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.
e. The two objects rise to the same height.

Option E is correct.

The two objects rise to the same height.

Explanation:

Let the mass, velocity and height the bigger object will rise to be M, V and H respectively.

Let the mass, velocity and height the object will rise to be m, v and h respectively.

Note that V = v

Using the work energy theorem

The change in kinetic energy of a body between two points is equal to the work done on the body between those two points.

Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(M)(V²) = (-MV²/2)

(The final kinetic energy = 0 J because the object comes to rest at the final point)

Work done on the bigger object = work done by all the forces acting on the body

But the only force acting on the body is the force of gravity (since the inclined plane is frictionless)

Workdone on the body = work done by the force of gravity in moving the body up a height of H = - MgH

(-MV²/2) = - MgH

H = (V²/2g)

For the small body,

Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(m)(v²) = (-mv²/2)

Workdone on the body = work done by the force of gravity in moving the body up a height of H = - mgh

(-mv²/2) = - mgh

h = (v²/2g)

Since V = v as given in the question (Both bodies have the same speeds)

H = h = (V²/2g) = (v²/2g)

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