Problem 1A-12****: Bouncing Balls without Air Resistance

Both the position of an object tossed up vertically and a free falling object can be written in a general form

``` ```
where g is a given positive constant, the meaning of which will become clear later. The constants A and B are to be determined by initial conditions like the position and the speed of the object at a certain given time.

Suppose a ball is released gently from a height H at time t = 0. The ball will bounce many times as it repeatedly hits the ground. Let the n-th bounce occur at time tn, and let the speed of the ball right before the n-th bounce be denoted as -vn-1 where the minus sign means that the ball is moving downward. The speed of the ball right after the bounce, un, is assumed to be
un = vn-1/2.

(a) Determine t1, the time of the first bounce, and t2, the time of the second bounce.

(b) Derive a general formula for the time of the (n+1)-th bounce, tn+1, in terms of tn, the time of the n-th bounce.

(c) How high will the ball reach between the n-th bounce and the (n+1)-th bounce?

(d) A second identical ball is tossed up from the ground level at time T with an upward speed of U, when and where will it collide with the first ball?

(e) If the second ball is released gently from the same height H as the first ball at time T instead, when and where will it collide with the first ball?

Scroll down for solution

Solution:

Eq.(1) only describes the motion of a free-falling object or a tossed up object, but does not describe the violent process of the ball's sudden reversal of direction when it bounces. Eq.(1) thus must be applied to each time interval tn ≤ t ≤ tn+1 separately, and the bouncing processes handled as initial conditions to those equations. The motion of the ball during the above mentioned time interval can be written as

``` ```
where the constants An and Bn are determined by the initial conditions of the ball at t = tn.

(a) During the first time interval 0 ≤ t ≤ t1, we have

``` ```
At t = 0 the ball is at H with speed 0, the gentle release condition, so y(0) = H and v(0) = 0. Applying those conditions to Eqs.(2), we get A0 = H and B0 = 0. Thus Eqs.(2) becomes
``` ```
At t = t1 when the ball hits the ground we must have y(t1) = 0, so the first equation of Eqs.(2) yields
``` ```
Define u0 as -u0 = v(t1), then the second equation of Eqs.(2A) and Eq.(3) combined say
``` ```
In the second time interval t1 ≤ t ≤ t2, we have
``` ```
According to the given condition of the bounce, at t = t1, right after the bounce, we have u1 = v(t1) = u0/2. The second equation of Eqs.(5), Eq.(3) and Eq.(4) lead us to
``` ```
Then from the first equation of Eqs.(5) we get
``` ```
Eqs.(5) become
``` ```
To find t2, we note that y(t2) = 0. The first equation of Eqs.(5A) says
``` ```
Comparing with Eq.(3), the first option leads to t2 = t1, so is not acceptable. Thus we must have
``` ```
In summary we have
``` ```

(b) Consider a time interval tn ≤ t ≤ tn+1. Right after the n-th bounce the ball has an upward speed of un. In this time period Eqs.(2) can be written as

``` ```
The conditions of bouncing imply that v(tn) = un, and y(tn) = 0, so we get from Eq.(7)
``` ```
Eqs.(7) become
``` ```
To find tn+1 we need to solve the equation y(tn+1) = 0. We can rewrite the first equation of Eqs.(7A) as
``` ```
Since tn+1 ≠ tn, we have
``` ```
From the second equation of Eqs.(7A) we have
``` ```
This means that when the ball falls back to the ground at t = tn+1, the magnitude of its downward speed is equal to the bouncing speed of this time interval. Thus when the ball bounce the next time, its upward speed is one-half of the bouncing speed at t = tn. This implies that the bouncing speed decreases by a factor of 2 at each bounce so that we have the relations
``` ```
Eq.(8) can then be written as
``` ```
Using Eq.(4), it becomes
``` ```
Combining Eq.(8B) with Eq.(3), we have the recursive relation
``` ```
We can go further and calculate tn+1 explicitly by applying Eq.(9) repeatedly as follows:
``` ```

