Ineleastic perpendicular collision question

[MomentumIsHard]

Homework Statement

Car 1: 2200kg
Initial direction: South
Car 2: 1800kg
Initial direction: East
Cars lock together and slide 11.25M (S 22.25 E)
At an acceleration that is 9.81 m (back) (it's a coincidence that this is gravity.)

Homework Equations

Who's fault was the collision if the speed limit was 50km/h?

I can't find momentum without a velocity so I'm very lost

The Attempt at a Solution

 

[haruspex]

Homework Statement

Car 1: 2200kg
Initial direction: South
Car 2: 1800kg
Initial direction: East
Cars lock together and slide 11.25M (S 22.25 E)
At an acceleration that is 9.81 m (back) (it's a coincidence that this is gravity.)

Homework Equations

Who's fault was the collision if the speed limit was 50km/h?

I can't find momentum without a velocity so I'm very lost

The Attempt at a Solution

Just invent variables to represent the three unknown velocities, the two before and the one after collision.
Momentum gives you two equations, stopping distance gives you a third.

 

[MomentumIsHard]

Just invent variables to represent the three unknown velocities, the two before and the one after collision.
Momentum gives you two equations, stopping distance gives you a third.

Which equations are those? We only got one equation for in elastic in class, what about the distance one too? What you said makes sense but I can't find 3 equations

 

[gneill]

Hi MomentumIsHard,

You need to make some attempt before help can be given. What details about the motion after the collision do you know?

 

[MomentumIsHard]

Hi MomentumIsHard,

You need to make some attempt before help can be given. What details about the motion after the collision do you know?

All the question states is that it moves 11.25 M [Back] (even though other directions were given in terms of east and south) at an acceleration of 9.81 m/s. The problem is I don't know where to start so I can't really make an attempt. Once I get one momentum I could pretty easily figure this out but I can't quite figure that out with no velocities given

 

[haruspex]

Which equations are those?

Conservation of momentum (one for each of two dimensions) and the SUVAT equation relating speeds, distance and (constant) acceleration.

 

[gneill]

All the question states is that it moves 11.25 M [Back] (even though other directions were given in terms of east and south) at an acceleration of 9.81 m/s. The problem is I don't know where to start so I can't really make an attempt. Once I get one momentum I could pretty easily figure this out but I can't quite figure that out with no velocities given

No idea what "[Back]" is supposed to mean, but it doesn't matter if you're given the rest of the motion details. You have a distance, an acceleration, and a final speed. What can you deduce from that?

 

[haruspex]

No idea what "[Back]" is supposed to mean, but it doesn't matter if you're given the rest of the motion details. You have a distance, an acceleration, and a final speed. What can you deduce from that?

I think it means the direction of the given acceleration is opposite to the motion.

 

[gneill]

I think it means the direction of the given acceleration is opposite to the motion.

That's quite likely. Of course deceleration is also implied by the fact that the combined mass travels a limited distance.