Question

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.

Answer 1

Answer:

They will not stop at same elevation

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

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

Explanation:

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:

$E=K+P$

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

$P=g*h*m$

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:

$K=1/2*m*v^{2}$

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

$E=1/2*m*v^{2}+g*h*m$

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

$E_{base} =1/2*m*v^{2}$

When the car stops moving we have v=0:

$E_{top} =g*h_{top}*m$

This two must be equal:

$E_{base} =E_{top}$

$1/2*m*v^{2} =g*h_{top}*m$

solving for h:

$h_{top} =\frac{v^{2}}{2*g}$

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.