Conserved momentum (car accident)

[mujadeo]

Homework Statement

A blue car with a mass of 2000 kg and a yellow truck with mass 9000 kg enter an intersection. The truck comes from the south at vtruck = 7 m/s, and the car comes from the west. When the two collide they stick together and slide at an angle of q = 43° north of east.

What is the speed of the pair after the collision?

Homework Equations

how can i do anything with question without speed of the blue car?
i know p=63000, and angle plays into it, but can't figure this out
please help

The Attempt at a Solution

 

[learningphysics]

Homework Statement

A blue car with a mass of 2000 kg and a yellow truck with mass 9000 kg enter an intersection. The truck comes from the south at vtruck = 7 m/s, and the car comes from the west. When the two collide they stick together and slide at an angle of q = 43° north of east.

What is the speed of the pair after the collision?

Homework Equations

how can i do anything with question without speed of the blue car?
i know p=63000, and angle plays into it, but can't figure this out
please help

The Attempt at a Solution

Use conservation of momentum... momentum is conserved along any axis... so momentum is conserved in the north/south direction... and momentum is also conserved in the east/west direction... Write the equations out for conservation of momentum in the north/south direction, and also in the east/west direction... you'll have to use some trigonometry... You'll have two equations with two unknowns...

 

[mujadeo]

OK, so north south is y and east west is x
p in y before collision is 63000
i don't understand p is conserved in both directions?
the combined mass can't chaneg (=11000kg) and it moves with only one speed and p is the same after collision as it was before.
So i don't understand this

 

[learningphysics]

OK, so north south is y and east west is x
p in y before collision is 63000
i don't understand p is conserved in both directions?
the combined mass can't chaneg (=11000kg) and it moves with only one speed and p is the same after collision as it was before.
So i don't understand this

Yes, the combined mass is 11000kg. And there's an unknown final velocity call it vf... and it's moving at an angle of 43 degrees. So the momentum after the collision has a magnitude of 11000*vf... Draw a picture of this vector... Can you divide up this vector into a horizontal(east/west) and vertical component(north south)? Use cos and sin...

 

[mujadeo]

yes i understand about getting the components of vectors, but what components are they, speed or momentum. and how can that tell me the speed?

 

[Dick]

Since you CAN get components of vectors, I would suggest you start doing so. They are velocity AND momentum. And as learningphysics points out momentum is conserved along both axes. What do you get from the north-south axis? Your doubts you will get anything are stopping you from continuing.

 

[learningphysics]

Since you CAN get components of vectors, I would suggest you start doing so. They are velocity AND momentum. And as learningphysics points out momentum is conserved along both axes. What do you get from the north-south axis? Your doubts you will get anything are stopping you from continuing.

Yes. The first step is writing out the equations mujadeo. If you're having trouble writing out the 2 conservation of momentum equations, then let us know where you're getting stuck.

 

[mujadeo]

K= 1/2mv^2
p=mv
i am sorry i am not getting what the components are?
should i get x and y components of the total momentum (63000) befor the collision and use that to someone get v??

 

[mujadeo]

what i mean is, take total momentum (which is same cause its conserved) and then multiply by cos43, which will give v in x dire
then mult by sin43 giving v in y
then use trig to find the hyp side (velocity at 43deg)
?

edit -- i forget the divide by the mass step
(i am using p/m = v )

 

[learningphysics]

what i mean is, take total momentum (which is same cause its conserved) and then multiply by cos43, which will give v in x dire
then mult by sin43 giving v in y
then use trig to find the hyp side (velocity at 43deg)
?

edit -- i forget the divide by the mass step
(i am using p/m = v )

Yes, that's the right idea... I wouldn't worry about the velocity vector... I'd just deal with the momentum vectors...

So the east-west componenet of momentum would be mf*vf*cos43 and the north-south component would be mf*vf*sin43... where mf is the total mass of the clump after the collision, and vf is the final velocity.

So the total momentum in the east west direction before the collision = total momentum in the east west direction after the collision... and the same idea with the north-south direction...

 

[mujadeo]

OK I got this one nailed
thank you for the help!