Sequental elastic and inelastic collisions

[Jay Sachar]

Homework Statement

A taxi car weighing 2000kg hits a stationary mini-van that has a mass of 2200kg. The taxi stops and the mini-van rolls and hits a stationary sports car with a mass of 1830kg. Their bumbers hit and they move together at 2.3m/s. What is the velocity of the taxi before the collision?

Part one of the problem would be an elastic collision as they don't stay together, so use the formula m1v1+m2v2=m1v3+m2v4
M1=2000
V1=?
M2=2200
V2=0ms
V3=0m/s
V4=?
Part 2 of the problem would be inelastic so use the formula m1v1+m2v2=v3(m1m2)
M1= 2200
V1=?
M2=1830kg
V2=0m/s
V3=2.3m/s

Homework Equations

m1v1+m2v2=v3(m1m2)
m1v1+m2v2=m1v3+m2v4

The Attempt at a Solution

I first tried to find the v4 for the first part of the problem using the equation m1v1+m2v2=v3(m1m2) isolating for v1 as that would be equal to v4 in the first part. (2200×v1)+(1830×0)=2.3(2200×1830) and after isolating for v1 i got 4209m/s. I then punched this in for v4 in the first part of the problem so it was (2000×v1)+(2200×0)=(2000×0) +(4209×2200) and after isolating for v1 i got 4629.9m/s which just seems too high. Any help wpuld be greatly appreciated!

 

[Alloymouse]

After collision 2, you have correctly pointed out that momentunm is conserved according to that equation.

Consider this: since initial velocity of the taxi is 0, that would mean that all momentum after the second collision initially belonged to the moving van.

Do you need more hints?

 

[Jay Sachar]

After collision 2, you have correctly pointed out that momentunm is conserved according to that equation.

Consider this: since initial velocity of the taxi is 0, that would mean that all momentum after the second collision initially belonged to the moving van.

Do you need more hints?

Uh yeah maybe another hint would be appreciated. Still fairly confused about it.

 

[Alloymouse]

Uh yeah maybe another hint would be appreciated. Still fairly confused about it.

For starters, do you understand what I'm trying to say when "all the momentum belongs to the van" before collision 2? It's using the principle of conservation of linear momentum.

Here's another hint:

If you can calculate the total momentum involved in collision 2 (total momentum of objects after collision), you now know how much momentum the van had before colliding with the stationary taxi.

From this, you can find the speed of van after collision 1 happened.

 

[haruspex]

Part one of the problem would be an elastic collision as they don't stay together

No, an elastic collision is one in which KE is conserved, which is not the case here.
In fact, the problem is a bit strange... KE appears to increase! (Are you sure you have the details right?)

Part 2 of the problem would be inelastic so use the formula m1v1+m2v2=v3(m1m2)

I assume you are thinking of v3(m1+m2). What you have written makes no sense dimensionally.
At first I thought this was just a typo, but I see you really did multiply the two masses and got a crazy speed as a result.
You would have spotted the error if you had kept track of units. Your result should read 4630 kg m/s.

Also, the reason this (corrected) equation applies is not merely that it is inelastic but that it is a coalescence. As collisions go, that is as inelastic as possible.