Mechanics - two boxes down a hill

[Hioj]

Homework Statement

The first box slides down a hill and gains speed and ends at v=1 \dfrac{m}{s}.

The second box has an initial speed of v_{0}=1 \dfrac{m}{s} and slides down the same hill. What speed does it end up with?

The boxes have equal masses.

Solve this without first solving for the height h. This is intuition practice.

Homework Equations

\dfrac{1}{2}mv^{2}=mgh

The Attempt at a Solution

Since \dfrac{1}{2}mv^{2}=mgh, then v=\sqrt{mgh}. The first gains a speed of 1 \dfrac{m}{s}, so the second must also gain 1 \dfrac{m}{s}. So it ends up at 2 \dfrac{m}{s}?

I remember the answer being \sqrt{2}.

 

[Spinnor]

You have,

v = (m*g*h)^.5

It should be v = (2*g*h)^.5

Why don't you solve for h just for fun?


The second box will have twice the energy of the first but as kinetic energy goes as v^2 the velocity won't be double.

 

[Hioj]

The following assumes g=10\dfrac{m}{s} for simplicity.

We have
\dfrac{1}{2}mv^{2}=mgh
and so
h=\dfrac{v^{2}}{2g}=\dfrac{1}{20} as v^{2}=1\dfrac{m}{s} for the first box.
For the second box, it must be true that the kinetic energy with the end speed v equals the kinetic energy it starts off with (\dfrac{1}{2}mv_{0}^{2}) plus the gained potential energy. Then
\dfrac{1}{2}mv^{2}=\dfrac{1}{2}mv_{0}^{2}+mgh
\Longrightarrow v^{2}=v_{0}^{2}+2gh=1+\dfrac{2g}{20}=1+1=2
\Longrightarrow v=\sqrt{2}.

Correct?

Could this be done any easier?