SET UP Because momentum is a vector quantity, we need coordinate axes. We draw a sketch (Figure 1), labeling the car A and the truck B . SOLVE We need to find (1) the components of momentum of each vehicle, (2) the components of the total momentum, and then (3) the magnitude and direction of the total momentum vector (Figure 2). The components of momentum of the two vehicles are pA,xpA,ypB,xpB,y=mAvA,x=mAvA,y=mBvB,x=mBvB,y=(1000kg)(0)=0=(1000kg)(15m/s)=1.5104kgm/s=(2000kg)(10m/s)=2.0104kgm/s=(2000kg)(0)=0 There are no z components of velocity or momentum. The components of the total momentum P are PxPy=mAvA,x+mBvB,x=0+2.0104kgm/s=2.0104kgm/s=mAvA,y+mBvB,y=1.5104kgm/s+0=1.5104kgm/s The total momentum P is a vector quantity with these components. Its magnitude is P=(2.0104kgm/s)2+(1.5104kgm/s)2=2.5104kgm/s Its direction is given by the angle in (Figure 2), where tan=pypx=1.5104kgm/s2.0104kgm/s=34=36.9 Part A - Practice Problem: If the car is moving at 17 m/s , how fast must the truck move for the total momentum vector to make a 30 angle with the +x axis (and 60 with the +y axis)? Express your answer with the appropriate units. Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate u

(north ) V = 10 m/S X (east) mp = 2000 kq A = 15 m/S MA = 1000 k