A runaway railroad car,with mass 45.0 × 10⁴ kg,coasts across a level track at 3.80 m/s when it collides with a spring-loaded bumper at the end of the track.If the spring constant of the bumper is 3.80 × 10⁶ N/m,what is the maximum compression of the spring during the collision? (Assume the collision is elastic. )

Question

Quizplus · Accepted Answer

The maximum compression of the spring can be found using the conservation of energy principle. The initial kinetic energy of the railroad car is converted into the potential energy stored in the spring at maximum compression. The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the car and $v$ is its velocity. The potential energy (PE) stored in the spring at maximum compression is given by $PE = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the compression distance. Setting KE equal to PE and solving for $x$ gives:$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$Given $m = 45.0 \times 10^{14} \, \text{kg}$, $v = 3.80 \, \text{m/s}$, and $k = 3.80 \times 10^{16} \, \text{N/m}$, we can solve for $x$:$$x = \sqrt{\frac{mv^2}{k}} = \sqrt{\frac{(45.0 \times 10^{14} \, \text{kg})(3.80 \, \text{m/s})^2}{3.80 \times 10^{16} \, \text{N/m}}}$$Solving this equation gives $x = 1.31 \, \text{m}$, which corresponds to choice A.

A Runaway Railroad Car,with Mass 45