Question

How fast did the wagon move, with a mass of 20 t, if during the collision with the stop, each of its two systems was compressed by 10 cm. It is known that each spring is compressed 1 cm under the action of force 9800 N.

Answer 1

According to the energy conservation law, the kinetic energy of the mooving wagon was converted to the potential energy of spring elastic deformation:

$$\dfrac{mv^2}{2} =2\cdot \dfrac{kx^2}{2}$$

where $m = 20t = 20\cdot 10^3kg$ is the mass of the vagon, $v$ is its initial velocity, $k$ is the spring constant of each spring, $x = 10cm = 0.1m$ is the compression rate of each spring. Factor 2 appeares since there are two identical springs in the system.

The spring constant can be found from the Hook's law:

$$k = \dfrac{F}{x_0}$$

where $F = 9800N$ is the force, and $x_0 = 1cm = 0.01m$ is the compression under this force.

Thus, obtain:

$$v = x\sqrt{\dfrac{k}{m}} = x\sqrt{\dfrac{F}{x_0m}}\\ v = 0.1\cdot \sqrt{\dfrac{9800}{0.01\cdot 20\cdot 10^3}} = 0.7m/s$$

Answer. 0.7 m/s.