Essay about Phys 151

Submitted By Senou-Dah Tinde
Words: 1833
Pages: 8

Senou Dah tinde
Phys_151 L
Lab Report 6: Dynamics of Motion II.

Purpose: Tested the application of Newton’s laws of motion by study the accelerated motion applying different force on an object.

Resume: Used the fellow material: air track with accessory kit, air track cart with flag, slotted mass set,
2 infrared photogates, computer timer and interface, platform balance, string and paper clip to study the acceleration of an object under an applied force. record the result from each try of this experiment, an tell what happened the force applied to it increased or decreased.

Part I
Mass of the cart (0.1899 kg) + mass of the attach (plug­in = 0.0102 kg): 0.2001 kg
Mass of paper clip: 0.005 kg
Mass of knife edge: 0.01kg

Run #1
M 1 = 0.2506 kg

M 2 = 0.010 kg

t1 = 0.1854 s

t3 = P ulse = 0.4514 s

t2 = 0.1303 s

Δt1 = 7.1855 ∙ E − 3 s

Δt3 = 0.0111 s

Δt2 = 2.0763 ∙ E − 3 s

Run #2
M 1 = 0.2406 kg

M 2 = 0.020 kg

t1 = 0.1428 s

t3 = P ulse = 0.3391 s

Δt1 = 1.6357 ∙ E − 3 s

Δt3 = 2.3876 ∙ E − 3 s Δt2 = 4.1633 ∙ E − 4 s

t2 = 0.0963 s

Run #3
M 1 = 0.2306 kg

M 2 = 0.030 kg

t1 = 0.1159 s

t3 = P ulse = 0.2757 s t2 = 0.0787 s

Δt1 = 8.2630 ∙ E − 4 s Δt3 = 1.2318 ∙ E − 3 s Δt2 = 2.9155 ∙ E − 4 s

Run #4
M 1 = 0.2206 kg

M 2 = 0.040 kg

t1 = 0.1111 s

t3 = P ulse = 0.2546 s t2 = 0.0709 s

Δt1 = 1.5172 ∙ E − 3 s Δt3 = 2.0949 ∙ E − 3 s Δt2 = 5.7552 ∙ E − 4 s

Run #5
M 1 = 0.2106 kg

M 2 = 0.050 kg

t1 = 0.0952 s

t3 = P ulse = 0.2220 s t2 = 0.0626 s

Δt1 = 9.1068 ∙ E − 4 s Δt3 = 1.7539 ∙ E − 3 s Δt2 = 4.7889 ∙ E − 4 s

∂a
∂d

=

1−1 t2 t1 t3+ 1 t1+t2
2

(

)

∂a
∂t3

=

(

−d

1−1 t2 t1

)
2

(t3+ 1 (t1+t2))
2

a=

d−d t2 t1 t3+ 1 t1+t2
2

(

)

MT=M1+M2

∂a
∂t1

(

2

2

t1 t2(t3+ 1 (t1+t2))
2

Δa =

)

(

2

d − 1 t1 +t1t2 +t2t3+ 1 t2
2
2

=



2

∂a
( ∂d ∙ Δd)

2

+

(

∂a
∂t2

∂a
∂t1

2

) +(

∙ Δt1

∂a
∂t2

2

2

t2 t1(t3+ 1 (t1+t2))
2
2

) +(

∙ Δt2

)

2

d − 1 t1 −t1t2 −t1t3+ 1 t2
2
2

=

∂a
∂t3

2

2

)

∙ Δt3

Table 1
− t1

t1t2

t1t3

t2t3

2
1
2 t1

2
1
2 t2

t1 t2

t1t2

1

2.2809

0.02416

0.08369

0.05882

0.01719

0.00849

0.00448

0.00315

2

3.3814

0.01375

0.04842

0.03266

0.01020

0.00464

0.00196

0.00132

3

4.0784

0.00912

0.03195

0.02170

0.00672

0.00310

0.00106

0.00072

4

5.1035

0.00788

0.02829

0.01805

0.00617

0.00251

0.00088

0.00056

5

5.4702

0.00596

0.02113

0.00453

0.00453

0.00196

0.00057

0.00037

1 t2 Run #

1

2

2

Table 2

(

)

[

(

2

)]

2

[

(

[

2

(

Run #

t3 + 1 t1 + t2
2

1

0.60925

0.37119

0.00166

0.00117

2

0.45865

0.21036

0.00041

0.00028

3

0.37300

0.13913

0.00015

0.00010

t3 + 1 t1 + t2
2

t1 t2 t3 + 1 t1 + t2 t1t2 t3 + 1 t1 + t2
2
2

4

0.34560

0.11944

0.00011

0.00007

5

0.30090

0.09054

0.00005

0.00003

Table 3
2

Run #

2

d(− 1 t1 + t1t2 + t2t3 + 1 t2 d(− 1 t1 − t1t2 − t1t3 + 1 t2
2
2
2
2

−d

(

1 t2 )

− t1
1

1

0.00992

­ 0.01166

­ 0.22809

2

0.00409

­ 0.00677

­ 0.33814

3

0.00272

­ 0.00447

­ 0.40784

4

0.00223

­ 0.00398

­ 0.51035

5

0.00173

­ 0.00297

­ 0.54702

Table 4
Run #

∂a
∂d

∂a
∂t1

∂a
∂t2

∂a
∂t3

Δa

1

3.7438

5.9625

­ 9.9684

­ 0.61448

0.09

2

7.3725

9.9078

­ 24.398

­ 1.6074

0.15

3

10.934

18.453

­ 44.645

­ 2.9314

0.22

4

14.767

21.189

­ 59.626

­ 4.2729

0.30

5

18.179

33.508

­ 88.802

­ 6.0417

0.37

Table 5
Run #

M1 (kg)

M2 (kg)

MT=

M2g (N)

a (m/s2)

Δa (m/s2)

M1+M2
1

0.2506

0.010

0.2606…