A classical mechanics problem by Stuti Malik

Important facts:

1) As there is no extra input of energy during the process, the time duration of subsequent bounces decreases exponentially, tending towards zero (when the ball stops bouncing). So, the total time taken before the ball finishes bouncing involve sum to infinity.

2) The vertical displacement of a ball when it bounce ( rebound and hit the ground again) is zero.

3) The initial speed of ball after rebounding is given by: u*e, where u is the speed with which the ball initially hit the ground (before rebounding) and e is the coefficient of restitution.

4) Useful SUVAT equations for this question: s= (u)(t) + (1/2)(a)(t^2) and v= u + at, where s is the vertical displacement, u is the initial speed, v is the final speed, t is the time taken for a certain displacement, a is the constant acceleration.

Important results:

*The time taken for the ball to reach the ground = √2,

The time duration of 1st bounce = √2,

The time duration of 2nd bounce = (1/2) √2

The time duration of 3rd bounce = (1/4) √2

Time duration of subsequent bounces forms a geometric series with common ratio- 1/2 or 0.5

*Total duration of the ball bouncing = √2 + (1/2) √2 + (1/4) √2 + .....

(Sum to infinity) = (√2) / ( 1- 0.5) =2 √2

Final result:

Total time taken before the ball finishes bouncing = Time taken for the ball to reach the ground + total duration of the ball bouncing = √2 + 2√2 = 3 √2

Quick method:

It can be shown that the total time taken before the ball finishes bouncing is given by:

(1+e) / (1-e) multiplied by √(2h/g),

Where e is coefficient of restitution, h is the height from which the ball falls and g is acceleration due to gravity.

As e is 1/2 and h= g= 9.8,

The answer = 3 √2

If coefficient of restitution is $e$ and the ball is dropped freely from a height $h$ , then,

• Total displacement till it comes to rest $= -h$

• Total distance covered till it comes to rest $= \left(\dfrac{1+e^{2}}{1-e^{2}}\right)h$

• Total time taken by ball to come to rest $= \left(\dfrac{1+e}{1-e}\right)\sqrt{\dfrac{2h}{g}}$

• Total linear momentum delivered by ball to the surface $= mv_{0}\left(\dfrac{1+e}{1-e}\right)$ where $m = \text{mass of ball}$ and $v_{0} = \text{velocity with which the ball hits the ground for the first time}$

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$\begin{aligned} \therefore \text{Total time} & = \left(\dfrac{1+e}{1-e}\right)\sqrt{\dfrac{2h}{g}} \\ & = \left(\dfrac{1+0.5}{1-0.5}\right)\sqrt{\dfrac{2\times9.8}{9.8}} \\ \\ & = 3\sqrt{2} = k\sqrt{2} \\ & \implies k=\boxed{3} \end{aligned}$