Question

A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumper at the end of the track. If the spring constant of the bumper is 2x10^6 N/m, what is the maximum compression of the spring during the collision? (Assume collision is elastic)

Answer 1

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

$U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where we have

$k=2\cdot 10^6 N/m$ is the spring constant of the bumper

x is the maximum compression of the bumper

$m=30\cdot 10^4 kg$ is the mass of the car

$v=2.0 m/s$ is the speed of the car

Solving for x, we find the maximum compression of the spring:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m$

Answer 2

Answer:

0.6 m

Explanation:

The law of conservation of energy states that:

$\Delta E_m=0$

The mechanical energy ($E_m$) is the sum of the kinetic energy and the potential energy:

$\Delta K+\Delta U=0\\K_f-K_i+U_f-U_i=0\\\frac{mv_f^2}{2}-\frac{mv_i^2}{2}+\frac{kx_f^2}{2}-\frac{kx_i^2}{2}=0$

$U_i$ is zero since the spring is not initially compressed and $K_f$ is zero since all kinetic energy becomes potentital energy:

$\frac{kx_f^2}{2}=\frac{mv_i^2}{2}$

Finally, we solve for x and replace the given values:

$x_f^2=\frac{mv_i^2}{k}\\x_f=\sqrt{\frac{mv_i^2}{k}}\\x_f=\sqrt{\frac{(30*10^4kg)(2\frac{m}{s})^2}{2*10^6\frac{N}{m}}}\\x_f=0.6 m$