A rubber ball is dropped from a height of 5 m on a planet where the value of acceleration due to gravity is not known. On bouncing, it rises to 1.8 m. Find the factor by which the ball loses its velocity.

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Solution

By conserving energy, we get,
$m g \times 5 = \frac{1}{2} m v_{1}^{2}$ ......(1)
$m g \times 1.8 = \frac{1}{2} m v_{2}^{2}$ (2) ......(2)
Dividing equation (1) by (2) and taking square root,
$\frac{v_{1}}{v_{2}} = \frac{5}{3}$
$∴ v_{2} = \frac{3}{5} v_{1}$
and $\frac{v_{1} - v_{2}}{v_{1}} = \frac{2}{5}$

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A rubber ball is dropped from a height of 5 m on a planet where the value of acceleration due to gravity is not known. On bouncing, it rises to 1.8 m. Find the factor by which the ball loses its velocity.

By conserving energy, we get, mg×5=12mv21 ......(1) mg×1.8=12mv22 (2) ......(2) Dividing equation (1) by (2) and taking square root, v1v2=53 ∴v2=35v1 and v1−v2v1=25