A car moving at 10 m/s approaches a hill. If the car were put in neutral, it would roll up the hill to a stop at a certain elevation. If the car were approaching the hill instead at 20 m/s, would it roll to stop at the same elevation? You may ignore the effects of friction in your calculations.

If we neglect the effects of friction in the calculations the energy if the system must be conserved. The car energy can be described as a combination of kinetic energy and potential energy:

[tex]E=K+P[/tex]

The potential energy is due to the gravitational forces and can be describes as:

[tex]P=g*h*m[/tex]

Where g is the gravitation acceleration, m the mass of the car, and h the elevation. This elevation is a relative quantity and any point of reference will do the work, in this case we will consider the base of the hill as h=0.

The kinetic energy is related to the velocity of the car as:

[tex]K=1/2*m*v^{2}[/tex]

As the energy must be constant E will be always constant, replacing the expressions for kinetic and potenctial energy:

[tex]E=1/2*m*v^{2}+g*h*m[/tex]

In the base of the hill we have h=0:

[tex]E_{base} =1/2*m*v^{2}[/tex]

When the car stops moving we have v=0:

[tex]E_{top} =g*h_{top}*m[/tex]

This two must be equal:

[tex]E_{base} =E_{top}[/tex]

[tex]1/2*m*v^{2} =g*h_{top}*m[/tex]

solving for h:

[tex]h_{top} =\frac{v^{2}}{2*g}[/tex]

Lets solve for the two cases:

for v=10m/2 => h=5.1m

for v=20m/2 => h=20.4m

As you can see, when the velocity is the double the height it reaches goes to four times the former one.