# Two balls falling and bouncing one off the other

1. Nov 5, 2007

### black_squirrel

I'm not sure how to do this.

The Problem:

We have two balls one with mass M and another with mass m. M>>>>>m both balls are dropped at the same time, the heavier ball below the lighter ball from a height of 10 meters. (even though the lighter is on top and heavier on bottom, for the purposes of this problem, you can assume they are falling from the same height). So both balls fall. When the heavier ball hits the ground, it bounces straight back up and immediately collides with the lighter ball falling down. THe question is how high will the lighter ball rise assuming all collisions are elastic.

My Take:
Both balls start at a potential energy. Bigger ball has M*g*10m and smaller ball has m*g*10m.
For big ball, PE turns into KE so we have:
Mgh = .5*M*v^2 so velocity when it hits the ground is 14.14m/s (g = 10 m/s^2) which is the velocity with which it would hit smaller ball.
THe smaller ball would hit the ground at the same velocity but i don't know how to figure out what it's velocity is going to be after it collides with the bigger ball. I'm thinking something involving conservation of momentum to find that velocity and then just figure out max height of a ball of mass "m" with that initial velocity...

2. Nov 5, 2007

### rl.bhat

By using conservation of momentum and energy, we can get the velocities of both the bodies by using
V1 = [(m1 - m2)/(m1 + m2)]u1 + [2m2/(m1+m2)]u2
V2 = [2m1/(m1 + m2)]u1 + [(m2-m1)/(m1+m2)]u2
Use proper approximation to find the velocity of m and hence find the height to which it rebounds

3. Nov 5, 2007

### andrevdh

You might try this:

Assume initial velocities

$$v_1 \ and \ V_1$$

before the collision - which can be evaluated with your formula assuming that they fell through 10 meters. Assume the velocities after the collision

$$v_2 \ and \ V_2$$

(which are unknowns). Set up energy and momentum equations using conservation of both. Eliminate $$V_2$$ from the energy conservation equation using the momentum conservation equation. Use the simplification that terms with

$$\frac{m}{M}$$

in them can be ignored. Solve for $$v_2$$. I must say this approach did not work for me - I got that it will rise to the same height - 10 meters! Maybe I made a mistake?

4. Nov 5, 2007

### black_squirrel

Sorry I don't understand your equations. What equations are these and what does u1 and u2 correspond to? thanks for the help

5. Nov 5, 2007

### rl.bhat

u1, u2 are the initial velocities of M and m and v1,v2 are the velocities of M and m after impact.

6. Nov 6, 2007

### black_squirrel

Hmm...what did you derive that equation from? I don't follow it.

Conservation of momentum states:

how did u derive V'a and V'b from that?

7. Nov 6, 2007

### GTrax

Er.. I am still working through this one, so allow me to be foolish for awhile, but my first reaction is you might not need to mess with actually calculating velocities. All collisions were elastic, so what kinetic energies were gained in falling would turn back into potential energies at the rebound. Is it just possible that the two balls would fall together, and rise together, and end up at (censored for now) metres high?

8. Nov 6, 2007

### black_squirrel

i'm not sure. The way they ask the question makes it seem that the heavier ball bounces and collides with the lighter ball, launching it upward. even if you consider it a combined system of mass M+m, with all collisions being elastic, would that mean it would bounce back to the same height?

9. Nov 6, 2007

### rl.bhat

The equation you have written is conservation of energy.
Conservation of momenum = MaUa + MbUb = MaVa + MbVb. By solving these two equations we can get the values of Va and Vb. To get the final velocities neglect m compared to M in the final expression.

10. Nov 6, 2007

### GTrax

Consider the 'Newton's Momentum Balls" toy of a row shiny ball bearings suspended. Let go two together, and they swing..and smack.. and they stop dead.. and two together!! take off from the other end .. and rise to a level of potential energy. They return, and the original two take a trip together!! to end up somewhere near where they started.

The masses of the balls is arbitrary. The collisions being perfectly elastic, and assuming no other losses, they just have to end up where they started, ready for an indefinite repeat performance.

Even if you let one go first, delaying the drop of the second, so it meets the big one on its rebound..

Even, even if you let the big one go first, and dropped the smaller one from a greater height.

11. Nov 6, 2007

### rl.bhat

Since both the balls start from the same point , they will have the same velocity u.At the instant when the body of mass M rebound, its velocity will be u( since collision is elestic). So the M is moving up with velocity u and m is moving down with velocity u. Since their directions are different,one will be +ve and another will be -ve. Substite these values in the equations which I have given, neglecting m, you will get the answer.

