A pool ball moves at 2.1 m/s to the right (+x direction) and hits an identical, stationary ball. After the first ball strikes the 2nd ball, the first ball's velocity is 1.961 m/s, at an angle of +21. Assuming an elastic collision and ignoring friction and rotational motion, find the struck ball’s (a) speed and (b) direction (in unit circle notation) after the collision.
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Answer:
0.75 m/s
-68.9° or 291.1°
Explanation:
A pool ball moves at 2.1 m/s to the right (+x direction)
momentum = M * 2.1 in x direction
After collision first ball's velocity is 1.961 m/s at angle 21°
so Momentum in x direction = M * 1.961Cos21 = 1.83 M
Momentum in y Direction = M * 1.961Sin21 = 0.7 M
Let say second ball speed = V at angle α
Momentum in x direction = M * VCosα
Momentum in y Direction = M * VSinα
2.1M = 1.83 M + M * VCosα
=> VCosα = 0.27
& 0 = 0.7 M + M * VSinα
=> VSinα = -0.7
√V²Cos²α + V²Sin²α = V = 0.75
α = -68.9° or 291.1°
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