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.

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

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.

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