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You are playing pool. You hit a striped ball with the white ball which is moving with a velocity of 8 m/s before the collision. After the collision the striped ball goes at an angle of 30 degrees to the left of the line of the original direction of the white ball and the white ball goes at an angle of 60 degrees to the right of the line of the original direction it came in with. What velocity does the striped ball go off with

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Answer:

The striped ball has a final velocity of 6.93 m/s

Explanation:

The initial velocity of the white ball = 8 m/s

The initial velocity of the striped ball = 0 m/s

let the mass of the ball = m

Momentum = mass * velocity

Initial momentum of the white ball = 8m kgm/s.............(1)

Initial momentum of the strip ball = 0 kgm/s...................(2)

After the collision:

let the final velocity of the white ball be [tex]v_{w}[/tex]

and the final velocity of the strip ball be [tex]v_{s}[/tex]

The final momentum can be resolved in the x and y directions

[tex]P_{xf} = mv_{s} cos30 + mv_{w} cos60[/tex].....................(3)

[tex]P_{yf} = mv_{s} sin30 + mv_{y} sin60[/tex]......................(4)

From the conservation of momentum:

Final momentum = Initial momentum

that is, equating (2) and (4)

[tex]mv_{s} sin30 + mv_{w} sin60 = 0\\0.5v_{s} + \frac{\sqrt{3} }{2} v_{w} = 0\\v_{s} = \sqrt{3} v_{w}\\[/tex]

[tex]v_{w} = v_{s}/\sqrt{3}[/tex]....................(5)

Equating (1) and (3) and inserting (5)

[tex]mv_{s} cos30 + mv_{w} cos60 = 8m\\\frac{\sqrt{3} }{2} v_{s} + \frac{v_{w} }{2} = 8\\\sqrt{3} v_{s} +v_{w} = 16\\\sqrt{3} v_{s} + \frac{v_{s} }{\sqrt{3} } = 16\\3 v_{s} + v_{s} = 16\sqrt{3} \\4 v_{s} = 16\sqrt{3}\\ v_{s} = 4\sqrt{3}\\ v_{s} = 6.93 m/s[/tex]

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