Angle between velocity vectors of 2D billiard ball collision?

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naianator
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Homework Statement


During a game of billiards, a white cue ball traveling at speed v strikes a green ball that was initially at rest. The green ball's speed after the collision is twice the speed of the white ball after the collision. The billiard balls have equal mass.

What is the angle between the cue ball's final velocity vector and the cue ball's initial velocity vector? (Enter an angle between 0 and 90 degrees.)

Homework Equations


v^2 = v_1^2+v_2^2

v = v_1+v_2

The Attempt at a Solution


I have determined that the speed of the cue ball after hitting the green ball is v/sqrt(5) making the speed of the green ball 2v/sqrt(5). From what I can tell cos(theta) = speed of cue ball after striking green ball/initial speed = v/sqrt(5)/v = 1/sqrt(5). But obviously this isn't correct. What am I doing wrong?
 
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What is the exact question ? What kind of collision takes place ?

Hint : Does this even seem possible ( Compare KEf and KEi ) ?
 
Qwertywerty said:
What is the exact question ? What kind of collision takes place ?

Hint : Does this even seem possible ( Compare KEf and KEi ) ?
I'm not sure what you mean... v^2 = v^2/5+4v^2/5 = v^2
its an elastic collision implying that the balls take off perpendicular to each other so I can construct a triangle
 
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naianator said:
I'm not sure what you mean... v^2 = v^/5+4v^2/5 = v^2
its an elastic collision implying that the balls take off perpendicular to each other so I can construct a triangle
naianator said:
The green ball's speed after the collision is twice the speed of the white ball after the collision.
Sorry , this seemed a bit confusing .

Okay , anyways , your answer is correct . However , what do you mean by ' constructing a triangle ' ?
 
Qwertywerty said:
Sorry , this seemed a bit confusing .

Okay , anyways , your answer is correct . However , what do you mean by ' constructing a triangle ' ?
a right triangle of the velocity vectors - initial velocity being the hypotenuse... Is that correct? Anyhow, I'm not sure how I got the wrong answer before - must have been a calculator error. Thanks!
 
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