KING MONGKUT’S INSTITUTE OF TECHNOLOGY LATKRABANG SCHOOL OF ENGINEERING GROUP OF KMIDS 01006702/01006724– PHYSICS -I LABORATORY LINEAR MOMENTUM & COLLISIONS INSTRUCTED BY: SUPUN DISSANAYAKA, KASUN HETTIHEWA NAME: WANTANA, NAPHAT, SINGHANART, PAWARIT, KRITTAPAS ID NUMBER: 1707070, 1707023, 1707054, 170704, 1707015 GROUP NO: KMIDS, 2 DATE OF SES.: 9/11/2022 DATE OF SUB.: 03/12/2022 Purpose: 1) 2) 3) 4) 5) To understand the concepts of momentum and how to apply it. Understand how to classify and differentiate different collisions. Understand the concept of impulse and how to apply it. Know the relationship between kinetic energy and momentum. Understand how momentum is conserve and apply it to the real-world. Theory: Linear Momentum • Momentum is always conserved. • The momentum formula is defined as πππ π ∗ π£β πππππ‘π¦ Or π =π∗π£ • The unit of momentum is equal to kilogram meter per second ππ ∗ π π Impulse and Average Force • The amount of time requires to the change the momentum πΌ = ππ π£π − ππ π£π • Average force is defined by the force multiply by time equal to the change in momentum πΉβ = ππ π£π − ππ π£π πΌ = π₯π‘ π₯π‘ Collisions • In all types of collisions momentum is conserved. π1π + π2π = π1π + π2π Or • If we express, it in respect of masses and velocities. π1π π£1π + π2π π£2π = π1π π£1π + π2π π£2π Elastic collision • The total kinetic energy and momentum are conserved. 1 1 1 1 2 2 2 2 π1 π£1π + π2 π£2π = π1 π£1π + π2 π£2π 2 2 2 2 Perfectly inelastic collision • The momentum is conserved however the total energy is not conserved. Pre-Lab Exercise: EX: 1) What will happen to the system’s momentum and what is it? Find the u velocity. - The momentum will be conserved since momentum is conserved in all collisions. The type of collision is an explosion. Since in all collision ππ π£π = ππ π£π We can infer that ππ π£π₯π = ππ π£π₯π πππ ππ π£π¦π = ππ π£π¦π Now, the formula for our horizontal momentum is: ππ π£π₯π = π3.8 (π£3.8π₯ ) + π4.2 (π£4.2π₯ ) + π2 (π’ πππ π) Which mean our π’ πππ π is equal: π’ πππ π = ππ π£π₯π − π3.8 π£3.8π₯ − π4.2 π£4.2π₯ π2 Thus, π’ πππ π = 10(0) − 3.8 ∗ (−21πππ 20) − 4.2 ∗ (12πππ 30) 2 π’ πππ π = 15.669 π/π Now, the formula for our vertical momentum is: ππ π£π¦π = π3.8 (π£3.8π¦ ) + π4.2 (π£4.2π¦ ) + π2 (π’ π ππ π) Which mean our π’ π ππ π is equal: π’ π ππ π = ππ π£π¦π − π3.8 π£3.8π¦ − π4.2 π£4.2π¦ π2 Therefore, π’ π ππ π = 10(0) − 3.8 ∗ (21π ππ20) + 4.2 ∗ (12π ππ30) 2 π’ π ππ π = −1.0466 π/π To find π we must find our theta first by find π‘ππ π π π ππ π 1.0466 =− = π‘ππ π π πππ π 15.669 π = π‘ππ−1 (− 1.0466 ) 15.669 π = −3.821° 15.669 = 15.703 πππ π π= π = 15.703 π/π EX: 2) What was the original speed of the bullet? From the law of conservation of momentum ππ π£π = ππ π£π The momentum of our system is defined as: πππ’ππππ‘ π£π = (πππ’ππππ‘ + ππ€πππ )π£π The formula for our initial velocity is: π£π = (πππ’ππππ‘ + ππ€πππ )π£π πππ’ππππ‘ Thus π£π = (0.010 + 5)0.600 0.010 π£π = 300.6 π/π Experiment: Elastic Collisions (Fixed Object) Procedure: 1. 2. 3. 4. 5. 6. Set up two carts on the track. Set up end stop with holders of magnets on the end of the track. Starting each trial with the reference poles to track the result. Move the two carts to each of the end sides on the track. Push one of the carts to other end side of the track. Record the results of collision by video camera. Observation: The moving cart hit the fixed cart and gradually moving back with a low speed. Calculations: π = 527.5 π ππ 0.53 ππ Calculating the initial and final velocity: Trial 1 2 3 Avg. d1 (cm) 72.6 72.6 72.6 t1 (s) 2.46 2.64 3.45 Vi (cm/s) 29.5 27.5 21.04 26.01 0.26 m/s d2 (cm) 33 43 22.8 π£π = 0.26 m/s π£π = 0.13 m/s Momentum: Equation used: ππ = ππ ππ£1 = ππ£2 Therefore 0.53 × 0.26 = 0.53 × 0.13 0.14 ππ × π ππ × π = 0.069 π π t2 (s) 1.83 3.5 2.66 Vf (cm/s) 18.03 12.29 8.57 12.96 0.13 m/s Kinetic Energy: Equations used: πΎπΈπ = πΎπΈπ 1 1 ππ£12 = ππ£22 2 2 Therefore 1 1 × 0.53 × (0.26)2 = × 0.53 × (0.13)2 2 2 0.018 π½ = 0.0045 π½ Results: Momentum: 0.14 ππ × π ππ × π = 0.069 π π Momentum is not conserved. Kinetic Energy: 0.018 π½ = 0.0045 π½ Kinetic Energy is not conserved Conclusion: According to the theory, momentum is always conserved as well as kinetic energy for elastic collision. However, the results above show that this is not true. This might be because in the real world there are factors like friction, human error, energy transfer, impulse, etc. These factors might cause the result to be inaccurate. Elastic Collisions (Moving Object) Procedure: 1) Set up two carts on the track. 