# Can Physics Confirm the Height of a Carnival Ride Start?

• PascalPanther
In summary, the car starts from a height of around 10 feet, and the block of cushioned material compresses about 15 feet from equilibrium after the collision.
PascalPanther

## Homework Statement

Oscillations: You and some friends are waiting in line for "The Mixer", a new carnival ride. The ride begins with the car and rider (150 kg combined) at the top of a curved track. At the bottom of the track is a 50 kg block of cushioned material which is attached to a horizontal spring whose other end is fixed in concrete. The car slides down the track ending up moving horizontally when it crashes into the cushioned block, sticks to it, and oscillates at 3 repetitions in about 10 seconds. Your friends estimate that the car starts from a height of around 10 feet. You decide to use your physics knowledge to see if they are right. After the collision, you notice that the spring compresses about 15 ft from equilibrium.

## Homework Equations

E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
omega = Sqrt(k/m)
f = omega/(2pi) = 1/(2pi)*Sqrt(k/m)

## The Attempt at a Solution

m = 150kg
M = 150kg + 50kg = 200kg
x = 15ft = 4.57m
h1 = 3.04 m
h_real = ?
x = A?

mgh = (1/2)Mv^2 + (1/2)kx^2
Since there is no velocity given, I use:
mgh = (1/2)kA^2

I need to find k:
3 cycles every 10 seconds
f = 0.3 cycles/s

f = (1/2pi)Sqrt(k/M)
0.3 = (1/2pi)Sqrt(k/M)
k = 710 N/m

(150kg)(9.8m/s^2)(h) = (1/2)(710N/m)(4.57m)^2
h = 5.04m

h_real > h1

Did I do that correctly?

Almost: you have a rounding error in the final answer.

PascalPanther said:
mgh = (1/2)Mv^2 + (1/2)kx^2
Since there is no velocity given, I use:
mgh = (1/2)kA^2
You have assumed that mechanical energy is conserved during the collision. Rethink that assumption, given that the car and block stick together.

I need to find k:
3 cycles every 10 seconds
f = 0.3 cycles/s

f = (1/2pi)Sqrt(k/M)
0.3 = (1/2pi)Sqrt(k/M)
k = 710 N/m
Good.

Hmm...
Before the collision, the block would be:
mgh = (1/2)mv^2
Then
mv = Mv_2
Is there a way to use this relationship without velocity data (actual height)?

Sure. Start with the SHM. That should enable you to find the post-collision speed, then use that relationship (conservation of momentum) to find the pre-collision speed. Then you can deduce the height.

(1/2)Mv^2 = (1/2)kA^2 right?

v = Sqrt(k/M)A
v = Sqrt(710/200kg) * (4.57m)
v = 8.6 m/s

mv = Mv
(150kg)v = (200kg)(8.6 m/s)
v= 11.5 m/s

mgh = (1/2)mv^2
(150kg)(9.8)h = (1/2)(150kg)(11.5m/s)^2
h = 6.75 m?

Looks good.

## 1. What is oscillation?

Oscillation is a repetitive movement or fluctuation around an equilibrium point. It can occur in various forms, such as sound waves, electromagnetic waves, and mechanical vibrations.

## 2. What is conservation of oscillation?

The conservation of oscillation is the principle that states that the total energy of a system undergoing oscillatory motion remains constant over time, despite the fluctuations in potential and kinetic energy.

## 3. How does conservation of oscillation apply to real-life systems?

Conservation of oscillation applies to a wide range of real-life systems, such as pendulums, springs, and musical instruments. It helps to predict the behavior and stability of these systems and is essential in fields such as engineering and physics.

## 4. What factors can affect the conservation of oscillation?

The conservation of oscillation can be affected by factors such as friction, air resistance, and damping. These external forces can cause a loss of energy in the system, leading to a decrease in the amplitude or frequency of oscillation.

## 5. How is the conservation of oscillation related to the laws of thermodynamics?

The conservation of oscillation is closely related to the first and second laws of thermodynamics. The first law states that energy cannot be created or destroyed, only transferred or converted. The second law states that entropy, or disorder, in a closed system will always increase. The conservation of oscillation aligns with these laws as the total energy of a system oscillating between potential and kinetic energy remains constant, but the system will eventually reach equilibrium due to energy dissipation caused by entropy.

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