Question

A runaway railroad car, with mass $3.0\times {10}^5\text{kg}$, coasts across a level track at$2.0\text{m/s}$when it collides elastically with a spring-loaded bumper at the end of the track. If the spring constant of the bumper is, what is the maximum compression of the spring during the collision? (a) (b) $0.58m$(c)$0.34\text{m}$(d)$1.07\text{m}$(e)$1.24\text{m}$.

Short answer

Option (a) is correct. The maximum compression of the spring during the collision is$x=0.77\text{m}$

Step-by-step solution

Step 1: The total energy of a simple harmonic oscillator

The total energy of a simple harmonic oscillator is a constant of the motion and is given by

$E=\frac{1}{2}k{A}^2$

$E=$Total energy

$L=$Length of pendulum

$K=$Constant

Find when it collides elastically with a spring-loaded bumper at the end of the track

The car will continue to compress the spring until all of the car’s original kinetic energy has been converted into elastic potential energy within the spring, i.e., until

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$Or

$$\begin{gathered} x=v_i\sqrt{\frac{m}{k}} \\ x=\left(2.0m/s\right)\sqrt{\frac{3.0\times {10}^{5}kg}{2.0\times {10}^{6}N/m}} \\ x=0.77m \end{gathered}$$

Option (a) is correct. The maximum compression of the spring during the collision is$x=0.77\text{m}$