# Homework Help: Linear momentum head on collision

1. Jun 20, 2009

### desixcutie04

1. The problem statement, all variables and given/known data
Two cars, one a compact with mass 1200 kg and the other a large with mass 3000 kg, collide head on at 60 mi/hr.
a. Which car has a greater magnitude or momentum change? Which car has a greater change in velocity?
b. Which car's occupant's would you expect to sustain greater injuries?

2. Relevant equations
P=mv
m1v1i + m2v2i = m1v1f + m2v2f

3. The attempt at a solution
P(small car)= m1v= 1200 X 96.56 km/hr = 115, 872
P (large car) = m2v= 3000kg x 96.56 km/hr= 289, 680
I'm having trouble understanding whether this collision is inelastic of elastic. So I can't figure out whether the momentum is going to be conserved or not.

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

2. Relevant equations

3. The attempt at a solution

2. Jun 20, 2009

### LowlyPion

Welcome to PF.

I think you can pretty much be sure that it is an elastic collision.

Assume that the final velocity is determined by the combined crumpled mass. These are not Six Flags bumper cars at 88 ft/sec.

By the way you may want to work in more useful units like m/s as opposed to km/hr.

3. Jun 20, 2009

### Staff: Mentor

It's best to use standard units for speed: m/s, not km/hr. (But you don't really have to worry about units to answer the questions.)
Generally, cars get tangled up together when they crash, so I would assume that the collision is perfectly inelastic unless told otherwise.
When is momentum conserved in a collision?

Edit: LowlyPion beat me to it.

4. Jun 20, 2009

### desixcutie04

so momentum is conserved when there are no external forces acting on the collision. In this case, it would be conserved because the two cars are the only thing in the collision. (the prompt doesnt mention whether the collision is inelastic or elastic)
The momentum is conserved in a collision because kinetic energy is transferred .
This would mean that that both cars have the same velocity after the collision.

m1v1 + m2v2= (m1+m2) V
V= (m1v1+m2v2)/ (m1+m2)
V= (1200kg(26.82m/s) + 3000kg(26.82m/s))/ (1200kg+ 3000kg)
V= 26.82 m/s

So if their velocity is the same after the collision, they experience the same change in velocity.
Would this mean that passengers in both cars are likely to experience similar injuries?

5. Jun 20, 2009

### Staff: Mentor

Good.

Careful. Since momentum is a vector, directions--and thus signs--matter. If one vehicle has an initial velocity in the positive direction, the other must be going in the negative direction.

You'll have to redo the above, but take into account the direction of the velocities when finding the change.

6. Jun 20, 2009

### desixcutie04

m1v1 + m2v2= (m1+m2) V
V= (m1v1+m2v2)/ (m1+m2)
V= (1200kg(-26.82m/s) + 3000kg(26.82m/s))/ (1200kg+ 3000kg)
V= 11.49 m/s

change in velocity:
large car: 26.82-11.49= 15.33 m/s
small car: -26.82-11.49= -38.31 m/s

so the direction is what influences which car has a greater change in velocity. Since I have made the small car in the negative direction, it has a greater change in velocity.
I'm not sure if i did the calculation correctly this time around.

7. Jun 20, 2009

### Staff: Mentor

Looks good.

Change in anything is usually final minus initial. But what really matters is the magnitude of the change, since the sign is arbitrary. (You could have chosen the opposite directions for the cars.)

What determines the magnitude of the change in velocity is the relative momentum of the cars. The more massive car starts with the most momentum, so ends up with the smallest change in speed. (It doesn't matter which car starts in the positive direction; the sign is arbitrary.)
Looks OK to me, except for the sign of your velocity changes, but what matters there is the magnitude of the change, not the sign.

8. Jun 20, 2009

### desixcutie04

thank you so much for your help!
So for part b, do the passengers in the car that has the larger change in speed sustain greater injuries?

9. Jun 20, 2009

### Staff: Mentor

That's what I would say (all else being equal).