Answer:
The acceleration at lowest point is 19.62 m/s^2
Explanation:
Conservation of energy is an concept in which it is stated that the energy of an isolated object remains the same. Energy changes from one form to another.
Lets Assume
Constant of string is K
By using the conservation of energy we will have the following equation
1/2 x 80^2 x K = m x 9.81 x 120
3200 K = 1177.2 m
K = 1177.2 m / 3200
K = 0.368 m
At the lowest point we will have
a = ( K x X - m x g ) / m
a = ( 0.368 m x 80 - m x 9.81 ) / m
a = 19.62 m / s^2
So, the acceleration at lowest point is 19.62 m/s^2
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.
Learn more here: brainly.com/question/7694106
God is good man what can you say but its 18.66x30301=362728
Answer:
0.66 s
Explanation:
∆x = v∆t → 2 × 115 = 350 ∆t → ∆t = 230/350 = 0.66 s