12. Nov 6, 2007

### black_squirrel

thank you sooo much for helping me through this.
solving for V2 using the conservation of momentum equation gave me:

the velocity of the smaller ball is dependent on the velocity of the bigger ball after the collision which is also unknown. Also, i'm not sure how to take care of the M/m ratio there.

13. Nov 7, 2007

### black_squirrel

okay my final answer is that it will rebound back to the original 10 meters. Is this right rl.bhat?

14. Nov 7, 2007

### rl.bhat

No. It is not right. If you put u1 = u2 = v1 = u ( because they are falling from the same height and collision is elastic) and m2 = 0 ( because M>>>>m) you get v2 = 3u and h = 90 m. Surprised? But it is true! When you find the expression for v1 and v2 by using both conservstion of energy and momentum the term M/m does not appear. The expression which I have given are standerd expressions.

15. Nov 7, 2007

### black_squirrel

Okay, I accept that it's true but I still can't derive the expression you have written. I found the same expression listed on wikipedia on this page: http://en.wikipedia.org/wiki/Momentum#Relating_to_mass_and_velocity

I'd really appreciate it if you could explain the derivation of the expressions. I've wasted a lot of paper trying to derive it from momentum and energy equations.

16. Nov 8, 2007

### rl.bhat

From conservation of momentum we have m1u1 + m2u2 = m1v1 +m2v2.....(1)
From conservation of energy.. 1/2[m1u1^2 + 1/2m2u2^2 ]= 1/2[m1v1^2 + m2v2^2]..(2)
Rearranging both the equations we get, m1(u1 - v1) = m2(v2 - u2) ...(3) and
m1(u1^2 - v1^2) = m2(v2^2 - u2^2). ..(4).
..(4)/(3) gives u1 + v1 = u2 + v2. To find v2 put v1 = u2 + v2 - u1 in equ. (1) and solve for v2.

17. Nov 8, 2007

### GTrax

I will settle for the qualitative approach for the present.

The original question stressed that one mass was very large compared to the other, but I will not be distracted by this. Strictly, the two masses are in contact and perfectly elastic. The elastic compression and rebound of the smaller off the larger exactly equals the compression and rebound off the unyielding landing surface. Even more strictly, we have a bouncing oscillatory system with the planetary mass also participating!

The velocities upward immediately after collision are the same for both masses (given that we are artificially permitted to take the fall distance of both masses as equal and being 10m). They had better be, or the collision was not elastic. They rise together, back to the start, and may as well be regarded as one mass.

.... or can they? Is this really about thinking of a dual elastic collision so perfect that the masses can be considered as stuck together, behaving as one? Are there assumptions hidden in this scenario?

Imagine them separated by a finite vanishingly small amount. Upon arrival, instead of a connected hard surface, small mass meets large mass coming the other way, and at the same speed. Large mass gets slowed a little, ending up somewhat short of its start point, while small mass gets bounced to a new height different to its start point. This last bit can be calculated easily from momentums.

It also hides the little lie. That collision, although a infinitesimal gap away, was a new scenario, where different conditions were sneaked in. Small mass did not get to experience an elastic collision with a planetary mass showing it a hard elastic surface. This scene treats the large mass still in contact with it, and still in elastic compression, as if it were a connected part of the planet. Instead, it gets to have a completely different collision with a large mass that had a previous collision adventure of its own.

The point here is that if they shared the collision simultaneously, then no part of the large mass and its momentum gets the opportunity to bounce the small mass anywhere new.

If they start in contact, and the collisions are elastic, and simultaneous, they stay in contact, and end up where they started. A very artificial and narrow condition maybe, but its what you get when you pick it apart.

18. Nov 9, 2007

### rl.bhat

As the figure shows the two balls are one upon the other. They start in contact. Until the large ball does not touch the ground, small ball does not experiance any force. When they collide with the ground, the force of action is (M + m)g. The force of reaction must be the same. And each ball must experiance the same force. Since the balls are not attached firmly, the smaller ball will experiance the force (M+m)g. In that case will it reach the same height or more? Now imagin the other way round. Smaller ball reaches the ground first. In that case what happens?

19. Nov 9, 2007

### aq1q

u1 + v1 = u2 + v2. works for all elastic collisions right?

20. Nov 9, 2007

### rl.bhat

u1 + v1 = u2 + v2. works for all elastic collisions right? No.
It is true only during elastic head on collision of two identical bodies. In that case v1 becomes u2 and v2 becomes u1 and move in the opposite direction.