Physics SL

Sample Extended Essays

Table of content

Research question

Rationale

Background information

Variables

Hypotheses

Considerations

Procedure

Qualitative observations

Data collection and processing

Data analysis

Scientific Justification

Evaluation of hypotheses

Conclusion

Evaluation

References

Appendix

21 mins Read

4,019 Words

How does the terminal velocity in cm s^{-1} of an object falling through a liquid depends on the density of the liquid it is falling through, determined using velocity-time graph?

Physics has always been a subject of interest for me. I am always fascinated by the real-life application of this subject especially in industrial, technological, electrical and mechanical sectors. Real-life observations make us curious towards scientific inquiries. As a swimmer, I have often been asked to remove my body hairs before swimming competitions. When I have asked this thing to my coach, I was informed that having smooth body surface allows a person to swim at a faster speed. I was more intrigued to understand what is the science behind this. After some research, I came to know that this is related to motion of objects through fluid. As the surface of the body becomes smooth, the drag force that restricts the motion of an object through a fluid also decreases. This made me interested to know what are the various factors that can influence the motion of an object through a fluid column. Thus, I decided to explore how the speed of an object falling through a fluid depends on the density of the fluid.

When a body is allowed to fall through a fluid (liquid or gas), the velocity of the body keeps on increasing as the body keeps falling down. At a point, the velocity of the body attains the maximum value and does not increase further. This velocity attained by the body is known as the terminal velocity.

When a body is moving through a fluid, there is a mechanical force acting on the body against the direction of motion of the body. This originates from the fact that there is a difference in the velocity with which the body moves and the fluid moves. This is a kind of frictional force that opposes the motion of the object. It is essential for the object to be in physical touch with the fluid for a drag force to act on it. The mathematical expression of drag force is given as:

\(F=\frac{1}{2}×ρ×ϑ^2×C_d×A\)

Here, F = drag force

*ρ = density of the object*

*ϑ = velocity of the fluid relative to the object*

*C _{d} = drag coefficient*

*A = projected area*

When an object is partly or totally submerged in a fluid, the fluid exerts a force on that object which pushes it upwards towards the surface of the fluid. This is known as upthrust or buoyant force5. Mathematically, the buoyant force is given by the expression:

*F _{b} = ρ × V × g*

*F _{b} = buoyant force*

*ρ = density of the fluid*

*V = volume of fluid displaced by the object*

*g = acceleration due to gravity*

At a point where the object attains terminal velocity, there are three forces acting on the body- the drag force and the weight of the object acting downwards. The drag force acts upwards (against the direction of motion) and the weight acting downwards along the direction of motion.

At any point, the net force acting on the object (F) is the difference of the weight of the object acting downwards and the drag force acting upwards.

*Net force acting on object (F) = Weight of the object (W) - Drag force (F _{d})*

When the body starts to move, the drag force is smaller than the weight of the object acting downwards and that makes the body keep moving downwards. As the body keeps moving, the velocity of the body keeps increasing and thus the magnitude of the drag force also increases. This in turn decreases the net force acting on the object along the direction of motion. As a result, the acceleration on the object decreases because

*Force (F) = mass (m) × acceleration (a)*

At a point, after the object has travelled downward for a distance, the the forces acting upwards (drag force) and the force acting downwards (weight of the object) is equal. As a result, the net force that is acting on the object is zero. If the net force becomes zero, the acceleration is also zero. This means that the velocity of the object is not changing and thus it attains a maximum constant value.

At terminal velocity,

*F _{d} = Weight of the object*

*\(\frac{1}{2}\)×ρ × ϑ × C _{d} × A = mg*

*ϑ = \(\frac{2mg}{ρC_dA}\)*

Mathematically, the expression of terminal velocity is given as-

*ϑ _{t} = \(\sqrt{\frac{2mg}{ρCdA}}\)*

*ϑ _{t} = terminal velocity*

*m = mass of the object*

*g = acceleration due to gravity*

*ρ = density of the fluid*

*C _{d }= drag coefficient*

*A = projected area of the object*

In CGS system, the unit of terminal velocity is cm s^{-1} and in SI system it is m s^{-1}.

