Head-on Collision: Solving for Ball A & B Velocities

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In a perfectly elastic head-on collision between two equal mass billiard balls, the initial velocities are 1.9 m/s for ball A (in the +x direction) and 3.9 m/s for ball B (in the -x direction). To find the final velocities after the collision, both conservation of momentum and conservation of kinetic energy must be applied, resulting in two equations. The first equation represents the conservation of kinetic energy, while the second represents the conservation of momentum. Solving these equations simultaneously allows for the determination of the final velocities of both balls. Proper algebraic manipulation is essential to find the unknown velocities accurately.
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Can anyone help me with this problem?


Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball (ball A) is initially moving at 1.9 m/s in the +x direction, and the other (ball B) is initially moving at 3.9 m/s in the -x direction, what will be the velocities of (a) ball A and (b) ball B after the collision? (Indicate the direction by the sign of your answer, taking a positive velocity as being in the +x direction.)

I've tried using the conservation of momentum and energy... doesn't seem to work.
 
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First I tried mvA+MvB=MvA'+MvB' but that didn't work... then I got some long formulas from class like V(B)'=2M(A)(VA)/(MA+MB)+(M(B)-(MA)(VB)/(MA+MB)
 
Hi Jaykawk1,

Ok this kind of problem requires more algebra than physics...

Ok you have to use two principles two solve this

1) (1/2)M(a)(V^2) + (1/2)M(b)(V^2) = (1/2)M(a)(V'(a)^2) + (1/2)M(b)(V'(b)^2)
[ Conservation of total kinetic energy for elastic collisions ]

2) M(a)V(a) + M(b)V(b) = M(a)V'(a) + M(b)V'(b)
[ Conservation of Momentum ]


Now what you need to see is that here you have two equations.

And...the problems asks:

"what will be the velocities of (a) ball A and (b) ball B after the collision?"

So there are TWO velocities that need to be found thus two unkowns.

Therefore we have two equations;

1) (1/2)M(a)(V^2) + (1/2)M(b)(V^2) = (1/2)M(a)(V'(a)^2) + (1/2)M(b)(V'(b)^2)

2) M(a)V(a) + M(b)V(b) = M(a)V'(a) + M(b)V'(b)

And also two unkowns, V'(a) and V'(b), Now from here its just algebra.

Solve for one of the unkowns, V'(a) or V'(b), using equation (2) and substitute that into equation (1), and then solve for the other unkown velocity (you may have to use the quadratic formula) and than once you get the velocity.

Then go back to what you came up with earlier when you were first trying to solve for one of the unkowns and substitute what you found from above into that equation and solve for the final one.

And your set.

Regards,

-Jose Antonio Zalles II

P.S. just reply if its still giving you issues.
 
Last edited:
Thanks so much!
 
No prob, best of luck with your studies.

-Jose Antonio Zalles II
 
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