# Homework Help: Car Collisions

1. Mar 6, 2014

### Jimmy87

1. The problem statement, all variables and given/known data

Hi, could someone help me with some physics homework for a head on collision between a truck and a car. Truck and car are going the same speed (50mph) but truck weighs 5 times more than car.

Q.1. What is the force on each car?
Q.2. What is the impulse?
Q.3. Explain in detail using work-energy theorem why it is worse to be in the car
Q.4. If the vehicles collide at 50mph how would the impact compare for both vehicles to that of a collision with a solid wall?

2. Relevant equations

I = Ft = change in momentum
F = ma

3. The attempt at a solution

Think I know Q's 1 and 2. The force of collision is equal and opposite and so is the time of the collision therefore I think both impulse and force will be the same. Using F =ma the acceleration will be much greater on the smaller car.

My questions:

How does this bigger acceleration cause a greater kinetic energy of the car as I thought KE is proportional to momentum and the momentum change is the same for both vehicles? I'm a bit unsure how you relate forces and accelerations to kinetic energy.
For the last part I'm not sure - I think the truck will experience an impact which is not as bad as colliding with a wall. Does the car experience an impact worse than colliding with a wall? not quite sure why though in terms of KE, impulse etc.

2. Mar 6, 2014

### HallsofIvy

We keep getting this kind of question- typically from people who want to prove that the accident was the other guys fault. Were you actually given this problem? You are given the speeds and (relative) masses of the vehicles so you can determine the (relative) kinetic energies and momenta at the instant of collision. But the force depends upon how long the collision takes and how much each vehicle "collapses". And that information is not given.

3. Mar 6, 2014

4. Mar 7, 2014

### BvU

Soothe soothe
So what did you get for the third question ?

$d T = \vec F \cdot d\vec s \quad$ if F is constant, and $T = {\vec p^2 \over 2m}$

Your relevant equations should include something about momentum, in particular: momentum conservation. There is something about the momentum of the center of mass and the momentum w.r.t. the center of mass you can write down.

I think it is safe to assume the car doesn't bounce back too much. That gives some hold to do calculations. In particular that the car driver dies and the truck driver doesn't, so that's why it's better to be in the truck (which is not the answer to Q 3!).

Concerning the brick wall: Gives a few scratches on the truck and a bit more damage to the car, unless it is a bit sturdier than a single 3 5⁄8" . Just joking:
The intention here is that the collision is inelastic and all speed is killed. This time the truck driver dies for sure and the car driver has a small chance.

5. Mar 7, 2014

### Jimmy87

Thanks BvU. The understanding of why the car is worse off if it collides with the truck is because the car receives a greater change in acceleration since they both receive the same force. I mean I kind of know that a greater acceleration is worse during a collision but can't put any physics to it to explain why in terms of kinetic energy. So how do you then relate kinetic energy to a changing acceleration? With the brick wall the question we have to answer is that we need to compare the effect of a car hitting a brick wall and then a car hitting a truck in a head on collision where both vehicles are going the same speed. Which is worse for the car? Hitting a brick wall or having a head on collision with a truck? I think hitting the truck would be worse because I looked it up and if two cars of equal mass collide at the same speed teach will feel a collision which is the same as hitting a wall, therefore hitting the truck must be worse for the car than hitting a wall?

6. Mar 7, 2014

### BvU

Yes. For two identical cars, same speed, the center of mass is at rest.

Since there are no external horizontal forces, momentum of c.o.m. is conserved: $\vec F_{1\rightarrow 2} = - \vec F_{2\rightarrow 1} \Leftrightarrow {d\vec p_1\over dt} = {-d\vec p_2 \over dt} \Leftrightarrow {d\over dt} \left ( \vec p_1 + \vec p_2 \right ) = 0$

This conservation of c.o.m. momentum is also true for the car-truck collision. If the car has mass m, truck 5 m, c.o.m. has 4 mv before and after collision. So speed of the combined wreck is $4 mv/6m = {2 \over 3}v$ or 33 mph. So car changes from -50 mph to +33 mph truck from 50 to 33. A factor 5 (no coincidence 5m/m !) worse for the car driver.

Note that I state all this in terms of momentum. With F = dp/dt, which with constant mass becomes F = ma, the change in momentum is important. And the more spread out in time, the better. Therefore cars are designed to have a maximum crumple zone, plus airbags and soft internals. Everything to spread out the change in momentum as much as possible.

For trucks there is no point: if you carry 45 tonnes of cargo you would need an immense crumple zone.

Now for hitting the proverbial brick wall (think of a medieval castle wall or something). No difference with the case identical cars at same speed and head on: Forces same, Δp same, crumpling equally long, etc. Our car driver would havve a(very) small chance. Not so our truck driver: he is between the wall and the cargo and his minimal crumple zone is useless.