2D Collision Momentum Lab Report
(gr 12 assignment, got a 89% on it)
Introduction:
This lab involves calculating the velocity of 2 marbles that have a collision. This occurs when the first
marble rolls down a ramp colliding with a second marble, launching both off the ramp and a raised
platform. A number of physics concepts are demonstrated in this experiment.
Momentum is a force that can impact the movement of objects. Momentum is the force that
transfers between objects when they collide with each other. It is affected by the mass of an object and
the velocity of an object, the variables (m) and (v). At its simplest form momentum can use the equation
P = mv with P being momentum. Momentum is directly proportional to the mass and velocity meaning
as either variable increases or decreases momentum does the same. Like any other force it is measured
in Newtons, and is a vector quantity, meaning it has a direction as well as a magnitude. One of the
properties of momentum is that it is always conserved, meaning that the force of momentum is
preserved, regardless of whether a collision occurs in a closed or open system. There are 2 main types of
collisions, both where momentum is conserved. Elastic collisions conserve kinetic energy, whereas
Inelastic collisions don’t conserve kinetic energy. The perfect version of an elastic collision is where
both momentum and kinetic energy are fully conserved meaning no energy is lost. A perfectly inelastic
collision is where both objects which are colliding, stick together, whereas momentum is conserved but
kinetic energy is not. All 4 scenarios can utilize the equation 𝑚1𝑣𝑓1 + 𝑚1𝑣𝑓2 = 𝑚1𝑣𝑖1 + 𝑚1𝑣𝑖2 because
the total final momentum must equal the total final momentum in any collision. Because elastic
collisions also conserve energy, the kinetic energy equation for momentum
1
2
2
𝑚1𝑣𝑓1 +
1
2
2
𝑚2𝑣𝑓2 =
1
2
2
𝑚1𝑣𝑖1 +
1
2
2
𝑚2𝑣𝑖2, also applies. In inelastic equations velocities are added
together when the 2 objects collide leading the the equation 𝑣𝑓 =
(𝑚1𝑣𝑖1+𝑚2𝑣𝑖2)
(𝑚1+𝑚2)
. Impulse is the final
piece to momentum. Impulse can be seen as the change in momentum in a given time frame, it can be
represented by the equation ∆𝑝 = 𝑚𝑣, where (𝚫p) is impulse.
These equations and concepts will be used in the lab to determine the type of collisions experimented
on.
Below is a diagram of what the experiment will look like for reference (Fig 1).
Fig 1.
Purpose: To walk through the conservation of momentum in a 2D collision between 2 marbles and find
if the collision is elastic or inelastic.
Hypothesis: The marbles will form an elastic collision when they hit as they are smooth, and will not
attach together. Though in theory they should have a perfect elastic collision, but since this is an “open
system” experiment it will just be a regular elastic collision.
Materials:
2 Marbles
Mechanical Weighing Scale
Clear tape
Masking tape
3 sheets of Carbon paper
2 Photogates + 2 wood planks
Measuring tape
Black clip
Procedure:
1. The mass of the 2 marbles was measured as well as their combined mass.
2. The ramp was lined up with the edge of the desk.
3. The photgates were set up above the ramp, supported using tape and 2 pieces of wood.
4. The initial velocity was then measured by using the photogates to record the speed of the ball,
when it passed between the 2 sensors. The incident ball was dropped from the screw at the top of
the ramp and this was repeated 5 times. The values were then recorded in a table and averaged
out.
5. 3 pieces of carbon paper were taped in front of the 2D collision ramp, lined up with the edge of
the desk on the ground, so that the point in which the target ball is resting, is at the base of the
paper. The point where the ball is resting was marked on the carbon paper.
6. The height of the desk was then measured where the target ball would be, along with the height
of the ramp where the incident ball would start.
7. The incident ball was then dropped a total of 3 times, with the target ball resting on the screw at
the end of the metal slide/ramp.
8. Using the markings on the carbon paper that the incident ball and target ball leave after the
collision, a circle was drawn to determine the average landing point of each ball.