(c) Let us assume that the maximum height of the ball is reached at t = τn in the time interval tn ≤ t ≤ tn+1. It means that
v(τn) = 0. Using the second equation of Eqs.(7A), we get

``` ```
The maximum height that the ball reaches after the n-th bounce, hn, can be calculated from the first equation of Eqs.(7A) as
``` ```
With the help of Eq.(8A) and Eq.(4) the maximum height of the ball after the n-th bounce can be written as
``` ```

(d) This problem is not well defined and does not have a simple answer. Suppose that the time of the toss of the second ball, T, is such as tn ≤ T ≤ tn+1, then we know that the two balls will collide in the time interval tn ≤ t ≤ tn+1. However, before they collide, the second ball may undergoes multiple bounces. If we could have written down a master function for the bouncing second ball, then we can equate this master function with the position function of the first ball in the time interval and get a precise time of collision. Unfortunately we do not have such a master function for the second ball, just as we do not have such a master function for the original ball. Therefore, this problem is not solvable unless all the numerical values of all the constants are given and all the bouncing intervals are numerically calculated to pin down to see in which intervals of bouncing that two balls will collide. The reason that such an ill-defined problem is given is probably due to the naiive thinking that two balls will collide before the second ball falls back to the ground. We can go ahead to assume this hidden condition and reveal what kinds of constraints need to be imposed on the time of the toss of the second ball, T, and the toss up speed of the second ball, U.

Let the position and the speed of the second ball be denoted as Y(t) and V(t) respectively. From Eqs.(7A) with tn replaced by T and un replaced by U, and then with the help of Eq.(8), they can be written as

``` ```
Apparently if tn+1 ≤ T1, two balls will collide before the second ball falls back to the ground, and thus this is the constraints needed to be imposed on T and U to make the hidden condition true. We can show that this inequlity is also necessary for the two balls to collide before the second ball touches the ground again. Let us assume that this inequality is not satisfied so that the second ball is tossed up at most at the same time as the bounce of the first ball or later, but landed before the first ball. This means that the first ball's bounce speed must be larger than the toss-up speed U of the second ball. Thus there is no chance the collision can take place when the first ball is ascending. This means that if the collision takes place, it is when the first ball is descending. This also means that at the moment of the collision the downward speed of the first ball must be larger than the downward speed of the second ball, so tn+1 must be smaller than T1 and thus we reach a contradiction. Once this condition is satisfied, then we can proceed to calculate the time of collision and the position of collision. First we equate y(t) of Eqs.(7A) with Y(t) of Eqs.(11), but with t replaced by the colliding time τ to get
``` ```
This equation can be solved for τ to give
``` ```
Substituting this τ into Y(t) of Eq.(11) gives the position of the collision, Yn as
``` ```
where tn and un are gien in Eqs.(9) and (8A) respectively. We will not go ahead to perform the rather meaningless exercise to simplify Eq.(13) further.

(e) This problem is as ill defined as part (d) discussed before. The problem is that after the release of the second ball and before the two balls collide, the first ball can go through many more bounces. Since we do not have a master function to describe the bouncing first ball, we can not equate the position function of the second ball with that non-existing master equation of the first ball to find the exact time of collision. If all the numerical values of the given constnts are included in the problem, then we will be able to solve each interval explicitly and find out in which bouncing interval of the first ball that two balls will collide. The reason that such an ill defined problem is given is again due to the naiive hidden assumption that two balls will collide before the first ball will complete another bounce after the second ball is released at t = T. We will go ahead to analyse further the property of this problem.

We know from Eq.(3) of part (a) that if the first ball is not there, the second ball will touch the ground at t = T1 and it is given as

``` ```
Apparently the given T must fall into some time interval such as

tn ≤ T ≤ tn+1,

where tn+1 is given in terms of tn in Eq.(9) of part (b). Eq.(9) tells us that the largest time interval between tn+1 and tn is for n = 1, that is, between t1 and t2. But this largest time interval is just equal to the difference between T1 and T. This means that if T = t1, then two balls will collide at t = t2. If T falls into other time intervals, it is not possible for two balls to collide within the same time interval. If T falls within a time interval of large n, then the first ball will make many bounces before two balls will collide. We will consider the simplest case that T is chosen such that the collision will take place between the first bounce and the second bounce of the first ball. That means T must be equal to or smaller than t1. This condition can also be obtained from more direct but more tedious calculations as shown below:

The position and the speed of the second ball, Y(t) and V(t), can be written as
``` ```
The initial condition V(T) = 0 implies that D = gT, and the initial condition Y(T) = H implies that C = H - (g/2)T2. Thus we have
``` ```
Equating y(t) of Eqs.(5A) and Y(t) of Eqs.(15) and replacing t by τ, we have
``` ```
This can be simplified to
``` ```
To make this solution meaningful we must have t1 ≤ τ ≤ t2. Eqs.(3), (6) and (16) then lead us to
``` ```
The last inequality is not possible, so the option
``` ```
Eq.(17) can be rewritten as
``` ```
Thus we have
``` ```
Since the inequality of Eq.(17C) is always satisfied, the only constraint is Eq.(17B) that supersedes the inequality of Eq.(17A). Eqs.(3) and (17B) imply that T ≤ t1. This means that the second ball must be dropped before the first ball touches the ground first time for two balls to collide before the first ball touches the ground second time. The time of collision is given in Eq.(16) so that the position of the collision is obtained by substituting τ of Eq.(16) into Y(t) of Eq.(12) to give the value hcollide as
``` ```