Two skiers travel two different frictionless paths. The two paths start at the same place and finish at the same place. Skier A is heavier than Skier B. With no more information, which of the skiers has the greater speed at the finish? Ignore air resistance.

A. Skier A has the same speed as Skier B at the finish.

B. Skier A has the greater speed at the finish.

C. Skier B has the greater speed at the finish.

D. There is no way to know which skier has the greater speed at the finish.

D. There is no way to know which skier has the greater speed at the finish.

Explanation::

Total energy is conserved because there is no friction

(Et) ₀ = (Et) f

(Et) ₀ = Initial total energy (J)

(Et) f = Final total energy (J)

(Et) ₀ = K₀ + U₀

(Et) f = Kf + Uf

K₀ : Initial kinetic energy

U₀ : Initial potential energy

Kf : final kinetic energy

Uf : final potential energy

The formulas to calculate the kinetic energy (K) and potential energy (U) are:

K = (1/2) m*v²

U = m * g*h

m : mass (kg)

v: speed (m/s)

h: hight (m)

Problem development

Skier A

(Et) ₀ = (Et) f

K₀ + U₀ = Kf + Uf

(1/2) mA * (v₀A) ² + mA*g*h₀ = (1/2) mA * (vfA) ² + mA*g*hf,

We divide by mA on both sides of the equation

(1/2) * (v₀A) ² + g*h₀ = (1/2) (vfA) ² + g*hf

(1/2) * (v₀A) ² + g*h₀ - g*hf = (1/2) (vfA) ²

We multiply by 2 both sides of the equation

(v₀A) ²+2g (h₀ - hf) = (vfA) ²

(vfA) ² = (v₀A) ²+2g (h₀ - hf) Equation (1)

Skier B

(Et) ₀ = (Et) f

K₀ + U₀ = Kf + Uf

(1/2) mB * (v₀B) ² + mB*g*h₀ = (1/2) mB * (vfB) ² + mB*g*hf

We perform the same procedure above:

(vfB) ² = (v₀B) ²+2g (h₀ - hf) Equation (2)

Comparison of equation (1) with equation (2)

The term 2g (h₀ - hf) is the same in both equations because the paths of the two skiers start in the same place and end in the same place.

The final speed (vf) of skiers depends on their initial speed (v₀).

Because the initial speed of the skiers is unknown, it cannot be determined which has the highest final speed