# Finding the velocity after a collision?

1. Nov 27, 2013

### PhyIsOhSoHard

1. The problem statement, all variables and given/known data
A car is driving and hits the brakes for l1 distance before colliding with another car at rest who has the brakes on. After the collision they slide distance l2 together while braking.
The mass of both cars are known, and the friction between the road and the wheels is known as well.

Find the velocity of the car driving right before it hits the brakes.

2. Relevant equations
Something with momentum and impulse perhaps?

3. The attempt at a solution
Originally, I thought about using the total momentum equation, however I do not know any of the velocities before or after the collision. I only know the velocity of the car at rest before the collision.

$m_Av_{A1x}+m_Bv_{B1x}=(m_A+m_B)v_{2x}$

Where A denotes the moving car and B denotes the car initially at rest. 1 is before the collision and 2 is after.
I want to find $v_{A1x}$ and I know that $v_{B1x}=0$ but that still leaves me with $v_{2x}$.

Besides I don't even think that this equation is correct because I haven't taken the friction force into consideration.

Can somebody lead me to the right direction?

2. Nov 27, 2013

### Dick

You know what the problem is. Take frictional force into account. Use conservation of energy to find the velocity of the first car when they collide taking friction into account. Use conservation of momentum at the moment they collide to find the velocity of both cars after they collide. Then use conservation of energy again.

3. Nov 27, 2013

### PhyIsOhSoHard

So I use the law of conservation of energy:
ΔK + ΔU + ΔUint = 0

ΔK = K2 - K1 = 1/2mv22 = 1/2mv12

Already there I'm stuck. I don't know what the velocity of the car is before or after so how do I find an expression for the velocity before the collision?

4. Nov 27, 2013

### Dick

You aren't taking the frictional force into account. If the first car is traveling at velocity v1, then its initial kinetic energy is (1/2)m1*v1^2. If coefficient of friction is μ then what's the force of friction? If the car travels a distance l1, then how much energy is lost to friction? What's the final velocity at the point of collision?

5. Nov 28, 2013

### PeroK

It won't help, but it might be worth noting that when a car brakes, it doesn't slow down due to friction between the tyres and the road (unless you lock the brakes and it skids). Instead, the braking mechanism increases the "rolling resistance" on the motion of the wheels.

After the collision, it's possible that the brakes would be locked and the cars would skid.

The problem really ought to say that the braking forces are known.

6. Nov 28, 2013

### PhyIsOhSoHard

But how do I implement the friction force into the equation?
I know that the friction force is equal to the normal force multiplied by the coefficient of friction:
fk = μkn

And I know from my FBD that the normal force is equal to the gravity force:
fk = μkmg

But how can I implement this into my conservation of energy?

I can transform it into work done by friction. The work that friction does is:
Wfric = -fkl1 = -μkmgl1

Would it be collect to just subtract the kinetic energy with the friction energy? So I get:

$\frac{1}{2}mv^2_1-\mu_kmgl_1$

So this velocity is right before the collision. But what are they equal to?

7. Nov 28, 2013

### Dick

$\frac{1}{2}m_1 v^2_1-\mu_k m_1 g l_1$ will be the kinetic energy of the first car just before collision. So it's equal to $\frac{1}{2}m_1 v_c^2$ where $v_c$ is the velocity just before they collide. PeroK makes a good point about braking, but I think this is what they expect you to do.

8. Nov 28, 2013

### PeroK

That force would apply whether or not you are braking. The brakes don't turn on gravity, nor increase the mass of the car, nor cause friction on the road surface.

Yes, that would be correct for a block sliding along a road. Not for a car! But, the maths is correct.

The answer here will be a complicated function for v in terms of all the other variables. You're not going to get v = 10 m/s or anything like that.

You just have to keep going now and do the same for the second deceleration when the two cars are together. The answer will be a complicated function like you're working out.

PS Yes, I agree I think you're doing the problem as expected.

Last edited: Nov 28, 2013
9. Nov 28, 2013

### PhyIsOhSoHard

So I have from my conservation of mechanical energy that:
$K_1+U_1=K_2+U_2$

Since the potential energies are zero that leaves me with:
$K_1=K_2$
$\frac{1}{2}m_1v^2_1-\mu_km_1gl_1=\frac{1}{2}m_1v^2_c$

But my problem is here that I don't know either of the two velocities $v^2_1$ and $v^2_c$

I assume the first velocity is the velocity of the car before it brakes, correct?

I then have my conservation of momentum:
$m_Av_{A1x}+m_Bv_{B1x}=(m_A+m_B)v_{2x}$

Just before the collision, the car B is still at rest which means that $v_{B1x}$ is zero.

$m_Av_{A1x}=(m_A+m_B)v_{2x}$

From my conservation of energy I can isolate $v^2_1$ which is the velocity of the car before it brakes. But if I insert that into my conservation of momentum then I still do not know the velocity of both cars after collision $v_{2x}$

10. Nov 28, 2013

### PhyIsOhSoHard

Ya, I thought a long time about the braking force versus friction force and it does make sense that the breaking force plays a much bigger importance.
But I tried to draw a FBD with the braking force implemented and no matter what i tried, I could find any expression for the braking force.

11. Nov 28, 2013

### Dick

I think the idea here is just to write the initial velocity, which I called $v_1$, as a function of all of the 'known' variables in the problem $\mu_k, m_1, m_2, l_1, l_2$. So no, you don't know any of those quantities as a number. It is kind of complicated, but you can do it. Next solve the collision problem with $v_c$ as the initial velocity of $m_1$.

