Answer:
Part a)

Part b)

So this speed is independent of the mass of the rider
Explanation:
Part a)
By force equation on the rider at the position of the hump we can say

now we will have


now we have



Part b)
At the top of the loop if the minimum speed is required so that it remains in contact so we will have

at minimum speed




So this speed is independent of the mass of the rider
Explanation:
If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.
Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.
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