Conservation of energy and momentum Lab PDF

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Energy conservation and momentum conservation laws are particularly useful for solving mechanical problems and also arise in other fields of physics, such as electromagnetism, thermodynamics, and optics. You will discuss these laws in this experiment across a number of examples where non-conservativ...

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Falling balls (15 pts) Calculation 1a: (1 pts) Bocce ball: • 3.0 m: (0.77 + 0.78 + 0.76) = 0.770s • 5.0 m: (1.00 + 1.02 + 0.98) = 1.000s • 7.5 m: (1.22 + 1.26 + 1.22) = 1.233s • 10.0 m: (1.39 + 1.40 + 1.43) = 1.407s Lead ball: • 3.0 m: (0.78 + 0.78 + 0.77) = 0.777s • 5.0 m: (1.01 + 1.00 + 0.99) = 1.000s • 7.5 m: (1.24 + 1.26 + 1.22) = 1.240s • 10.0 m: (1.44 + 1.43 + 1.43) = 1.433s Calculation 1b: (1 pts) 2ℎ 𝑉𝑎𝑣𝑔 = 𝑡 Bocce ball: • • • •

3.0 m: 𝑉𝑎𝑣𝑔 =

5.0 m: 𝑉𝑎𝑣𝑔 =

7.5 m: 𝑉𝑎𝑣𝑔 =

• • •

0.770 2(5.0)

1.000 2(7.5)

3.0 m: 𝑉𝑎𝑣𝑔 =

5.0 m: 𝑉𝑎𝑣𝑔 =

7.5 m: 𝑉𝑎𝑣𝑔 =

= 7.792 m/s

= 10.000 m/s

= 12.165 m/s

1.233 2(10.0)

10.0 m: 𝑉𝑎𝑣𝑔 =

Lead ball: •

2(3.0)

1.407

2(3.0)

0.777 2(5.0)

1.000 2(7.5)

= 7.722 m/s

= 10.000 m/s

= 12.097 m/s

1.240 2(10.0)

10.0 m: 𝑉𝑎𝑣𝑔 =

= 14.215 m/s

1.433

= 13.957 m/s

Table 1: (2 pts) Type of Item and its mass

Bocce ball (0.95 ± 0.05) kg

Lead ball (5.40 ± 0.10) kg

Graph 1: (3 pts)

v2 (m2/s2)

0.770

Average Maximum Speed (m/s) 7.792

0.98

1.000

10.000

100.000

1.26

1.22

1.233

12.165

147.987

1.39

1.40

1.43

1.407

14.215

202.066

3.0

0.78

0.78

0.77

0.777

7.722

59.629

5.0

1.01

1.00

0.99

1.000

10.000

100.000

7.5

1.24

1.26

1.22

1.240

12.097

146.337

10.0

1.44

1.43

1.43

1.433

13.957

194.798

Height (m)

Time 1 (s)

Time 2 (s)

Time 3 (s)

Average Time (s)

3.0

0.77

0.78

0.76

5.0

1.00

1.02

7.5

1.22

10.0

60.715

Calculation 1c: (1 pts) Using the slope of the linear regressions for Graph 1, calculate the experimental value for the acceleration due to gravity exp (and its uncertainty) for both the bocce ball and lead ball data. Bocce Ball • Slope = 20.10 • Acceleration due to gravity = 20.10/2 = 10.05 m/s2 • Uncertainty Calculation (calculated only for one value) ∆g = |g|√22

∆t2 t

+ 12

∆h2 h

→ ∆g = |10.05|√4

0.0062

0.770

0.1 2

+ 3.0 → ∆g = 0.37

• g = (10.05 ± 0.37) m/s2 Lead Ball • Slope = 19.22 • Acceleration due to gravity = 19.22/2 = 9.61 m/s2 • Uncertainty Calculation (calculated only for one value) ∆g = |g|√22 •