9. The length from the center of each circle to the base of the target ball was then measured and
recorded (with T1 being the target ball and I1 being the incident ball).
Analysis:
1. Draw a line and measure the distance from point X to point T1 and from point X to point I1.
T1(Marble 2 (T1)) Hyp lengths I1 (Marble 1 (I1))
0.2175 +/- 0.0005 m
0.176 +/- 0.0005 m
2. Determine the angle between the line XT1 and the x-axis as well as between XI1 and x-axis.
Mention the method by which you determine the angles and show work if necessary. [A 1]
T1(Marble 2 (T1))
Hyp lengths
I1 (Marble 1 (I1))
Hyp lengths
0.176 +/- 0.0005
0.211 +/- 0.0005 m (x)
0.2175 +/- 0.0005 m
0.045 +/- 0.0005 m (y)
m
Sinθ = opp/hyp
Sinθ = XT1x/XT1
θ = Sin(0.211/0.2175)-1
θ = 90 - 75.95
θ = 14.042°
+/- (0.0005/0.211)*100% + (0.0005/0.2175)*100%
+/- 0.46%
Sinθ = opp/hyp
Sinθ = XT1x/XT1
+/- (0.0005/0.045)*100% + (0.0005/0.176)*100%
+/- 1.39%
θ = Sin(0.045/0.176)-1
θ = 90 - 14.813
θ = 75.186°
3. Calculate the final speed of the marbles just after the collision. Show all proper equations and
manipulations for full marks. [A 5]
T1(Marble 2 (T1))
Hyp lengths
I1 (Marble 1 (I1))
Hyp lengths
0.211 +/- 0.0005 m (x)
0.2175 +/- 0.0005 m
0.045 +/- 0.0005 m (x)
0.176 +/- 0.0005 m
0.054 +/- 0.0005 m (y)
0.169 +/- 0.0005 m (y)
XT1fy
d = 0.78 m +/-0.0005 m a = 9.8 +/- 0.1 m/s2
find t
d = 1/2 at2
2(0.78)
(9.8)
=𝑡
t = 0.399 s
= +/- (0.0005/0.045)*100%
= +/- 0.0641%
XT1fx
d = 0.176 +/- 0.0005 m
t = 0.399 +/- 0.0641% s
find v
v = d/t
v = (0.176)/0.399
v = 0.44 m/s
= +/- (0.0005/0.176)*100% + (0.0005/0.045)*100%
= +/- 0.28% + 0.0641%
= +/- 0.3441%
XI1fx
t = 0.399 +/- 0.0641% s
d = 0.2175 +/- 0.0005 s
v = d/t
(0.2175) = v/(0.399)
v = 0.545 m/s
= +/- (0.0005/0.176)*100% +/- 0.0641%
= +/- 0.28% +/- 0.0641%
= +/- 0.294%
4. Using your knowledge of the principle of conservation of momentum determine the initial speed
of the incident ball just before the collision (A 6) Note, separately, the speed of the ball at impact
that was found using the photogates.
Use 1 as incident marble, 2 as target marble.
