Posted by Shaila on Sunday, September 26, 2010 at 2:52pm.
We do not know if it is an elastic collision, but momentum is always conserved.
Your calculations seem to be based on the conservation of momentum in both directions, but you had a cos(θ) for both the "horizontal", or x-direction, and the "vertical", or y-direction.
Assuming the original direction of the first ball is along the y-axis, then the deviation of its course, θ, is -29.7°, and the new velocity, v1=9.56 m/s.
The stationary ball now moves with a velocity v2, and at an angle φ with the y-axis.
Equate momentum in the x-direction:
v1*sin(θ)+v2*sin(φ)=0
Equate momentum in the y-direction:
v1*cos(θ)+v2*cos(φ)=12.5
Solving the equations, I get v2=6.3 and φ=48°, different from your answer.
After that, you equate initial KE and final KE to determine if the collision was elastic.
How is it possible that mass which is 0.155kg is not taken into consideration?
You solve it as a 2D collision and thereafter you obtain the two equations, two variables and you solve you will get V'
Thereafter you find if it is elastic or not by equating the kinetic energy
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