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A white billiard ball with mass mw = 1.3 kg is moving directly to the right with

ID: 1344788 • Letter: A

Question

A white billiard ball with mass mw = 1.3 kg is moving directly to the right with a speed of v = 3.44 m/s and collides elastically with a black billiard ball with the same mass mb = 1.3 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 = 62° and the black ball ends up moving at an angle below the horizontal of ?b = 28

What is the final speed of the white ball?

2)

What is the final speed of the black ball?

3)

What is the magnitude of the final total momentum of the system?

4)

What is the final total energy of the system?

INITIAL: FINAL I W

Explanation / Answer

m1 = mass of white ball = 1.3 kg

m2 = mass of black ball = 1.3 kg

Along X-direction :

V1ix = velocity of white ball before collision = 3.44 m/s

V2ix = velocity of black ball before collision = 0 m/s

V1fx = velocity of white ball after collision = V1f Cos62

V2fx = velocity of black ball after collision = V2f Cos28

Along Y-direction :

V1iy = velocity of white ball before collision = 0 m/s

V2iy = velocity of black ball before collision = 0 m/s

V1fy = velocity of white ball after collision = V1f Sin62

V2fy = velocity of black ball after collision = - V2f Sin28

Using conservation of momentum along X-direction ::

m1 V1ix + m2 V2ix = m1V1fx + m2 V2fx

1.3 x 3.44 + 1.3 x 0 = 1.3 V1f Cos62 + 1.3 V2f Cos28

0.47 V1f + 0.88 V2f = 3.44                      eq-1

Using conservation of momentum along Y-direction ::

m1 V1iy + m2 V2iy = m1V1fy + m2 V2fy

1.3 x 0 + 1.3 x 0 = 1.3 V1f Sin62 - 1.3 V2f Sin28

0.88 V1f = 0.47 V2f

V1f = 0.534 V2f                                   eq-2

Using Eq-1 and eq-2

0.47 (0.534 V2f) + 0.88 V2f = 3.44

V2f = 3.04 m/s

V1f = 0.534 V2f   = 0.534 x 3.04 = 1.62 m/s

1)

V1f = 1.62 m/s

2)

V2f = 3.44 m/s

3)

final Total momentum = initial total momentum = 1.3 x 3.44 + 1.3 x 0 = 4.47

4)

final total energy = (0.5) (m1V21f + m2 V22f) = (0.5) (1.3) (1.622 + 3.442) = 9.4 J

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