Laboratory 3
Forces
Michael Akopov
PH1004 Section E2
Abstract
The objective of this lab was to explore and analyze general collisions between two
items. A In this experiment, an air track was used to investigate the conservation of momentum
and kinetic energy for elastic collisions, and the conservation of momentum for inelastic
collisions. A collision is an isolated event in which each colliding particle exerts a force on the
other particle for a relatively short period of time. Collisions happen every time one object comes
into contact with another and physical energy is altered. Collisions can be grouped into elastic
collisions, inelastic collisions, and perfectly inelastic collisions. Elastic collisions are such
collisions that the kinetic energy and the momentum of the system are conserved. Inelastic
collisions are such collisions where only momentum is conserved. A perfectly inelastic collisions
is one after which the objects stick together and move with one velocity as one unit.
The two types of collisions observed in this experiment were elastic and perfectly
inelastic collisions. In the first part we had to see if the collisions that happened were really
elastic. For the elastic collision we used mathematics to prove that the kinetic energy and
momentum were conserved. For the perfectly inelastic collision we had to show that there was
no conservation in the kinetic energy but only momentum conservation. For part A and part B,
the results proved the above implication, and the experiment was thereby successful.
Data
Variables
Symbol
xb1
xub1
xb2
xub2
x

Description
Position of blocked photogate for flag 1
Position of unblocked photogate for flag 1
Position of blocked photogate for flag 2
Position of unblocked photogate for flag 2
Distance measurement uncertainty
Measurement tool
Ruler
Ruler
Ruler
Ruler
Ruler
m1
m2
m1
m2
Glider1 Mass
Glider 2 Mass
Glider 1 Mass measurement uncertainty
Glider 2 Mass measurement uncertainty
tb1i
tb1f
tb2f
Initial time when Glider 1 passes photogate 1
Final time when Glider 1 passes photogate 2
Final time when Glider 2 passes photogate 3
Final time when Glider 1 + 2 pass photogate 2 (B)
Time measurement uncertainty
tb(1&2)f
t
xf1
xf2
xf1
xf2
Digital Scale
Digital Scale
Digital Scale
Digital Scale
Computer Clock
Computer Clock
Computer Clock
Computer Clock
Given (0.001s)
Flag 1 width
Flag 2 width
Flag 1 Width measurement uncertainty
Flag 2 Width measurement uncertainty
Ruler
Ruler
Ruler
Ruler
Part A
Table 1: Data from part a of the experiment
Trial
tb1i
1
0.068 s
2
0.068 s
3
0.070 s
4
0.068 s
5
0.069 s
6
0.071 s
7
0.069 s
8
0.069 s
9
0.070s
10
0.069s
tb1f
0.244 s
0.228 s
0.240 s
0.230 s
0.235 s
0.224 s
0.220 s
0.191 s
0.247 s
0.221 s
tb2f
0.057 s
0.057 s
0.057 s
0.056 s
0.056 s
0.058 s
0.057 s
0.059 s
0.058 s
0.057 s
xb1
xub1
xb2
xub2
x
m1
m2
m
t
50.0 cm
52.5 cm
50.0 cm
52.5 cm
0.05 cm
406 g
205 g
0.05 g
0.001 s
Table 2: flag measurement and instrumental uncertainty
xb1
50.0 cm
xub2
52.5 cm
xf1
2.5cm
xf1
0.05 cm
xb2
50.0 cm
xub2
52.5 cm
xf2
2.5cm
Part B
Table 3: Data from part b of the experiment
Trial
1
2
3
4
5
6
7
8
9
10
tb1i
0.069 s
0.069 s
0.070 s
0.069 s
0.069 s
0.070 s
0.070 s
0.069 s
0.069 s
0.069 s
tb(1&2)f
0.117 s
0.114 s
0.111 s
0.111 s
0.111 s
0.113 s
0.112 s
0.113 s
0.116 s
0.119 s
m1
m2
m1
m2
t
406 g
205 g
0.05 g
0.05 g
0.001 s
xf2
0.05 cm
Analysis
Part A:
1. The width of the flags was determined by subtracting the Block from the Unblock of
the measurement taken to determine the width shown in table 2.
xb1 + xub1 = xf1 = 50.0cm = 52.5cm = 2.5 cm
xb2 + xub2 = xf2 = 50.0cm = 52.5cm = 2.5 cm
xf1 = 0.05 cm
xf2 = 0.05 cm
2. To calculate the best estimates for the times it took the flag to pass at each gate the
average of the 10 trials were taken using Excel. The averages were calculated to be as follows:
tb1i = 0.069 s
tb1f = 0.228 s
tb2f = 0.057 s
The standard deviation of the times was calculated in Excel to be as follows:
σ tb1i = 0.000994s
σ tb1f = 0.016028s
σ tb2f = 0.000919s
To find ∆tb the following formula was used
tb   2  t ins
2
We know that ∆tins = 0.001s
t b1i  0.0009942  0.0012 = 0.001410 s
t b1 f  0.0160282  0.0012 = 0.016059 s
t b 2 f  0.0009192  0.0012 = 0.001358 s
3. The velocity for each glider passing in the appropriate gate was calculated according to
w
the formula: vinst  x .
t
2.5
cm
 36.23
v1i =
0.069
s
2.5
cm
 10.96
v1f =
0.228
s
2.5
cm
 43.86
v2f =
0.057
s
4. The uncertainty in velocity is given by the formula:
v 
w
1
v
v
tb 
w = 2 tb +
∆w.
tb
tb
w
tb
and calculated to be the following:
cm
2.5
1
0.001410 
0.1 = 2.1897
2
0.069
0.069
s
cm
2.5
1
0.016059 
0.1 = 1.2109
∆v1f =
2
0.228
0.228
s
cm
2.5
1
0.001358 
0.1 = 2.7993
∆v2f =
2
0.057
0.057
s
∆v1i =
5. The initial and final momentums were calculated according to the formulas:
g  cm
cm
= 14709
s
s
g  cm
cm
cm
Pf  m1v1 f  m2 v 2 f = 406 g * 10.96
+ 205 g * 43.86
= 13441
s
s
s
6. The uncertainty in the initial momentum was calculated by the error propagation rule:
Pi  m1v1i = 406 g * 36.23
pi 
pi
p
m1  i v1i = v1i  m1  m1  v1i
m1
v1i
And calculated to obtain the following result:
g  cm
.
s
The uncertainty in the final momentum was calculated by the error propagation rule:
pi = 36.23 0.5  406  2.1897 = 907
p f 
p f
m1
m1 
p f
v1 f
v1 f 
p f
m2
m2 
p f
v2 f
v2 f = v1 f m1  m1v1 f  v2 f m2  m2 v2 f
And calculated to be the following:
p f = 10.96 * 0.5 + 406 * 1.2109 + 43.86 * 0.5 + 205 * 2.7993 = 1093
g  cm
s
Fractional uncertainties were found to be as follows:
∆ Pi / Pi = 907 / 14709 = 0.06
∆ Pf / Pf = 1093 / 13441 = 0.08
7. see conclusion
8. The Initial and final kinetic energies were calculated according to the following
formulas:
m1v12i 406  36.23 2
g  cm 2
=
= 266460
2
2
s2
m2 v22 f
g  cm 2
405 10.96 2 205  43.86 2
=
= 221503
.


