Using conservation of momentum to find final velocity in a collisionby elsternj Tags: collision, conservation, final, momentum, velocity 

#1
Mar3011, 07:11 PM

P: 42

1. The problem statement, all variables and given/known data
Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle [tex]\phi[/tex]south of east (as indicated in the figure). After the collision, the twocar system travels at speed v_{final} at an angle [tex]\theta[/tex] east of north. Find the speed v_{final} of the joined cars after the collision. Express your answer in terms of v and [tex]\phi[/tex] . 2. Relevant equations p=mv p_{i}=p_{f} 3. The attempt at a solution first i tried to break this down in terms of its components in the x direction: m_{1}v_{1i}cos[tex]\phi[/tex] =(m_{1}+m_{2})V_{final}sin[tex]\theta[/tex] in the y direction: m_{1}v_{1}sin[tex]\phi[/tex]+m_{2}2v = (m1+m2)v_{final}cos[tex]\theta[/tex] Now here is where I am starting to get mixed up. I have both of my components. (They may be wrong so please help me with those equations) Do I just solve both for v_{final} and then square them and take the square root? Any insight is much appreciated. 



#2
Mar3011, 07:30 PM

P: 51

"in the x direction:m1v1icos + m2=(m1+m2)Vfinalsin"
I think maybe your equation got problem because direction x for v2 is zero so momentum for m2 also be zero. If i m wrong please check it out ! thanks ! 



#3
Mar3011, 07:34 PM

P: 42

ah yes, i actually omitted that mass when i originally wrote it and then for some reason when looking at my paper i brought it back for some reason.. that was more or less a typo. i am aware that the second mass does not have momentum in the x direction, i will edit that.




#4
Mar3011, 07:55 PM

P: 51

Using conservation of momentum to find final velocity in a collision
If no going wrong my answer maybe like this:
First take it down then square both sides propose is remove the delta , (m1v1icos/(m1+m2)V_{final})^{2} =sin^{2 }1 [(m1v1sin+m22v)/(m1+m2)v_{final}]^{2} = cos^{2}2 1+2 : (m1v1icos/(m1+m2)V_{final})^{2}+[(m1v1sin+m22v)/(m1+m2)v_{final}]^{2}=1 After that take up [(m1+m2)V_{final}]^{2} shift it to right then remove the square so that my answer . 


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