Enough blogging about personal stuff like our cats. Here is a neat little physics puzzle instead.

Solving this question requires nothing more than elementary high school physics (assuming you were taught physics in high school; if not, shame on the educational system where you grew up). No tricks, no gimmicks, no relativity theory, no quantum mechanics, just a straightforward application of what you were taught about Newtonian physics.

We have two parallel rail tracks. There is no friction, no air resistance, no dissipative forces.

On the first track, let’s call it *A*, there is a train. It weighs 10,000 kilograms. It is accelerated by an electric motor from 0 to 10 meters per second. Its kinetic energy, when it is moving at \(v=10~{\rm m/s}\), is of course \(K=\tfrac{1}{2}mv^2=500~{\rm kJ}\).

Next, we accelerate it from 10 to 20 meters per second. At \(v=20~{\rm m/s}\), its kinetic energy is \(K=2000~{\rm kJ}\), so an additional \(1500~{\rm kJ}\) was required to achieve this change in speed.

All this is dutifully recorded by a power meter that measures the train’s electricity consumption. So far, so good.

But now let’s look at the *B* track, where there is a train moving at the constant speed of \(10~{\rm m/s}\). When the *A* train is moving at the same speed, the two trains are motionless relative to each other; from *B*‘s perspective, the kinetic energy of *A* is zero. And when *A* accelerates to \(20~{\rm m/s}\) relative to the ground, its speed relative to *B* will be \(10~{\rm m/s}\); so from *B*‘s perspective, the change in kinetic energy is \(500~{\rm kJ}\).

But the power meter is not lying. It shows that the *A* train used \(1500~{\rm kJ}\) of electrical energy.

Question: Where did the missing \(1000~{\rm kJ}\) go?

First one with the correct answer gets a virtual cookie.

Expand for my answer.

I guess there’s no missing kinetic energy because it is dependent on the frame of reference. Electrical energy recorded on the power meter is not a function of speed.

1. Kinetic energy is not relative like potential energy.

2. If you put the A train on the B train and accelerates there the necessary energy is just 500 kJ. But it’s very difficult before the length of train B. ;-)

Let me know when you sort it out. I am following with much interest.

My understanding is that you are accelerating the train over a longer distance as the speed increases, so higher energy.

You integrate over m * v.

The 1000 kj is not lost anywhere. 1500 kj reading is recorded when A accelerated from 10 to 20. If similar recording was recorded when A accelerated from 0 to 10 it should be 500 kj. So in relative to B when A moves from 0 to 10 ( actually 10 to 20) one would expect the reading to be 500kJ but since the recordigs are absolute not relative you have 1500 in machine. Velocity to Kinetic energy relationship is not linear. (Try plotting V vs Ke for 5, 10, 15 and 20 )So 0-10 brings 500 while 10 – 20 brings 1500 difference.

Of course this is assuming the second train is also 10 000 kg

I got the cookie ðŸ˜‹

To whom it may concern: I now posted the solution in my blog. See https://spinor.info/weblog/?p=8789

3 guys go into a hotel & rent room for $25, giving clerk 3 ten dollar bills. clerk sends bellhop with change. he pockets 2 & returns a dollar to each . 9 x 3 = 27, *+* 2 = 29. where is the extra dollar ?