Problem — Impact speed and coefficient of restitution

Problem

A ball is dropped from $h_0 = 1{,}0$ m; it bounces up to $h_1 = 0{,}64$ m. Calculate the impact speed at the ground and the coefficient of restitution $e = \sqrt{h_1/h_0}$, justifying the formula.

Solution

Impact speed. During free fall from rest, conservation of energy (or $v^2 = 2gh$) gives the speed at which the ball reaches the ground: $v = \sqrt{2\,g\,h_0} = \sqrt{2 \cdot 9{,}8 \cdot 1{,}0} \approx 4{,}43 \ \mathrm{m/s}$

Justification of the coefficient of restitution. The coefficient of restitution is defined as the ratio between the magnitude of the velocity immediately after the collision and immediately before: $e = \frac{v_{\text{dopo}}}{v_{\text{prima}}}$ Both before (falling) and after (rising) the relationship between speed and height $v^2 = 2gh$ holds, i.e. $h \propto v^2$. So for the rising speed $v_{\text{dopo}} = \sqrt{2 g h_1}$ and for the impact speed $v_{\text{prima}} = \sqrt{2 g h_0}$, giving $e = \frac{\sqrt{2 g h_1}}{\sqrt{2 g h_0}} = \sqrt{\frac{h_1}{h_0}}$ This explains why the coefficient of restitution can be obtained simply by measuring the two heights, without directly measuring the speeds.

Numerical calculation. $e = \sqrt{\frac{0{,}64}{1{,}0}} = \sqrt{0{,}64} = 0{,}80$

$\ev{v \approx 4{,}4\ \mathrm{m/s} \qquad e = 0{,}80}$

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Topics: Experimental method Skills: Using significant figures