- #1

- 33

- 0

A ball of mass m = 2.2 kg moving with a speed of v0 = 20.1 m/s strikes an identical ball which was initially at rest. After the collision, the incoming ball (ball 1) goes off at q1 = 24° relative to its original direction and the struck ball (ball 2) moves off at q2 = 34° as shown in the diagram.

*I can't post the diagram, but ball 1 shoots off 24 degrees above the x-axis and ball 2 shoots 34 degress below the x-axis.*

--------------------------------------------------------------------------------

a) What is the magnitude of the initial momentum of the two ball system?

pi =

**44.22 kg m/s**

--------------------------------------------------------------------------------

b) What is the ratio of the balls' final speeds?

|v1|/|v2| =

**1.37**

--------------------------------------------------------------------------------

c)

**What is the final speed of the incoming ball?**

|v1| = m/s

Okay, I have another problem that I'm stuck on! I was able to figure out part A & B (answers above), but I can't seem to get part C.

I tried setting the horizontal momentum of the two objects after the collision equal to the initial horizontal momentum (44.22) and then solving for one variable.

44.22 = m1v1 + m2v2 or 44.22 = m (v1+v2)

Then I thought I would have to use the ratio of v1/v2 in order to solve for a variable. So I did the following:

v1/v2 = 1.37 ... v2 = v1/1.37

I then plugged that into the momentum equation above and solved for the horiztonal velocity and got 11.619m/s for v1.

Then I took that number and divided it by COS(24) to get the Vf since the ball shoots off at an angle. I got 12.72m/s, but it isn't right. Where have I gone wrong? :(