Mathematical proof of statement

[Juraj]

There is an article on Wikipedia gravity assist and in the explanation of the term, author gives an analogy:

A close terrestrial analogy is provided by a tennis ball bouncing off the front of a moving train. Imagine standing on a train platform, and throwing a ball at 30 km/h toward a train approaching at 50 km/h. The driver of the train sees the ball approaching at 80 km/h and then departing at 80 km/h after the ball bounces elastically off the front of the train. Because of the train's motion, however, that departure is at 130 km/h relative to the train platform; the ball has added twice the train's velocity to its own.

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.

 

[Qwertywerty]

Isn't this violating the law of conservation of energy and momentum?

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 .

 

[Juraj]

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 .

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

 

[A.T.]

Isn't this violating the law of conservation of energy and momentum?

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:

 

[Juraj]

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:

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.

 

[A.T.]

but how can we say that the velocity of the ball will be 130 m/s

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.

 

[Qwertywerty]

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.

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

Use formula for e → e = v_sep/v_app . 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 ) .

 

[Juraj]

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

Use formula for e → e = v_sep/v_app . 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 ) .

I think that the analogy assumes the collision is perfectly elastic. I don't quite follow your math here.

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.

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...

 

[Qwertywerty]

I think that the analogy assumes the collision is perfectly elastic.

This is answered by -

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.

I don't quite follow your math here.

Which part ?

 

[Juraj]

Which part ?

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

 

[A.T.]

I don't quite follow your math here.

https://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

 

[Qwertywerty]

I am using the definition of the coefficient of restitution ( e ) - velocity of separation 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 v_approach = v_seperation .

 

[Dale]

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.

 

[A.T.]

So the small object bounces back with twice the speed

Twice which speed in which frame?

 

[russ_watters]

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

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.

 

[A.T.]

When you bounce a ball off a wall, the velocity change is 30 - (-30) = 60 m/s.

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).

 

[russ_watters]

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).

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.

 

[russ_watters]

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.

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.

 

[Juraj]

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.

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

 

[A.T.]

Since this is a 1d problem, speed change and velocity change are the same

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).

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.

The ball's momentum isn't conserved.

 

[Dale]

Twice which speed in which frame?

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.

 

[Juraj]

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

Use formula for e → e = v_sep/v_app . 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 ) .

How actually did you get

v1 - v = u + v

this?

 

[A.T.]

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

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..

 

[Qwertywerty]

How actually did you get this?

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

 

[Juraj]

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

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.

 

[russ_watters]

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.

Yes. Momentum (and energy) is conserved here, but the larger the more massive body, the smaller its speed change and the larger the smaller body's speed change. As the larger body's mass tends toward infinity, its speed change tends toward zero and the smaller body's speed change is double the initial speed (or the same speed, in the opposite direction). Since the masses are so different from each other, we just ignore the larger object's momentum gain and assume the smaller object keeps all its momentum (in the opposite direction) in the bounce.

As a counterexample, in billiards, since the masses of the balls are the same, the speed change of each ball is the same (in opposite directions). So a moving ball hits a stationary ball and stops, whereas the stationary ball moves away with the initial velocity the moving ball previously had.

 

[Juraj]

Yes. Momentum (and energy) is conserved here, but the larger the more massive body, the smaller its speed change and the larger the smaller body's speed change. As the larger body's mass tends toward infinity, its speed change tends toward zero and the smaller body's speed change is double the initial speed (or the same speed, in the opposite direction). Since the masses are so different from each other, we just ignore the larger object's momentum gain and assume the smaller object keeps all its momentum (in the opposite direction) in the bounce.

As a counterexample, in billiards, since the masses of the balls are the same, the speed change of each ball is the same (in opposite directions). So a moving ball hits a stationary ball and stops, whereas the stationary ball moves away with the initial velocity the moving ball previously had.

Thanks for the explanation, it's much clearer now.

Thank you everyone

 

[A.T.]

Thanks for the explanation, it's much clearer now.

As a follow up you should try this: Drop a well inflated basketball, with a small rubber bouncy ball on top, onto a hard surface. How high can the small ball jump compared to the initial drop height?

 

[A.T.]

As a follow up you should try this: Drop a well inflated basketball, with a small rubber bouncy ball on top, onto a hard surface. How high can the small ball jump compared to the initial drop height?

@Juraj Also demonstrated here:

 

[FactChecker]

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.

"Unchanged" is probably the wrong word to use. "Negligible change" is probably better. The change of the train velocity would be tiny because it is millions of times more massive than a tennis ball. It would change. More accurate would be something like the velocity of the train changes to 49.99999999 km/hour and the velocity of the tennis ball would be 80 + 49.99999999 = 129.99999999. But that is getting picky, right?