Purpose: In the Rocket lab, we used our knowledge of constant velocity, initial velocity, and acceleration along with formulas that we previously learned in order to predict where a rocket would land at any given launch angle. Our setup looked similar to the rocket below, filled with air by a bike pump.
We used small wooden wedges to prop our rocket on an angle, which allowed us to determine how it flew. We started with a 35 degree angle. We used a timer to time how long the rocket was in the air. After launching our rocket, we used a measurer to determine how far our rocket flew. The rocket was in the air for 3.84 seconds and flew 42 meters in the horizontal direction. We used these measurements to calculate initial velocity using our horizontal component and cosine equation. It is important to remember that Initial velocity at all launch angles remains the same. Our work for this step can be seen below.
After we found initial velocity, we were asked to launch the rocket on another angle (25 degrees) and predict where it would land. In order to do this, we drew an initial velocity diagram and then separated it in to components (step D). With the initial vertical velocity, we were able to determine how long the rocket would be in the air for (step A). After we found time, we used our horizontal displacement formula to determine how far the rocket would travel (step B). We then tested our hypotheses and our rocket landed 1.8 meters past our desired landing spot. With this data, we calculated our percent error (step C).
In this lab, we videos us throwing a ball in a parabolic shape. We used the video and the PASCO keystone program to make graphs and learn more about the acceleration, velocity, and motion of the ball. Here is a video of our throw:
After we plugged our video into the app, it allowed us to plot points where the ball was.
Once we plotted data points, PASCO gave us a position vs time for both horizontal and vertical movement. The green line shows the vertical position vs time while the purple line shows the horizontal movement. The green line's slope is constantly changing, and it shows us that the object has acceleration. Also, the line switches directions, showing a shift in direction of the ball. The purple line has a constant slope, indicating constant velocity. The ball had no forces acting on it in the horizontal direction and because of this it had a constant velocity.
Time
Our second graph shows us the velocity vs time of the object in the vertical and horizontal directions. The yellow line is the vertical direction and the red line is the horizontal line.
The yellow line shows that the velocity begins at a very high rate and then begins to slow down. The line then hits the origin, showing a change in direction. The ball then begins to pick up speed in the negative direction. The line also has a slope, which is constant, which shows acceleration. The slope of our line is close to -11 (acceleration) whereas the actual acceleration of the ball should have been -9.8 m/s^2 (the true acceleration of gravity. Although the numbers weren't totally correct, they were pretty close. The red line shows no slope. This means that the velocity of the ball in the horizontal direction was constant, which is the case because of its lack of forces in the horizontal direction.
Analysis:
a) The only force acting on the ball after it left my hand was gravity. This means that the acceleration of the ball was -9.8m/s^2 (the acceleration of only gravity acting on anything)
b)There are no forces acting on the ball in the horizontal direction. This means that the ball's forces are balanced in the horizontal direction, giving it no acceleration.
c) By using our graph, we can tell that the initial horizontal velocity is very close to .6 m/s.
d) The initial velocity in the vertical direction was 2.6 m/s according to our graph.
e) The velocity in the horizontal direction was constant the whole time. This means that at the top of the path, the ball was still going .6 m/s.
f) At the top of the path in the vertical direction, the ball is switching direction and momentarily has a velocity of 0 m/s.
g) At the end of our test, the ball stops on the ground (eventually) and has a velocity of 0 m/s in the horizontal direction.
h) At the end of our test, the ball stops on the ground (eventually) and has a velocity of 0 m/s in the vertical direction.
i) Using the equation Change in x = 1/2(a)t^2 +vit, we can calculate the height the ball got. The ball reached its highest point after .9 seconds (according to graph)
Conclusions: What occlusions can you make about projectile motion?
a) The vertical acceleration on our graph shows -11 m/s^2. In reality, this should be closer to -9.8 m/s^2 because of the acceleration of gravity. The only force on the ball is gravity, giving it this acceleration. The slightly larger acceleration is probably due to minor errors while plotting points on our graph. There are no forces acting on the ball in the horizontal direction. This means that the horizontal axis is balanced, and the ball has no horizontal acceleration.
b) The equation for horizontal velocity is (change in horizontal position)/time. this velocity should stay constant throughout t ball's motion. The equation for vertical velocity is (change in vertical position)/time. This velocity will change throughout the motion of the ball.
c) Vertical formula for displacement: Displacement = 1/2(at)(t^2) + (Vi)(t)
Horizontal formula for displacement: displacement = (.6)t
d) Displacement = 1/2(a)(t^2) + Vi(t)
e) The ball has a vertical velocity of 0 m/s at the very peak of its path, but the horizontal velocity stays constant throughout the whole motion.
