PHYs 207 Lab 5 - Physics 20700- GH3 – Lab5 - Linear Momentum PDF

Preview

Summary

Physics 20700- GH3 – Lab5 - Linear Momentum...

Description

Samuel Obadina

Physics 20700- GH3 – Lab5 - Linear Momentum Introduction In our fifth lab experiment, my group was tasked to prove the law of conservation of linear momentum. The goal was to prove that this motion conserves the momentum if there is not an external force acting on the moving object. In order to prove this, we rolled a steel ball down a ramp and had it roll into a wooden block. We then compared the momentum before and after the system. We expect the momentum to be equal in our result. A benefit for participating in this lab was we were also able to learn more about the concepts of momentum, and energy.

Procedure Part I: Determination of the Velocity of the Sphere before Impact: We put the sliding ball clamp at any point in the ramp and secure it to the ramp. We then place the block on the platform. From the box, the slider guide and slider was removed. We then let the steel ball roll down the track from the highest point and have a group member make sure an imprint is left on the wax paper when the ball hits the surface. We measure the height of the ball’s fall and the distance which the sphere falls after rolling the track from a fixed position. This procedure was repeated 9 times or more to lower the chance or error. We then calculated the average of the distance. We calculated the uncertainty of the height and the distance that was measured. We also calculated the mean of distance and use the standard deviation of the mean to find another

value of uncertainty. We finally calculated the velocity and time of flight of the ball using the data we procured. Part II: Determination of the Velocity of Sphere and Block after Impact: We first measured the mass of the steel sphere and wooden block using the weight scale and recorded the values. We then mounted the slider guide and the slider in the box. We have suspended the block at an even level so that it is perpendicular to the table. We then made sure that there is about a 1/8th of an inch gap between the block and the ramp. We measured the length of the string and calculated the uncertainty. A group member had to make sure the block barely touches the sliding pointer or not be in contact at all after the ball collides. We redo this experiment 9 more times and calculated “x” based on the sliding pointer position. With the data acquired, we also calculated the linear momentum before and after the impact.

Data/ Calculations x(t) = d (horizontal travel distance) y(t) = -y (ball drop distance)

Conservation of Energy Equation: P E + KE (initial) = P E + KE (final) 0 + 21 (m(ball) + m(block) ) v 2 = (m(ball) + m (block) ) gy v (ball and block ) =

√2gy = √2g(h − √h

2

− x2 )

Loss of Kinetic Energy Equation: KE (final)−KE (initial) KE (initial)

=

1 (m 2

(ball)+m (block ))v2 (ball and mass)− 12 m(ball)v2 (ball) 1 2m

(ball)v2(ball)

Without Collision (Free Fall): From the first part of the experiment, we obtained the following values, The height, y from which the ball dropped: 0.665m Uncertainty in the height which ball dropped: 1mm/ 2= 0.5mm = 0.0005m 0.0005m/0.665m = 0.0007m

Distance (m)

Uncertainty (cm)

.245

.0007

.285

.0007

.27

.0007

.29

.0007

.26

.0007

.26

.0007

.225

.0007

.225

.0007

.225

.0007

.225

.0007

Average distance value in meters: 0.251 m Uncertainty in distance in meters: 1mm/2 = 0.0005m 0.0005m/0.251m = 0.00199m

Standard Deviation For standard deviation, we used the Excel program to calculate our standard deviation. We input the data into Excel and the results show the standard deviation to be: 0.025690m. Standard Deviation Error: 0.121774 S tandard deviation error =

standard deviation √number of trials

=

0.121774 √10

= 0.0385

Standard Deviation = 0.121774 +- 0.0385

We can now calculate the time of flight. Height= .665m g= 9.8m/s^2 Height = initial distance + velocity x time + (acceleration x time^2)/2 v = √2gh Uncertainty in velocity: S tandard deviation error =

standard deviation √number of trials

=

0.121774 √10

= 0.0385

The standard deviation of velocity is the same standard deviation error because it is directly proportional to one and the other. Uncertainty in time of flight: t=

