Extended Essay – Investigating the limitations of the law of conservation of momentum – – To what extent does changing the weight of an toy car (ranging from 20 to 100 grams) going on a wooden surface affect the loss in momentum (in kgms −1 ) with respect to the law of conservation of momentum, given that the surface, temperature and force applied on the object remains the same? – Vishank Rughwani Subject: Physics Supervisor: Ms. Sarah Siddiqi School: Dubai International Academy Candidate Number: Words: ​ ​3512 1Introduction: The law of conservation of momentum states that momentum before a collision equals the momentum after the collision but only in an isolated system. This research paper aims to investigate how momentum changes with respect to external forces – not an isolated system. With this experiment, as common sense would tell you, the object would lose momentum due to forces opposing the direction of the motion of both objects, hence momentum after would be less. However, despite having a general impression of this, there is no actual relation that represents this relationship. This makes it hard and inaccurate to use these concepts in the real world, where a number of external forces are present. Take friction as an example. Frictional force will oppose the motion of the object at the points of contact between two surfaces; hence, the object will lose momentum as it will be losing energy to overcome these frictional forces. The fact that these forces are present in the real world and there being no relationship between the two has incited me to look into how momentum changes when the normal force of the objects changes hence, frictional force as well. This gives rise to the research question “To what extent does changing the weight of an toy car (ranging from 0 to 100 grams) going on a wooden surface affect the loss in momentum (in kgms −1 ) with respect to the law of conservation of momentum, given that the surface, temperature and force applied on the object remains the same?” This experiment will only take into account friction as methods to measure other external forces are not within the scope of this research paper. Background Information: The conservation of momentum is a fundamental concept of physics. Momentum is defined to be the mass of an object multiples by the velocity of the object and is equivalent to the force required to bring the object to a stop in a unit length of time. ​For any array of several objects, the total momentum is the sum of the individual momenta. ​ The conservation of momentum states that, in an isolated system, the amount of momentum is remains constant; momentum is neither created nor destroyed, but can only change through the action of forces as described by Newton’s 2laws of motion. Momentum is a vector quantity, having both a magnitude and a direction. Momentum is conserved in all three physical directions at the same time. Friction is the force exerted by the surface of an object when another object moves against it. Friction occurs in the direction opposite to motion, and because of this, it is a force that affects the motion of objects. When a box is pushed across a surface, friction works against the box in the direction opposite to the box’s motion. This happens because when two objects rub together, it sets off attractive forces between the molecules of the objects. For this reason, friction can occur between any two objects. As friction results from attractive forces, the amount of friction depends on the materials of those two interacting. For this reason, all two surfaces have different coefficients of friction. Frictional force is calculated by multiplying the coefficient of friction and the normal force in newtons. However, for every two objects there are two coefficients of friction: static and kinetic. Static friction is that that occurs when the object is at rest it prevents motion instead of slowing it down. In all cases the force needed to overcome static friction is much larger than the force needed to overcome kinetic friction. Kinetic friction is that that occurs when the object is moving and works to reduce the motion. Due to the fact the law of conservation of momentum only holds true when the