12. Nov 28, 2013

### Dick

Braking force would have to be yet another given variable, you can't really derive it from anything else here. I think we have enough variables. I'd just assume the brakes lock the wheels and the braking force is the road friction.

13. Nov 28, 2013

### PhyIsOhSoHard

Alright, so I tried the best that I could and this is what I came up with:

Conservation of mechanical energy before and after the collision gives me (no potential energy):
$K_1=K_2$

K1 is the kinetic energy of the driving car before the collision which when taking the work of the friction force which is opposite into consideration yields:
$K_1=\frac{1}{2}m_{driving}v^2_{driving}-\mu_km_{driving}gl_1$

K2 is the kinetic energy after the collision (as they collide their masses are added together):
$K_2=\frac{1}{2}(m_{driving}+m_{rest})v^2_{collision}$

Equal to each other as per the conservation of mechanical energy:
$\frac{1}{2}m_{driving}v^2_{driving}-\mu_km_{driving}gl_1=\frac{1}{2}(m_{driving}+m_{rest})v^2_{collision}$

The expression for the velocity after the collision is:

Conservation of momentum before and after the collision:
$m_{driving}v_{driving}+m_{rest}v_{rest}=(m_{driving}+m_{rest})v_{collision}$

Since the other car was at rest:
$m_{driving}v_{driving}=(m_{driving}+m_{rest})v_{collision}$

Now I can insert the expression of the velocity after the collision that I found earlier and find the expression for the velocity of the driving car before the collision, vdriving

Is this the correct method?

14. Nov 28, 2013

### Dick

Your are doing some stuff right. For one thing you should spell out clearly what your variables mean, particularly the v's. There are three interesting values of v, the initial velocity of the first car, which I think is your $v_{driving}$. Then there is the velocity of that car immediately before the collision and the velocity of both cars immediately after the collision. One thing that is definitely wrong is to relate those two by conservation of energy. Kinetic energy is not conserved in the collision, it's inelastic. You need to use momentum conservation alone at the collision time.

Last edited: Nov 28, 2013
15. Nov 28, 2013

### PhyIsOhSoHard

Alright. So my variables are:
vdriving=First car driving before the collision
vimpact=First car's velocity right before the collision
vcollision=Both of the car's velocities together after the collision

m1=First car that drives
m2=Second car at rest

My goal is to find vdriving.

Momentum conservation between the speed of the first car and the speed of both after collision gives:
$m_1v_{driving}=(m_1+m_2)v_{collision}$

Then I have energy conservation just before the impact:
$\frac{1}{2}m_1v^2_{imapct}=\frac{1}{2}m_1v^2_{impact}-\mu_km_1gl_1$

Do we agree on those two equations? :) Am I missing anything else?

Last edited: Nov 28, 2013
16. Nov 28, 2013

### Dick

Great job on defining variables. That's always a good start. But for your equations don't you mean:
$m_1v_{impact}=(m_1+m_2)v_{collision}$ and $\frac{1}{2}m_1v^2_{driving}=\frac{1}{2}m_1v^2_{impact}-\mu_km_1gl_1$?

17. Nov 28, 2013

### PhyIsOhSoHard

Hmm, let me see if I understand this correctly.

Right before the collision and after the collision, we have an inelastic collision.
The energy is conserved between the scenarios for when the car is driving and when the car is breaking and ends at right before the collision. This is not when we can use the momentum conversation because kinetic energy is not conserved.

Ya, that makes sense now. :)

But from what I can see, we are still missing an extra equation, right?

We still cannot fully express our $v_{collision}$ in terms of those two equations, so what if we use the conservation of energy right after the collision and until both cars are at rest?

$\frac{1}{2}(m_1+m_2)v^2_{collision}=\frac{1}{2}(m_1+m_2)v^2_{rest}-\mu_k(m_1+m_2)gl_2$

Since after the collision their length when they're braking is l2 and their mass is now added together.
We also know that at the last scenario where the energy is still conserved and where both cars are at rest, the velocity $v_{rest}=0$

So we end up with:
$\frac{1}{2}(m_1+m_2)v^2_{collision}=-\mu_k(m_1+m_2)gl_2$

Now we can find an expression for $v_{collision}$ and then isolate this and insert it in our conservation of momentum to find $v_{impact}$ and then we can insert the expression of $v_{impact}$ and finally find an expression for $v_{driving}$

Did I understand this correctly? :)

18. Nov 28, 2013

### Dick

Yes, you are getting it. You start with an initial amount of kinetic energy $\frac{1}{2}(m_1)v^2_{driving}$. You lose some of it due to friction before the collision. Then you also lose some during the collision because it's inelastic. Then you lose the rest while both cars coast to a halt. So I think your last equation should be $\frac{1}{2}(m_1+m_2)v^2_{collision}=\mu_k(m_1+m_2)gl_2$ without the minus sign.

Last edited: Nov 28, 2013
19. Nov 29, 2013

### PhyIsOhSoHard

Great, thanks!

I have another question. How energy was used for the deformation of the cars?

Does it have something to do with the kinetic energies before and after the collision?

20. Nov 29, 2013

### Dick

Well, sure. The difference between the kinetic energies is the energy consumed by the collision.