Physics - Elastic Collision: Solving for Unknowns

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In the discussion about an elastic collision involving two balls of equal mass, participants focus on calculating the angle of motion and velocity ratios post-collision. The initial velocity of ball B is twice that of ball A, and the angle θ is given as 56°. The equations derived for the final velocity components indicate that the total mass cancels out, simplifying the calculations. The final angle above the horizontal is determined to be approximately 26.3 degrees. Key questions include how to handle the unknowns and whether to consider magnitudes in the calculations.
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Homework Statement



Two balls of equal mass collide and stick together as shown in the figure. The initial velocity of ball B is twice that of ball A. (Take θ = 56°.)

(a) Calculate the angle above the horizontal of the motion of mass A + B after the collision.

(b) What is the ratio of the final velocity of the mass A + B to the initial velocity of ball A, vf/vA?

(c) What is the ratio of the final energy of the system to the initial energy of the system, Ef/Ei?

The Attempt at a Solution



a) x= m ( 2vcos(56) + vcos(56) ) = 2mVfx

2vcos(56) + vcos(56)= mvfx

y = m ( 2vsin(56) + vsin(56) ) = 2mVfy

2vsin(56) + vsin(56) = 2mVfy

Am I suppose to take the magnitude of this ? How do I solve for two unknowns ?
Do all the m's and v's cancel out?

Thanks
 

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y = m ( 2vsin(56) + vsin(56) ) = 2mVfy

2vsin(56) + vsin(56) = 2mVfy

There is something bad in the setup of the first equation.

Am I suppose to take the magnitude of this ? How do I solve for two unknowns ?
Do all the m's and v's cancel out?

Divide by the total mass of the body.
v remains.
 
2vcos(56) + vcos(56) = 2vfx => 3vcos(56)

2vsin(56) + -vsin(56) = 2vfy => vsin(56)

tan-1 ( y/x) => (1/3) => 26.3 degree
 
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