Momentum and Elastic Collisions

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In an elastic collision scenario involving a cue ball and an eight ball, the cue ball initially travels at 8.0 m/s while the eight ball is at rest. The eight ball moves at a 30° angle post-collision, leading to the establishment of momentum and energy conservation equations. The equations derived include momentum in both x and y directions, along with the conservation of kinetic energy. A suggested method for solving the complex equations involves squaring and adding the momentum equations to simplify the calculations. This approach allows for the determination of the cue ball's velocity and the angle of deflection after the collision.
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Homework Statement


In a pool game, the cue ball, which has an initial speed of 8.0 m/s, make an elastic collision with the eight ball, which is initially at rest. After the collision, the eight ball moves at an angle of 30° to the original direction of the cue ball.


Homework Equations


V8 = Velocity of 8-ball
Vc = Velocity of cue ball

(1)Epx = m*V8*cos(30) + m*Vc*cos(x) = m*8
(2)Epy = m*V8*sin(30) + m*Vc*sin(x) = 0
(3).5*m*Vi^2 = .5*m*V8^2 + .5*m*Vc^2


The Attempt at a Solution


While these equations are technically solvable, they are nearly impossible by hand. Solving (3) for one velocity and using substitution twice I get:
sin(x)^2 -cos(30)*cos(x) + sin(30)*cos(30)*cos(x) - sin(30)*cos(x)^2 = -1/8

Is there an easier way to solve this since I know I will not be able to solve this equation myself?
 
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(1)Epx = m*V8*cos(30) + m*Vc*cos(x) = m*8
(2)Epy = m*V8*sin(30) + m*Vc*sin(x) = 0

Rewrite these two equations as
m*V8*cos(30) = m*8 - m*Vc*cos(x) -------(1)
m*V8*sin(30) = - m*Vc*sin(x) ----------(2)
Square both sides of eq.1 and 2 and add them. After simplification you will get the value of vc*cos(x)
From the conservation of energy equation, find the value of vc. Then you can find the angle x.
 
Awesome, thanks so much!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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