A white billiard ball with mass m w = 1.57 kg is moving directly to the right wi

Question

A white billiard ball with mass mw= 1.57 kg is moving directly to the right with a speed of v = 2.85 m/s and collides elastically with a black billiard ball with the same mass mb= 1.57 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° and the black ball ends up moving at an angle below the horizontal of ?b= 61°.

1) 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?

Explanation / Answer

After a zero-friction collision of a moving ball with a stationary one of equal mass, the angle between the directions of the two balls is always 90 degrees. This is an important fact that professional billiards players take into account, although it assumes the ball is moving frictionlessly across the table rather than rolling with friction. Consider an elastic collision in 2 dimensions of any 2 masses m1 and m2, with respective initial velocities u1 and u2 = 0, and final velocities V1 and V2. Conservation of momentum gives m1u1 = m1V1+ m2V2. Conservation of energy for an elastic collision gives (1/2)m1|u1|2 = (1/2)m1|V1|2 + (1/2)m2|V2|2. Now consider the case m1 = m2: we obtain u1=V1+V2 and |u1|2 = |V1|2+|V2|2. Taking the dot product of each side of the former equation with itself, |u1|2 = u1

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