Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?

A) Object 1 rises to the greater height because it weighs less.

B) Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.

C) Object 2 rises to the greater height because it contains more mass.

D) Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.

The two objects rise to the same height.

E) The two objects rise to the same height : h₁=h₂ = (v²) / (2g)

Explanation:

With coefficient of kinetic friction, μk = 0:

We apply the principle of energy conservation:

E₀ = Ef Formula (1)

K₀+U₀ = Kf + Uf Formula (2)

K = (1/2) * m*v² Formula (3)

U = m*g*h Formula (4)

Where:

E₀:

Initial total energy (J)

Ef: Final total energy (J)

K₀: Initial kinetic energy (J)

U₀: Initial potential energy (J)

Kf: Final kinetic energy (J)

Uf:

Final kinetic energy (J)

v : speed (m/s)

m: mass (kg)

h : hight (m)

Known data

v₁=v₂=v

m₁=m

m₂ = 2m

μk = 0 : coefficient of kinetic friction

Problem development

We apply formulas (1), (2), (3) and (4) for the first object, (1):

E₀₁ = Ef₁

K₀₁+U₀₁ = Kf₁ + Uf₁ U₀₁=0, Kf₁ = 0, m₁=m

(1/2) * m*v²+0 = 0+m*g*h₁ : We eliminate m, then,

(1/2) * v²=g*h₁

h₁ = (v²) / (2g)

We apply formulas (1), (2), (3) and (4) for the second object, (2):

E₀₂ = Ef₂

K₀₂+U₀₂ = Kf₂ + Uf₂ U₀₂=0, Kf₂ = 0, m₂=2m

(1/2) * 2m*v²+0 = 0+2m*g*h₂ : We eliminate m, then,

(1/2) * 2*v² = 2*g*h₂ : We divide by 2 on both sides of the equation, then,

(1/2) * v²=g*h₂

h₂ = (v²) / (2g)