The velocity of an object falling through a fluid if plotted against time will yield a graph as given below:

As the graph shows that initially, the velocity of the object keeps on increasing with the increase in time. At a point, the velocity attains a maximum value and becomes constant after that. This is the point where, the drag force and the weight of the object balances each other making the object at a value of zero acceleration. Thus, if the velocity of the object is measured at different instant of time and plotted graphically along time, the terminal velocity can be obtained by drawing a perpendicular to the y axes from the inflexion point (where the graph becomes a straight line parallel to the x axes).

The velocity of an object is defined as:

*Velocity = *\(\frac{Distance}{Time}\)

The object was allowed to fall through a graduated cylinder so that the vertical distance covered by the object can be measured using the graduations of the cylinder. A stop- watch was used to take a record of the time. Thus, as raw data the distance covered by the object at different point of time was recorded and the velocity of the object was calculated. Following this, a graph was obtained of velocity against change in time and at a point where the graph shows an inflexion point with the line becoming parallel to x axes, the terminal velocity was obtained.

**Density of the liquid:**

The terminal velocity of the freely falling object will depend on the viscosity of the liquid through which it is falling. Thus, the aim was to use liquids of different viscosities and thus to do so liquids of various densities were used. The values of densities as reported on the packaging were used. The five different liquids and their densities as reported in the packets are mentioned in the table below:

Serial number

Type of liquid used

Density (× 10^{-1} g cc^{-1})

1

Coconut oil

9.25

2

Castor oil

9.69

3

Paraffin oil

8.00

4

Lamp oil

8.20

5

Lubricant

8.75

The dependent variable is terminal velocity of the object measured in cm s^{-1}. The metal ball will be allowed to fall through the column of fluid and the distance travelled by the ball will be recorded at different intervals of time. This will be used to obtain a velocity- time graph using which the terminal velocity will be deduced.

- Temperature:

Temperature has an effect on the density of liquids. With the increase of temperature, the average kinetic energy of the molecules of the liquid increases and thus they can overcome the intermolecular forces between each other to increase the intermolecular space and thus occupy more volume while the mass remains the same. This in turn decreases the density of the liquid. To control this, all the trials were conducted at room temperature.

- Value of acceleration due to gravity:

The magnitude of acceleration due to gravity changes from one location to another. The value of terminal velocity depends on the value of acceleration due to gravity (g). As the value of g increases, the magnitude of the force (mg) pulling the object downward also increases and thus the terminal velocity also increases. To control this, the experiment was performed at one particular location which is the school laboratory.

- Mass of the object:

The value of terminal velocity will depend on the mass of the object falling down. As the mass of the object increases, the pulling force (mg) also increases and thus the acceleration of the object increases. Though in this investigation, three different objects were used yet they were chosen in such a way that they are of the same mass. The mass of the one cent coin, metal ball and the dice used was 2.27 ± 0.01 g. A digital mass balance was used to determine the mass of the object in use.

- Projected area (A):

The projected area is the area of contact between the object and the fluid. It depends on the shape of the object. As the projected area increases, the magnitude of drag force also increases. In all cases, the object used was a spherical metal ball so that the projected area remains constant in all cases.

**Trend - 1: **As the density of the liquid through which the liquid is falling increases, the value of terminal velocity will decrease.

**Justification: **As the density of the liquid increases, the viscous force increases and the drag coefficient too. Thus, there is an internal friction of larger magnitude that resists the motion of the liquid and this in turn decreases the value of the terminal velocity.

Apparatus

Quantity

Least count

Uncertainty

Digital mass balance

1

0.01 g

± 0.01 g

A graduated measuring cylinder of 1000 cc

1

1.00 cc

± 0.05 cc

A meter scale

1

0.10 cm

± 0.05 cm

A stand and clamp

1

NA

NA

Digital Stop-watch

1

0.01 s

± 0.01 s

- Use a gloves and mask to avoid direct spillage of the liquids on hand and face.
- Use a laboratory coat.
- Do not carry food items inside the workstation.
- Do not touch any electrical devices in wet hands.

