How Does Drop Height Affect The Amount Of Water Drop In Water
The splash is cause by sudden disturbance in liquid. For example if the ball is dropped into water, its energy will affect the water, causing some amount to splashed out. When the ball is released from a height, its initial potential energy combined with the gravitational energy form a kinetic energy, which transmitted to the water. So I will investigate the relationship of drop height of the ball and the amount of water splashed out.
Research question:
How does the volume of water splashed out depend on drop height of the ball?
Explanation for controlled variable:
The change in mass in ball will affect the kinetic energy as KE = mv2/2, so will use the same ball for all experiments.
Initial volume of the water will be also controlled using …show more content…
the same beaker, as change in volume of water will change the splash.
Hypothesis
I assume that the higher drop height of the ball will result in larger volume of water splashed out. Because if the ball falls from the higher location, its final velocity will be higher and the kinetic energy with which it falls into the water will be more. So it will eventually result in more water splashed out.
I conducted my experiment based on the following method.
Methods of experiment:
1. I attached ruler to the base in order to make it easier to measure drop height
2. I made a ball using plasticine. I tried several types of ball, but the plasticine was the best material. The iron has too high density, so even a small ball is too heavy, so that it breaks the beaker. The wooden ball, on the contrary, is not dense, so that even big ball have no significant impact on splash. Plasticine is dense enough to create big splashes, also it is not likely to break the glass.
3. I filled a beaker with water. It is important to fill it to the brim. If there is a space between the surface of water and edge of the beaker, there is possibility that water will not splash out of the beaker, so it is impossible to measue amount of splash.
4. Using the measuring glass, I estimated initial volume of the water.
5. I placed the beaker next to the ruler.
6. Using a ruler, I regulated the location of the ball, attached to the wooden holder, so that distance between the ball and top of the beaker was 5cm. The major difficulty was to place the ball vertically to the beaker, so that it could fall exactly to the center of the beaker.
7. I made the wooden holder loose to release the ball.
8.
Using measuring glass, I measured the volume of the water after the splash. Before pouring water into the measuring glass, I removed the ball from the beaker.
9. Repeat 1-9 steps changing the drop height
I have done five trials for each drop height and considered the average value.
I subtracted from the initial volume of water in beaker the volume of water after dropping a ball in order to find out the volume of water splashed out.
Water splashed out = initial V – final V
To measure the drop height I used a common metric ruler, so uncertainty is ±0.05cm, which is indicated as horizontal bars in Graph 1.
To measure the volume of water, I used measuring glass with smallest division 5ml, so uncertainty is ±2.5ml
All points on the Graph seem to lie on a straight line. And the correlation coefficient of best-fit line is r = 0.9959. So we can assume that there is relationship between drop height of the ball and volume of water splashed out is very close to linear.
The slope of the best fit line is m = 1.423 ml/cm, which means that 1 cm increase in drop height will approximately increase 1.423 ml of water splashed out.
The minimum slope is mmin = 1.194 ml/cm and the maximum slope is mmax = 1.722 ml/cm.
Δm = (mmax = mmin)/2 = (1.722 – 1.194)/2 = 0.264
ml/cm
So if the ball is dropped 1 cm higher the volume of water splashed out will increase 1.423 ± 0.264 ml
Evaluation:
My hypothesis was right: as the drop height of the ball rises, the volume of the water splashed out will increase. And this relationship is approximately linear, because the best-fit line is located close to all points and intersects all error bars. The largest deviation from best fit is at drop height = 25 cm. Based on the slope of best-fit line, I determined that 1 cm increase in drop height leads to 1.423 ± 0.264 ml increase in volume of water splashed out.
During the experiment I could not avoid different types of errors, which create limitations of my investigation.
1. I could not locate the ball exactly above the beaker, so it might dropped not always in the center of beaker
2. The water in beaker was not perfectly motionless. There might be slight waves, which could affect the splash
3. When I took out the ball from the beaker after the splash, some amount of water could rest on the surface of the ball. Also the loss of water could occur when pouring the water into the measuring glass
4. I held the wooden holder with my bare hands. So because of human error, there might be inaccuracy in drop height
Methods of improvement:
1. Let the water in the beaker to rest for a while, to make it motionless
2. To fix the location of the ball using extension clamp attached to the stand
3. Use pincette while taking the ball out of the beaker to minimize the water loss
Research question:
How does the volume of water splashed out depend on drop height of the ball?
