What is the speed of the silver sphere at the moment it hits the ground? (Assume that energy is conserved during the fall and that 100% of the sphere’s initial potential energy is converted to kinetic energy by the time impact occurs.)

Consider the two spheres shown here, one made of silver and the other of aluminum. The spheres are dropped from a height of 2.1 m. (composition of sphere : density is 10.49 g/cm³, volume is 196 cm³), (composition of aluminum : density is 2.7 g/cm³, volume is 196 cm³).

Answer:

The speed of the silver sphere at the moment it hits the ground is 6.416 m/s

Explanation:

Potential energy of the silver sphere = mgh

where;

g is acceleration due to gravity = 9.8 m/s²

h is height above the ground = 2.1m

m is mass of the silver sphere =?

Density = mass/volume

mass of the silver sphere = density X volume = (10.49 g/cm³) X (196cm³ )

mass of the silver sphere = 2056.04 g = 2.056 kg

Potential energy of the silver sphere = (2.056 X 9.8 X 2.1) J

Potential energy of the silver sphere = 42.31248 J

If energy is conserved and 100% potential energy is converted to kinetic energy, the speed of the silver sphere will be calculated as follows:

Kinetic Energy (K.E) = 1/2mv²

where;

m is mass of the sphere in kg = 2.056kg

v is the velocity of the silver sphere = ?

If energy is conserved; Potential energy = Kinetic energy

42.31248 = 1/2mv²

v² = (42.31248 X 2)/m

v² = (42.31248 X 2)/2.056

v² = 41.16

v = √41.16

v = 6.416 m/s

Therefore, the speed of the silver sphere at the moment it hits the ground is 6.416 m/s