Experiment 3
Forces
Aleksandr Shpiler
Section E2
Abstract
The objective of this lab was to analyze the forces which occur during collisions.
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 either kinetic or potential are 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 proved 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.
Data
Part A (General raw data)
Part A
xb1
xb2
m1
Delta x
Delta t
65.9
66.2
305.3
0.05
0.001
cm
cm
g
cm
s
xub1
xub2
m2
Delta m
68.3
68.6
204.8
0.05
cm
cm
g
g
0.683
0.686
0.2048
0.00005
m
m
kg
kg
General data converted into SI units.
Part A
xb1
xb2
m1
Delta x
Delta t
0.659
0.662
0.3053
0.0005
0.001
m
m
g
m
s
xub1
xub2
m2
Delta m
Part A Obtained Data
Trial
tb1i
tb1f
tb2f
1
2
3
4
5
6
7
8
9
10
0.043 s
0.045 s
0.043 s
0.047 s
0.047 s
0.045 s
0.045 s
0.047 s
0.047 s
0.047 s
0.266 s
0.264 s
0.256 s
0.259 s
0.248 s
0.256 s
0.258 s
0.264 s
0.254 s
0.256 s
0.041 s
0.041 s
0.041 s
0.041 s
0.041 s
0.041 s
0.041 s
0.041 s
0.041 s
0.041 s
Part B (General raw data)
Part B
m1
Delta m
305.5
0.05
g
g
m2
Delta t
204.8
0.001
g
s
0.2048
0.001
kg
s
General Data converted to SI units.
Part B
m1
Delta m
0.3055
0.00005
kg
kg
m2
Delta t
Obtained Data
Trial
1
2
3
4
5
6
7
8
9
10
tb(1i), s
0.073
0.074
0.075
0.072
0.073
0.073
0.072
0.072
0.072
0.072
tb(1&2 f), s
0.123
0.131
0.129
0.125
0.127
0.125
0.131
0.127
0.129
0.127
Analysis
1) Flag width
flag wid.1
flag wid.2
xub1 - xb1
xub2 - xb2
0.683 – 0.659 =
0.686 – 0.662 =
0.024 m
0.024 m
2) tbs, standard deviation, and uncertainties in tbs
w =
tb1i
xb1+ xb2 =
0.0005+
0.0005 =
0.001 m
0.0739s
σ tb1i
0.000994s
deltaTB1I
t b1i  0.000994 2  0.0012
0.00141s
tb1f
0.33s
σ tb1f
0.005598s
deltaTB1F
t b1 f  0.0055982  0.0012
0.005686s
tb2f
0.0644s
σ tb2f
0.000516s
deltaTB2F
t b 2 f  0.000516 2  0.0012
0.001125s
3) Velocities
v1i 
v1i =
w1
t b1i
0.024
m
 0.324763
0.0739
s
v1 f 
v1f =
w1
v2 f 
tb 2 f
0.024
m
 0.072727
0.33
s
v2f =
w2
tb2 f
0.024
m
 0.372671
0.0644
s
4)
v 
uncertainties in the velocity
w
1
v
v
tb 
w = 2 tb +
∆w
tb
tb
w
tb
 V=1/tb1i*  W1+W1/tb1i^2*  tb1i
deltaTB1I = Squareroot(uncert^2+.001^2)
dv1i
dv1f
dv2f
0.019729
0.004283469
0.022041
∆v1i =
0.024
1
m
0.00141 
0.001 = 0.019729
2
s
0.0739
0.0739
∆v1f =
0.024
1
m
0.005686 
0.001 = 0.00428347
2
0.33
s
0.33
∆v2f =
0.024
1
m
0.001125 
0.001 = 0.022041
2
0.0644
s
0.0644
5) Initial and Final momentum
Pi  m1v1i
Pf  m1v1 f  m2 v 2 f
0.09915
0.098527
Pi  m1v1i = 0.3053 kg * 0.324763
Pf  m1v1 f  m2 v 2 f = 0.3053 kg * 0.072727
m
kg  m
= 0.09915
s
s
m
m
kg  m
+ 0.2048 kg * 0.372671 = 0.098527
s
s
s
6) Uncertainty in the momentums
partial derivative of p
delta Pi=V1i*dM1+M1*dV1i
Delta Pf=V1f*dM1+M1*dVf1+V2f*dM2+M2*dV2f
delta Pi =
delta Pf =
0.006039642
0.005843968
Pi +/- dPi
Pf +/- dPf
(+)
0.10519
0.104371
(-)
0.093111
0.092683
pi 
pi
p
m1  i v1i = v1i  m1  m1  v1i
m1
v1i
pi = 0.324763  0.00005  0.3053  0.019729 = 0.006039642
p f 
p f
m1
m1 
p f
v1 f
v1 f 
p f
m2
m2 
p f
v 2 f
kg  m
s
v 2 f = v1 f m1  m1v1 f  v2 f m2  m2 v2 f
p f = 0.072727 * 0.00005 + 0.3053 * 0.004283469 + 0.372671 * 0.00005 + 0.2048 * 0.022041
= 0.005843968
kg  m
s
The results clearly indicate that momentum (p) was conserved during these collisions.
