Solving an Inelastic Collision: Finding the Angle of Motion for Two Cars

[alexito01]

Homework Statement

A red car and a blue car collide at an in intersection. Prior to the collision the red car with mass 1000kg was heading North at 15m/s. The blue car with mass 1500 kg was heading East at 20m/s. The collision is completely inelastic with the cars sticking together and moving as one. At what angle measured North of East do the cars move off?

Homework Equations

I used Vf = √(Vfx2) + (Vfy2)

I alsoUsed m1vf1x+ m2vf2x= m1vi1x + m2vix for x component and
m1vf1y+ m2vf2y= m1vi1y + m2viy for y component

and then θ = tan-1(vfy/vfx) to findthe angle but i keep geting it wrong

The Attempt at a Solution

I ended up with an angle of around 27 degrees

 

[PhanthomJay]

This is a conservation of momentum problem, not conservation of energy as your title suggests. However, your method looks good and I agree with your answer. What's wrong with it?

 

[Mthees08]

That is correct about 27 degrees north of east. at a velocity about 13.5 m/s (13 with 2 sig figs)

 

[visharad]

Cars stick after collision. So they have common velocity. Let velocity = vf
Take east as +ve x and north as +ve y

For x component:-
(m1+m2)vfx = m1vi1x + m2vix
(1000+1500)vfx = 1000*0 + 1500*20
2500 vfx = 30000
vfx = 30000/2500 = 300/25
vfx = 12 m/s

For y component
(m1+m2)vfy = m1vi1y + m2viy
(1000+1500)vfy = 1000*15 + 1500*0
2500 * vfy = 15000
vfy = 15000/2500 = 150/25 = 6 m/s

theta = tan-1(vfy/vfx) = tan-1(6/12) = tan-1(1/2) = 26.565 deg
So, as approximation, this is same as your answer. Why do you think your answer is wrong?