A 1.00 kg sphere M, suspended by a string from point P, is lifted to a height h. The sphere is released and passes through the lowest point in its swing at a speed of 10.0 meters per second. [Neglect friction.] a. Calculate the height from which the sphere was released. [Show all work including equation and substitution with units]

b. Compared to the sphere’s speed through the lowest point of its swing when released from h, the sphere’s speed through the lowest point when released from 2h would be

(a) h = 5.1 m

(b) v = 14.13 m/s

Explanation:

(a)

We will use the law of conservation of energy. For this situation it states that:

Loss in Potential Energy of Sphere = Gain in Kinetic Energy of the Sphere

mgh = (1/2)mv²

h = v²/2g   equation (1)

where,

h = height = ?

v = speed at lowest point = 10 m/s

g = 9.8 m/s²

Therefore,

h = (10 m/s)²/(2)(9.8 m/s²)

h = 5.1 m

(b)

using the equation (1)

h = v²/2g

v = √2gh

where,

v = velocity = ?

g = 9.8 m/s²

h = height = 2(5.1 m) = 10.2 m

Therefore,

v = √[2(9.8 m/s²)(10.2 m)]

v = 14.13 m/s

b would be the answer

a.   mass number from atomic number