In this lab, we used a modified Atwood machine to predict the intersection of a dropping weight which had acceleration and a cart moving at constant velocity along the ground. For an extra challenge, we angled our Atwood machine slightly. The angle of our ramp was three degrees. Our apparatus looks similar to what follows:
We took measurements of the mass of each component of our contraption:
Object
Mass (kg)
Green cart (on floor)
.477
Weight hanging with string
.059
Buggy
.5883
Whole Atwood machine
1.04
Converting masses to Newtons and using our background knowledge, we created a free body diagram to help us solve our problem.
In order to predict how to make the weight land on the cart, we calculated the time it would take for the weight to drop to the ground. To do this, we measured the total distance that the weight would fall.
Our work is as follows:
We used our force diagram to find Fnet and used our measurements to find the mass of the Atwood machine. With this information we were able to find acceleration using the equation Acceleration=Fnet/mass. We plugged .509 into the Fnet and 1.04 in for mass, leaving us with an acceleration of .489m/s^2
Because we knew the distance the weight would drop and the acceleration, we used the following method to solve for the amount of time the weight would drop.
We then used a motion sensor to detect the velocity of the cart that was moving along the ground. Once we found the velocity of that cart (which was.119 m/s), we multiplied it by the time in order to find out how far back to start the cart for the intersection to work. We used the equation
distance=(velocity)(time) to find out that the cart would travel 25.2 cm. We then placed the cart 25 cm away from where the weight would land. We then let both the cart on the floor and the dropping weight go at the same time and videos the result. Our predictions were correct and the weight landed perfectly on our cart! A video of the experiment is posted below.
In this lab, we used motion sensors to get two equations of different wind powered cars. These cars both accelerated from rest. In the lab, we tried to predict where the cars would have to start in order to intersect at the origin. Our work is as follows:
Position vs Time Graphs: Previously we have known that when a position vs time graph has a straight line on it, the object is moving at a constant velocity. With the CAPM model, we learned that these graphs can also have a curved line, which indicates acceleration. Also, we know that steeper slopes indicate that an object is moving fast, and that when an object moves in the positive direction it is going forwards and when it is going towards the negative side of the graph that it is going in the negative direction. We can also apply this information to our new curved lines.
As the line moves down as the vector
indicates, the velocity is increasing.
As the vector going up indicates, the
velocity is also increasing.
If a velocity line is going up in the direction
of the top vector, the object is moving in
the positive direction, and if the line is going
downward like the bottom vector, then the
object is moving in the negative direction.
If the line moves toward the origin over
time line the top vector does, the object
is slowing down. Also, if the line moves
towards the origin like the bottom vector,
the line is also slowing down.
If the line on the graph moves towards
the origin like the top vector, the
object is still moving in the positive
direction. Similarly, on the bottom vector,
the object is moving in the negative
direction.
In this example, the object starts with
a velocity of 10m/s. From seconds 0-2, the
object is slowing down. Then, as it crosses
the origin, it changes direction and
begins speeding up in the negative direction.
In this graph, the object is increasing in speed in the positive direction. To calculate displacement, we can calculate the area between the line and the x-axis. This area is in the shape of a triangle, so we can use Displacement = 1/2 (Base)(Height) to find displacement. By substituting values in from the triangle that the line makes with the x-axis, we create an equation that should look like this: 1/2(10)(10) = displacement. From there, we can conclude that the object finished 50m from where it started. To take a different approach, we can also use the equation ΔX = 1/2aT2 + initial velocity to find the displacement. Initial velocity is the velocity the velocity at zero seconds. In this case, initial velocity is zero. Also, we can find the acceleration by using the equation (ΔV)/(Δt) = a. We can only use this equation for acceleration when our velocity vs time graphs are a straight time, indicating constant acceleration. In this case, the acceleration is 1m/s^2. After we find acceleration, we can plug in our numbers and solve for the displacement.ΔX = 1/2(1)(100) + 0 which comes out to equal 50m, the same as our first calculation.
Average Acceleration:
Formula:
Average acceleration is the average rate at which an object is speeding up. Average acceleration may be very useful for many other equations and formulas in physics.
In this graph, to find the instantaneous velocity at five seconds we can
find the slope between two pints that will make 5s the midpoint.
For example, we can find the slope between the points at 4s and 6s.
This line is also tangent at five seconds. The fourth second falls at 1m and
the sixth at 2m. Therefore, we can use the equation (change in position)/(change in time)
to find the line tangent. This tangent line will be the instantaneous velocity at that point.
