Question

In: Physics

A white billiard ball with mass mw = 1.33 kg is moving directly to the right...

A white billiard ball with mass mw = 1.33 kg is moving directly to the right with a speed of v = 2.96 m/s and collides elastically with a black billiard ball with the same mass mb = 1.33 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of ?w = 29

Solutions

Expert Solution

mw = mass of white ball = 1.33 kg

mb = mass of black ball = 1.33 kg

Vwix = velocity of white ball before collision along X-direction = 2.96 m/s

Vwiy = velocity of white ball before collision along Y-direction = 0 m/s

Vbix = velocity of blaack ball before collision along X-direction = 0 m/s

Vbiy = velocity of blaack ball before collision along Y-direction = 0 m/s

Vwf = velocity of white ball after collision

Vwfx = velocity of white ball after collision in X-direction = Vwf cos29

Vwfy = velocity of white ball after collision in Y-direction = Vwf Sin29

Vbf = velocity of black ball after collision

Vbfx = velocity of black ball after collision in X-direction = Vbf cos61

Vbfy = velocity of black ball after collision in Y-direction = -Vbf Sin61   (since it is along negative y-direction)

Using conservation of momentum along X-direction ::

mwVwix+ mbVbix= mw Vwfx + mbVbfx

inserting the values

2.96 + 0 = Vwf cos29 + Vbf cos61

Vwf (0.875) + Vbf (0.485) = 2.96                         eq-1

Using conservation of momentum along Y-direction ::

mwVwiy+ mbVbiy= mw Vwfy + mbVbfy

inserting the values

0 + 0 = Vwf Sin29 -Vbf Sin61

  Vwf Sin29 = Vbf Sin61                    

Vwf = 1.804 Vbf                             eq-2

Using eq-1 and eq-2

(1.804 Vbf ) (0.875) + Vbf (0.485) = 2.96  

Vbf = 1.4345 m/s          --------------------final speed of black ball

Vwf = 1.804 Vbf   = 1.804 x 1.4345

Vwf = 2.588 m/s          ----------------------- final speed of white ball

final total momentum = initial total momentum = mw Vwi= 1.33 x 2.96 = 3.937 kgm/s

final total energy = (0.5) mw V2wf + (0.5) mb V2bf = (0.5) (1.33) (2.5882 + 1.43452)

Final total energy = 5.822 J


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