Momentum, vector addition question

[ConfusedPhysics1234]

Homework Statement

A car with a mass of 1400 kg is westbound at 50 km/h. It collides at an intersection with a northbound truck having a mass of 2000 kg and traveling at 40 km/h. What is the initial common velocity of the car and trucks after collision if it's an inelastic collision?

Homework Equations

p = mv

The Attempt at a Solution

So I got this question wrong, but I can't seem to get the right answer no matter how I look at it. I think maybe I lost marks because I didn't use the method the teacher wanted (which is fine with me)?? Let me know guys.

m₁v₁ + m₂v₂ = (m₁ + m₂)v_f

(1400kg)(13.888m/s) + 0 = (1400kg + 2000kg)(v_fx)
vf = 5.7189... m/s

0 + (2000 kg)(12.5 m/s) = (1400kg + 2000kg)(v_fy)
vf = 7.35294.. m/s

(5.7)² + (7.3)² = r²
vf = 9.3 m/s

tan-1(7.3 / 5.7 m/s)
= 52 N of W
= 128 degrees

I tried a different approach and found P(car) and P(truck), then P(combined), solved for Vf, but I got the same answer:

(1400 kg)(13.888 m/s) = 19443.2 kg m/s West
(2000 kg)(12.5 m/s) = 25000 kg m/s North
pythagorean theorem...
= 31670.77559 kg m/s
momentum is conserved so
31670.78 = (2000 + 1400) vf
vf = 9.3 m/s!

 

[NascentOxygen]

Hi.

Your method looks right, though I haven't checked your arithmetic.

 

[Simon Bridge]

Your method is correct - your answer is different from mine.
Note: do algebra before plugging numbers in, check for rounding errors.

 

[ConfusedPhysics1234]

Hey, redid the algebra, got the same answer. What was your answer Simon Bridge?


v_fx = (1400 kg)(13.888m/s) / (1400kg + 2000kg)
=5.7189 m/s
v_fy = (2000 kg)(12.5 m/s) / (1400kg + 2000kg)
= 7.35294 m/s

(5.7189² + 7.35294²)^1/2 = v_f

 

[ConfusedPhysics1234]

Never mind, I converted 40km/hr to 12.5 m/s, should have been 11.1 m/s. That sucks, lost a lot of marks for it.

I got 8.7 m/s as my answer.

 

[Simon Bridge]

... and that was my answer.
I did the components in kmph and converted at the end... the fewer conversions you do the better.