P
0.006039642
∆ Pi / Pi = ∆ i =
= 0.06091
0.09915
Pi
Pf
0.00584396 8
∆ Pf / Pf = ∆
=
= 0.0593134
0.098527
Pf
10) Kinetic Energy
Ki=1/2M1V1i^2
0.0161
Ki 
Kf 
m1v12f
2

m2 v22 f
2
=
Kf=1/2M1V1f^2+1/2M2V2f^2
0.015029
m1v12i 0.3053  0.324763 2
kg  m 2
=
= 0.0161
2
2
s2
0.3053  0.072727 2 0.2048  0.3726712
kg  m 2
= 0.015029

2
2
s2
11) Uncertainty in Kinetic Energy
partial derivative of k
delta Ki=dM1(V1i^2)/2+dV1iM1V1i
delta Kf=dM1(V1f^2)/2+dV1fM1V1f+dM2(V2f^2)/2+dV2fM2V2f
(+)
delta Ki= 0.001958817
Ki+/- dKi= 0.018059
delta Kf= 0.001780932
Kf+/- dKf=
0.01681
(-)
0.014319
0.013248
v
K i
K i
K i 
m1 
v1i = 1i  m1  m1v1i  v1i
2
m1
v1i
2
Ki = (0.3247632 * 0.00005)*0.5 + (0.019729 * 0.3053 * 0.324763) = 0.001958817
K f 
K f
m1
K f
v1 f
v2 f
Kf
Kf
v1 f 
m2 
v2 f =
 m1  m1v1 f  v1 f 
 m2  m2 v2 f  v2 f
v1 f
m2
v2 f
2
2
2
m1 
kg  m 2
s2
2
K f = (0.00005* 0.0727272 )* 0.5 + (0.004283469*0.3053*0.072727) + (0.00005*0.3726712)*0.5 +
kg  m 2
(0.022041*0.2048*0.372671) = 0.001780932
s2
The results clearly indicate that kinetic energy was conserved during these collisions.
Fractional Uncertainties
K
0.001958817
∆ Ki / Ki = ∆ i =
= 0.1217
0.0161
Ki
Kf
0.00178093 2
∆Kf /Kf = ∆
=
= 0.1185
Kf
0.015029
Part B
1)
w =
tb1i
xb1+ xb2 =
0.0005+
0.0005 =
0.001 m
0.0728s
σ tb1i
0.001033s
deltaTB1I
t b1i  0.0010332  0.0012
0.001438s
tb(1+2)f
0.1274s
σ tb1f
0.002633s
deltaTB1F
t b1 f  0.0026332  0.0012
0.002817s
2)
v1i 
v1 =
w1
t b1i
0.024
m
 0.32967
0.0728
s
v1 f 
v2 =
w1
tb 2 f
0.024
m
 1.8838
0.1274
s
Uncertainties in the velocity
w
1
v
v
tb 
w = 2 tb +
∆w
tb
tb
w
tb
v 
 V=1/tb1i*  W1+W1/tb1i^2*  tb1i
deltaTB1I = Squareroot(uncert^2+.001^2)
dv1i
dv1f
0.02025
0.01201
∆v1 =
m
0.024
1
0.001438 
0.001 = 0.02025
2
0.0728
0.0728
s
∆v2 =
m
0.024
1
0.002817 
0.001 = 0.01201
2
0.1274
0.1274
s
Initial and Final momentums
Pi  m1v1i
Pf  (m1  m2 )v 2
0.10065
0.9609
Pi  m1v1 = (0.3053) 0.32967
m
kg  m
= 0.10065
s
s
Pf  (m1  m2 )v 2 = (0.3053 + 0.2048) 1.8838
m
kg  m
= 0.9609
s
s
6) Uncertainty in the momentums
delta Pi =
delta Pf =
0.006198
0.006
Pi +/- dPi
Pf +/- dPf
(+)
(-)
0.106848
0.9669
0.094452
0.9549
pi  v1  m1  m1  v1 = 0.32967  0.00005  0.3053  0.020246 = 0.006198
kg  m
s
p f  v 2 m1  v 2 m2  (m1  m2 )v 2 = 0.32967  0.00005  0.188383  0.00005  0.5101 0.012014 =0.006
The results clearly indicate that momentum (p) was conserved during these collisions.
