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Question

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? 

I know how to do part (a) and I got 7.5x10^4 J but how to do parts (b) and (c)? 
For (b), I got 2100(14.17 m/s-11.39 m/s)=5838 and I know that's not the answer.

debydete · Accepted Answer

The change in momentum "dP" is the vector eq;
dP = Pf - Pi

You find this by adding "tip to tail" the arrows representing the two momentums.
You first draw arrow Pf to the right (East). Then at the tip of the Pf arrow you draw the -Pi arrow, which will be South (negative North).
The arrow from the tail of Pf to the tip of -Pi is the vector dP.
Since Pf & -Pi are perpendicular, dP is just the hypotenuse of the right triangle.
The magnitude is;
IdPI = Sqrt[Pi^2 + Pf^2]

The angle is () measured down from the East direction and found from;
Tan() = Pi/Pf
() = 38.8 deg South of East

Where in SI units 
Pi = (2100)(41)(1000/3600) kgm/s
Pf = (2100)(51)(1000/3600) kgm/s

Apparao V · Answer

It is easy one . why do not you calculte
Vector difference ,should be taken Root of v1 square +v2 square - 2X v1X v2 is in magnituge
directionN-S.

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Urgent! Help me with AP Physics?

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