# Headon inelastic collsion between small car and big truck

1. May 31, 2013

### bkraabel

1. The problem statement, all variables and given/known data
Which head-on collision between a small car and a large truck
causes a larger conversion of kinetic energy: one in which their
initial momenta are the same magnitude, or one in which their
initial kinetic energies are the same? Assume the same kinetic
energy for the two-vehicle system in both cases.

2. Relevant equations

Notation:
$v_{1,i}^{(1)}$ is the initial speed of small car.
$v_{1,f}^{(1)}$ is the final speed of small car.
$v_{2,i}^{(2)}$ is the initial speed of large truck.
$v_{2,f}^{(2)}$ is the final speed of large truck.

Initial kinetic energy $K_i$ is the same in both cases,which gives
$\frac{1}{2}m_1(v_{1,i}^{(1)})^2\approx m_1(v_{1,i}^{(2)}) \equiv m_2(v_{2,i}^{(2)})$
and
$v_{1,i}^{(1)}\approx\sqrt{2}v_{1,i}^{(2)}$

3. The attempt at a solution
Assume same coefficient of restitution in both cases. Because $m_{1} << m_{2}$, this leads to $\frac{v_{1,f}^{(1)}}{v_{1,i}^{(1)}}\approx\frac{v_{1,f}^{(2)}}{v_{1,i}^{(2)}}$

For initial momentum the same (case 1), I get
$K_{f}^{(1)}\approx\frac{1}{2}m_{1}({v_{1,f}^{(1)}})^2\approx \frac12m\left[\frac{v_{1,f}^{(2)}}{v_{1,i}^{(2)}}v_{1,i}^{(1)}\right]^2= K_i\left[\frac{v_{1,f}^{(1)}}{v_{1,i}^{(1)}}\right]^2$

For initial kinetic energy the same (case 2), I get
$K_{f}^{(2)}\approx\frac12m_1(v_{1,f}^{(2)})^2+\frac12K_i$

Calculating the amount of kinetic energy converted into other forms of energy for both cases and taking the difference gives
$\Delta K^{(1)}-\Delta K^{(2)}=\frac12m_1(v_{1,f}^{(2)})^2[\sqrt{2}-1]-\frac12K_i$

But I can't determine whether the second term is less than or greater than the first term. Another approach is to note that the $K_f^{(2)}$ is greater than one half the initial kinetic energy, so if we could show that
$\left[\frac{v_{1,f}^{(1)}}{v_{1,i}^{(1)}}\right]^2<\frac12$
then we would know that $K_f^{(1)}< \frac12K_i$

Any ideas? Other approaches? Mistakes in my algebra?
thanks!

2. May 31, 2013

### haruspex

I feel you're making this much harder than necessary. Suppose the KE of the system before collision is E. If the two vehicles have equal and opposite momenta, how are they moving after collision? What will the KE of the system be then?

3. Jun 1, 2013

### bkraabel

First, let me clarify the notation that I've used:
$v_{k,i}^{(n)}$ is the initial speed of vehicle k for case n.
$v_{k,f}^{(n)}$ is the final speed of vehicle k for case n.

n = 1 is when initial momentum of vehicles is the same.
n = 2 is when initial kinetic energy of vehicles is the same.
k = 1 is car
k = 2 is truck

With that said, I'm afraid I don't quite see the point of the previous reply. When the initial momentum of each vehicle is the same, I know how the final speeds and kinetic energies are related. But I want to know how the difference in kinetic energy (pre vs post collision) compares between cases 1 and 2. Which gives a greater difference in kinetic energy?

thanks!

4. Jun 1, 2013

### haruspex

It's an inelastic collision. I would take it that the two vehicles coalesce. If the two vehicles have equal and opposite momenta, that allows you to write down immediately their final speeds.

5. Jun 2, 2013

### bkraabel

Yes, if the vehicles stick together it's quite simple, I agree. My take on is that the two vehicles do not stick together. I think the only thing that we can assume is that there is the same coefficient of restitution in both cases.

6. Jun 2, 2013

### haruspex

Ok, then assuming there is some 'bounce'...
This is unclear. I'll assume the lower index denotes vehicle while the upper index denotes scenario.
Don't you mean $\frac{1}{2}m_1(v_{1,i}^{(1)})^2+\frac{1}{2}m_1(v_{2,i}^{(1)})^2=\frac{1}{2}m_1(v_{1,i}^{(2)})^2+\frac{1}{2}m_1(v_{2,i}^{(2)})^2$?
I don't see the need to be making approximations.
How about starting with the equation that says the initial momenta are the same, no more and no less?

7. Jun 2, 2013

### bkraabel

I think you need to know the coefficient of restitution to solve this. Consider the case in which the two cars have the same initial momentum. The final kinetic energy can range from 0 to $K_i$ depending on the coefficient of restitution. The same argument applies to the case in which the cars start with the same kinetic energy, although the limits of final kinetic energy are not so clear. So the best I could do is the following:

$\Delta K\approx \frac14u_{1,f}^2-\frac12K_i$

where $u_{1f}$ is the final speed of the car in the case where the momenta of the car and truck are the same (I've changed to using u for case 1 and v for case 2). To relate $u_{1f}$ to $K_i$ you need to know the coefficient of restitution.

8. Jun 2, 2013

### haruspex

It might only be necessary to use the given fact that the coefficient of restitution is less than 1. (I'm always a bit unsure what people mean by elastic and inelastic. I understand completely elastic and completely inelastic, but which term is used to cover the realistic case, where it's somewhere in between?)
If so, you will not get there using approximations - it will have to be exact.

9. Jun 2, 2013

### bkraabel

The coefficient of restitution covers the case between elastic and inelastic collisions. For $C_R=1$ the collision is elastic. For $C_R=0$ the collision is totally inelastic. I pushed the analysis a bit farther, again assuming $m_1\ll m_2$, and found
$\Delta K\approx \frac14m_1C_R^2u_{1i}^2-\frac12K_i$
which shows explicitly the dependence on the coefficient of restitution. But if you let
$\epsilon^2=1-C_R^2$
and recall that $K_i=\frac12m_1u_{1,i}^2+\frac12m_2u_{2,i}^2$ you get
$\Delta K\approx -\frac14\epsilon^2u_{1,i}^2-\frac14m_2u_{2,i}^2 < 0$
So it would appear that the change in kinetic energy is greater for the case in which both vehicles have the same initial kinetic energy.