Calculate the velocity of the system After the collision

[MrJoseBravo]

1. The problem statement, all variables and given/own data
http://prntscr.com/661d5d

Homework Equations

m1v1 + m2v2 = m1v1p +m2v2p

The Attempt at a Solution

I asumed that the momentum in the y direction is constant, and that in the x direction it is conversved.
Also, i thought of it as 2 parts to this problem. The momentum of the first cart plus that of the rock has to equal the combined momentum.
This combination plus the negative momentum of the 2nd cart, must equal the desired velocity of the entire system.
I don't think this logic is correct since i didnt get the right answer, so clarification on that should help me solve the problem.
anyway here's the math
(x direction)
Vrx = 1.20cos30
Mc1Vc1 + MrVr = McVc1f + MrVrf * here i assume that the final velocity is going to be the same for both of them.
plugging in the values i get this
Vf=300/(Mc1 + Mr)
Vf = .75m/s

Now, the the momentum of the entire system should be equal to the rock and the first cart minus the second.

(Mc2+Mr)Vf - Mc2Vc2 = M(system)V(system)

I get V = .359
Ans : .205 m/s

Any help is appreciated, thanks!

 

[BvU]

Is it clear to you what all these symbols stand for ? A list would be helpful.

First collision: what is the 300 ? Fill in the correct momentum to get a better value for v_f.

Second collision: if you don't show what you are doing, only give the result you get, it is very difficult to guess what goes wrong. Most physicists have very limited telepathic capabilities. The expression is correct and the answer I get is indeed 0.2049 m/s.

I tried some possible errors but couldn't reverse-engineer the .359 whatever I tried !

 

[AlephNumbers]

Mc1Vc1 + MrVr = McVc1f + MrVrf

Are you sure that is correct? That expression would apply in the case of a perfectly elastic collision. I would assume that the rock collides with the cart and stays in the cart.

 

[AlephNumbers]

Oh. Sorry. If you assumed that the velocities of the cart and the rock are the same after the collision then that should be correct.

 

[MrJoseBravo]

BvU, my apologies for not expressing things cleary enough

so i did (300)(.6) +(100)(1.2) for the initial momentum of the first cart and the rock, which gives
300 = Mc1Vc + MrVr
so 300=(Mc1+Mr)Vf
300/(Mc1+Mr) = Vfx
Vfx=.75m/s

As for the equation, aleph, are you saying that the coefficient of restitution comes into play here? if so, I am a little confused on how to implement that, however i do know how to find it.

The second calculation i did this

400(.75) - 400(.3) = MfVf

(300-120)/800 = .225 m/s

Looks like i did the calculation wrong at the end there BvU and for any grief that might have caused.
Regardless the answer is still wrong!

 

[AlephNumbers]

I got the same answer as BvU. It would appear that you rounded a little too heavily. The value for the velocity of cart 1 and the rock that I got is .709808. Other than that everything is correct.

 

[BvU]

Dear Jose,

No need for apologies. I now understand the .75 m/s, and it's not good. Reason: momentum is a vector. To get (horizontal) speed of car + rock, you can add horizontal momentum of car and only the horizontal component of momentum of rock. I though that's what you intended to do, because you did write Vrx = 1.20 cos30 .
The vertical component of the momentum of the rock is somehow absorbed by the car, either in shock absorbers or by denting the floor.

So that settles the first part. Forgot the numbers, but you get some horizontal momentum (in fact you don't need the speed of the first car after rock has settled an before the connecting collision with car 2). That one you did just fine. IF you re-do it, you'll get the book (?) answer