Kinetic Energy/Momentum from change in direction and speed

[threewingedfury]

From a calculus based physics course:

A 2100 kg truck that travels north at 41 km/hr is turning east and accelerating to 60 km/hr
-What are the magnitudes and direction of change in the trucks momentum?
- What is the change in the trucks kinetic energy?

I need some help on steps - because I'm lost

[Max Eilerson]

momentum p = mv. Use this to work out the magnitudes of the vectors, then resolve the vector system.
K.E. = 1/2mv^2

[threewingedfury]

would the change in KE be:
KE = .5mv^2
KE1 = .5(2100)(41km/hr * 0.277777778 m/s)^2
KE2 = .5(2100)(60km/hr * .277777778 m/s)^2
KE1-KE2 = 136192.132-291666.671
KE=-155474.54

[threewingedfury]

and the momentum would be
p =mv
p1 = 2100*41km/hr
p2 = 2100*60km/hr
p1-p2 =-39900 kg*km/hr
Are those the right units? Or does it need to be in m/s?
Would the direction change be east? thats where I'm lost, or would it be the negative direction?

[threewingedfury]

so am I right, or really wrong?

[Max Eilerson]

KE is right since it's just a scalar. Momentum is a vector so you need to resolve it.

[threewingedfury]

but how exactly would you do that unless you multiply the mass to the velocity?

I tried the average of the 2 velocities, but thats not right