- All the used materials were returned back to the laboratory for further use.
- No harmful or abandoned chemicals were used.

**Part-A: Determining the mass of the objects used:**

- Take a digital mass balance.
- Use the “Tare” button to adjust the reading of the balance to 0.00 ± 0.01 g.
- Place the object on the top pan of the digital mass balance.
- Record the reading displayed on the screen of the mass balance and note it down.

**Part-B: Determining the terminal velocity of the object:**

- Take a clean and long 1000 cc graduated measuring cylinder.
- Align a 100 cm meter scale along it.
- Use a marker to mark graduations on the measuring cylinder as it is there on the meter scale.
- Clean the measuring cylinder and dry it properly using a dryer.
- Fill in the cylinder till the mark of 0.00 cm at the top using coconut oil.
- Take the metal ball and drop it slowly and vertically into the liquid inside the cylinder.
- Start your stop-watch.
- Note down the distance travelled by the ball after every 10.00 ± 0.01 s.
- Keep noting the distance travelled until the stop-watch reads 90.00 s.
- Repeat this step 1-9 for two more times.
- Repeat all of the above steps with other liquids-paraffin oil, lubricating oil, lamp oil and castor oil.
- Repeat all of the above steps with the other two objects – dice and coin.

- Castor oil was most sticky and viscous in nature as far as appearance is considered.
- The objects did not fall with a constant velocity.
- The time taken for the object to reach the bottom of the cylinder was not the same in all cases.

Formulas used:

Mean = \(\frac{Trial-1+Trial-2+Trial-3}{3}\)

Standard deviation = \(\frac{\sum^{i=3}_{i=1}(Trial-Mean)^2}{3}\)

Time (in ± 0.01 s)

Distance travelled in ± 0.05 cm

Velocity in * 10^{-2} cm s^{-1}

10.00

0.20

2.00

20.00

0.47

2.70

30.00

0.80

3.30

40.00

1.40

6.00

50.00

2.00

6.00

60.00

2.60

6.00

70.00

3.20

6.00

80.00

3.80

6.00

90.00

4.40

6.00

As indicated in the scatter graph above, the velocity increases as the time increases.

However, from 50.00 ± 0.01 s, the velocity becomes constant as the trend line becomes parallel to the x axes. The terminal velocity is 6.00 × 10^{-1} cm s^{-1}. Refer to Appendix for raw data and processing for other oils.

Serial no.

Type of oil used

Density in × 10^{-1} g/cc

Terminal velocity in × 10^{-1} cm s^{-1}

1

Coconut oil

9.25

6.00

2

Castor oil

9.69

4.00

3

Paraffin oil

8.00

22.00

4

Lamp oil

8.20

17.00

5

Lubricating oil

8.75

13.00

**Error propagation:**

For coconut oil,

At time (t) = 60.00 ± 0.01 s, the velocity becomes maximum and the terminal velocity is attained.

Distance travelled at time t = 60.00 ± 0.01 s = 2.60 ± 0.05 cm

Terminal velocity (Vt) = \(\frac{s}{t}\)

\(=\frac{2.60±0.05\ cm}{60.00±0.01\ s}\)

\(=\frac{2.60}{60.00}±\bigg(\frac{0.005}{2.60}+\frac{0.01}{60}\bigg)\)

Fractional uncertainty in terminal velocity \(\bigg(\frac{∆V_t}{V_t}\bigg)=±1.93×10^{-2}\)

Percentage uncertainty in terminal velocity

= \(\frac{∆V_t}{V_t}\) × 100 = ± 1.93 × 10^{-2} × 100 = ± 1.93

The scatter graph shown above depicts the variation of the terminal velocity of the freely falling object against the density of the oil through which it is falling in g cc^{-1}.

The terminal velocity is plotted along the y axes as it is the dependent variable and the density of the oil is plotted along the x axes as it is the independent variable.

As the graph shows, the terminal velocity of the object decreases from 2.20 cm / s to 0.40 cm/s as the density of the oil increases from 0.800 g/cc (paraffin oil) to 0.969 g/cc (castor oil). This indicates that as the oil becomes more dense, the viscosity of the oil increases and thus the terminal velocity also decreases.

