A billiard ball moving at 5 m/s strikes another ball which is initially at rest. After the collision, the first ball moves at a velocity of 4.35 m/s at an angle of 30o below its original motion. Find the velocity and angle of the second ball after the collision. Assume that the collision is perfectly elastic.

The velocity of the second ball is approximately 2.588 m/s

The angle direction of the second ball is 75° counterclockwise from the horizontal

Explanation:

The initial velocity of the first billiard ball = 5 m/s

The initial velocity of the billiard ball the first billiard ball strikes = 0 m/s

The final velocity of the first billiard ball = 4.35 m/s

The final direction of motion of the first billiard ball = 30° below its original motion

For perfectly elastic collision, whereby the target is at rest initially, by conservation of momentum, we have;

m₁ × = m₁· + m₂·

Which gives;

m₁ × 5·i = m₁·((√3)/2×5·i - 2.5·j) + m₂·

∴ m₂· = m₁ × 5·i - m₁·((√3)/2×5·i - 2.5·j)

m₂· = m₁ × 5·(1 - √3/2)·i + m₁·2.5·j = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

Therefore, given that the mass of both billiard balls are equal, we have, m₁ = m₂, which gives;

m₂· = m₁·  = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

∴ = 2.5·(2 - √3)·i + 2.5·j

The magnitude of the velocity of the second ball is = √((2.5·(2 - √3))² + 2.5²) ≈ 2.588 m/s

The direction of the second ball, θ = arctan(2.5/((2.5·(2 - √3))) = 75° counterclockwise from the horizontal.

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.