Physics, asked by maga8, 1 year ago

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.​

Answers

Answered by amitnrw
0

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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