A cannon is on the edge of a cliff 5.2 x 102 m above the ground. A cannon ball leaves the cannon at 1.5 x 102 m/s. If the cannon
1 answer:
You might be interested in
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 × = 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.