# A frame-of-reference riddle

1. Apr 20, 2014

### bobie

Suppose a bowler is standing on the edge of platform of a train speeding at 100 m/s, a bowl in his hand (m=1Kg) has KE 5000 J.
The bowler throws his ball at 10 m/s giving the ball 50 J of KE , so the balls has total KE 5050 J, right?.

Suppose now that the ball falls outside the platform on the earth, hitting a ball of identical mass, this should fly off at 110 m/s, is this right?..
..but, if so, the second ball has 6050 J of KE, where does the missing 1000 J of KE come from?

2. Apr 20, 2014

### Staff: Mentor

It came from the kinetic energy of the train, which was very slightly reduced when the bowler threw the ball forward and therefore pushed slightly backwards against the train. When you said that "the bowler throws his ball at 10 m/sec", you were ignoring this very small effect in the train frame.

Last edited: Apr 20, 2014
3. Apr 20, 2014

### sophiecentaur

There is nothing "missing". You just have to be careful about what you compare with what. The KE of the rest of the train is not relevant because the momentum change, when the ball is thrown can be as little as you like (just choose the mass of the train to be large enough wrt the ball).
It's all down to how KE is defined and relates to the amount of damage the ball can do when it hits something. If the ball hits a target on the train then it can only do 50J worth of damage (mv2/2 where v is the relative velocity). If the ball is dropped from the train is will just do 5000J of damage. But if the ball has a velocity of 110m/s, it can do 6050J of damage. 6050J is the amount of energy that would be needed to accelerate the ball from rest to 110m/s. The figure of 50J, although it seems relevant, represents no more than the energy needed to accelerate the ball relative to the train. It's the velocity relative to the ground that counts when calculating KE in the Earth frame.

Arm waving argument, but you could flow it up with sums if you wanted to:
If you wanted to accelerate the ball from 100m/s to 110m/s with an externally applied force (not a moving hand on the train), then the Force times distance (work done) would involve a distance (during the acceleration time), appropriate to speeds of 100 up to 110 m/s and not 0 up to 10m/s (the speeds of the hand). That's what accounts for your "missing KE" - but it's not actually missing.

4. Apr 20, 2014

### dauto

The energy transferred to the ball is frame dependent someone in the train sees the ball gain 50 J someone off the trains sees the ball gain 1050 J. In physics lingo we say "energy is not an invariant" the amount of energy transferred between objects is relative. Welcome to relativity land.

5. Apr 20, 2014

### bobie

Thanks, sophiecentaur, the ball falling to the ground surely can do 6050J of damage, that is a fact. So I suppose it possesses that KE also before it leaves the platform, is that right?.
I realize that the bowler's arm is swinging at 100 m/s (relative to the ground) but, suppose the acceleration of 10 m/s is given not by the arm on the train nor by an external force, but by an independent force , say internal to the bowl, like a jet propulsion.
This force would be absolute, it would provide the same amount of thrust/work/KE of 50J in any possible frame or reference or not? Would we then know for sure that the ball gets 5050J?

6. Apr 20, 2014

### sophiecentaur

It doesn't matter what is providing the force. The Work done will be Force times average speed times time.
If the Force of the arm is F and the time is t, the Work done by the thrower will be 10Ft/2 but the work done ON the ball will be 110Ft/2, in the Earth frame. Whatever provides the force (rocket or string plus winch on the ground), the result is the same for the final KE referenced to the ground. The ball would have to end up with 6050J, either way. The 5050 J figure is not relevant to this problem; it's just a red herring number that has no physical meaning. (What is does represent is the energy of a collision with the floor of the train and the a collision with the ground if it subsequently fell of the train at 100m/s)

7. Apr 20, 2014

### Staff: Mentor

Then you'd have to keep track of the kinetic energy of the jet exhaust as well. The train and ground observer will agree about the amount of energy produced by the fuel burn, but will allocate it to the ball and the exhaust in different ways.

8. Apr 20, 2014

### sophiecentaur

The wasted energy is not really an issue in this case. All that's necessary to consider is the Force produced and the distance it acts for. This tells you the work done on the ball and that is the KE added. (1050J).

9. Apr 20, 2014

### dauto

No, the work produced by a force is not an invariant. it will be different depending on the reference frame. There is no way around that.

10. Apr 20, 2014

### Vb123

It came from KE of train because if motion is analysed from some some who is on the ground he /she will take into account the kinetic Energy generated from the motion of train.

11. Apr 20, 2014

### Staff: Mentor

The amount of damage done is frame invariant, a fact as you call it. But that does not imply anything about the KE of the ball. Consider a ball at rest, with 0 KE, it can still cause damage if it collides with a fast moving vehicle.

12. Apr 21, 2014

### sophiecentaur

It just depends upon the relative velocities of the colliding objects. The whole point of this thread is that the definition of KE is, in fact, counterintuitive. If it were not, there would be no worry about 'missing energy'.

13. Apr 21, 2014

### D H

Staff Emeritus
Wrong.

Kinetic energy is a frame dependent quantity, proportional to the square of velocity. That means you cannot add kinetic energies computed in two different frames of reference. You're adding apples and oranges. The kinetic energy of the unbowled ball in the platform frame is ½*(1 kg)*(100 m/s)2 or 5000 J. The kinetic energy of the bowled ball in the bowler's frame is ½*(1 kg)*(10 m/s)2 or 50 J. That doesn't mean that the kinetic energy of the bowled ball in the platform frame is 5050 J. You can't add things that way. The kinetic energy of the bowled ball in the platform frame is ½*(1 kg)*(110 m/s)2 or 6050 J.

From the train. You forgot about Newton's third law.

Suppose the train has a mass of 1 million kg and that it rolls on a frictionless track. Suppose the train is initially standing still next to a platform. The bowler bowls the ball such that an observer on the platform sees the ball having a velocity of 10 m/s. The train is now moving backwards at a velocity of 10-5 m/s thanks to Newton's third law. The total energy (in the platform frame) of the train+ball system has changed from zero joules to 50.00005 joules.

Now suppose the train is moving at 100 m/s just before the bowler bowls the ball. The energy of the train+ball system prior to bowling the ball is ½*(1e6 kg + 1 kg)*(100 m/s)2 or 5.000005 GJ. (The energy of the ball itself is just 5000 J in the platform frame). The bowler now applies that same 50.00005 joules of energy needed to make the ball move 10 m/s faster than the train. The energy of the ball is not 5050 joules. It's 6050 joules.

So where did that extra 1000 joules come from? It came from the train. The train's kinetic energy prior to bowling the ball was 5 GJ. After bowling, it's ½*(1e6 kg)*(100 m/s - 10-5 m/s)2 or 4.999999 GJ. The net change in energy of the train+ball system is 50.00005 joules. No matter which inertial frame you choose to use, the kinetic energy of the train+ball system increases by 50.00005 joules as a result of the bowler bowling the ball.

14. Apr 21, 2014

### D H

Staff Emeritus
Aside: There's a similar effect in orbital mechanics, the Oberth effect. You get the biggest bang for your buck if you fire your jets at periapsis (closest approach).

15. Apr 21, 2014

### sophiecentaur

That neatly nails the problem I think. Total energy is the same.