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Inelastic collision, two cars approach each other at intersection.

Two cars approach each other at an intersection. One car has a mass of 928.4 kg and is travelling in the negative y direction with a velocity of 21.4 m/s. The second car has a mass of 951.2 kg and is travelling in the positive x direction with a velocity of 39.5 m/s. If the collision is totally inelastic, what is the magnitude of the resultant velocity (in m/s) after the collision?

V=9.41 m/s?


I dont have time to type my work right now. But will show work if answer is wrong to see where i went wrong.
 

Delphi51

Homework Helper
3,407
10
I get more than twice that value, but I often make mistakes in calculations. It will be interesting to see your method.
 
m1vo1 + m2vo2 = (m1 + m2)V

(928.4)(-21.4) + (951.2)(39.5)=(928.4 + 951.2)V

17704.64=1879.6V

V=9.41
 
your equation:
m1vo1 + m2vo2 = (m1 + m2)V
is for one dimensional case. But you have a 2D problem. Momentum is conserved, and so is its y and x components.
 
my teacher has not clearly presented the X & Y compents in this type of probelm. would i use something along the lines of that problem but have cos and sin in there somewhere?
 
Momentum is a vector quantity, just draw those two vectors. That collision is inelastic, means that two masses stick together and travel as one after collision. But because, momentum is conserved, both components of momentum will be conserved too (x and y components). You can simply spot right triangle here, with Pythagorean theorem crying to be used ;]
 

Delphi51

Homework Helper
3,407
10
North/south: mv before = mv after; solve for vy after
East/west: mv before = mv after; solve for vx after
Combine vx and vy with Pythagorean theorem to get the combined velocity after the collision.
 

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