Question :

two railroad boxcars weighing $50,000$ N each are approaching one another at $5$ m/s and $7$ m/s. After they collide and couple together, how fast will the combination be moving and in which direction?

From the problem I have the following setup:

enter image description here

Using $C_i$ as $car_i$ we have weight $C_1 = $ weight $C_2 = 50,000 N$ Then using $g = 10$ we have:

$$ \text{mass } C_1, C_1 = 5000 kg $$

Conservation of linear momentum states

$$ m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2 $$

As $m_1 = m_2$ this gives

$$ m (v_1 + v_2) = m (u_1 + u_2) $$

Using given values (and taking left to be negative) we have

$$ 5 - 7 = u_1 + u_2 $$

Then we know that, as the cars stick together after the collision they're moving in the left direction, with a velocity of $-2 m/s$

Assuming that the above working is correct - I'm not sure how I would have determined whether or not the cars would have stuck together if the problem hadn't stated that they had? Do I need information about the materials to answer that?

Hopefully the working, and reasoning, is clear and valid.

  • 1
    $\begingroup$ Call the final speed $u$ for both of the boxcars so the final momentum is $mu+mu=2mu$. $\endgroup$ – Farcher Apr 29 '17 at 15:39
  • $\begingroup$ Shouldn't the answer be -1 m/s ? $\endgroup$ – Not_Einstein Feb 1 at 14:01

Yes, in a 1D collision you would also need information about the elasticity of the collision (ie what proportion of the total KE is lost) as well as conservation of momentum to determine both final velocities. Elasticity of collision is quantified by the coefficient of restitution COR which is the ratio of the relative velocities of separation and approach. Like coefficients of friction, the COR is usually the same for two particular materials across a range of relative speeds.

In this case you are told that the objects stick together. This tells you that the COR=0.

In a 2D or 3D collision you would need further information. If the objects are circles or spheres you would need to know their radius and the distance between their centres along the line of approach. Point particles have no radius so it is impossible to find this vital information and impossible to predict the outcome. If there is friction the spheres could exchange rotational momentum as well as linear momentum. See How do linear momentum and angular momentum divide on collision?

| cite | improve this answer | |
  • $\begingroup$ thanks - so I would need to use have the KE lost to determine that without the information given in the problem? (And the other working makes sense?) $\endgroup$ – baxx Apr 29 '17 at 15:40
  • 1
    $\begingroup$ That is correct. If you were not told that the objects stick together (COR=0) then you would not have enough information to predict the outcome even for a 1D collision. $\endgroup$ – sammy gerbil Apr 29 '17 at 15:44
  • $\begingroup$ OK cool - I wasn't too sure whether there was a way of solving that. thanks $\endgroup$ – baxx Apr 29 '17 at 15:45

In the real world predicting such outcomes is not trivial. To begin with, the properties of the materials involved in the collisions and how they behave under various forces involved matter.

For example, putty balls stick together when they hit each other and lose kinetic energy as heat and work done to change the shape of the balls. Steel balls on the other hand usually collide elastically and waste little kinetic energy as heat (they lose some energy as sound). A typical demonstration of this is Newton's cradle.

In the case of rail road cars, the engineering details may come into question too. For example, when cars collide they may bounce back elastically or get locked at impact using a hook.

The bottom line is that figuring this out is not trivial and that is why most undergraduate level problems instead state that as one of the given facts.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.