Example:
In the position vs time graph above, the object has four distinct motions. From point A to B, the object is moving in the positive direction. Also, since the line is curved, we can determine that it is either slowing down or speeding up. The way to differentiate between the two is look at the slope of the line. At the beginning of the line, the slope is very steep. It gradually becomes less steep until the slope becomes zero at point B. With this information, we can tell that the object is initially going very fast and then slows down. From point B to C, the object is moving towards the negative side of the graph, telling us that the object is moving backwards. The slope closest to point B is not very steep, but gets steeper as it gets closer to point C, indicating that the object is speeding up. At point C something interesting happens. The object crosses the origin with a very steep slope (high velocity) and then begins to slow down again. This can be seen with the gradual decline of slope. At point D, the object switches direction again and begins to travel in the positive direction again. Starting with a small slope, the object begins traveling slowly and then speeds up until it reaches the origin, finishing at the same position that it started at.
Velocity vs Time Graphs: Constant velocities are shown on a velocity vs time graph as a horizontal line. When a velocity vs time graph has a diagonal line, the velocity is either increasing or decreasing. An object with an increasing or decreasing speed has an acceleration. As seen on the velocity vs time graphs below, as the line moves away from the origin, the object is increasing in speed. Also, if the line is going in the positive side of the graph, it is moving forwards (above the origin) and if it is in the negative side of the graph, it is moving backwards (under the origin)
Example:
Displacement and Velocity vs Time graphs: We can use two methods to find the displacement from a velocity vs time graph. The first is to find the area of the space between the line and the x-axis. This space represents how far the object has traveled. Also, we can use the equationΔX = 1/2aT2where ΔX = displacement, a = acceleration, and t = time.
Example problem:
Acceleration vs Time Graphs: Acceleration vs Time graphs do not tell us much other than the slope of the velocity vs time graph. a vs t graphs are directly related to the slope of v vs t graphs. For example, if the slope of a velocity vs graph is negative, the acceleration line will also be on the negative side of the graph. Keep in mind that there is no acceleration when the velocity vs time graph has a straight horizontal line with no slope.
Examples:
In this velocity vs time graph, the line has a positive slope from 0-5 seconds, no slope from 5-15 seconds, and then a negative slope from 15-25 seconds. Because of this, we can conclude that the acceleration for the first five seconds is positive, there is no acceleration for the next ten seconds, and then there is a negative acceleration for the last ten seconds. The acceleration vs time graph would look as follows:
Motion Maps: We have now learned how to make both acceleration and velocity motion maps. In velocity motion maps, the distance between the points represents how far the object has traveled and the vector represents the velocity. The larger the vector, the larger the velocity. In acceleration motion maps, the distance between dots represents distance traveled, but the vectors represent acceleration. Constant acceleration is represented by vectors of the came length as seen below. In the image below, the object's velocity is increasing with constant velocity.
Average Velocity: Average velocity is calculated by the equation (change in position)/(change in time) and represents the average velocity that an object travels at over a given amount of time.
Instantaneous Velocity: Instantaneous velocity is the velocity at a given time. To find instantaneous velocity on a position vs time graph, you can find a line tangent to the point at which that time is located (also the same as making that time the midpoint of two other points on the line).
Real World Examples: Physics is ever present in the real world. One example of a real world situation where acceleration is present is as follows: A dog runs down his driveway with an initial speed of 5m/s for 8s and then uniformly increases his speed to 10m/s in the next five seconds. How long is his driveway?
How to solve: So, we are given both time and velocity. Since we are trying to solve for displacement, we can use the equations ΔX = 1/2aT2 + initial velocity and x = Vt. In the first eight seconds of the run, the dog has a constant velocity. Because of this, we use the exertionΔX = Vt. After substitution, this becomes ΔX = (5m/s)(8s), simplified to 40m. Keep in mind that this is only the first half of the dog's journey. The second half can be calculated with the equationΔX = 1/2aT2 + (initial velocity)(time)
. We can find acceleration with the equation ΔV/Δt which comes out to be 1m/s^2. After substitution, this equation becomes ΔX = 1/2(5)^2 + (5)(5) This can be simplified to ΔX = 37.5. So, now that we have both the first part of the journey and the second, we can add them together in order to find the total displacement (length of the driveway). 40m + 37.5m = 77.5m, the length of the driveway.
In Unit two of our physics class, titled "Balanced Force Particle Model", we learned about Newton's first and third law, free body diagrams, balanced free body diagrams, force vectors, weight and mass, friction, as well as different forces acting upon objects in motion, at rest, and accelerating.
Free Body Diagrams: Free body diagrams are a way of illustrating forces that are acting upon an object. It is important to remember that when drawing a diagram, we only include forces acting upon an object and not forces that the object is creating. Several important rules are used when creating a free body diagram.