√

2h g

=

√

2(0.251 m ± 0.0019m) 9.8m/s2

=0.2263s +- .0008s

Uncertainty in height from block to top of the apparatus: 1 cm/2 = 0.5 cm 0.5cm / 47cm = 0.0106 cm = 0.000106 m Report Questions 1. Now we have two ways to calculate the uncertainty of an experimental measurement. The first simply looks at the average of the uncertainties over a repeated measurement, as you

did in the first lab. The new method uses the standard deviation. How do multiple measurements of d change the uncertainty? Compare the uncertainties in the velocity of the ball using these two different methods. The more trials done, the less value the uncertainty is. This is because it gave us a better average distance. With more trials being done, there is an increase in confidence in our experimental data because there is a smaller chance of data being affected by errors being a big part of the overall experiment. There was also uncertainty in the velocity of the sphere using the standard deviation method is lower than the other experiment. By calculating the standard deviation of the mean, we get more accurate values as the number of measurements increases. The denominator will increase because of this but the numerator will be more accurate due to the number of trials Uncertainty of h :

0.0005m 0.665m

* 100% = 0.0751%

Uncertainty of d :

0.0005m 0.251m

* 100% = 0.1992%

2. Within the limits of your experimental accuracy, is momentum conserved during the collision? Initial x = .205m Final x = .251m Change in x = .251m - .205m = 0.046m Before Collision p = m * v = 0.0668 kg * 3.58m/s = 0.239144 v (ball and block ) =

√2gy = √2g(h − √h

2

kg *m s

− x2 )

=

√

2 * 9 .81m/s2

(

0.47m −

√(0.47m)

2

− (0.046m)

2

)

= 0.2103m/s2

After Collision p (block and ball) = (M (block) + m (ball))v (block and ball) = (0.0668 kg + 0.0865 kg) (0.2103m/s) = 0.03223kg x m / s It can be seen that momentum is somewhat conserved but there is a noticeable difference before and after the collision. This may be because of human errors or frictional forces.

3. Derive equation (1), starting from general physics principles. The law of conservation of energy applies in this case. This means initial energy equals final energy. We will need to calculate the initial kinetic energy before the sphere hits the block and final energy will be potential energy because that is when the sphere and block reached the highest height. This means potential energy will equal kinetic energy. This formula will be used to find velocity. mgh = .5mv^2 After adjusting the equation to make it more suitable for what we are trying to find, we get v = √2gh

4. From your results, compute the fractional loss of kinetic energy of translation during impact. Disregard rotational energy of the sphere.

In order to find the loss of kinetic energy, we need to create an equation that can calculate this using the initial kinetic energy as the KE (F inal) −KE (Initial) KE (Initial)

=

1 2 (0.0668

2

2

kg +0.0865 kg )( 1.42m/s) −12 (0.0668 kg )(3.58 m/s) 1 2

2

(0.0668 kg )(1.42 m/s)

=− 0.24 J

5. Derive an expression for the fractional loss of kinetic energy of translation in terms only of m and M, and compare with the value calculated in the preceding question. Consider the collision as a totally inelastic one. mv= (m+ M) x v^2 v^2 = v(m/(m+M)) Ef =.5 ( m + M ) v^2’ Ef = (M^2 + m^2)/ (2(m+M)) Ef/ Ei = m/ (m+ M) Loss in kinetic energy = 0.0668kg/ (0.0668kg+ 0.0865kg)= 0.437 Conclusion The purpose of this lab was to confirm if there was a conservation of momentum in our experiment. To confirm this, we were tasked to prove that the experiment followed the law of conservation of momentum in a linear system. Based on the results obtained, we concluded that there was no conservation of momentum based on what we did in our experiment. This is because our data shows a noticeable difference in momentum of what it should be and what we obtained in value, giving us a decent loss of kinetic energy. This loss of energy was most likely caused by sources of errors such as human error, giving estimations based what we see on the

ruler and other issues.We do believe that momentum is conserved, it is just there is loss in energy from the sound of the collision and there is still friction involved. In a perfect experiment, these values would be added into our calculation....