Question

A child has two red wagons, with the rear one tied to the front by a (non-stretching) rope. If the child pushes on the rear wagon, what happens to the kinetic energy of each of the wagons and the two-wagon system?

Answer 1

Answer:

The rear wagon gains the kinetic energy, but the front wagon will remain at rest.

The two-wagon system will gain a kinetic energy $\dfrac{1}{16}$ of the kinetic energy gained by the rear wagon.

Explanation:

Let's consider that the masses of the wagons to be 'M'. When the child pushes the rear wagon let's assume that the velocity of the rear wagon be 'v'.

Therefor the kinetic energy gained by the rear wagon be $K_{r} = \frac{1}{2}Mv^{2}$.

Now let's assume that the velocity of the centre of mass (C), as shown in the figure, be 'V'. So from momentum conservation law we can write,

$&& Mv = (M + M)V\\\\&or,& V = \frac{v}{2}$

Now the centre of mass ($M_{C}$) is given by

$M_{C} = \frac{M \times M}{M + M} = \frac{M}{2}$

So the kinetic energy ($K_{C}$) of the system will be

$K_{C} = \frac{1}{2}M_{C}V^{2} = \frac{1}{2}\frac{M}{2}(\frac{v}{2})^{2} = \frac{1}{16}Mv^{2}$