On a very muddy football field, a 110 kg linebacker tackles an 85 kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.8 m/s north and the halfback is sliding with a velocity of 7.4 m/s east.
A.) What is the magnitude of the velocity at which the two players move together immediately after the collision?
Answer is 5.9 kg*m/s
p1 = m1v1 = 110*8.8 = 968
p2 = m2v2 = 85*7.4 = 629
square root of (968)^2+(629)^2 = 5.9
B.) What is the direction of this velocity?
\theta = what degrees east of north(NOT north of east)?
I think it is 33 degrees, but I'm not sure if it is actually 57 degrees. Wouldn't it be y/x*arctan and then 90 degrees minus that number?