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

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SUMMARY

In a perfectly elastic head-on collision involving two billiard balls of equal mass, ball A, initially moving at 1.9 m/s in the +x direction, and ball B, initially moving at 3.9 m/s in the -x direction, the velocities after the collision can be determined using the conservation of momentum and kinetic energy principles. The equations governing the system are: (1) conservation of kinetic energy: (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), and (2) conservation of momentum: M(a)V(a) + M(b)V(b) = M(a)V'(a) + M(b)V'(b). Solving these equations yields the final velocities for both balls.

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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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