Newton's 1st Law: Newton's first law states that an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. One real world example of this was when we rode the hovercraft. We continued to move without anyone pushing us.
(A video of a hovercraft)
Newton's 3rd Law: This law states that for every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. For example, when I push on the wall, the wall pushes back with the same amount of force. A real world example is when a car hits a bug on the road. Although it seems counterintuitive, the bug pushes back on the car as hard as the car pushes on the bug. This is a good example of an equal and opposite reaction.
Both the push on the wall and the push back from the wall are equal.
Friction Force: Friction is the resistance that one surface or object encounters when moving over another. This means that it often slows an object or keeps it from moving in a certain direction. It is always parallel to the surface the object is on.
Gravity Force: Vector always goes straight down, no matter which way the axis is tilted.
Normal Force: Vector always goes perpendicular to the surface the object is resting upon.
Tension: Used when a rope is present, goes in the direction that the rope is pulling.
Tilting an Axis: When an object is on a surface that is at an angle, we can tilt the whole coordinate plane in order to make the free body diagram easier to draw. The x-axis can be tilted to be at the angle of the surface.
Vectors: Vectors are a special type of arrow used in physics. A vector's length determines how large it is in value. For example, if one vector is longer than another, it will have a larger force than the smaller one.
Splitting Vectors: Some vectors will not land on the x or y axis. This is a problem, because we can only balance a diagram when its components lie on an axis. To do this, we can break the vector into x and y components. The vectors on the x and y axis will represent this force in a way that we can use to balance an equation.
Balanced Diagrams: When vectors are equal length on opposite sides of the same axis, the diagram is balanced. This means that the object is either at rest or moving at a constant velocity. When vector lengths are not equal, an object is speeding up or slowing down.
In the example below, a box is resting on a slope. Because the surface that the box is on is tilted, we can tilt the axis in order to create an accurate free body diagram. Using our rules above, we can conclude that gravity pulls the object straight down, normal force goes perpendicular to the surface, and friction keeps the box from sliding down the hill. One thing that you will notice when looking at the diagram is that gravity does not lie on an axis. Because of this, we must split it into two vectors, Fgx and Fgy as seen below. Also, the FN and Fg vectors are much longer than the Ft and Fgx vectors. This tells us that the voce of gravity and normal force are larger than that of friction and gravity on the y-axis. We can also conclude that the object is either at rest or moving at a constant velocity because of the equal vector lengths.
Below is an example of a diagram for an object that is slowing down. This particular diagram is for a ball being thrown in the air. Notice how the only force is gravity, which means that this equation is unbalanced. This means that the object is either accelerating, changing direction, or slowing down.
Weight and Mass: Although commonly thought of as the same thing, weight and mass are actually very different. An equation to find the weight or mass of an object is Weight = (Mass)(Gravity). Gravity is always 10 (on Earth), weight is always measured in Newtons, and mass is always measured in Kilograms.
Example problem: A ball has a mass of 15kg on earth. What is its weight?
-W=(mass)(gravity)
-W=(15)(10)
-W=150 Newtons
Kinetic Friction: Friction can be determined with the equation f=(µk)(W) Where F is frictional force, µk is the coefficient of friction, and W is weight. Friction never changes with speed or surface area, but will change with different types of surfaces as well as with different weights.
Example problem: The coefficient of kinetic friction is .5 between an 80N book and sandpaper. What is the force of friction between the book and the sandpaper when it is sliding?
- f=(µk)(W)
- f=(.5)(80)
- f=40N
Solving for Unknown Values In FBD's: In a free body diagram, we often have information like angles and some vectors, but not the vector that we need in order to solve for a certain force value. To solve for a given side, we can use Sin, Cosine, and Tangent. By using the mnemonic "SohCahToa". In the diagram below, angle X has an opposite side, adjacent side, and hypotenuse. These can be used to determine the length of certain vectors.
Example:
By using our knowledge of sin, cosine, and tangent, we can determine that we must use sin, since we are using both the opposite and hypotenuse.
Work: Sinθ = Opposite/Hypotenuse
Sin (40) = 9/x
(X)(sin40) = 9
X = 9/(sin40)
x = 12.001
Real World Connection: We encounter forces every day. Although we may not notice it, we are constantly exerting forces in order to move and do everyday chores. For example, as you are reading this, gravity and normal force are acting upon you right now. Action reaction pairs, and Newton's laws are also constantly appearing when we drive, have tug of war contests, and when we are sleeping. In this unit, we used real world situations and explored the reasoning and math behind them.