objects are in an isolated system, when friction is taken into consideration, momentum is no longer conserved. This is because friction is an external force and when it is present energy is lost due to the frictional forces which the object will have to overcome and since the formula for momentum is the product of the object’s mass and velocity the momentum before will no longer equal the momentum after. Research Question: To what extent does changing the mass of an toy car (ranging from 0 to 100 grams) going on a wooden surface affect the loss in momentum (in kgms −1 ) with respect to the law of conservation of momentum, given that the surface, temperature and force applied on the object remains the same? 3Body – Theoretical Hypothesis: It is hypothesised that as the mass on the car’s increase, the force of friction will increase, hence the loss in momentum will be greater. This is fairly obvious. In order to get into the details, it is important to look at the equations pertaining to this experiment. The equations pertaining to this include: ● F f = ​μ ​k​N : Force of friction, where is the ​μ ​k​ coefficient of friction and N is the normal force exerted by the object it is on in Newtons. ● F = ma : Newton’s second law of motion, where m is mass in kilograms and a is acceleration in ms −2 . ● p = mv : Formula for momentum, where m is mass in kilograms and v is velocity in ms −1 . With these formula’s, it is possible to to devise another formula that could tell us the loss in momentum with regard to friction. By equating the formulas for Newton’s second law of motion and the force of friction you get: ma = ​μ ​k ​N By rearranging for a (acceleration, or in this case deceleration): a = (​μ ​k​N)/m With this formula, we can now use one of the suvat equations. For the sake of convenience, the formula v2 = u2 + 2as where v is the final velocity, u is the initial velocity (ms −1 ) a is acceleration or in this case deceleration (ms −2 ) and s is the distance in meters. 4By substituting a in the equation, we get: v = √u2 + 2(− UkN/m)(s) With this, we can calculate the velocity after travelling a certain distance with respect to the coefficient of friction and frictional force. Hence, by substituting v in the p=mv formula we arrive at the formula for momentum after: p = (m)(√u2 + 2(− UkN/m)(s)) Hence, by substituting in the values for the different masses, initial speed and the distance between the light gates, a graph can be formulated for this hypothesis. With research, it has been found that the coefficient of kinetic friction is 0.65. The distance between the two light gates will be 33 centimeters (+-0.05cm) or 0.33 m (+-0.0005m). The mass will increase from 0 to 100 grams (+-0.01g) with increments of 20 grams, while the toy cars A and B weigh 250.99 grams and 570.01 grams (+-0.01) respectively. The initial speed will be assumed to decrease over time as the force applied will remain constant. Here, normal force equals mass x gravity where gravity is 9.81 ms −2 . 5Table for hypothesis (2 d.p.) Mass of Toy Car A (grams [+-0.01g]) Normal Force from surface onto toy car A(N ewtons) Assumed Speed (ms −1 ) Momentum Before (p=mv) ( kgms −1 ) Momentum After (using formula stated above)( kgms −1 ) Momentum lost (​ ​kgms −1 ) 250.99 2.46 3.00 0.75 0.74 0.01 270.99 2.66 2.50 0.68 0.65 0.03 290.99 2.86 2.00 0.58 0.55 0.03 310.99 3.05 1.50 0.47 0.42 0.05 330.99 3.25 1.00 0.33 0.25 0.08 350.99 3.44 0.75 0.26 0.13 0.13 Hypothesis Graph 6As shown by the graph, it is hypothesised that the loss in momentum increases exponentially as the normal force increases. This is so du
