Jan 272018

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.

 Posted by at 9:54 am

  10 Responses to “Kinetic energy”