Explanation for controlled variable:
The change in mass in ball will affect the kinetic energy as KE = mv2/2, so will use the same ball for all experiments.
Initial volume of the water will be also controlled using …show more content…
the same beaker, as change in volume of water will change the splash.
Hypothesis
I assume that the higher drop height of the ball will result in larger volume of water splashed out. Because if the ball falls from the higher location, its final velocity will be higher and the kinetic energy with which it falls into the water will be more. So it will eventually result in more water splashed out.
I conducted my experiment based on the following method.
Methods of experiment:
1. I attached ruler to the base in order to make it easier to measure drop height
2. I made a ball using plasticine. I tried several types of ball, but the plasticine was the best material. The iron has too high density, so even a small ball is too heavy, so that it breaks the beaker. The wooden ball, on the contrary, is not dense, so that even big ball have no significant impact on splash. Plasticine is dense enough to create big splashes, also it is not likely to break the glass.
3. I filled a beaker with water. It is important to fill it to the brim. If there is a space between the surface of water and edge of the beaker, there is possibility that water will not splash out of the beaker, so it is impossible to measue amount of splash.
4. Using the measuring glass, I estimated initial volume of the water.
5. I placed the beaker next to the ruler.
6. Using a ruler, I regulated the location of the ball, attached to the wooden holder, so that distance between the ball and top of the beaker was 5cm. The major difficulty was to place the ball vertically to the beaker, so that it could fall exactly to the center of the beaker.
7. I made the wooden holder loose to release the ball.
8.
Using measuring glass, I measured the volume of the water after the splash. Before pouring water into the measuring glass, I removed the ball from the beaker.
9. Repeat 1-9 steps changing the drop height
I have done five trials for each drop height and considered the average value.
I subtracted from the initial volume of water in beaker the volume of water after dropping a ball in order to find out the volume of water splashed out.
Water splashed out = initial V – final V
To measure the drop height I used a common metric ruler, so uncertainty is ±0.05cm, which is indicated as horizontal bars in Graph 1.
To measure the volume of water, I used measuring glass with smallest division 5ml, so uncertainty is ±2.5ml
All points on the Graph seem to lie on a straight line. And the correlation coefficient of best-fit line is r = 0.9959. So we can assume that there is relationship between drop height of the ball and volume of water splashed out is very close to linear.
The slope of the best fit line is m = 1.423 ml/cm, which means that 1 cm increase in drop height will approximately increase 1.423 ml of water splashed out.
The minimum slope is mmin = 1.194 ml/cm and the maximum slope is mmax = 1.722 ml/cm.
Δm = (mmax = mmin)/2 = (1.722 – 1.194)/2 = 0.264
ml/cm
So if the ball is dropped 1 cm higher the volume of water splashed out will increase 1.423 ± 0.264 ml
Evaluation:
My hypothesis was right: as the drop height of the ball rises, the volume of the water splashed out will increase. And this relationship is approximately linear, because the best-fit line is located close to all points and intersects all error bars. The largest deviation from best fit is at drop height = 25 cm. Based on the slope of best-fit line, I determined that 1 cm increase in drop height leads to 1.423 ± 0.264 ml increase in volume of water splashed out.
During the experiment I could not avoid different types of errors, which create limitations of my investigation.
1. I could not locate the ball exactly above the beaker, so it might dropped not always in the center of beaker
2. The water in beaker was not perfectly motionless. There might be slight waves, which could affect the splash
3. When I took out the ball from the beaker after the splash, some amount of water could rest on the surface of the ball. Also the loss of water could occur when pouring the water into the measuring glass
4. I held the wooden holder with my bare hands. So because of human error, there might be inaccuracy in drop height
Methods of improvement:
1. Let the water in the beaker to rest for a while, to make it motionless
2. To fix the location of the ball using extension clamp attached to the stand
3. Use pincette while taking the ball out of the beaker to minimize the water loss