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Nadusha1986 [10]

3 years ago

11

billiardballsnew2 A white billiard ball with mass mw = 1.49 kg is moving directly to the right with a speed of v = 3.09 m/s and

collides elastically with a black billiard ball with the same mass mb = 1.49 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 29° and the black ball ends up moving at an angle below the horizontal of θb = 61°. 1)What is the final speed of the white ball? m/s 2)What is the final speed of the black ball?

1 answer:

Kobotan [32]3 years ago

7 0

Answer:

Final velocity of white ball is 0m/s

Final velocity of black ball is 3.09m/s

Explanation:

An elastic collision is one that conserves internal kinetic energy

An internal kinetic energy is the sum of kinetic energies of objects in the system

Initial kinetic energy of white ball is Vi1 = 3.09m/s

Final kinetic energy of white ball is Vf1 = ?

Initial kinetic energy of black ball is Vi2 = 0m/s

Final kinetic energy of black ball is Vf2 = ?

m1 = 1.49kg mass of white ball

m2 = 1.49kg mass of black ball

The formula to calculate internal kinetic energy is

1/2m1Vf1^2 + 1/2m2Vf2^2 = 1/2m1Vi1^2

Solving the equation

1.Vf1 = (m1 - m2)Vi1/m1+m2

Vf1 = (1.49-1.49)*3.09/1.49+1/49

Vf1 = 0m/s

2. Vf2 = 2m1Vi1/m1+m2

Vf2 = 2*1.49*3.09/1.49+1.49

Vf2 = 3.09m/s

N:B following the general principle of collision when 2 bodies of same masses collide in elastic collision they exchange velocities.

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Answer:

a) t=1s

y = 10.1m

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b) t=1.5s

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Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

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y: The vertical distance the ball moves at time t

vi: Initial speed

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v= Speed the ball moves at time t

Known information

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Vi=15 m / s

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t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

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