Answer to Question #128093 in Mechanics | Relativity for John

Question

A 500-ton train and collides with a buffer stop while traveling at 2 m/s. A buffer stop is a safety device and is designed to stop trains in case of brake failures. The mass of the buffer stop is 1.5 tons. It acts as a spring with a spring constant k = 1 x 106 N/m.

a. Estimate the mean deceleration, in [m/s2], experienced by the train as it is halted.

b. Estimate the distance, in [m], by which the buffer stop spring is compressed.

Answer 1

The kinetic energy of the train got stored as elastic potential energy of the spring as a result of stopping it. So, if the spring gets compressed by a distance of x and it's force constant is k, its elastic potential energy stored will be $0.5kx^{2}$. And if the train of mass m was moving with a speed of v, then its kinetic energy was $0.5mv^{2}$ . So, equating both, we get:

$0.5mv^{2} = 0.5kx^{2}$

$0.5\times500\times10^{3}\times2^{2}=0.5\times1\times10^{6}\times x^{2}$

x=1.41m

Answer b) The distance, in [m], by which the buffer stop spring is compressed = 1.41m

Then we can calculate the deceleration of the train:

$v^{2}=u^{2}+2as$ , where v - final speed, u - initial speed, a - acceleration, s - distance.

$0^{2}=2^{2}+2\times a\times 1.41$

a=-$1.41\frac{m}{s^{2}}$

Answer a) Deceleration $1.41\frac{m}{s^{2}}$