SYPT2024 CatB Q4: Ping Pong Rocket. A ping pong ball is placed in a container of water. When the container is dropped, the ping pong ball will get launched to a great height. What maximum height can you reach with up to 2 liters of water?

 SYPT2024 CatB Q2: Ping Pong Rocket: A ping pong ball is placed in a container of water. When the container is dropped, the ping pong ball will get launched to a great height. What maximum height can you reach with up to 2 liters of water?

https://www.youtube.com/shorts/Uy4_CT4aFts

gpt4o

Let's delve deeper into the physics of this situation to verify the expected maximum height:

1. Buoyant Force and Acceleration:

   • When the container is in free fall, the water and the ping pong ball experience effective weightlessness.
   • Upon impact with the ground, the water inside the container pushes upwards with a significant force, which launches the ping pong ball upwards.
   • The upward acceleration $𝑎$ imparted to the ping pong ball depends on the force of the water jet and the mass of the ping pong ball.

2. Force Analysis:

   • The buoyant force ${𝐹}_{𝑏}$ exerted on the ping pong ball is given by:
     $${𝐹}_{𝑏}=𝜌\cdot 𝑉\cdot 𝑔$$
     where $𝜌$ is the density of water, $𝑉$ is the volume of the ping pong ball, and $𝑔$ is the acceleration due to gravity.
   • The net upward force ${𝐹}_{𝑛𝑒𝑡}$ acting on the ping pong ball when the water is expelled is:
     $${𝐹}_{𝑛𝑒𝑡}={𝐹}_{𝑏}-𝑚\cdot 𝑔$$
     where $𝑚$ is the mass of the ping pong ball.

3. Acceleration and Velocity:

   • The acceleration $𝑎$ of the ping pong ball is:
     $$𝑎=\frac{{𝐹}_{𝑛𝑒𝑡}}{𝑚}=\frac{(𝜌\cdot 𝑉\cdot 𝑔-𝑚\cdot 𝑔)}{𝑚}=𝑔(\frac{𝜌\cdot 𝑉}{𝑚}-1)$$
   • The ping pong ball will be propelled with this acceleration over a very short distance during the impact, giving it an initial upward velocity ${𝑣}_0$.

4. Calculating Initial Velocity ${𝑣}_0$:

   • The exact velocity ${𝑣}_0$ can be complex to derive analytically due to rapid dynamics but can be estimated based on energy considerations.
   • The kinetic energy imparted to the ping pong ball is related to the potential energy of the water column:
     $$\frac{1}{2}𝑚{𝑣}_0^2=𝜌\cdot 𝑉\cdot 𝑔\cdot ℎ$$
   • Solving for ${𝑣}_0$:
     $${𝑣}_0=\sqrt{\frac{2𝜌\cdot 𝑉\cdot 𝑔\cdot ℎ}{𝑚}}$$

5. Maximum Height Calculation:

   • The maximum height ${𝐻}_{𝑚𝑎𝑥}$ can be calculated using the kinematic equation:
     $${𝐻}_{𝑚𝑎𝑥}=\frac{{𝑣}_0^2}{2𝑔}$$
   • Substituting ${𝑣}_0$ into the equation:
     $${𝐻}_{𝑚𝑎𝑥}=\frac{{\left(\sqrt{\frac{2𝜌\cdot 𝑉\cdot 𝑔\cdot ℎ}{𝑚}}\right)}^2}{2𝑔}=\frac{2𝜌\cdot 𝑉\cdot 𝑔\cdot ℎ}{2𝑔𝑚}=\frac{𝜌\cdot 𝑉\cdot ℎ}{𝑚}$$

Example Calculation

Given typical values:

• Density of water, $𝜌\approx 1000{\text{kg/m}}^3$
• Volume of a ping pong ball, $𝑉\approx 3.35\times 10^{-5}{\text{m}}^3$
• Mass of a ping pong ball, $𝑚\approx 2.7\times 10^{-3}\text{kg}$
• Drop height, $ℎ=10\text{m}$

Let's compute the maximum height:

$${𝐻}_{𝑚𝑎𝑥}=\frac{1000\cdot 3.35\times 10^{-5}\cdot 10}{2.7\times 10^{-3}}=\frac{0.335}{2.7\times 10^{-3}}\approx 124\text{m}$$

This result indicates that, under ideal conditions, the ping pong ball could potentially reach a height of approximately 12.4 meters when dropped from a height of 1.0 meters, considering the energy imparted by the buoyant force of the water.

This height is a theoretical maximum and practical constraints like air resistance and imperfect energy transfer would likely reduce the actual height reached.

In a real-world scenario, factors like air resistance and energy losses might slightly reduce this height. For maximum accuracy, experimental validation would be necessary to account for these factors.

Theory of COE

To predict the maximum height of the ping pong ball using the conservation of energy, we can consider the energy transformation from potential energy to kinetic energy and then back to potential energy.

