A billiard player shoots a ball with an initial speed of 1.5 m/s. It strikes ano

Question

A billiard player shoots a ball with an initial speed of 1.5 m/s. It strikes another billiard ball and moves off at an angle of 75.4° from its original path. The target ball was initially stationary. Assume an elastic collision and neglect effects of friction. What is the component of velocity of the second ball perpendicular to the initial direction of the first ball? (Assume that the first ball is initially travelling in the positive i and deflects such that it has a non-negative j velocity component after the collision. )

Explanation / Answer

Let the y component of final velocity of first ball be v1y and second ball be v2y, and corresponding x components be v1x and v2x.

Since there was no y component of momentum initially, v1y = -v2y

and v1x = v1y / tan 75.4 degree = 0.2605 v1y

v2x = u - v1x = 1.5 -  0.2605 v1y.

since collision is elastic,

0.5 m u^2 = 0.5 mv1^2 + 0.5 mv2^2

1.5^2 = [v1y ^2 + (0.2605 v1y)^2] + [(-v1y)^2 + (1.5 - 0.2605 v1y)^2]

v1y = 0.366

v2y = -0.366 m/s answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.