Consider a perfectly elastic collision of two objects (1 and 2) of the same mass in the x − y plane. Initially object 1 has velocity ⃗v1i = v1iˆi and object 2 is stationary. After the elastic collision, object 2 is heading at a nonzero angle θ with respect to the x axis; i. e., the collision is not a head-on collision. In this problem, you will develop a geometric proof to show that, after the collision, the angle between the trajectories of objects 1 and 2 is always 90◦, for any angle θ in the range 0 < θ < 90◦. If you play pool (billiards), you are probably aware of this result.
(a) Express conservation of momentum for this particular collision as a vector equation. Now use the fact that object 1 and object 2 have the same mass to draw a vector relationship between three velocity vectors. (Hopefully, you have drawn a triangle for which each side is a clearly labeled velocity vector.)
(b) Write down an expression for conservation of kinetic energy for this collision. Again use the fact that object 1 and object 2 have the same mass. Does the equation look familiar?
(c) Use the vector diagram in (a) and the equation in (b) to prove the final result.
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Consider a perfectly elastic collision of two objects (1 and 2) of the same mass in the x − y plane....
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