# Conservation of momentum 1-D problem: track is not level

This problem is a standard kind of 1-D problem. You have 2 equal carts approaching each other and collide on linear air track. BUT the person setting up the track did NOT do a good job at making the track Level(or NOT perfectly horizontal). THE QUESTION is : HOW "UNLEVEL" is the track?

The collision took t= 0.750 seconds. And the data collected is below:

## DATA

m1=m2= 1kg

Initial velocities:

v1i= +0.450m/s

v2i= -0.505m/s

Final velocities:

v1f= -0.510m/s

v2f= +0.420m/s

From the data we can easily calculate the momentum before and after the collision and see that the momentum of this system is NOT Conserved.

To me this makes sense because the track is unlevel, we have made the track to be like an inclined plane which means that gravity now has influence on this system! SO we have now gone into 2-D space, and we have acceleration happening due to gravity.

I was able to figure out this much. BUT how do I calculate, based on the given information, say the angle that the track makes with respect to the horizontal.(This is what I guess the question is asking).

BUT, I also don't understand how they can measure the velocities(in the data given) if we have acceleration, which means that the velocity is changing(velocity is no longer constant). SO I see some contradictions here in this question.

I found this question in an old book, and there are no answers to the questions. BUT this one got me interested.

Hope somebody has seen this kind of question before. Your help is greatly appreciated.

• Typically collision experiments like this use photogates and simply measure the time between sensing and the known width of the object to arrive at the velocity. If multiple points are taken, then lines of best fit are used to create a velocity. – Kyle Kanos Nov 19 '14 at 19:09
• OK, thanks Kyle. So you are saying that with these devices, this measurement process, in the presence of acceleration, this would then measure the average velocity. BUT, if am not mistaken, isn't momentum all about the velocities all being constant velocities, except for the collision aspect where the velocities change, but still remain constant afterwards. – Palu Nov 19 '14 at 19:15
• Yes, it still measures the average velocity. Momentum for an object can clearly change, though in a force-free region the net momentum will be constant. – Kyle Kanos Nov 19 '14 at 19:21
• In the general types of momentum questions, i believe it is always done with constant velocity. I just read that if velocity is changing(ie, acceleration) then the momentum is changing from one point to the next. In this situation if momentum is changing from t1,t2,t3 etc, how would one be able to apply conservation of momentum in this case. I only know how to apply it when velocity before and after a collision is constant. SO that is why this problem is confusing, is the given velocities constant, afterwards, or is it just a snapshot of a particular time, with acceleration being present. – Palu Nov 19 '14 at 19:35
• If there is an acceleration, then you do not have a force-free region and thus the conservation of momentum does not apply. – Kyle Kanos Nov 19 '14 at 19:36

You always have to satisfy the momentum equations, which is only the linear momentum equation for this one dimensional case:

$$m_1v(t_1)-m_1v(t_0) = \int_{t_0}^{t_1} F dt$$

Assume the collision is completely elastic and all is conservative, so no plastic deformation, drag or any kind of damping. Then the only force which acts is the gravity:

$$F = sin(\theta)mg$$

where $\theta$ is the angle the track is making with respect to 'levelness'. Now we know that the momentum equation should be valid for each individual body, but also for the entire system. Again, the only external force is gravity. I assume that the velocities which are given are at the instant before and after the collision, as they are also a function of time and no time 'after collision' is given.

The momentum before the collision is:

$$L(t_0) = m_1 v_1(t_0) + m_2 v_2(t_0)$$

And after the collision:

$$L(t_1) = m_1 v_1(t_1) + m_2 v_2(t_1)$$

We know the collision time, and the force integral is not a function of time. So we obtain:

$$L(t_1) - L(t_0) = sin(\theta)(m_1 + m_2)g t_c$$

with $t_c$ as collision time. Solve for $\theta$, which now is the only unknown.

$$sin(\theta) = \frac{L(t_1) - L(t_0)}{(m_1 + m_2)g t_c}$$

• Thanks Rhino, i believe that did it. The question does not make clear that the velocities are immediately before and after the collision. BUT assuming this, then I agree with you that this would be the answer. Thanks again. – Palu Nov 20 '14 at 0:44