First, find the magnitude of v, that is, the speed v of the two-car unit after the collision.?

Question

In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic.

Two cars of masses m1 and m2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of V1, and car 2 was traveling northward at a speed ofV2 . After the collision, the two cars stick together and travel off in the direction shown.

Express v in terms of m1 ,m2 , and the cars' initial speeds v1 and v2.

Answer 1

let me get you started...

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 m1v1 since the other car has no x motion.

Before the collision, the total y momentum is m2v2.

After the collision, the total momentum in the x direction is m1v1, and we have the equation:

m1v1 = (m1+m2)v(x)

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

obviously, v(x) = m1m1/(m1+m2)

similarly in the y direction, v(y)=m2v2/(m1+m2)

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

tan(theta)=v(y)/v(x) = [m2v2/(m1+m2)]/[m1v1/(m1+m2)]=m1v1/m2v1

since the tan of 45 degrees is one, you know that if the object moves off at a 45 degree angle that:

1=m2v2/m1v1 or m2v2=m1v1 or that p2=p1