P
0.006198
∆ Pi / Pi = ∆ i =
= 0.06158
0.10065
Pi
Pf
0.006
∆ Pf / Pf = ∆
=
= 0.006244
Pf
0.9609
) Kinetic Energy
Ki
0.01659
Kf
0.
00905
m1v12i 0.3053  0.32967 2
kg  m 2
=
= 0.01659
Ki 
2
2
s2
(m1  m2 )v22 0.5101  0.188383 2
kg  m 2
=
= 0.00905
.
Kf 
2
2
s2
11) Uncertainty in Kinetic Energy
delta Ki=
delta Kf=
0.002
0.00116
Ki+/- dKi=
Kf+/- dKf=
(+)
0.01859
0.01021
(-)
0.01459
0.00789
The results clearly indicate that kinetic energy was not conserved during these collisions.
kg  m
s
v1
kg  m 2
 m1  m1v1  v1i = 0.329672x 0.5x0.00005 + 0.32967 x 0.3053 x 0.02025 = 0.002
2
s2
2
K i 
2
2
v
v
K f  2  m1  2  m2  (m1  m2 )v 2  v 2 =0.1883832x 0.5x0.00005+0.1883832x 0.5x0.00005 + 0.5101 x
2
2
kg  m 2
0.188383 x 0.012014= 0.00116
s2
Fractional Uncertainties
K
0.002
∆ Ki / Ki = ∆ i =
= 0.12055
0.01659
Ki
Kf
0.00116
∆Kf /Kf = ∆
=
= 0.1282
Kf
0.00905
Results
Part A
Velocity
m
m
 0.019729
s
s
m
m
v1f = 0.072727  0.004283469
s
s
m
m
v2f = 0.372671  0.022041
s
s
v1i = 0.324763
Momentum
kg  m
kg  m
.
 0.006039642
s
s
kg  m
kg  m
P f  0.098527
 0.005843968
s
s
Pi  0.09915
∆ Pi / Pi = 0.06091
∆ Pf / Pf = 0.0593134
Kinetic Energy
kg  m 2
kg  m 2
0.001958817

s2
s2
kg  m 2
kg  m 2
K f = 0.015029
 0.001780932
s2
s2
K i = 0.0161
Fractional Uncertainties:
∆ K i / K i = 0.1217
∆ K f / K f = 0.1185
Part B
Velocity
m
m
 0.02025
s
s
m
m
v2 = 1.8838  0.01201
s
s
v1 = 0.32967
Momentum
kg  m
g  cm
.
 0.006198
s
s
g  cm
kg  m
Pf = 0.9609
 0.006
s
s
Pi = 0.10065
Fractional Uncertainties:
∆ Pi / Pi = 0.06158
∆ Pf / Pf = 0.006244
Kinetic Energy:
kg  m 2
kg  m 2
K i  0.01659
 0.002
s2
s2
kg  m 2
kg  m 2
K f  0.00905
0.00116

s2
s2
Fractional Uncertainties:
∆ K i / K i = 0.12055
∆ K f / K f = 0.1282
Conclusion
The momentum was conserved in part A of this experiment because there is an
intersection between the two ranges of values. Likewise, the kinetic energy is also
conserved because there is an intersection between the two ranges of values. 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 there is an intersection between the two
values and therefore the momentum is conserved. The kinetic energy obtained showed
that there is no intersection between the two answer sets and therefore kinetic energy was
not conserved. This proves that the collisions performed in part B are non-elastic and
contrariwise.