The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
The given parameters;
- mass of the ball, m₁ = 0.8 kg
- speed of the ball, u₁ = 2.5 m/s
- mass of the object at rest, m₂ = 2.5 kg
- final velocity of the object at rest, v₂ = 1 m/s
Let the final velocity of the 0.8 kg ball immediately after collision = v₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)
2 = 2.5 + (0.8)v₁
-0.5 = (0.8)v₁

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
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Answer:
v = 31.3 m / s
Explanation:
The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.
Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane
Highest point
Em₀ = U = m g y
Lowest point
= K = ½ m v²
As there is no friction, mechanical energy is conserved
Em₀ =
m g y = ½ m v²
v = √ 2 g y
Where we can use trigonometry to find and
sin 30 = y / L
y = L sin 30
Let's replace
v = RA (2 g L sin 30)
Let's calculate
v = RA (2 9.8 100.0 sin30)
v = 31.3 m / s
Complete Question
The Complete Question is attached below
We have that the Cartesian components of tension
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Answer:
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