System of particles, linear momentum (easy problem)

[lemonpie]

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]

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

 

[lemonpie]

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]

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?

 

[lemonpie]

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]

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.