Collision Lab Report 2 PDF

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Boom, Collisions

Lab partner: PHYS181-007

18 March 2020, 3:00:00PM

Abstract The goal of the experiment was to demonstrate both elastic and inelastic collisions of carts by using different methods of conservation of linear momentum, which can show the loss or gain of energy in the system. Introduction Newton’s 3rd Law states that for every action there exists an equally opposite reaction, meaning momentum (P) is conserved. Momentum is also equal to mass time velocity (P=m*v). Transfer of energy is applicable to momentum, meaning if Object1 collides with Object2, the momentum from 1 transfers to 2. Kinetic Energy (KE) takes place in the lab because it accelerates an object from rest to maximum velocity. Potential Energy (PE) is also the energy of the system of an object dependent on its position. In the experiment, Elastic and Inelastic. Elastic occurs when no energy loss occurs in the system during collision, and Inelastic is when energy transfer does occur. Figure1 shows the schematic for the track and carts. Since linear momentum is a vector, the net force of the system is zero so the momentum is constant. The lab examines both elastic and inelastic collisions and explosion. Figure1. Schematic for Track

Theory

The Linear momentum for an object with mass (m) moving with a velocity (v) is expressed as: 𝑝 = 𝑚𝑣

(1)

𝑝: linear momentum m: mass of object (g) 𝑣: 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (m/s) If a change in linear momentum occurs, the change in linear momentum with respect to time is given by the net force of that object. Net force is zero and momentum is constant 󰇍󰇍 ∆𝑝 ∆𝑡

𝑚 𝑎: 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 ( 2 ) 𝑠 𝐹𝑛𝑒𝑡 : net force acting on an object (N) ∆𝑡: change in time (s)

󰇍󰇍𝑣 = 𝑚 𝑎 = 󰇍󰇍󰇍󰇍󰇍󰇍󰇍 𝐹𝑛𝑒𝑡 = 𝑚 ∆𝑡

(2)

Initial momentum of carts with elastic collision is Equation1. Cart 2 is initially at rest, so initial momentum of the system before collision is expressed as: pi=m1v1,i+m2v2,i=m1v1,i

(3)

m1: mass of cart 1 (g) m2: mass of cart 2 (g) v1,i: initial velocity of cart 1 (m/s) v2,i: initial velocity of cart 2 (m/s) After the collision, the total momentum of the carts can be found by using Equation3 Pf=m1v1,f+m2v2,f

(4)

v1,f: final velocity of cart 1 (m/s) v2,f: final velocity of cart 2 (m/s) Calculating KE before and after a collision. Since the second cart is at rest, the first moving before the collision, so the initial kinetic energy is: 1

(5) KEi=2m1v21,i Final KE is calculated the same as the equation in 5. Final KE should be equal to the initial kinetic energy because this is an elastic collision 1

KEf= m1v21,f+ m2v22,f 2

(6)

Second lab is creating an inelastic collision between a stationary cart and a moving one. The final momentum is a change: 𝑣1,𝑓 =𝑣2,𝑓 = 𝑣𝑓

(7)

Given Equation7, the final momentum is expressed: 𝑝𝑓 = (𝑚1 + 𝑚2 )𝑣𝑓

(8)

Simplifying to: 1

𝐾𝐸𝑓 = (𝑚1 + 𝑚2 )𝑣𝑓2

(9)

2

Final KE is supposed to be less than the initial KE since this setup is modeled for an inelastic collision. No external forces are applied to the system, meaning momentum is constant. The final part of the lab considers momentum when both carts are moving. There exists initial velocity and initial momentum: pi=m1v1,i+m2v2,i

(10)

KE also has initial values of: 1

KEi= m1v21,i+ m2v22,i 2

(11)

Procedure Part 1: 1-Dimensional Elastic Collision 1: set track according to Figure1 2: make sure Cart1’s plunger is against wall so it pushes the cart and gives high initial velocity 3: position Cart2 near middle of track to remain stationary 4: Use PASCO Capstone on linear velocity 5: Vary masses on the cart 6: Trigger the plunger so Cart1 collides with Cart2 (record velocities (3x) and track in Excel sheet) 7: Calculate initial and final momentum of KE 8: Add 50g mass to Cart2 and run 3 trials for initial and final velocities, repeat on Excel 9: Add 50g mass to Cart1 (repeat step 8 with Cart1) Part 2: 1-Dimensional Inelastic Collision 1: set track according to Figure1, except Velcro sides facing each other 2: make sure Cart1’s plunger is against wall on left side of track 3: position Cart2 near middle of track to remain stationary 4: Trigger the plunger so Cart1 collides with Cart2 (record initial and final velocities and enter data in Excel file, and repeat for 3 trials) 5: Add 50g to Cart2 (collect for 3 trials and enter data in Excel) 6: Add 50g to Cart1 (collect for 3 trials and enter data in Excel

