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

They will not stop at same elevation

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

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

Step-by-step 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) =(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.