Two Rubber Ball Collision Physics Questions

Question

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

 

1. At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .

 

2. What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?

 

3. For what value of h does the collision occur at the instant when the first ball is at its highest point?

Answer 1

Hi Kundi!

 

Kinematics, the subject of this problem, is the study of motion using five key quantities: acceleration (a), initial velocity (v_o), final velocity (v), displacement (x-x_o, where x is final position and x_o is initial position), and time (t). Usually, a problem asks you to solve for one of these quantities. You should have four equations at your disposal, relating these five quantities. Each equation involves four of the five quantities. Therefore, if you want to use the equations to solve for one of the kinematic quantities, you have to know at least three others for the given situation. The challenge of kinematic problems is usually finding out what information you are given.

The way I usually work these problems is to ask, first, which of the five kinematic quantities is being asked for? This is often easiest to pick out, because the question has to directly state it.  In this problem, we have two separate objects, and are asked to come up with an expression for their displacement.  So we need to come up with expressions for each ball's displacement in terms of quantities we are allowed to write the answer in terms of:  v_o, h, and g (where h is the initial position of the, x_o, of the dropped ball).

 

So let's take a look at the dropped ball.  In that case, we have:

 

to find:  x-x_o

 

The question is, what three things do we "know" about the dropped ball (at least, in terms of being able to write the answer with them)?

 

First, for any object in vertical motion under the influence of only gravity (often confusingly called "free fall," even though it may be rising vertically), we always immediate know one kinematic quantity:  acceleration, which equals g, downward (which I usually default to being the negative direction).  So, a = -g.

 

Second, we are told the ball is dropped from rest.  Therefore, we know its initial velocity, v_o, must be 0.

 

In this case, we do not technically "know" a third item, so any kinematic equation we write will have two unknowns in that one equation.  But we do know that the final positions of the two balls must be the same, and the final time when they collide must be the same.  So if we use a kinematic equation that involves time, we will have options to write one kinematic equation for each ball with two unknowns, but those two unknowns will be the same, so we will have a system of two equations with two unknowns, which we can solve.

 

So, let's count our knowns as

 

know:  a, v_o, t

 

So, we need a kinematic equation that connects what we want to find with the quantities we know.  The one I would suggest is:

 

(x-x_o) = v_ot + (1/2)at²

 

For the dropped ball, this becomes:

 

x-h = 0(t)+(1/2)(-g)t² = -(1/2)gt²

 

It turns out that, for the ball shot upward, we want to find and know the same quantities, though they have different values (x_o = 0, taking the ground to be zero height; v_o is non-zero).  But the acceleration is the same, because this ball is also moving vertically and, once it leaves the launcher), is only being accelerated by gravity.

 

So we have, for the upward-moving ball:

 

to find: x-x_o

know: a, v_o, t

 

And we can set up the same kinematic equation as above:

 

(x-x_o) = v_ot + (1/2)at² --> (x-0) = v_ot + (1/2)(-g)t² --> x = v_ot-(1/2)gt²

 

Because the x's (final position) and the t's (final time) in both of these expressions are the same, we have a system of two equations with two unknowns:

 

x-h = -(1/2)gt²

x = v_ot - (1/2)gt²

 

There are multiple ways to solve this, but I think the one that is easiest algebraically is to solve for t first and then plug it back in to get an expression for the position x.  This is because that -(1/2)gt² term in each expression is identical, which simplifies the algebra.

 

x = h-(1/2)gt²   (dropped ball)

x = v_ot - (1/2)gt²   (upward ball)

 

Equating the right sides of these equations will quickly get you an expression for t in terms of v_o in terms of h.  Then plug that expression back in for t in either equation and solve for an expression for x in terms of v_o, h, and g.  For comparison, I obtained, as a result:

 

x = h - (1/2)g(h/v_o)²

 

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If the balls are to collide before the ball projected upwards returns to the ground, they must collide before x = 0.  If we want to solve for a value for h at which this occurs, we set x = 0 in the above expression and solve for h.  The above expression for x in terms of h in that case becomes:

 

0 = h -(1/2)g(h/v_o)² = h - (1/2)(g/v_o²)(h²) = h[1-(g/2v_o²)h] = 0

 

Since this is a quadratic expression in h, there are two solutions for h.  One is the (trivial) h = 0 solution;  if the dropped ball is dropped right at he ground as the second ball is launched, the two balls will (of course) meet right at the ground.  This is the "minimum" h solution.

 

The other solution, when the quantity in brackets = 0, provides the solution for the maximum h at which this condition of colliding above the ground occurs.  This is what you want to solve for.  Let me know if you want to look at the algebra of this solution in more detail.

 

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For the third problem, we have to come up with an expression for the position of the upward-projected ball when it is at its maximum.  In this case, we can set up, just for the upward ball, our kinematic knowns, and what we want to find:

 

to find: x-x_o

know: a, v_o

 

This is all as before.  But now, for the upward ball, we know a third kinematic quantity.  Any time you see the words "maximum height" regarding vertical free fall or projectile motion, you know one thing:  the object must have a final velocity of 0.  If it does not, it is either still going toward its maximum height, or it must be coming back from its maximum height.  So we can add v=0 to our list of knowns for the upper ball, and we have:

 

to find:  x-x_o

know: a, v_o, v   (with a = -g)

 

There is a standard kinematic equation that connects these four quantities:

 

v² = v_o²+2a(x-x_o)

 

which becomes, in this case:

 

0 = v_o²+2(-g)(x-0) = v_o²-2gx

 

So, the condition on the position of the upward ball so that it is at its maximum height is:

 

x = v_o²/(2g)

 

The dropped ball must also be at this position at the same time if the two balls are to meet there.  We already solved for the position x where the two balls meet, in terms of h and other allowed quantities, in the first part.  So we just have to use that expression to solve for h when x = v_o²/(2g)

 

v_o²/(2g) = h - (g/2v_o²)h²

 

This can be rearranged into a quadratic expression in h, which I solved for h using the quadratic formula  It turned out there was one real solution (the discriminant was zero).  If you want to go into the algebra details of that, just let me know.

 

But I hope this gets you on track for solving these problems.  If you have any other questions about this, or similar problems, just let me know!

Answer 2

At t seconds after this scenario begins,

 

Ball projected upward is at vertical location s(t) = v₀t - ½gt² 

 

Ball dropped from height h is at vertical location D(t) = h -  ½gt²

 

Equate these two expression to solve for the time of collision.