e to the fact that as the normal force increases the energy lost to overcoming the frictional force is also increased hence more speed is lost; therefore, the momentum lost increases. One major issue with this however, is the fact that it does not take into account the coefficient of static friction, which is usually higher than coefficient of kinetic friction and hence some energy would be lost overcoming the static friction force. In addition to this, since the collisions are elastic, the sound the cars would make would also reduce energy and as the momentum of the first car increases, the sound it would make with the second car would also increase; hence, higher loss in energy. Therefore, this graph would most likely shift downwards due to the coefficient of static friction and become slightly more steep due to the energy lost due to sound. Experimental Variables: Dependant Variable Loss of momentum (kgms −1 ) Independent Variable Mass (0 to 100 grams) (+-0.01g) Fixed Variables Surface the car was on (Cardboard) in order to keep the friction between the two surfaces the same. Using different surfaces is not within the scope of this research paper. Temperature by conducting the experiment in a controlled environment with the AC at a constant temperature. Force applied to the car before. This was 7kept the same for consistency within each trial and to increase accuracy. It was kept the same by making a slingshot with rubber bands that propelled the first toy car. Uncontrolled Variables Wear and tear on the tyres of the car. This would cause the friction to increase and hence cause greater inaccuracies within the results. Experiment Apparatus: For the experiment to be conducted these are the materials required: ● Two toy cars with a large, flat body to add weights onto. ● Balance in order to measure how much the toy cars weigh. ● Two Vernier light gates with cables in order to measure the velocity of both the toy cars. ● Clamps in order to hold the light gates high. ● 10 masses of 20 grams to add to the cars. ● Wooden ramp of 2 meters for the toy cars to run on. ● Rubber bands to make a slingshot to pull the first toy car. ● Vernier Logger Pro with all cables to record the data ● Vernier Logger Pro Software to transfer the data to the computer. ● Marker in order to mark the start lines for the cars. 8Experimental Methods: In order to gather the results, I first placed the wooden ramp on a flat surface so that the normal force and force due to gravity cancel out completely. Then I setup the data logger: I connected the in the two light gates into DISC1 & DISC2 respectively and plugged in the logger pro to the power and used a USB cable to plug it into a laptop. I started up the logger pro software, clicked ‘Open’ from ‘File’ and opened the ‘Photogates’ package and then chose ‘Two gate timing’. 9After this page opened up, I added the length of each of the cars 0.07m and 0.11m for toy car A and B respectively. I made sure that the data collection was set on time based and 18 seconds. After this, I set up the light gates by using stands with clamps. I made sure that the two light gates had enough space between them so that the toy cars had enough space to collide; hence, I set them 33 centimeters (+-0.01cm) apart as shown below. The light gates were also placed 8.5 centimeters (+-0.01cm) above the ramp Then I measured the weight of both toy cars A and B, 250.99 and 570.01 grams (+-0.01g) respectively and placed them on the ramp before marking their positions. After this, I started to set-up the slingshot. I used two rods a few centimeters in front of the first toy car and rubber bands so that the each trial the car would get the same applied force. After the complete set up I started the experiment. I started off with 0 grams extra on each car while recording the data onto a paper for toy car’s A velocity and toy cars B velocity and continued to do so 5 times and moved on to 10adding 20 grams onto each toy car. I repeated these steps till I came to 100 grams on each toy car. Procedure – Collecting Data Raw Data Collection (2 d.p.) Mass, Normal Force and Frictional Force of The Toy Tars Extra Mass Added (grams [+-0.01g] ) Mass of Toy Car A (grams [+-0.01g] ) Normal Force from surface onto toy car A (Newtons ) Frictional Force on toy car A (Newtons ) Mass of Toy Car B (grams [+-0.01g] ) Normal Force from surface onto toy car B (Newtons ) Frictional Force on toy car B (Newtons ) 0 250.99 2.46 1.60 570.01 5.59 3.63 20 270.99 2.66 1.73 590.01 5.80 3.77 40 290.99 2.86 1.86 610.01 5.98 3.89 60 310.99 3.05 1.98 630.01 6.18 4.03 80 330.99 3.25 2.11 650.01 6.38 4.14 100 350.99 3.44 2.24 670.01 6.57 4.27 11Raw Data (2 d.p.) Velocity of toy car A vs. Velocity of toy car B (ms ) −1 Extra Mass Added (grams [+-0.01g] ) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Average 0 1.43 0.25 1.47 0.28 1.48 