Question

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.

Answer 1

Answer:

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 × $\underset{v_1}{\rightarrow}$ = m₁·$\underset{v'_1}{\rightarrow}$ + m₂·$\underset{v'_2}{\rightarrow}$

Which gives;

m₁ × 5·i = m₁·((√3)/2×5·i - 2.5·j) + m₂·$\underset{v'_2}{\rightarrow}$

∴ m₂·$\underset{v'_2}{\rightarrow}$ = m₁ × 5·i - m₁·((√3)/2×5·i - 2.5·j)

m₂·$\underset{v'_2}{\rightarrow}$ = 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·$\underset{v'_2}{\rightarrow}$ = m₁·$\underset{v'_2}{\rightarrow}$  = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

∴ $\underset{v'_2}{\rightarrow}$ = 2.5·(2 - √3)·i + 2.5·j

The magnitude of the velocity of the second ball is $\underset{v'_2}{\rightarrow}$ = √((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.