Mechanics - two boxes down a hill

[Hioj]

Homework Statement

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

The second box has an initial speed of [itex]v_{0}=1 \dfrac{m}{s}[/itex] 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 [itex]h[/itex]. This is intuition practice.

Homework Equations

[itex]\dfrac{1}{2}mv^{2}=mgh[/itex]

The Attempt at a Solution

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

I remember the answer being [itex]\sqrt{2}[/itex].

 

[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 [itex]g=10\dfrac{m}{s}[/itex] for simplicity.

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

Correct?

Could this be done any easier?