2
2
2
s2
Ki 
Kf 
m1v12f
2
9. The uncertainties in the initial kinetic energy were xcalculated using the error
propagation rule:
2
v1i
K i
K i
 m1  m1v1i  v1i
K i 
m1 
v1i =
2
m1
v1i
and was calculated to obtain the following result:
g  cm 2
K i = 656 * 0.5 + 406 * 36.23 * 2.1897 = 32537
s2
The uncertainties in the final kinetic energy was calculated using the error propagation rule:
K f
K f
Kf
Kf
K f 
m1 
v1 f 
m2 
v2 f =
m1
v1 f
m2
v2 f
=
v1 f
2
2
 m1  m1v1 f  v1 f 
v2 f
2
2
 m2  m2 v2 f  v2 f
K f = 30.03 + 5388 + 480 + 28147 = 30694
Fractional uncertainties were found to be as follows:
∆ K i / K i = 32537 / 266460 = 0.12
∆ K f / K f = 30694 / 221503 = 0.14
10. see conclusion
g  cm 2
s2
Part B
1. To calculate the best estimates for the times it took the flag to pass at each gate the
average of the 10 trials were taken using Excel. The averages were calculated to be as follows:
tb1i = 0.0693 s
tb(1&2)f = 0.1137 s
The standard deviation of the times was calculated in Excel to be as follows:
σ tb1i = 0.000483s
σ tb(1&2)f = 0.002791s
To find ∆tb the following formula was used
t b   2  tins
2
We know that ∆tins = 0.001s
t b1i  0.0004832  0.0012 = 0.000493 s
t b(1&2) f  0.0027912  0.0012 = 0.002965 s
The velocity for the glider before the collision is given by v1 and after the collision by v2
was calculated according to the formula:
w
vinst  x .
t
And found to be the following:
2.5
cm
 36.08
v1 =
0.0693
s
2.5
cm
 21.99
v2 =
0.1137
s
The uncertainty in velocity is calculated using the formula derived in part A and found to be
1
w
∆v = 2 t b + ∆w.
tb
tb
cm
2.5
1
0.000493 
0.1 = 1.700
2
0.0693
s
0.0693
cm
2.5
1
0.002965 
0.1 = 1.453
∆v2 =
2
0.1137
0.1137
s
∆v1 =
2. The initial and final momentums were calculated according to the formulas:
g  cm
cm
= 14648
s
s
g  cm
cm
Pf  (m1  m2 )v 2 = (204+406) 21.99
= 13435
s
s
Pi  m1v1 = (406) 36.08
The uncertainty in the momentum was calculated by the error propagation rule derived in part A
and used to obtain the following results:
g  cm
.
pi  v1  m1  m1  v1 = 36.08  0.5  406 1.700 = 708
s
g  cm
p f  v 2 m1  v 2 m2  (m1  m2 )v 2 = 21.99  0.5  36.08  0.5  6111.453 = 917
s
Fractional uncertainties were found to be as follows:
∆ Pi / Pi = 708 / 14648= 0.048
∆ Pf / Pf = 917 / 13435= 0.068
3. The Initial and final kinetic energies were calculated according to the following
formulas:
m v 2 406  36.08 2
g  cm 2
= 264259
K i  1 1i =
2
2
s2
(m  m2 )v 22 611 21.99 2
g  cm 2
=
= 147728
.
Kf  1
2
2
s2
The Uncertainties in the initial and final kinetic energies were calculated using the error
propagation rule:
v1
g  cm 2
 m1  m1v1  v1i = 650.1 x 0.5 + 406 x 36.08 x 1.700 = 25227
2
s2
2
2
v2
v2
K f 
 m1 
 m2  (m1  m2 )v 2  v 2 =
2
2
g  cm 2
= 241.8 x 0.5 + 241.8 x 0.5 + 611 x 21.99 x 1.453= 19764
s2
2
K i 
Fractional uncertainties were found to be as follows:
∆ K i / K i = 25277/ 264259 = 0.10
∆ K f / K f = 19764 / 147728= 0.13
Results
Part A
Velocity
cm
cm
 2.1897
s
s
cm
cm
v1f = 10.96
 1.2109
s
s
cm
cm
v2f = 43.86
 2.7993
s
s
v1i = 36.23
Momentum
g  cm
g  cm
.
 907
s
s
g  cm
g  cm
P f  13441
 1093
s
s
Pi  14709
Fractional Uncertainties:
∆ Pi / Pi = 0.06
∆ Pf / Pf = 0.08
Kinetic Energy
g  cm 2
g  cm 2
K i = 266460
 32537
s2
s2
g  cm 2
g  cm 2
K f = 221503
30694