Part 3: 1: set track according to Figure1 except Cart1’s plunger is facing against Cart2 in center of track 2: Trigger plunger so carts disperse against each other and enter data into Excel, for 3 trials 3: Add 50g to Cart2 and repeat procedure for 3 trials and enter data into Excel Part 4: 1: set track according to Figure1, but position carts so magnets are facing each other towards middle (each track on opposite sides of track) 2: Push carts towards the center (with equal force) (record and enter data in Excel) 3: Add 50g to Cart2 and repeat procedure, entering data in Excel 4: Add 50g to Cart1 and repeat procedure, entering data in Excel 5: Repeat measurements, but push both carts in the same direction (Cart1 on left side of track and Cart2 in center, meaning Cart1 needs a higher velocity than Cart2 so they collide) 6: Repeat 3 collisions with additional 50g mass on Cart2, and additional 50g mass on Cart1 7: Repeat all measurements under inelastic conditions Sample Calculation/ Results Part 1 consisted of carts undergoing elastic collisions, meaning total momentum and KE were conserved. Masses varied between the moving and stationary carts and their momentum and KE were conserved. Calculation for momentum and KE for 2 massless carts: Equation3 pi= (0.26g) (1.118m/s) +(0.26g)(0 m/s)=0.2907 Post collision, total momentum of the system is calculated from Equation4 Pf= (0.26g) (0.142 m/s)+(0.26g)(0.967 m/s)=0.2883 Solving for KE, Equation5: KEi= (0.26g) (1.1183 )2=0.1629 2 s 1

m

Final KE is calculated with Equation6: 1

Discussion

KEf= (0.26g) (1.42 m/s)2+ (0.26g) (0.967 m/s)2=0.1246 2

This lab uses concepts of linear momentum and vector’s in relation to motion. When 2 objects collide (carts) and a force occurs. Newton’s 3rd Law states that if an Object1 exerts force on Object2, an equal force occurs from Object2 onto Object1, so the net force is 0. Total momentum in the system has to remain constant. The energy in the system has to remain constant but the KE is subject to change. If it remains constant, its Elastic. If it decreases, its inelastic.

Part 1: this part of the experiment is elastic collision, meaning momentum and KE are conserved. Initial and final values are equal (or should be). Since momentum remains constant, the change of p to determine how much momentum was conserved. Table1 shows across varying masses momentum conservation because initial and final momentum are equal. Part 2: this part of the experiment is inelastic collision, meaning final KE is (or should be) less than initial KE. Momentum should remain constant since no external forces are acting on the system. Table2 displays the final KE is lower than initial, so a decrease in KE shows inelastic collision. Part 3: this part of the experiment is both carts starting at rest, so only final values are recorded. Table3 shows momentum conserved in the system, so both carts are from rest, so the momentum is 0. After explosion, momentum of the system must still be 0 because the vector sum is 0. Both carts travel in opposite directions. Part 4: this part of the experiment shows both carts moving. In the first part, it focused on collisions towards each other and the second part showed a collision from behind. Both measures were recorded in elastic and inelastic collisions. Table4 shows inelastic, since there is a decrease in KE across all masses. Values from elastic show energy is conserved because the initial and final values are equal. Tables Table 1: Elastic Collision with Stationary Cart M1 (g)

M2 (g)

0.26 0.26 0.26 0.26 0.26 0.26 0.76 0.76 0.76

0.26 0.26 0.26 0.76 0.76 0.76 0.76 0.76 0.76

V1i (m/s) 1.07 1.10 1.185 1.102 1.149 1.05 0.712 0.703 0.672

V1f (m/s) 0.131 0.14 0.156 -0.176 -0.316 -0.077 0.179 0.243 0.211

V2f (m/s) 0.933 0.95 1.019 0.46 0.461 0.396 0.527 0.456 0.470

P1i

P1f

P2f

0.2907

0.037

0.251

0.2861

0.0337

0.5287

0.1604

∆𝒑

∆𝑲𝑬

KEi

KEf

0.0023

0.1629

0.1247

0.0384

0.3336

0.0138

0.1576

0.0759

0.0816

0.2490

0.1193

0.1840

0.07869

0.1053

Table 2: Inelastic Collision with Stationary Cart M1 (g)

M2 (g)