The mathematical relationship between terminal velocity and density has been expressed according to the equation: y = 2.5274 x^{2} – 55.064x + 300. The equation suggests that the terminal velocity of the object decreases polynomial as the density of the oil increases.

*y* = 2.5274 x^{2} - 55.064x + 300

Differentiating the equation with respect to x,

\(\frac{dy}{dx}=2.5274\frac{d}{dx}(x^2)-55.064\frac{d}{dx}(x)+\frac{d}{dx}(300)\)

\(\frac{dy}{dx}\) = 2.5274 × 2x - 55.064

\(\frac{dy}{dx}\) = 5.0548 x - 55.064

Differentiating with respect to x again,

\(\frac{d}{dx}\bigg(\frac{dy}{dx}\bigg)=\frac{d}{dx}(5.0548x)-\frac{d}{dx}(55.064)\)

\(\frac{d^2y}{dx^2}\) = 5.0548

As the value of double derivative is positive it indicates that there is a minima.

At minima,

\(\frac{dy}{dx}\) = 5.0548 x - 55.064 = 0

5.0548 x − 55.064 = 0

5.0548 x = 55.064

x = \(\frac{55.064}{5.0548}\) = 10.89

It means that when the density of the oil is 10.89 × 10^{-1} g/cc, the terminal velocity of the freely falling object would be minimum.

y = 2.5274 x^{2} − 55.064 x + 300

y = 2.5274 × (10.89)^{2} − (55.064 × 10.89) + 300

y = (2.5274 × 118.59) − 599.64 + 300

y = 299.72 − 599.64 + 300

y = 0.084

At the minima, using an oil of density 1.089 g/cc, the velocity of the object will be 0.0084 cm/s.

As the density of the oil (the fluid medium) increases, the magnitude of drag force increases. The drag force acts opposite to the motion of the object and against the weight of the object which pulls it downwards. This in turn decreases the net force acting on the object. As a result, the velocity of the object also decreases.

At a molecular level, this can be better explained using friction and viscosity. As the density of the fluid increases, the mass of the fluid per unit area also increases which in turn means that the number of particles per unit volume also increases. This results in more amount of friction between the particles in the fluid and the object. As a result, the viscous force opposing the motion of the object is also greater. This eventually reduces the maximum velocity that can be attained by the object.

In Figure -1, a polynomial trend line is obtained that slopes downward and reaches a minimum. The same graph is plotted below using a linear trend line.

Figure - 2: Terminal velocity versus density with linear trend line

y = −10.441x + 104.05

Here, the gradient is -10.441 which is a negative value and that indicates that there is a negative relationship between density of the fluid and the density of the medium. At the same time, the graph also shows the value of R^{2} is 0.9655. This claims that there is a 96.55% of association ship between the density of the medium and the density of the medium. Thus, both the gradient and the value of the R^{2} clearly indicates a relationship between the density of the medium and the terminal velocity and identifies that as an indirect relationship.

How does the terminal velocity in cm s^{-1 }of an object falling through a liquid depends on the density of the liquid it is falling through, determined using velocity-time graph?

- The terminal velocity of the object decreases from 2.20 cm / s to 0.40 cm/s as the density of the oil increases from 0.800 g/cc (paraffin oil) to 0.969 g/cc (castor oil). This indicates that as the oil becomes more dense, the viscosity of the oil increases and thus the terminal velocity also decreases.
- The mathematical relationship between terminal velocity and density has been expressed according to the equation: y = 2.5274 x2 – 55.064x + 300. The equation suggests that the terminal velocity of the object decreases polynomial as the density of the oil increases.
- It means that when the density of the oil is 10.89 × 10
^{-1}g/cc, the terminal velocity of the freely falling object would be minimum. - As the density of the oil (the fluid medium) increases, the magnitude of drag force increases. The drag force acts opposite to the motion of the object and against the weight of the object which pulls it downwards. This in turn decreases the net force acting on the object. As a result, the velocity of the object also decreases.

- The density of the oils used is the independent variable. The values of density has been obtained from a secondary source. The fact that the secondary source used is reliable and free of bias makes the data used more reliable and accurate.

