# Conservation of Momentum Exercise

2015-11-12# Problem Statement

In a train yard, train cars are rolled down a long hill in order to link them up with other cars as shown. A car of mass 4000. kg starts to roll from rest at the top of a hill 5.0 m high, and inclined at an angle of 5.0° to the horizontal. The coefficient of rolling friction between the train and the track is 0.050. What velocity would the car have if it linked up with 3 identical cars sitting on flat ground at the bottom of the track?

Hint: The equation for rolling friction is just like the one for sliding friction.

# Illustration

# My Solution

The speed of the car after leaving the hill is equal to

```
v = sqrt(kinetic_energy * 2 / m)
```

And its kinetic energy will be equal to the work done on it:

```
kinetic_energy = F * d
```

Getting the distance is a matter of simple geometry:

```
d = csc (5°) * 5 m ~= 57.37 m
```

The force is a bit more complicated, note that I am using the net force, taking friction into account

```
F = tan (5°) * 4 * 10^4 N - 0.05 * sec (5°) * 4 * 10^4 N
= 4 * 10^4 N * (tan (5°) - 0.05 * sec (5°)) ~= 1,492 N
```

Throwing everything together

```
v = sqrt(F * d * 2 / m) ~= 6.542 m / s
```

As the velocity will be equally divided between the cars

```
v = (6.542 m / s) / 4 = 1.635 m / s
```

Thus, the answer is **1.64 m / s**.

Note that I used `10 m s^-2`

for the gravitational acceleration. If you want to
be more precise, use `9.80665 m s^-2`

, the the standard acceleration due to
gravity.