2) Set up the magnets on the cart so that both have the same pole facing each other. 3) Starting each trial with the reference poles to track the result. 4) Push one of the cart towards the other to make them collide. 5) Record the results of collision by video camera. Observation: Calculations: π1 = 527.5 π ππ 0.53 ππ π1 = 521 π ππ 0.52 ππ Calculate the initial and final velocity for both carts Trial 1 2 3 Avg. d1 (cm) 30 30 30 t1 (s) 1.87 1.35 1.3 v1 (cm/s) 16.04 22.2 28.7 22.31 0.22 m/s d2 (cm) 19.55 49.3 39.7 t2 (s) 3.31 4.21 3.9 π£1π = 0.22 m/s π£1π = 0 m/s π£2π = 0 m/s π£2π = 0.13 m/s Momentum: Equation used: ππ = ππ π1 π£1π + π2 π£2π = π1 π£1π + π2 π£2π Therefore 0.53 × 0.22 + 0.52 × 0 = 0.53 × 0 + 0.52 × 0.13 0.12 + 0 = 0 + 0.068 v2 (cm/s) 16.24 11.71 10.18 12.71 0.13 m/s 0.12 ππ × π ππ × π = 0.068 π π Kinetic Energy: Equation used: πΎπΈπ = πΎπΈπ 1 1 1 1 2 2 2 2 π1 π£1π + π2 π£2π = π1 π£1π + π2 π£2π 2 2 2 2 Therefore 1 1 1 1 × 0.53 × (0.22)2 + × 0.52 × (0)2 = × 0.53 × (0)2 + × 0.52 × (0.13)2 2 2 2 2 0.013 + 0 = 0 + 0.0044 0.013 π½ = 0.0044 π½ Results: Momentum: 0.12 ππ × π ππ × π = 0.068 π π Momentum is not conserved Kinetic Energy: 0.013 π½ = 0.0044 π½ Kinetic Energy is not conserved Conclusion: According to the theory, momentum is always conserved as well as kinetic energy for elastic collision. However, the results above show that this is not true. This might be because in the real world there are factors like friction, human error, energy transfer, impulse, etc. These factors might cause the result to be inaccurate. Inelastic Collisions (Moving Object) Procedure: 1) Set up two carts and place them on the track. 2) Set up a reference pole 3) Set up the magnets on the cart, so that they have opposite pole so that they could attract when they collide. 4) Push one cart towards the other so that they could collide 5) Use a recording device to record the collision. Observation: When two carts hit each other. It attract and continues moving forward at a lower speed in the same direction. Calculations: π1 = 527.5 π ππ 0.53 ππ π2 = 521 π ππ 0.52ππ π3 = 0.53 + 0.52 = 1.05 ππ Calculate the initial and final velocity of the carts Trial 1 2 3 Avg. d1 (cm) 30 30 30 t1 (s) 1.2 1 1.53 v1 (cm/s) d2 (cm) 25 39.9 30 39.5 19.6 19.55 24.87 0.28 m/s π£1π = 0.28 m/s π£2π = 0 m/s π£3π = 0.0079 m/s Momentum: Equation used: ππ = ππ t2 (s) 4.25 4.8 3.2 v2 (cm/s) 9.4 8.2 6.1 7.9 0.079 m/s π1 π£1π + π2 π£2π = π3 π£3π Therefore 0.53 × 0.28 + 0.52 × 0 = 1.05 × 0.0079 0.15 + 0 = 0.0083 0.15 ππ × π ππ × π = 0.0083 π π Kinetic Energy: Equation used: 1 1 2 2 π1 π£1π + π2 π£2π 2 2 1 2 πΎπΈπ = π3 π£3π 2 πΎπΈπ = Therefore πΎπΈπ = 1 1 × 0.53 × (0.28)2 + × 0.52 × (0)2 2 2 πΎπΈπ = 0.021 π½ πΎπΈπ = 1 × 1.05 × (0.0079)2 2 πΎπΈπ = 0.000033 π½ π²π¬π ≠ π²π¬π Results: Momentum: 0.15 ππ × π ππ × π = 0.0083 π π Momentum is not conserved Kinetic Energy: πΎπΈπ = 0.021 π½ πΎπΈπ = 0.000033 πΎπΈπ ≠ πΎπΈπ Kinetic Energy is not conserved Conclusion: According to the theory, momentum will always be conserved, but kinetic energy is not with inelastic collisions. In this experiment, the results show that both momentum and kinetic energy are not conserved. This shows that kinetic energy is not conserved with inelastic collisions. Kinetic energy is either converted or transferred into other types of energy, thus decreasing. However, momentum is not conserved in this experiment. This might be because of factors like friction, impulse, and human error. Discussion: There might be many factors that affect the experiment. One of the factors is friction. In the real-world scenarios, there will always be friction that will occur when there is movement. Thus, we cannot really neglect the fact that friction will always occur in our experiment. The other factor is the use of initial velocity to calculate the momentum. However, during the experiment we couldn't find the initial velocity, so we decided to use the average velocity instead, so the calculations might not be accurate. Another factor might be energy, the possibility of conserving all the energy might be difficult because there will always be energy conversion or transfer of energy, whether it is heat, sound, etc. Finally, there might be human errors while conducting the experiment that might impact the results.
