A billiard player shoots a ball with an initial speed of 4.8 m/s. It strikes ano

Question

A billiard player shoots a ball with an initial speed of 4.8 m/s. It strikes another billiard ball and moves off at an angle of 72.6° 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

using law of conservation of momentum

along the original direction

momentum before collision = momentum after collision

(m*4.8)+(m*o) = (m*V1*cos(72.6))+(m*V2x)

4.8 - (V1*cos(72.6)) = V2x

V2x = 4.8-(0.3*V1)

along the perpendicular to the initial direction of first ball

0 + 0 = (m*V1*sin(72.6)) + (m*V2y)

V2y = -V1*sin(72.6) = -0.954*V1

V2^2 = (4.8-(0.3*V1))^2 + (0.954*V1)^2

using law of conservation of kinetic energy

0.5*m*4.8^2 = (0.5*m*v1^2)+(0.5*m*v2^2)

0.5*4.8^2 = (0.5*V1^2)+(0.5*(4.8-(0.3*V1))^2 + (0.954*V1)^2)

V1 = 0.989 m/sec

V2y = -0.954*0.989 = -0.943 m/sec

in vecor form V2y = -0.943 j hat m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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