2D momentum ( in understanding, but have the answer)

[Elbobo]

Homework Statement

Assume an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 3.6 m/s strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 1.9 m/s .

Find the queue ball’s angle θ with respect
to its original line of motion. Answer in units
of ◦.

Homework Equations

p1x + p2x = p1x' + p2x'
p1y + p2y = p1y' + p2y'
p = mv

The Attempt at a Solution

OK I got the right answer (really I just copied my teacher's work which was too confusing for me to repeat, I only plugged in numbers to get the right answer), which was about 58.14 degrees. Really don't feel like typing out the whole process, but she (the teacher) used the sin^2 theta + cos^2 theta = 1 identity during the process.

THEN! My friend shows me how she did it, and all she did was arccosine (final speed of ball 1/initial speed of ball 1), which got the exact same answer.

I don't get the physics behind this. Why did my friend's solution work?

 

[tiny-tim]

Assume an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 3.6 m/s strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 1.9 m/s .

Find the queue ball’s angle θ with respect
to its original line of motion. Answer in units
of ◦.
…
THEN! My friend shows me how she did it, and all she did was arccosine (final speed of ball 1/initial speed of ball 1), which got the exact same answer.

I don't get the physics behind this. Why did my friend's solution work?

Hi Elbobo!

(btw, it's a cue ball, not a queue ball!)

I think her method only works for the special case of equal masses.

It's a geometry thing …

Hint: if you draw the vector triangle representing the momentums, you should notice that the energy equation immediately gives you one of the angles of the triangle.

 

[Elbobo]

Ah I see! Didn't think of using the resulting momentums as the components of the initial momentum in constructing a vector diagram.

Why wouldn't this method work when the masses are different?

 

[tiny-tim]

Why wouldn't this method work when the masses are different?

As a momentum diagram, it will still work fine.

But if you've tried it, you should have found that the geometry won't work conveniently for the energies.

 

[Elbobo]

Oh, right, because energy is a scalar quantity.

(BTW I didn't make up "queue"; I thought that was strange too)