0.29 1.46 0.34 1.43 0.25 1.45 0.28 20 1.22 0.22 1.15 0.13 1.38 0.20 1.26 0.25 1.14 0.27 1.23 0.21 40 0.98 0.19 1.11 0.16 0.95 0.16 1.00 0.21 1.05 0.23 1.01 0.19 60 1.10 0.20 1.14 0.20 0.92 0.05 0.96 0.16 0.88 0.12 1.00 0.15 80 0.81 0.18 0.53 0.20 0.73 0.15 0.71 0.12 0.77 0.11 0.71 0.15 100 0.68 0.08 0.61 0.09 0.71 0.19 0.63 0.15 0.64 0.13 0.65 0.13 Average Speeds of different masses with Uncertainties (2 d.p.): Average Velocity ​(ms −1 ) Uncertainty ​(ms −1) Toy Car A Toy Car B Toy Car A Toy Car B 1.45 0.28 0.02 0.03 1.23 0.21 0.09 0.05 1.01 0.19 0.06 0.03 1.00 0.15 0.09 0.12 0.71 0.15 0.10 0.03 0.65 0.13 0.40 0.04 12Sample Calculation for Uncertainty: √ ((1.42−1.43)2+(1.46−1.43)2+(1.40−51.43)2+(1.45−1.43)2+(1.42−1.43)2) = ​uncertainty = 0.02 ms −1 Processed Data: Table listing the momentum before, after and loss in momentum Mass Added Momentum of Toy Car A Momentum of Toy Car B Momentum Lost 0 0.364 0.160 0.204 20 0.333 0.124 0.209 40 0.323 0.116 0.207 60 0.311 0.095 0.216 80 0.310 0.091 0.219 100 0.318 0.087 0.231 Processed Data Graph 13Data Analysis: The graph shows a consistent increase in the loss in momentum as mass increases. However, when mass added equals 40, the trend is broken. This could be said to be an anomaly in the data, however the standard deviation for the speed of the car is only 0.03 ms −1 meaning that all the trials done for this mass were more or less consistent; hence, there is no reasonable conclusion for this distrend. Conclusion: Compared to the graph in the hypothesis: In this graph, as shown the red line is the experimented results and the blue line is the hypothesised results. As represented, there seems to be a large difference between the two lines however, this is only due to the fact that I did not know the values of the initial speed for the hypothesis. Therefore, I assumed the values it 14would take and formed a graph. However, if formula stated in the hypothesis ( p = (m)(√u2 + 2(− UkN/m)(s)) ) is used to solve for any one of the values, the answer is found. For instance, for the first value: p = (250.99)(√1.452 + 2(− 0.65(2.46)/250.99)(0.33) ) = 0.364 ​kgms −1 Evaluation and further study: In order for the accuracy of this experiment to be greater, I believe that I would have to calculate the centre of mass of both the toy cars in order to place the weights exactly on that area. This would result in no change in the centre of mass hence, will not cause more force to be towards the end or towards the back. The fact that I did not calculate the centre of mass means that one set of wheels could have been experiencing more friction than the other and that calls for great inaccuracy. Some systematic errors that were present was the use of a ruler to measure the distances and width of the car. Compared to a digital measurer, a ruler has a much higher uncertainty, hence making the experiment inaccurate. Additionally, I feel that, even with the use of the slingshot to push the first car forward, there could have been discrepancies with the force applied on the car. Even the slightest differe
nce could have caused inaccuracies and since rubber bands were used, the tension could be so great that it caused deformation of the rubber band. This could help explain the dis trend within the data collected. If permitted, I would like to study this topic however, going down a ramp. With so many external forces on the object, the study would be very complex and would 15intrigue me. I would also like to calculate the coefficient of friction of both kinetic and static in order to make my predictions stronger and have for stronger analysis. This is a big factor because as stated in the hypothesis, energy would have to be used to overcome the static frictional force, which reduces speed and hence momentum. By investigating this, a formula could be devised with both static and kinetic friction. 16Works Cited “Conservation Laws – Real-life Applications.” ​Science Clarified ​. N.p., n.d. Web. Friedl, Sarah. “Friction: Definition and Types.” ​Study.com ​. Study.com, n.d. Web. PhysLink.com, Anton Skorucak. ​Coefficients of Friction ​. N.p., n.d. Web. “Science EE IB DP.” (n.d.): 1-13. Print. “Vernier Logger Pro.” ​Vernier Logger Pro ​ (n.d.): 1-13. Print. 17