Solving Collision of Billiard Balls: Speed and Direction

[aligass2004]

Homework Statement

Similar to the other one...
On billiard ball is shot east at 2.6m/s. A second, identical billiard ball is shot west at .8m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90 degrees and sending it north at 1.58m/s. What are the speed and direction of the first ball after the collision?

Homework Equations

The Attempt at a Solution

I tried using (P1x)f + (P2x)f = (P1x)i + (P2x)i ----> (m1xv1x)f + (m2xv2x)f = (m1xv1x)i + (m2xv2x)i. I tried solving for (v1x)f. I assumed since the billiard balls are identical that the mass cancels out.

 

[learningphysics]

yes, the masses cancel. Looks like you have the right idea. can you show your calculations?

 

[aligass2004]

V1f = V1i + V2i - V2f
= 2.6 + .8 - 1.58 = 1.82

 

[learningphysics]

V1f = V1i + V2i - V2f
= 2.6 + .8 - 1.58 = 1.82

but 1.58 is not in the x-direction... what is the Vxf of the second ball?

 

[aligass2004]

I didn't realize that. The Vxf of the 2nd ball is zero since it isn't moving in the x direction anymore.

 

[learningphysics]

I didn't realize that. The Vxf of the 2nd ball is zero since it isn't moving in the x direction anymore.

exactly.

 

[aligass2004]

Ok, so I tried V1f = 2.6 + .8 = 3.4, but that wasn't right.

 

[learningphysics]

Ok, so I tried V1f = 2.6 + .8 = 3.4, but that wasn't right.

Oh, sorry I didn't notice... you should have -0.8, not 0.8.

use east positive. west negative. north positive. south negative.

 

[aligass2004]

Alright, this time I tried V1f = 2.6 - .8 = 1.8, but it again wasn't right.

 

[learningphysics]

Alright, this time I tried V1f = 2.6 - .8 = 1.8, but it again wasn't right.

that's only in the east west direction... you also need to consider the north south direction.

I thought the question was asking for the components separately. I just noticed now it only asks for speed. I'm sorry... wish I realized before you submitted again...

get the north south component, same way you got the east/west component... then get the speed using pythagorean theorem.

 

[aligass2004]

Blah...freaking components. Ok, for the north/south component, V1f = V1i + V2i - V2f -----> V1f = 0 + 0 -1.58. There's only one problem with using the pythagorean theroem. With the last problem, I figured out the angle really easily since both components were the same. I don't know the angle with this one.

 

[learningphysics]

Blah...freaking components. Ok, for the north/south component, V1f = V1i + V2i - V2f -----> V1f = 0 + 0 -1.58. There's only one problem with using the pythagorean theroem. With the last problem, I figured out the angle really easily since both components were the same. I don't know the angle with this one.

you can get the angle using arctan(vy/vx). be careful before you submit though... how do they ask for the angle? Do they want the angle from the east axis etc..?

 

[aligass2004]

They ask for degrees in the south east direction.

 

[aligass2004]

So the angle would be -41.276 degrees, but it would be positive since they want it in the SE direction, right?

 

[learningphysics]

So the angle would be -41.276 degrees, but it would be positive since they want it in the SE direction, right?

yeah, but it is East 41.276 degrees South...

Do they just say SE... because it could be east of south or south of east... that makes a difference...

 

[aligass2004]

It says south of east

 

[learningphysics]

It says south of east

Cool. Then 41.276 is right.

 

[aligass2004]

Ok so the angle was right, but I don't know the velocity...and I'm out of tries. I tried the -1.58 = vcos(theta).

 

[learningphysics]

Ok so the angle was right, but I don't know the velocity...and I'm out of tries. I tried the -1.58 = vcos(theta).

The 1.58 is the sin. so vsin(41.276) = 1.58. v = 2.395m/s.

But it's best to just use the pythagorean theorem when you need the speed... you don't need to worry about the angles. sqrt(1.58^2 + (1.8)^2) = 2.395m/s

 

[aligass2004]

Alright.