s2
s2
Fractional Uncertainties:
∆ K i / K i = 0.12
∆ K f / K f = 0.14
Part B
Velocity
cm
cm
 1.700
s
s
cm
cm
v2 = 21.99
 1.453
s
s
v1 = 36.08
Momentum
g  cm
g  cm
.
 708
s
s
g  cm
g  cm
Pf = 13435
 917
s
s
Pi = 14648
Fractional Uncertainties:
∆ Pi / Pi = 0.048
∆ Pf / Pf = 0.068
Kinetic Energy:
g  cm 2
g  cm 2
K i  264259
 25227
s2
s2
g  cm 2
g  cm 2
K f  147728
19764

s2
s2
Fractional Uncertainties:
∆ K i / K i = 0.10
∆ K f / K f = 0.13
Conclusion
For part A there was conservation of Momentum because Pi = 14709  907
Pf = 13441  1093
g  cm
. And
s
g  cm
. There is an intersection between the two values and therefore the
s
momentum is conserved. The kinetic energy is also conserved because K i = 266460  32537
g  cm 2
g  cm 2
K
and
=
221503
30694
. There is an intersection between the two values

f
s2
s2
and therefore kinetic energy is conserved. From both the conservation of momentum and kinetic
energy we come to the conclusion that the collisions in part A were elastic collisions.
For part B the Momentum was conserved because: Pi = 14648  708
13435  917
g  cm
and Pf =
s
g  cm
. There is an intersection between the two values and therefore the
s
momentum is conserved. The kinetic obtained was K i = 264259  25227
147728  19764
g  cm 2
and K f =
s2
g  cm 2
. There is no intersection between the two answer sets and therefore
s2
kinetic energy was not conserved. This proves that the collisions performed in part B are nonelastic and contrariwise.