Momentum - two cars crashing at an angle

[XxphysicsxX]

Homework Statement

A 2000 kg car strikes a staitionary 1200 kg car and rebounds at 65.0 km/h@ 20 (deg) Sof E. After collision, the 1200 kg car moves at 40.00 km/h @ 35(deg) N of E. Determine the velocity of the 2000 kg car before the collision.

m1= 2000 kg
V1(before) = ?
V1(after) = 18.05 m/s

M2= 1200 kg
V2(before) = 0 m/s
V2 (after) = 11.11 m/s

Homework Equations

MV(before) = MV(after)
Momentum.

The Attempt at a Solution

In the X direction:
momentum of object 1 (before)= M1V1a + M2V2a - M2V2b
= (2000)(18.5)(cos 70) + (1200)(11.11)(cos 35) - (0)
= 23267. 855
In the Y direction
momentum of object 1 (before)= M1V1a + M2V2a - M2V2b
= (2000)(18.05)(sin 70) + (1200)(11.11)(sin 35)
= 41569.82

C^2 = A^2 + B^2
C =ROOT [(23267. 855^2) + (41569.82 ^2)]
C = 47638.67 = momentum

momentum / mass = velocity
47638.67 / 2000 kg = 23.81 m/s

23.81 m/s= 85.71 km/h <<<<<<<<<<<< which is wrong... its supposed to be 81.2 km/h

... and i don't get the right angle either.

Ive been at this for two days any help will be greatly appretiated

 

[Delphi51]

You have a mistake in the y direction; missing a minus sign on one of the mv's. One is north, the other south. Then partially cancel, reducing the answer. It works out to 81.2 for me.

Probably not a good idea to convert to m/s since the question and the answer are in km/hr.

 

[XxphysicsxX]

thankyou so muchhh!