- The distance travelled by the metal ball has been measured from the gradations of the graduated measuring cylinder. There can be a parallax error in this case which will make the reading of the distance travelled inaccurate. To avoid this, the graduated measuring cylinder must be kept at the eye level. Moreover, the distance has also been measured using three different trials and an average value is taken.
- There is a chance of systematic error in the investigation. The speed at which the object would travel through the fluid will depend on the mass of the object. If the object has a significantly low mass then the speed would be too less and that would cause the object to attain terminal velocity soon. On the contrary, if the object is heavier, the weight of the object pulling it down will also increase and thus the object would fall with a greater speed. This will make the measurement difficult as the object would take very less time to attain terminal velocity and reach the bottom of the beaker. Thus, it is essential that the object chosen has a mass which is neither too low nor too high making it convenient to track the change in distance travelled by the object against time.
- There is a source of methodical error in the investigation. The speed attained by the ball will also depend on the way the ball is thrown into the column of fluid. If the ball is not thrown vertically the motion of the ball will be along a line at an angle with the wall of the tube instead of along a straight line through the fluid. To improve this, a magnet can be used. The ball can be held by a magnet where the ball is attached to a magnet and the release of the magnet is used to project the ball into the fluid column.
- There is a confounding variable in the experiment. The terminal velocity also depends on the drag coefficient of the fluid medium. As the fluid changes, the drag coefficient also changes. This cannot be controlled as the fluid used has been changed.

To extend the investigation further, I would like to study how the terminal velocity depends on the temperature of the fluid medium used. A water bath can be used to monitor the temperature. The vertical column containing the fluid can be placed on a water bath and the distance travelled by the object can be measured against time to compute the terminal velocity.

- Charrondiere, U. Ruth, et al. FAO/INFOODS Density Database Version 2.0 (2012). https://www.fao.org/3/ap815e/ap815e.pdf
- Dey, Subhasish, et al. “Terminal Fall Velocity: The Legacy of Stokes from the Perspective of Fluvial Hydraulics.” Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences, vol. 475, no. 2228, Aug. 2019, p. 20190277. DOI.org (Crossref), https://doi.org/10.1098/rspa.2019.0277.
- Foote, G. B., and P. S. Du Toit. “Terminal Velocity of Raindrops Aloft.” Journal of Applied Meteorology, vol. 8, no. 2, Apr. 1969, pp. 249–53. DOI.org (Crossref), https://doi.org/10.1175/1520-0450(1969)008<0249:TVORA>2.0.CO;2.
- Langleben, M. P. “The Terminal Velocity of Snowflakes.” Quarterly Journal of the Royal Meteorological Society, vol. 80, no. 344, Apr. 1954, pp. 174–81. DOI.org (Crossref), https://doi.org/10.1002/qj.49708034404.
- LeClair, B. P., et al. “A Theoretical and Experimental Study of the Internal Circulation in Water Drops Falling at Terminal Velocity in Air.” Journal of the Atmospheric Sciences, vol. 29, no. 4, May 1972, pp. 728–40. DOI.org (Crossref), https://doi.org/10.1175/1520-0469(1972)029<0728:ATAESO>2.0.CO;2.
- Raymond, F., and J. M. Rosant. “A Numerical and Experimental Study of the Terminal Velocity and Shape of Bubbles in Viscous Liquids.” Chemical Engineering Science, vol. 55, no. 5, Mar. 2000, pp. 943–55. DOI.org (Crossref), https://doi.org/10.1016/S0009-2509(99)00385-1.
- Su, Dong, et al. “A Terminal-Velocity Model for Super-Ellipsoidal Particles.” Advanced Powder Technology, vol. 33, no. 12, Dec. 2022, p. 103882. DOI.org (Crossref), https://doi.org/10.1016/j.apt.2022.103882.
- Thorpe, Hugh Rankin. “II.—The Effect of Concentration on the Terminal Fall Velocity of Particles in Suspension.” Proceedings of the Royal Society of Edinburgh. Section A. Mathematical and Physical Sciences, vol. 67, no. 1, 1965, pp. 9–24. DOI.org (Crossref), https://doi.org/10.1017/S0080454100010700.
- Xu, Bin, et al. “Investigation on Terminal Velocity and Drag Coefficient of Particles with Different Shapes.” Journal of Physics: Conference Series, vol. 822, Apr. 2017, p. 012047. DOI.org (Crossref), https://doi.org/10.1088/1742-6596/822/1/012047.
- Yin, Qiu, and Ci Song. “A High Precision Model for the Terminal Settling Velocity of Drops in Fluid Medium.” Frontiers of Earth Science, vol. 15, no. 4, Dec. 2021, pp. 947–55. DOI.org (Crossref), https://doi.org/10.1007/s11707-020-0835-z.

