A 2.00 t van traveling at 50.0 km/h [S] collides with & remains locked to a 3.00 t truck traveling 60 km/h [W]

Calculate the resultant (total) momentum after the collision.
Answer should be ( 206 t km/h ) S 60.9 degrees W

Please please help me with this problem. Thank you so much

in order to find momentum after collision
use vector in south for van
use vector in west for train
so initial momentum of van ( in - x-axis) before collision will be = -100 t km/h
so initial momentum of train (in - y-axis) before collision will be = -180 t km/h

so final momentum will be in west-south direction which is -x and -y plane
SO final momentum will be resultant vector = sqrt{ (100)^2 + ( 180) ^2}
final momentum = sqrt {(10000) + (32400)}
= sqrt (42400)
final momentum = 205.9 t km/h
and angle will be tan@ = momentum of van/momentum of train
angle @ =tan inverse ( 100/180)

@ = tan inverse ( 5/9)
@ = 30 degree
as this is 29.05 degree from west to south
so 90- 29.05 = 60.9 degree from south to west

The momentum in southern direction (p_s) remains 2*50 t km/h = 100 t km/h. The momentum west (p_w) remains 3.00*60 = 180 t km/h. The sum is sqrt(100^2+180^2) = 206 t km/h.

For the direction use than tangent.

Gh