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Ball A (1.0kg) has an initial velocity of 1.8 m/s @ 0 (degrees). Ball A collides with Ball B (2.5kg). Ball B is initially at rest. At impact Ball A keeps going on with a velocity southeast @ 310 (degrees) and Ball B starts moving northeast @ 28 (degrees).

What is the final velocity of both balls?

2007-03-13 13:02:45 · 2 answers · asked by oscarjr1990 2 in Science & Mathematics Physics

Asume this collision is elastic. (KEi = KEf)

1/2 mA ViA^2= 1/2 mA VfA^2 + 1/2 mB VfB^2

2007-03-13 13:19:03 · update #1

I get answers for both final velocities but for some reason when I plug them into the equation for conservation of Kinetic Energy in an elastic collision the dont match up. KEi does not = KEf for me.

2007-03-13 13:20:53 · update #2

2 answers

I believe the problem as stated is overdetermined; that is, that there is a conflict in the data given. In any type of collision (inelastic, coefficient of restitution CR=0; partially elastic, 01) with these initial conditions, the bodies would have x-axis momenta summing to the total momentum, and equal and opposite y-axis momenta summing to 0. The kinetic energy required to achieve the specified CR would go to the y-axis momenta, which would change the path angles.
When the total momentum is known, the final x-axis and y-axis momenta are constrained to sum to a particular value. If the path angles are given, the y-axis momenta are also constrained and thus the CR is also. In this example, using "P" to represent momentum,
P1f(x) + P2f(x) = P1i(x) = Ptotal
P1f(y) + P2f(y) = 0
P1f(y) / P1f(x) = tan(-50)
P2f(y) / P2f(x) = tan(28)
These determine the final momentum x and y components of each body and therefore the CR. When I calculate this, I get KEi=1.62 and KEf = 0.771.
The ref. web page may be helpful in understanding this. If you explore the other pages associated with that page, you will find a collision calculator, but unfortunately it is strangely intolerant to the user entering the values he wants to enter. The collision of your example seems to be more complex than most web sites are willing to take on. It is 2-D and not "head-on", so the algebraic solution is more complex than the simple scalar (1-D) collisions.

2007-03-15 03:02:03 · answer #1 · answered by kirchwey 7 · 0 0

Draw a diagram and do some vector analysis.

2007-03-13 20:06:04 · answer #2 · answered by Anonymous · 0 0

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