# System of particles, linear momentum (easy problem)

## Homework Statement

A 2100 kg truck traveling north at 41 km/h turns east and accelerates to 51 km/h. (a) What is the change in the truck's kinetic energy? What are the (b) magnitude and (c) direction of the change in its momentum?

## Homework Equations

currently using delta P = mvf - mvi, but this is not working. i then tried p = mvcom, but this doesn't work either. please help.

## The Attempt at a Solution

first converted the speeds to 11.39 and 14.17 m/s.

a) delta K = 74510 J. this is correct.

b) Pi = (2100)(11.39) = 23919 kg m/s
Pf = (2100)(14.17) = 29757 kg m/s

delta P = Pf - Pi = 5838 kg m/s. this is incorrect.

c) i have no idea how to do this either.

LowlyPion
Homework Helper
Your problem with b) is that your momentum is changing in both x and y. Remember v and hence p are vectors.

wow, thanks so much! my new answer is 38178 kg m/s, which is correct.

any help with c) (from anyone) would be appreciated, but only if you have time and whatnot. thanks!

LowlyPion
Homework Helper
wow, thanks so much! my new answer is 38178 kg m/s, which is correct.

any help with c) (from anyone) would be appreciated, but only if you have time and whatnot. thanks!

c) is again a vector operation.

P_final - P_initial

Or more simply

P_final + (- P_initial)

In b) you found |ΔP|, now they want the angle right?

arctan(vy/vx) = 38.8 degrees um... i would have said northeast. but i guess that's not right (according to the back of the book). i mean, the magnitude of the angle is right. the direction isn't.

LowlyPion
Homework Helper
arctan(vy/vx) = 38.8 degrees um... i would have said northeast. but i guess that's not right (according to the back of the book). i mean, the magnitude of the angle is right. the direction isn't.

It was originally traveling north. Then it was going east. Hence it's "change" in direction north-south must be south and of course since it moved to the east then it is South and East at 38 degrees.