Experiment it!

Classical Mechanics Level pending

Imagine you are holding a ball at height 'h' metres above the ground. Then you leave the ball to drop without any force so that the ball falls freely to the ground under gravity. After the ball hits the ground with speed v, it rebounds with a lower speed and rises upwards to a lower maximum height as the ball's total energy (potential energy at height 'h') is used up as heat and sound energy when it hits the ground. The ball continues bouncing until all of its energy is used up.

The coefficient of restitution, e, between the ball and ground = (Speed with which the ball rebound) / (Speed with which the ball hit the ground). This ratio always stays constant and its magnitude is less than 1 as the ball rebounds with a lower speed.

Optional exercise: Show that the total time taken, T, since the ball is left free to accelerate downwards till it stops bouncing is:

T= √(2h/g) * (1+e)/(1-e),

Where e is the coefficient of restitution, h is the height from which the ball is dropped and g is the acceleration of free fall, 9.8 ms^-2.

Use this result in the following question:

Q. A ball is left free to accelerate downwards from a height of 19.6 metres above the ground. The time taken since the ball is left to drop and till it stops bouncing off the ground is recorded to be 12 seconds. Then the identical ball is dropped from the same height. To what height, in metres, will this ball rise after rebounding for the first time after being dropped from a height of 19.6 metres?

Assume that air resistance is negligible and take g= 9.8 m/s^2.

(Hint: Find e and use v^2 = u^2 + 2as, where u is the initial speed, v is the final speed, s is the vertical displacement and a is the acceleration).

The answer is 10.

1 solution

Tapas Mazumdar
Sep 18, 2016

$~~~~~~~~~~\text{Total time} = \left(\dfrac{1+e}{1-e}\right)\sqrt{\dfrac{2h}{g}} \\ \begin{aligned} \implies 12 & = \left(\dfrac{1+e}{1-e}\right)\sqrt{\dfrac{2\times19.6}{9.8}} \\ & = 2\left(\dfrac{1+e}{1-e}\right) \end{aligned} \\ \implies 6 = \left(\dfrac{1+e}{1-e}\right) \\ \implies e = \dfrac{5}{7}$

Now,

$\text{velocity of the body before collision}~~ v_{0} = \sqrt{2gh} = \sqrt{2\times19.6\times9.8} = -19.6ms^{-1} \text{(as velocity of body is in downward direction)}$

Let, $\text{velocity of body after collision} = v$

So,

$~~~~~~~~~~~~e = \left|\dfrac{v}{v_{0}}\right| \\ \implies \dfrac{5}{7} = \left|\dfrac{v}{-19.6}\right| \\ \implies v = 19.6 \times \dfrac{5}{7} = 14ms^{-1} \text{(as velocity of body is in upward direction)}$

Now,

Using, $v^{2}=u^{2}-2gh$ , where $v = 0~~, u=14ms^{-1}$ , we get,

$~~~~~~~~~~~~0 = {\left(14\right)}^{2} - 2(9.8)h \\ \implies h = \dfrac{196}{2\times9.8} = \boxed{10m}$

Experiment it!

The answer is 10.

1 solution

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