**Section-B: Raw data for castor oil:**

Formulas used:

Mean = \(\frac{Trial-1+Trial-2+Trial-3}{3}\)

Standard deviation = \(\frac{\sum^{i=3}_{i=1}(Trial-Mean)^2}{3}\)

Time (in ± 0.01 s)

Distance travelled in ± 0.05 cm

Velocity in * 10^{-2 }cm s^{-1}

10.00

0.10

1.00

20.00

0.20

1.00

30.00

0.40

2.00

40.00

0.60

2.00

50.00

0.90

3.00

60.00

1.30

4.00

70.00

1.70

4.00

80.00

2.10

4.00

90.00

2.50

4.00

As indicated in the scatter graph above, the velocity increases as the time increases.

However, from 60.00 ± 0.01 s, the velocity becomes constant as the trend line becomes parallel to the x axes. The terminal velocity is 4.00 × 10^{-1} *cm s*^{-1}.

Formulas used:

Mean = \(\frac{Trial-1+Trial-2+Trial-3}{3}\)

Standard deviations = \(\frac{\sum^{i=3}_{i=1}(Trial-Mean)^2}{3}\)

Time (in ± 0.01 s)

Distance travelled in ± 0.05 cm

Velocity in * 10^{-2} cm s^{-1}

10.00

0.90

9.00

20.00

2.10

12.00

30.00

3.60

15.00

40.00

5.40

18.00

50.00

7.50

21.00

60.00

9.70

22.00

70.00

11.90

22.00

80.00

14.10

22.00

90.00

16.30

22.00

As indicated in the scatter graph above, the velocity increases as the time increases. However, from 60.00 ± 0.01 s, the velocity becomes constant as the trend line becomes parallel to the x axes. The terminal velocity is 22.00 × 10^{-1} cm s^{-1}.

**Section-D: Raw data for lubricating oil:**

Formulas used:

Mean = \(\frac{Trial-1+Trial-2+Trial-3}{3}\)

Standard deviations = \(\frac{\sum^{i=3}_{i=1}(Trial-Mean)^2}{3}\)

Time (in ± 0.01 s)

Distance travelled in ± 0.05 cm

Velocity in * 10^{-1} cm s^{-1}

10.00

0.70

7.00

20.00

1.00

3.00

30.00

1.40

4.00

40.00

2.00

6.00

50.00

2.70

7.00

60.00

3.80

11.00

70.00

5.10

13.00

80.00

6.40

13.00

90.00

7.70

13.00

As indicated in the scatter graph above, the velocity increases as the time increases. However, from 70.00 ± 0.01 s, the velocity becomes constant as the trend line becomes parallel to the x axes. The terminal velocity is 13.00 × 10^{-1} cm s^{-1}.

**Section-E: Lamp oil:**

Time (in ± 0.01 s)

Distance travelled in ± 0.05 cm

Velocity in * 10^{-1} cm s^{-1}

10.00

0.70

7.00

20.00

1.00

3.00

30.00

1.60

6.00

40.00

2.30

7.00

50.00

3.30

10.00

60.00

4.80

15.00

70.00

6.50

17.00

80.00

8.20

17.00

90.00

9.90

17.00

As indicated in the scatter graph above, the velocity increases as the time increases.

However, from 70.00 ± 0.01 s, the velocity becomes constant as the trend line becomes parallel to the x axes. The terminal velocity is 17.00 × 10^{-1} cm s^{-1}.