# Mathematical proof of statement

1. Jul 30, 2015

### Juraj

There is an article on Wikipedia gravity assist and in the explanation of the term, author gives an analogy:
Isn't this violating the law of conservation of energy and momentum? Can someone make a mathematical proof of this? I tried, but the variables like the mass of the train and the tennis ball are missing - I'm interested how is it possible to predict such a movement without doing a calculation.

2. Jul 30, 2015

### Qwertywerty

No , it isn't .

Take a wall . Throw a ball at it . It will reflect back with nearly the same velocity ( e ≈ 1 ) , as wall will not have gained any velocity .

Now you have a moving wall . Throw the ball . What will happen ? The moving train case will be analogous to
this .

3. Jul 30, 2015

### Juraj

But why is the final velocity of the tennis ball 130 m/s? Why did it gain double train's velocity? I need a mathematical proof of this

4. Jul 30, 2015

### A.T.

No, because the train is slowed down a little by the collision, and transfers some of its kinetic energy and momentum. The same happens with a moving planet that you use for gravity assist. Or with the moving air mass that you use in dynamic soaring:

5. Jul 30, 2015

### Juraj

I understand that it transfers it's energy and momentum, but how can we say that the velocity of the ball will be 130 m/s without knowing the momentum of the train nor the tennis ball? We can't know how much energy and momentum will be transfered.

6. Jul 30, 2015

### A.T.

That's just an idealization, which assumes the collision is perfectly elastic and the mass of the ball is negligible compared to the train's mass. In reality it will always be less than that.

7. Jul 30, 2015

### Qwertywerty

Let velocity of train be v and that of ball initially be u .

Use formula for e → e = vsep/vapp . e ≈ 1 ( not exactly , as AT has said ) . This gives you → v1 - v = u + v ( where v1 is final velocity of ball - in opposite direction ) .

8. Jul 30, 2015

### Juraj

I think that the analogy assumes the collision is perfectly elastic. I don't quite follow your math here.
Are you sure? The explanation of this analogy says that the ball has gained twice the speed of train, this doesn't seem to me like idealization...

9. Jul 30, 2015

### Qwertywerty

Which part ?

10. Jul 30, 2015

### Juraj

Why are you defining e = vsep/vapp if the collision is perfectly elastic? I'm not familiar with the meaning of the indexes...

11. Jul 30, 2015

### A.T.

12. Jul 30, 2015

### Qwertywerty

I am using the definition of the coefficient of restitution ( e ) - velocity of seperation upon velocity of approach - between two bodies , which is valid for all collisions .

This is usually in the range of 0 to 1 → 0 for a perfectly inelastic collision and 1 for an elastic collision ( Why ? - See relative velocity ) .

Since this is assumed an almost elastic collision , I take e = 1 , and write vapproach = vseperation .

13. Jul 30, 2015

### Staff: Mentor

This is easiest to see in the center of momentum frame. In that frame we have:
$m v_0 + M V_0 = 0$ before the collision and
$m v_1 + M V_1 = 0$ after the collision due to conservation of momentum and
$v_0 = -v_1$ and $V_0=-V_1$ due to conservation of energy.

Solving for $\Delta v = v_1-v_0$ and $\Delta V = V_1-V_0$ and eliminating $v_1$, $V_1$, and $V_0$ we get:
$\Delta v = -2 v_0$
$\Delta V = 2 v_0 m/M$

In the limit as $m/M$ goes to 0 we have $\Delta V$ goes to 0, so the center of momentum frame is approximately the frame of the more massive object if the masses are very different. So the small object bounces back with twice the speed and the large object is (approximately) unaffected.

14. Jul 30, 2015

### A.T.

Twice which speed in which frame?

15. Jul 30, 2015

### Staff: Mentor

When you bounce a ball off a wall, the velocity change is 30 - (-30) = 60 m/s. Double the initial speed because you changed from 30m/s in one direction to 30 m/s in the other.

16. Jul 30, 2015

### A.T.

He isn't asking about the velocity change (which would be 160m/s not 60m/s is his example). He is asking about the speed change, which is double the wall(train) speed (not double the initial ball speed).

17. Jul 30, 2015

### Staff: Mentor

Since this is a 1d problem, speed change and velocity change are the same and whether its the ball or train speed he's concerned about, the choice of which object/frame is called "stationary", doesn't really change anything except the perspective. Ie, if the ball is moving toward a stationary wall at 30 m/s or the wall is moving toward the ball at 30 m/s, the speed change of the ball is the same 60 m/s. Adding a 3rd frame and adjusting the values doesn't change how that works. And from that one explanation, the OP should be able to figure out any other permutation.

18. Jul 30, 2015

### Staff: Mentor

There is also an implied assumption in these problems that the wall/train is truly unchanged by the collission. As such, only the ball's momentum is relevant, it is conserved and since the mass is constant, you can just use speed.

19. Jul 30, 2015

### Juraj

But the small object doesn't bounce back with the twice of it's initial velocity, according to the analogy. It bounces with velocity of 2V + v0

Last edited: Jul 30, 2015
20. Jul 30, 2015

### A.T.

No they aren't. In the given example the speed changes from 30 to 130 km/h (a change of 100 km/h), while the velocity changes from -30 to 130 km/h (a change of 160 km/h).

The ball's momentum isn't conserved.

Last edited: Jul 30, 2015
21. Jul 30, 2015

### Staff: Mentor

Aack! My description was horribly bad.

The change in velocity is equal to (negative) twice the initial velocity in the center of momentum frame. If M>>m then the initial velocity in the CoM frame is approximately equal to the relative velocity.

For other frames you need to add the velocity of M.

22. Jul 30, 2015

### Juraj

How actually did you get
this?

23. Jul 30, 2015

### A.T.

The ball's velocity change is twice the initial relative velocity. The ball's speed change is twice the train speed in the given frame..

24. Jul 31, 2015

### Qwertywerty

Velocity of approach = v + u ( Move towards each other ) .
Velocity of separation = v1 - u ( Move away from each other ) .

25. Jul 31, 2015

### Juraj

Oh, I get it now, but this is valid only if the momentum of the more massive body is taken as unchanged, right? We obviously can't apply this in every instance.