How do I solve elastic collisions?

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In an elastic collision involving a 2-kg ball moving at 3 m/s and a stationary 4-kg ball, both momentum and kinetic energy must be conserved. The conservation of momentum equation is 2(3) + 4(0) = 2vf1 + 4vf2, simplifying to 2vf1 + vf2 = 3. The total kinetic energy before the collision is calculated as 9, leading to the equation vf1^2 + 2vf2^2 = 9 after the collision. Solving these two equations simultaneously will yield the final velocities of both balls, vf1 and vf2. Understanding these principles allows for accurate calculations in elastic collision scenarios.
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A 2-kg ball is moving at 3 m/s toward the right. It elastically collides with a 4-kg ball that is initially at rest. Calculate the velocities of the balls after the collision.

I know that kinetic energy is conserved in elastic conditions, but I don't know how to use that to calculate this. I tried solving this using the conservation of momentum, but i ended up with 2(3) + 4(0) = 2vf1 + 4vf2 and I can't solve for both variables. I know PE = mgh and KE = (1/2) mv2
 
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Yes, in all collisions, in which there is no external force, momentum is preserved and that is what your equation, 2(3)+ 4(0)= 6= 4vf1+ 2vf2 or, dividing through by 2, 2vf1+ vf2= 3. And, as you say, in an elastic collision, kinetic energy is conserved. Since this doesn't involve change in height, the "potential energy" equation is irrelevant. And, as you say, kinetic energy is "(1/2)mv^2" the total kinetic energy before the collision is (1/2)(2)(9)+ (1/2)(4)(0)= 9. After the collision, with speeds vf1 and vf2, the kinetic energy is (1/2)(2)(vf1^2)+ (1/2)(4)(vf2^2)= vf1^2+ 2vf2^2. Since the kinetic energy does not change, vf1^2+ 2vf2^2= 9.

Solve the two equations 2v1+ vf2= 3 and vf1^2+ 2vf2^2= 9 for vf1 and vf2.
 
Thank you so much. I follow all of your work and understand it!
 
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