∆t2 t

+ 12

∆h2 h

0.006 2

0.1 2

→ ∆g = |9.61|√4 1.000 + 5.0 → ∆g = 0.22

g = (9.61 ± 0.22) m/s2

Question 1a: (2 pts) Compare the exp values obtained in Calc. 1c to = 9.8 m/s2. Which ball’s data provides a more accurate value? NB. You have used different methods in the past to compare experimental values to accepted/ theoretical values so we expect you to perform a similar analysis as you’ve done before. % 𝑒𝑟𝑟𝑜𝑟 = Bocce ball

|𝑞𝑎𝑐𝑐 − 𝑞𝑒𝑥𝑝 | 𝑥 100% |𝑞𝑎𝑐𝑐 |

|9.8 − 10.05| 𝑥 100% |9.8| % 𝑒𝑟𝑟𝑜𝑟 = 2.6% % 𝑒𝑟𝑟𝑜𝑟 =

Lead ball

|9.8 − 9.61| 𝑥 100% |9.8| % 𝑒𝑟𝑟𝑜𝑟 = 1.9% % 𝑒𝑟𝑟𝑜𝑟 =

The lead ball provides a more accurate value which is seen because the percent error for lead ball is 1.9% while bocce ball has 2.6% which is a higher error of percent.

Calculation 1d: (2 pts) Maximum Kinetic Energy 1

1

K = 2 𝑚𝑣 2 → K = 2 (5.40)(12.097)2 → K = 395.11 J Uncertainty of Average Time (calculated using excel)

Standard error for time at 7.5m. uncertainty= ±0.01s (Given) Standard error for initial height= ±0.1m Uncertainty of Average Max Speed ∆h 2

∆v = |v|√ h +

∆t2 t

→ ∆v = |12.097|√

0.102 7.5

+

0.01 2 1.24

→ ∆v = 0.189

Uncertainty of Kinetic Energy ∆K = |K|√𝐴2

∆v2 v

+ 𝐵2

∆m 2 m

0.189 2

→ ∆K = |395.11|√(2)2 12.097 + (1)2

0.10 2 5.4

→ ∆K = 14.35

K= (395.11 ± 14.35) J Calculation 1e: (2 pts) Initial Potential Energy Egp = mgh → Egp = mgh → Egp = (5.4)(9.8)(7.5) → Egp = 396.9 J m2

∆𝐸𝑔𝑝 = |𝐸𝑔𝑝 |√

m

+

∆h 2 h

0.10 2

→ ∆K = |396.9|√ 5.4 +

0.10 2 7.5

→ ∆𝐸𝑔𝑝 =9.1

Egp = (396.9 ± 9.1) J Question 1b: (1 pts) We can see that the values are fairly similar (395.11 J and 396.J) and from the work energy theorem we know that the net K-U=0 that is K=U for the conservation of energy to be true. We see that we have obtained very similar values for the kinetic and potential energy within experimental limits. This proves that energy is indeed conserved, but we can expect some discrepancy however due to the presence of friction forces (air drag).

Collisions in one dimension (13 pts)

There will not be any uncertainty calculations associated with Part 2 of the lab report so you do not need to show the uncertainty values on your graph’s linear regression parameters. Graph 2a: (3 pts)

Calculation 2a: (2 pts) Momentum Before the Collision pi = p1i + p2i → pi = m1 v1i + m2 v2i → pi = (0.1914)(0.2352) + (0.1919)(−0.0005740) pi = 0.044907129 pi = 0.0449 kg · m/s Momentum After the Collision pf = p1f + p2f → pf = m1 v1f + m2 v2f → pf = (0.1914)(0.004768) + (0.1919)(0.2307) pf = 0.045183925 pf = 0.0452 kg · m/s

Calculating % Change % change = % change =

|pi −pf | x pi

100%

|0.0449 − 0.0452| x 100% 0.0449

% 𝑐ℎ𝑎𝑛𝑔𝑒 = 0.67% As based on the theory of conservation of momentum, which states that total momentum before a collision is equal to total momentum after a collision without any external forces present in the system for both an elastic or inelastic collision, it is expected that the momentum be conserved in both elastic and inelastic collisions. The % change is very small (0.67%) so I believe that momentum is conserved which agrees with my expectations. Calculation 2b: (2 pts) Again using the data from Graph 2a, calculate the kinetic energy of the two gliders before and after the elastic collision. Compare the two kinetic energy values using the same “% change” calculation you used above. Does your result agree with your expectation based on conservation of energy?

