Question

A 500 tonne train collides with a station emergency buffer stop while travelling at 2 ms-1. The buffer stop has a mass of 1.5 tonnes and is designed to act as a spring with stiffness K = 1 MN/m. (i) Estimate the extent that the buffer stop spring is compressed (Hint: consider the velocity of the train and buffer system after the collision – is there an easy approximation?) [~1.4 m]; (ii) Estimate the mean deceleration experienced by the train as it is halted. Is this deceleration likely to be a danger to the passengers? [~1.4 ms-2]

Answer 1

Answer:

a) Δx=1.41 m; b) a=1.41 m/(s^2)

Explanation:

a) To solve this problem, we should consider two processes:

1) inelastic collision between the train and the buffer;

2) motion of the pair to reach complete stop.

As the collision between the train and the buffer is inelastic, we have some energy wasted. However, the velocity after the collision can be determined using Momentum Conservation Law:

mtVt=(mt+mb)*V, where mt- mass of the train, mb- mass of the buffer, Vt- initial velocity of the train, V- velocity after the collision.

V=mtVt/(mt+mb)=1.994 m/s.

After train and buffer have collided, the motion can be described using Energy Conservation Law, where kinetic energy of the buffer-train couple is fully transmitted to the potential energy of elastic deformation of the spring. From this idea, the following equation can be derived:

$\frac{(mt+mb)V^{2} }{2}=\frac{Kx^{2} }{2}$ , where x- displacement of the spring, K- spring stiffness.

From this equation, we can find: $x=\sqrt{\frac{(mt+mb)V^{2} }{k} } =1.41 m$

Regarding an approximation, we can assume, that the buffer is mass-less in comparison with the train and this assumption will yield very small error. If we consider mass-less buffer, there is no need to use mass of the buffer in calculations and there is no need to consider collision and finding velocity after the collision. To support this assumption, we can compare initial velocity of the train and final during the collision- train slowed down from 2m/s to 1.994 m/s- the change is insignificant, so we can directly apply Energy conservation law to the train-spring system and avoid buffer's mass.

b) To find mean acceleration, we can use the following known values: initial velocity is 1.994 m/s and the traveled length is 1.41 m. As we are looking for the mean value, we can assume, that the acceleration was constant and the following equation can be used: $x=\frac{V^{2} }{2a}$, where a- mean acceleration, V- velocity of the train and x- traveled distance.

From this equation, we can find acceleration as: $a=\frac{V^{2} }{2x} =1.41 \frac{m}{s^{2} }$

This acceleration is significantly smaller, than g (gravitational acceleration), so it is not dangerous to the passengers.