Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

v is the maximum speed

Hence, the maximum speed of the ball is 5.86 m/s.
The kinetic energy of the small ball before the collision is
KE = (1/2) (mass) (speed)²
= (1/2) (2 kg) (1.5 m/s)
= (1 kg) (2.25 m²/s²)
= 2.25 joules.
Now is a good time to review the Law of Conservation of Energy:
Energy is never created or destroyed.
If it seems that some energy disappeared,
it actually had to go somewhere.
And if it seems like some energy magically appeared,
it actually had to come from somewhere.
The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.
Answer:
Potential Energy to Kenetic Energy
Explanation:
When holding a ball in the air, the ball has potential energy. Once you drop the ball, the ball gains Kenetic Energy
Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.
In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.
Remember that the Force of Gravity is given under the principle

Where,
G = Gravitational Universal constant
M = Mass of the planet
m = mass of the object
r = Distance from center of the planet
When the radius grows considerably the gravitational force begins to decrease.