Question

A white billiard ball with mass m_w = 1.5 kg is moving directly to the right with a speed of v = 3 m/s and collides elastically with a black billiard ball with the same mass m_b = 1.5 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θ_w = 30° and the black ball ends up moving at an angle below the horizontal of θ_b = 60°.

1)

What is the final speed of the white ball?
m/s

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2)

What is the final speed of the black ball?
m/s

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3)

What is the magnitude of the final total momentum of the system?
kg-m/s

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4)

What is the final total energy of the system?
J

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Answer 1

Let the velocity of white ball is v(w) and velocity of black ball is v(b).

Apply the law of momentum conservation in X-axis:-

( 1.5 kg ). ( 3 m/s ) + 0 = ( 1.5 kg ) .v(w) . cos30* + (1.5 kg ).  v(b) . cos 60*

Simplify the equation

1.299 v(w ) + 0.75 v(b) = 4.5 ............... (1)

Along y- axis, the momentum conservation yields

( 1.5 kg ) . v(w) . sin30* = ( 1.5 kg ) . v(b) . sin60*

v(w) = 1.732 v(b) ........... (2)

Substitute (2) in equation (1)

1.299 [ 1.732 v(b) } + 0.75 v(b) = 4.5

Solve for velocity of black ball:

v(b) = 1.5 m/s

From the equation (2)

v(w) = 1.732{ 1.5 m/s} = 2.6 m/s

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Total momentum of the system:

P = mv = 1.5 x3 = 4.5 N-s

{as by the law of momentum conservation, momentum , P(final) = P(initial)}

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Total energy:

E = 1/2mv² = ( 1/2 ) ( 1.5 ) (3)² =6.75 J

as the collision is elastic=>E(final) = E(initial)}