Calculating KE before the collision 1 1 Ekin = m1 v1 2 + m2 v2 2 2 2 1 1 Ekin = (0.1914)(0.2352)2 + (0.1919)(−0.0005740)2 2 2 Ekin = 0.00529 J Calculating KE after the collision 1 1 Ekin = m1 v1 2 + m2 v2 2 2 2 1 1 Ekin = (0.1914)(0.004768)2 + (0.1919)(0. 2307)2 2 2 Ekin = 0.00511 J Calculating % Change % change = % change =

|Ekin(intial) −Ekin(final) | Ekin(initial)

x 100%

|0.00529 − 0.00511| x 100% 0.00529

% change = 3.40%

The % change is small (3.40%) so I believe that energy is conserved. An elastic collision is one in which the internal kinetic energy does not change (it is conserved) into other forms of energy due to internal friction. This does agree with my expectations for an elastic collision because I thought that kinetic energy would be conserved and the total kinetic energy of the objects before the collision remains in the form of kinetic energy afterwards. .

Graph 2b: (3 pts)

Calculation 2c: (3 pts) Momentum Before the Collision pi = p1i + p2i → pi = m1 v1i + m2 v2i → pi = (0.1914)(0.2039) + (0.1920)(0.00009871) pi = 0.039045412 pi = 0.0390 kg · m/s Momentum After the Collision pf = p1f + p2f → pf = m1 v1f + m2 v2f → pf = (0.1914)(0.08636) + (0.1920)(0.08767) pf = 0.033361944 pf = 0.0334 kg · m/s Calculating % Change % change = % change =

|pi −pf | pi

x 100%

|0.0390 − 0.0334| x 100% 0.0390

% change = 14.4% As based on the theory of conservation of momentum, which states that total momentum before a collision is equal to total momentum after a collision without any external forces present in the system for both an elastic or inelastic collision, it is expected that the momentum be conserved in both elastic and inelastic collisions. The % change is not that large (14.4%) in comparison to the % change for the kinetic energy in an inelastic collision (63.6%) so I believe that momentum is conserved but not perfectly conserved due to experimental errors. This mostly agrees with my expectations because I expected momentum to be conserved in an inelastic collision and in this case, it was for the most part and would have been completely conserved if not for the experimental errors. Potential errors may include human errors, material errors with the tools, measuring errors or the presence of unforeseen external forces within the system for instance uneven surface angle or higher friction in portions of the track. Calculating KE before the collision 1 1 Ekin = m1 v1 2 + m2 v2 2 2 2 1 1 Ekin = (0.1914)(0.2039)2 + (0.1920)(0.00009871)2 2 2 Ekin = 0.00398 J

Calculating KE after the collision 1 1 Ekin = m1 v1 2 + m2 v2 2 2 2 1 1 Ekin = (0.1914)(0.08636)2 + (0.1920)(0.08767)2 2 2 Ekin = 0.00145 J Calculating % Change % change = % change =

|Ekin(intial) −Ekin(final) | Ekin(initial)

x 100%

|0.00398 − 0.00145| x 100% 0.00398

% change = 63.6% The % change is very large (63.6%) so I believe that energy is not conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved) into other forms of energy due to internal friction. This does agree with my expectations for an inelastic collision because I thought that kinetic energy would not be conserved the total kinetic energy of the objects before the collision does not remain in the form of just kinetic energy afterwards....