First slide

Instantaneous velocity

Question

A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known, On bouncing it rises to 1 .8 m. The ball loses its velocity on bouncing by a factor of

Easy

A

2/5
B

3/5
C

9/25
D

16/25

Solution

When the ball strikes the surface of planet, the potential energy is changed into kinetic energy.
Thus
$\frac{1}{2} {mv}^{2} = mgh$
or v² =2gh=2 x g x 5=10g
After rebounce, v² = u² - 2 g h
or u²=v² + 2gh
$∴ u^{2} = 0 + 2 \times g \times 1 \cdot 8 = 3 \cdot 6 g$
Now $\frac{v^{2}}{u^{2}} = \frac{10 g}{3 \cdot 6 g} = \frac{100}{36} or \frac{v}{u} = \frac{10}{6} = \frac{5}{3}$
or $u = \frac{3}{5} v$
Thus the ball loses it velocity by a factor of
$1 - \frac{3}{5} = \frac{2}{5}$

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