Step-by-Step Calculation

1. Initial Potential Energy (PE_initial):

   • The potential energy of the water container at height $ℎ$ is:
     $$𝑃{𝐸}_{\text{initial}}={𝑚}_{\text{water}}\cdot 𝑔\cdot ℎ$$
   • For 2 liters of water (${𝑚}_{\text{water}}=2\text{kg}$), $𝑔=9.81{\text{m/s}}^2$, and $ℎ=1\text{m}$:
     $$𝑃{𝐸}_{\text{initial}}=2\cdot 9.81\cdot 1=19.62\text{J}$$

2. Energy Transfer to the Ping Pong Ball:

   • Assuming perfect energy transfer and negligible losses, the kinetic energy imparted to the ping pong ball $𝐾{𝐸}_{\text{pingpong}}$ is equal to the potential energy of the water container:
     $$𝐾{𝐸}_{\text{pingpong}}=19.62\text{J}$$

3. Kinetic Energy to Potential Energy:

   • The kinetic energy of the ping pong ball at launch is converted to potential energy at the maximum height.
   • The mass of the ping pong ball ${𝑚}_{\text{pingpong}}=2.7\times 10^{-3}\text{kg}$:
     $$𝐾{𝐸}_{\text{pingpong}}=𝑃{𝐸}_{\text{pingpong, max}}={𝑚}_{\text{pingpong}}\cdot 𝑔\cdot {𝐻}_{\text{max}}$$

4. Solving for Maximum Height ${𝐻}_{\text{max}}$:

   • Rearrange the equation to solve for ${𝐻}_{\text{max}}$:

     $${𝐻}_{\text{max}}=\frac{𝐾{𝐸}_{\text{pingpong}}}{{𝑚}_{\text{pingpong}}\cdot 𝑔}=\frac{19.62}{2.7\times 10^{-3}\cdot 9.81}$$

   • Perform the calculation:

     $${𝐻}_{\text{max}}=\frac{19.62}{2.7\times 10^{-3}\cdot 9.81}=\frac{19.62}{0.026487}\approx 741\text{m}$$

Result

Using the conservation of energy and assuming perfect energy transfer, the theoretical maximum height that the ping pong ball can reach is approximately 741 meters when the setup is dropped from a height of 1 meter.

This extremely high value is due to the assumption of perfect energy transfer, which is idealistic. In a real-world scenario, energy losses due to air resistance, imperfect energy transfer, and other factors would significantly reduce the actual height reached.

GPT4:  https://chat.openai.com/c/35cbd4d8-c302-42a1-81b6-40a3d3e64e67

The Singapore Young Physicists' Tournament (SYPT) problem you've described involves understanding the dynamics of a ping pong ball being launched upwards when a container filled with water is dropped. To solve this problem and predict the maximum height that can be reached by the ping pong ball with up to 2 liters of water, we'll need to analyze the forces involved and apply principles from fluid dynamics and mechanics.

### Key Principles:

1. **Buoyancy Force**: When the container is dropped, the buoyancy force will act on the ping pong ball. As the container accelerates downward, the effective gravitational field inside the container decreases, which affects the buoyancy force experienced by the ball.

2. **Drag Force**: As the ping pong ball moves through the water, it experiences a drag force proportional to the square of its velocity.

3. **Bernoulli's Principle**: The fluid dynamics involved, especially when considering the acceleration of the water and the ping pong ball, can be partly understood through Bernoulli's principle, indicating variations in pressure along the flow of the fluid.

4. **Newton's Second Law**: The acceleration of the ping pong ball can be analyzed by applying Newton's second law, taking into account the net force acting on the ball as a result of buoyancy and drag.

### Estimating Maximum Height:

To estimate the maximum height that the ping pong ball can reach, we can follow these steps:

#### Step 1: Initial Conditions

- **Container Size and Water Volume**: With up to 2 liters of water, the depth and surface area of the water will affect how the forces act on the ping pong ball. A deeper container might offer more potential for acceleration but also increases drag.

#### Step 2: Calculate Buoyancy Force

- The buoyancy force can be calculated using the formula: $$ (F_b = \rho V g) $$ where $$ \(\rho) $$ is the density of the water, $$(V)$$ is the volume of the ball submerged, and $$g$$ is the acceleration due to gravity.

#### Step 3: Calculate Drag Force

- The drag force experienced by the ball as it moves through water can be estimated using the formula: \(F_d = \frac{1}{2} C_d \rho A v^2\), where \(C_d\) is the drag coefficient, \(A\) is the cross-sectional area of the ball, and \(v\) is the velocity of the ball.

#### Step 4: Dynamics During Drop

- When the container is dropped, the acceleration of the container reduces the effective gravitational acceleration acting on the ball. This reduction affects the net force acting on the ball and its initial acceleration upwards.

#### Step 5: Maximum Height Calculation

- Once the initial upward velocity of the ping pong ball as it exits the water surface is determined, the maximum height can be calculated using the kinematic equation for motion under uniform acceleration: \(v^2 = u^2 + 2as\), where \(v\) is the final velocity (0 m/s at the maximum height), \(u\) is the initial velocity, \(a\) is the acceleration (negative due to gravity), and \(s\) is the distance (maximum height).

### Considerations:

- **Energy Losses**: Account for energy losses due to water resistance and air resistance once the ball is airborne.

- **Optimal Conditions**: Experimenting with different volumes of water, container shapes, and drop heights could help identify the optimal conditions for maximizing the launch height of the ping pong ball.

### Experimental Approach:

To accurately determine the maximum height, experimental validation is necessary. By dropping the container from various heights and measuring the resultant height reached by the ping pong ball (using high-speed video analysis), one can gather data to refine the theoretical predictions and account for real-world complexities not fully captured in the simplified model.

This problem is an excellent example of applying theoretical physics to a real-world scenario, requiring a blend of theoretical knowledge, experimental skill, and analytical thinking to solve.