0.26 0.26 0.26 0.26 0.26 0.26 0.76 0.76 0.76

0.26 0.26 0.26 0.76 0.76 0.76 0.76 0.76 0.76

V1i (m/s) 1.278 1.173 1.248 1.015 1.17 1.158 0.678 0.722 0.739

V1f (m/s) 0.61 0.557 0.589 0.251 0.287 0.283 0.343 0.352 0.353

V2f (m/s) 0.601 0.547 0.593 0.238 0.276 0.286 0.34 0.352 0.343

P1i

P1f

P2f

0.3214

0.1526

0.0995

0.2897

0.0712

0.5419

0.2655

∆𝒑

∆𝑲𝑬

KEi

KEf

0.0693

0.1984

0.0733

0.1251

0.1302

0.0884

0.1621

0.02659

0.1355

0.2622

0.0142

0.1934

0.0916

0.1018

Table 3: Explosion ∆𝒑

M1 (g)

M2 (g)

V1f (m/s)

V2f (m/s)

P1f

P2f

0.26 0.26 0.26 0.26 0.26 0.26

0.26 0.26 0.26 0.76 0.76 0.76

-0.312 -0.327 -0.269 -0.363 -0.322 -0.312

0.584 0.464 0.471 0.002 0.019 0.011

-0.0789

0.13165

0.0528

-0.0873

0.00811

-0.0792

Table 4: Collision with Two Moving Carts

Elastic moving opposite directions

Elastic moving same direction

Inelastic moving in opposite directions

Inelastic moving in same direction

M1 M2 (g) (g) 0.26 0.26

V1i (m/s) 0.17

V2i (m/s) 0.219

V2f (m/s) 0.168

P1i

p2i

P1f

P2f

0.0442

0.0569

-0.043

0.0437

0.051

0.0567

0.1368

0.0133

0.278

V1f (m/s) 0.167 0.305 -0.30

0.26 0.76

0.218

0.18

0.76 0.76

0.314

0.273

0.2386

0.2873

0.26 0.26

1.048

0.562

0.601

0.334

0.2725

0.26 0.76

0.871

0.544

0.106

0.15

0.76 0.76

0.602

0.586

0.154

0.26 0.26

0.396

0.065

0.26 0.76

0.456

0.76 0.76

0.334

0.26 0.26

0.555

0.305 0.407 0.137

0.26 0.76

0.705

0.76 0.76

0.638

∆𝒑

KEi

KEf

-0.101

0.001

0.0073

0.1368

-0.0434

0.0185

0.0131

-0.228

0.2075

-0.5464

0.0918

0.0625

0.1461

0.1563

0.0868

-0.1755

0.1838

0.0615

0.2265

0.0806

0.2265

0.114

0.0334

0.1029

0.0101

0.118

0.4575

0.4454

0.117

0.0897

-0.6962

0.2682

0.0143

0.285 0.054 0.013

0.065

0.103

0.0169

-0.074

0.0169

-0.1771

0.0209

0.0111

0.058 0.012

0.1186

-0.014

0.0522

0.0638

0.0017

0.0099

0.0441 0.0091

0.0745

0.1054

0.0001

0.554

0.13

0.1443

0.2316 0.3093 0.0356

0.1440

0.0338

-0.0021

0.0425

0.0421

0.11

0.267

0.266

0.1833

0.0836

0.0694

0.2022

0.0047

0.0691

0.0362

0.431

0.025

0.022

0.4849

0.3276

0.019

0.0167

0.77667

0.2252

0.004

0.2538

∆𝑲𝑬 0.0027 0.0054 0.0292 0.1224 0.0929 0.2539 0.0098 0.0607 0.1052 0.0004 0.0031 0.2248

Questions 1: Momentum conservation looks at the change of p for each part. Part1: momentum was conserved due to initial and final values are similar. Part2: energy conservation because of the decrease in KE final (supports inelastic theory). Part3: conserved because carts travel in equal vectors along opposite x direction, so the net force is 0. Part4: inelastic show decrease in KE final for elastic collisions and the final and initial values are equal. 2: Energy is conserved under elastic conditions if initial and final KE are equal. Table1 shows that Part1 conserved energy, same as Table4 for Part4. 3: Elastic collision occurs if energy is conserved in both momentum and KE. If both carts are equal during elastic collision, they release the same momentum and velocity (theoretically). The light cart should have a greater momentum because of greater velocity; therefore, the heavier cart will have a smaller velocity and momentum. 4: I think the energy from Part2 (inelastic) was lost due to the interaction of Velcro on the sides of the cart. Part4’s magnets played a role in the loss of energy. 5: Energy during explosion comes from energy released from plunger when it is launched from Cart1. 6: Possible error sources include inaccurate initial velocities, since the cart was moving slightly from the beginning, or pushing the carts at the same velocity in Part3 against each other. The experiment relied heavily on approximation....