If you experimented with the game you probably learned that it is more difficult to make the ball stay in the valley when the energy loss is low. This page uses an example to explain why this is true.

To simplify the analysis, we will assume that the ball is SLIDING and not rolling. By making this assumption we will not need to consider rotational kinetic energy and the analysis is easier.

Consider the figure and the following assumptions:

• Mass of ball m = 1 kg
• Acceleration of gravity g = 10 m/s/s
• Height of hill h = 10 m
• Distance from release point to hill d1 = 100 m
• Distance from hill to location on ramp where the height equals 10 m is d2 = 100 m.
• Ball slides (does NOT roll)

CASE 1: Loss = 0.01 J/m

We will first analyze the situation in which the energy loss is low. We will assume the loss is 0.01 J/m.

In order for the ball to make it over the hill when pushed from the release point, it will need to be given an energy equal to mgh + Loss where mgh is the potential energy of the ball on the hill:

mgh = 1(10)(10) = 100 J
Loss = (Loss/distance)Distance = 0.01(100) = 1 J

Therefore the ball needs to be given 101 J so that it just makes it over the hill.

The velocity that provides this amount of energy is found from the equation for kinetic energy: KE = 0.5mv²:

v = SQRT(2KE/m) = SQRT(2(101)/1) = 14.22 m/s

Therefore if the ball is released from the release point with a velocity of 14.22 m/s, it will just make it to the top of the hill.

Now we need to determine the maximum velocity which will ensure that the ball does not return back over the hill. We will make another simplifying assumption that the ball will travel a distance d2 after it passes over the hill so that it stops on the ramp at a height h. In reality, the ball will not make it to a height h on the ramp because the ball loses energy. However, this assumption simplifies the analysis without adding much error. Additionally we will assume that d2 equals 100 m.

Therefore the maximum energy that the ball can be given so that it does not pass over the hill on its return trip is found using:

Max Energy = PE_hill + Loss_d1 + 2(Loss_d2)

Max Energy = 1(10)10 + 0.01(100) + 2(0.01)(100) = 103 J.

The velocity that provides this amount of energy is found from the equation for kinetic energy: KE = 0.5mv²:

v = SQRT(2KE/m) = SQRT(2(103)/1) = 14.35 m/s

Therefore if the ball is released from the release point with a velocity of 14.35 m/s, it will just make it over the hill, go up the right ramp, and then return back to the top of the hill.

Therefore, for this example and based on the simplifying assumptions, if the energy loss is 0.01 J/m, the ball will need to be pushed with a velocity between 14.22 and 14.35 m/s.

CASE 2: Loss = 0.1 J/m

We will now analyze the situation in which the energy loss is higher. We will assume the loss is 0.1 J/m.

In order for the ball to make it over the hill when pushed from the release point, it will need to be given an energy equal to mgh + Loss where mgh is the potential energy of the ball on the hill:

mgh = 1(10)(10) = 100 J
Loss = (Loss/distance)Distance = 0.1(100) = 10 J

Therefore the ball needs to be given 110 J so that it just makes it over the hill.

The velocity that provides this amount of energy is found from the equation for kinetic energy: KE = 0.5mv²:

v = SQRT(2KE/m) = SQRT(2(110)/1) = 14.83 m/s

Therefore if the ball is released from the release point with a velocity of 14.83 m/s, it will just make it to the top of the hill.

Now we need to determine the maximum velocity which will ensure that the ball does not return back over the hill. We will make the simplifying assumption that the ball will travel a distance d2 after it passes over the hill so that it stops on the ramp at a height h. In reality, the ball will not make it to a height h on the ramp because the ball loses energy. However, this assumption simplifies the analysis without adding much error. Again we will assume that d2 equals 100 m.

Therefore the maximum energy that the ball can be given so that it does not pass over the hill on its return trip is found using:

Max Energy = PE_hill + Loss_d1 + 2(Loss_d2)

Max Energy = 1(10)10 + 0.1(100) + 2(0.1)(100) = 130 J.

The velocity that provides this amount of energy is found from the equation for kinetic energy: KE = 0.5mv²:

v = SQRT(2KE/m) = SQRT(2(130)/1) = 16.12 m/s

Therefore if the ball is released from the release point with a velocity of 16.12 m/s, it will just make it over the hill, up the right ramp, and then back to the top of the hill.

Therefore, for this example and based on the simplifying assumptions, if the energy loss is 0.1 J/m, the ball needs to be pushed with a velocity between 14.83 and 16.12 m/s.

SUMMARY

When the energy loss is low, the velocity must be more tightly controlled and so it is more difficult to make the ball stay in the valley then when the loss is higher.