Question

A billiard ball moving at 5.30 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.72 m/s, at an angle of θ = 27.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.

Answer 1

The velocity of the struck billiard ball after collision is 2.41 m/s at an angle of 62.8⁰.

Step-by-step explanation:

Given;

initial velocity of the billiard ball, u₁ = 5.3 m/s

initial velocity of the second ball, u₂ = 0

final velocity of the first ball, v₁ = 4.72 m/s at θ = 27.0°

final velocity of the second ball, v₂ = ?

Apply the principle of conservation of kinetic energy;

¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂m₁v₁² + ¹/₂m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

m₁u₁² + m₂(0) = m₁v₁² + m₂v₂²

m₁u₁² = m₁v₁² + m₂v₂²

m₁ = m₂

u₁² = v₁² + v₂²

v₂² = u₁² - v₁²

v₂² = (5.3)² - (4.72)²

v₂² = 5.812

v₂ = √5.812

v₂ = 2.41 m/s

The direction of the second ball is given by;

v₂sinθ = v₁sin27

sinθ = (v₁sin27) / (v₂)

sinθ = (4.72 x 0.454) / (2.41)

sinθ = 0.8892

θ = sin⁻¹ (0.8892)

θ = 62.8⁰

Thus, the velocity of the struck billiard ball after collision is 2.41 m/s at an angle of 62.8⁰.