity, $v = \sqrt{2 g h}$

Rebound velocity $v = - \sqrt{2 g h}$

Then the impulse

$I = m △ V I = m (\sqrt{2 g h} - (- \sqrt{2 g h})) I = 2 m \sqrt{2 g h}$

Force,

$F = \frac{m △ V}{△ t}$

$F = \frac{2 m \sqrt{2 g h}}{△ t}$

As given in the question

localid="1658076769643" $F_{avg} T = F △ t$

$F_{avg} T = F △ t F_{avg} T = \frac{2 m \sqrt{2 g h}}{△ t} \times △ t F_{avg} \frac{2 m \sqrt{2 g h}}{T}$

Thus, the value of average force becomes $F_{avg} \frac{2 m \sqrt{2 g h}}{T}$ .

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