**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 t_{n}, and let the speed of the ball right before the
n-th bounce be denoted as -v_{n-1} where the minus sign means that the
ball is moving downward. The speed of the ball right after the bounce, u_{n},
is assumed to be

u_{n} = v_{n-1}/2.

(a) Determine t_{1}, the time of the first bounce, and t_{2},
the time of the second bounce.

(b) Derive a general formula for the time of the (n+1)-th bounce, t_{n+1},
in terms of t_{n}, 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?

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**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 t_{n} ≤ t ≤ t_{n+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 A

(a) During the first time interval 0 ≤ t ≤ t_{1}, 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 A

At t = t

Define u

In the second time interval t

According to the given condition of the bounce, at t = t

Then from the first equation of Eqs.(5) we get

Eqs.(5) become

To find t

Comparing with Eq.(3), the first option leads to t

In summary we have

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

The conditions of bouncing imply that v(t

Eqs.(7) become

To find t

Since t

From the second equation of Eqs.(7A) we have

This means that when the ball falls back to the ground at t = t

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 t

(c) Let us assume that the maximum height of the ball is reached at
t = τ_{n} in the time interval t_{n} ≤ t ≤ t_{n+1}.
It means that

v(τ_{n}) = 0. Using the second equation of Eqs.(7A), we get

The maximum height that the ball reaches after the n-th bounce, h

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
t_{n} ≤ T ≤ t_{n+1}, then we know that the two
balls will collide in the time interval t_{n} ≤ t ≤ t_{n+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 t_{n} replaced by T and u_{n}
replaced by U, and then with the help of Eq.(8), they can be written as

Apparently if t

This equation can be solved for τ to give

Substituting this τ into Y(t) of Eq.(11) gives the position of the collision, Y

where t

(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 = T_{1} and it is given as

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

t

where t

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)T

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 t

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 ≤ t

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