Momentum and Kinetic Energy, Elastic Collision

AI Thread Summary
In an elastic collision between two identical balls, the first ball, moving at 3.0 m/s, strikes a stationary ball and moves away at a 30° angle. The conservation of momentum and kinetic energy equations are applied to determine the final speeds and direction of both balls. The calculations lead to the first ball's speed after the collision being approximately 2.6 m/s, while the second ball moves at 1.5 m/s at an angle of 60°. The discussion emphasizes the importance of keeping variables in symbolic form until the final calculations for clarity and insight. The approach and results are confirmed as correct by participants in the thread.
cassie123
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Homework Statement


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A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an
identical stationary ball as shown. If the first ball moves away
with angle 30° to the original path, determine:
a. the speed of the first ball after the collision.
b. the speed and direction of the second ball after the
collision.

Homework Equations


px: m1v1=m1v'1cosΘ1+m2v'2cosΘ2
py: 0=m1v'1sinΘ1+m2v'2sinΘ2
KE: 1/2m1v1^2=1/2m1v'1+1/2m2v'2

where I am using 1 and 2 to denote the first and second balls. and the "prime's" denoting final speed.
Θ1=30°
Since the balls have identical mass, I believe mass cancels out of the above equations.

The Attempt at a Solution


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What I tried to do was to find the components of the vectors of final motion, use the conservation of momentum and kinetic energy to create three equations to solve for the three unknowns (Θ2, v'1, v'2).

I canceled the mass out of all three equations, rearranged so that the Θ2 terms are on the same side of the momentum equations. I then squared the momentum equations and added them so that I could use the identity cos^2Θ+sin^2Θ=1 to get rid of the Θ2 terms and only have to solve for v'1 and v'2 using the KE equation and the added momentum equations. This is where I am getting stuck with the algebra:

(v1-v'1cos(30))^2+(-v'1sin(30))^2=(v'2)^2
and v'2=(v1)^2-(v'1)^2
so, subbing in for v'2:
(v1-v'1cos(30))^2+(-v'1sin(30))^2=(v1)^2-(v'1)^2

and at this point I should be able to solve for v'1, but i can't figure it out
 
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Just expand the terms and simplify (a lot).
 
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haruspex said:
Just expand the terms and simplify (a lot).

I gave that shot and ended up with v'1=2.598075 by using the quadratic equation.
This makes v'2=1.5 and Θ2=60°.

Does that make sense?
 
cassie123 said:
I gave that shot and ended up with v'1=2.598075 by using the quadratic equation.
This makes v'2=1.5 and Θ2=60°.

Does that make sense?
Looks right. You'll get more insight (and a heap of other benefits) if you resist the temptation to plug in numbers until the final step. Writing the original angle as theta, instead of 30 degrees, you find v1'=v1cos(theta), v2'=v1sin(theta).
 
Yes.
 
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