[د.امير]

بالنسبة للسؤال الثاني
2- Car A ( mass MA ) is traveling north at speed V0 and car B (mass 2MA/3 )is traveling at the same speed west .The two cars collide at an intersection ,sticking together .What is their common velocity just after collision ?
في هذا السؤال سيارتان تصدمان في تقاطع طريق ويلتصقان معا بعد التصادم
من قانون حفظ الزخم، حيث ان الزخم هو حاصل ضرب ألكتلة في السرعة
the point of this problem is to use conservation of momentum. The critical ideas are that momentum is a vector, and that momentum is conserved in collisions.

This means that you have to conserve both the x and y components of momentum.

Before the collision, the total x momentum is simply 2MA/3 vo since the other car has no x motion.

Before the collision, the total y momentum is MAvo.

After the collision, the total momentum in the x direction is (MA+2MA/3) v(x), which should equal to the total x momentum before collision and we have the equation:

2MA/3 vo = (MA+2MA/3)v(x)

where I use v(x) to mean the x component of velocity after collision

obviously, v(x) =2MA/3 vo / (MA+2MA/3)

similarly in the y direction, v(y)=MA vo / (MA+2MA/3)

the tangent of the angle after collision will be given by the ratio of v(y)/v(x):

tan(theta)=v(y)/v(x) = [MA vo / (MA+2MA/3)]/[2MA/3 vo / (MA+2MA/3)]=MA/2MA/3=3/2=1.5

since the tan of 56.3 degrees is 1.5, you know that after the collision both cars move at angle of 56.3 degree, at a velocity v1, that is
v1=v(x)/sin56.3