Answer:
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