Averaged Velocity From Photogates (Experimental): 0.035 m / 0.0236 s = 1.48 m/s +/- 3.55%
Uncertainties: (0.0005/0.035)*100% + (0.0005/0.0236)*100% = 3.55%
Mass of marble: 0.0055 kg
m1v1ix = m1v1fx + m2v2fx ← no momentum in the target marble (starts still)
(0.0055)v1ix = (0.0055)(0.44cos(14.042)) + (0.0055)(0.545cos(75.186))
v1ix = 0.001713 / 0.0055
v1ix = 0.6364 m/s
X uncertainties:
v2xf = 0.44 +/- 0.3441%
v1xf = 0.545+/- 0.294%
= 0.294% + 0.3441%
= 0.6381%
θ1 = 14.042° +/- 0.46%
θ2 = 75.186° +/- 1.39%
= 1.39% + 0.46%
= 1.85%
= 0.6381% * (0.44*0.3441%)+(0.545*0.294%)
= +/- 0.001611
1.85% * (75.186*1.39%)+(14.042*0.46%)
= +/- 0.08392
= +/- 0.001611 * 0.08392 / 0.6364 ← divide total uncertainty by velocity to get the % uncertainty
= +/- 0.085531/ 0.6364
= +/- 0.033381101%
v1ix = 0.6364 m/s +/- 0.033381101%
m1v1iy = m1v1fy + m2v2fy ← no momentum in the target marble (starts still)
(0.0055)v1iy = (0.0055)(0.0545sin(14.042)) + (0.0055)(0.44sin(75.186))
v1iy = 0.0013344 + 0.002897865 / 0.0055
v1iy =0.5389 m/s
Y uncertainties:
v2xf = 0.44 +/- 0.3441%
v1xf = 0.545+/- 0.294%
= 0.294% + 0.3441%
= 0.6381%
θ1 = 14.042° +/- 0.46%
θ2 = 75.186° +/- 1.39%
= 1.39% + 0.46%
= 1.85%
= 0.6381% * (0.44*0.3441%)+(0.545*0.294%)
= +/- 0.001611
1.85% * (75.186*1.39%)+(14.042*0.46%)
= +/- 0.08392
= +/- 0.001611 * 0.08392 / 0.6364 ← divide total uncertainty by velocity to get the % uncertainty
= +/- 0.085531/ 0.6364
= +/- 0.033381101%
v1iy = 0.5389 m/s +/- 0.033381101%
a2 + b2 = c2
v1iy2 + v1ix2 = v1i
√(0.5389)2 + (0.6364)2 = v1i
v1i = 0.8339 m/s
tanθ = opp/hyp
θ = tan-1(0.6364/0.5389)
θ = 49.742°
v1iy =0.5389 m/s +/- 0.033381101%
v1ix = 0.6364 m/s +/- 0.033381101%
0.033381101% + 0.033381101%
= 0.066762202
v1iy =0.5389 m/s +/- 0.033381101%
v1ix = 0.6364 m/s +/- 0.033381101%
= 0.033381101%*0.033381101%
= 0.00001114297%
= +/- 0.066762202 * 0.8339
= +/- 0.055673%
v1i = 0.8339 m/s +/- 0.055673[forward 49.742° +/- 0.00001114297% left]
5. Based upon your calculations (to two significant digits), is momentum conserved in the 2D
collision? Justify. Show all equations and work for full marks. [A 6]
Momentum should be conserved in both x and y axis
p1ix = p1fx + p2fx ← no momentum in the target marble (starts still no p2ix)
m1v1ix = m1v1fx + m2v2fx
m1v1ix = m1v1fx + m2v2fx
(0.0055)(0.6364) = (0.0055)(0.545cos(14.042)) + (0.0055)(0.44cos(75.186))
3.5x10-3 N ≐ 3.5x10-3 N
X uncertainties:
v2xf = 0.44 +/- 0.3441%
v1xf = 0.545+/- 0.294%
v1ix = 0.8339 m/s +/- 0.055673%
= 0.3441%/0.44 + 0.294%/0.545
= +/- 0.01439164528
3.5x10-3 N + 0.055673% ≐ 3.5x10-3 N +/- 1.439%
p1iy = p1fy + p2fy ← no momentum in the target marble (starts still no p2ix)
m1v1iy = m1v1fy + m2v2fy
(0.0055)(0.5389) = (0.0055)(0.545sin(14.042)) + (0.0055)(0.44sin(75.186))
2.9x10-3 N ≐ 2.9x10-3 N
By showing that the initial moment is equal to the final momentum in both the x and y components,
momentum is conserved even in 2D collisions regardless if it's a closed or open system.
6. Based upon your calculations (to two significant digits), is the collision (perfectly) elastic or
not? Justify. Show all equations and work for full marks. [A 2]
A perfect elastic collision would conserve all of the kinetic energy meaning that the initial kinetic energy
is equal to the final kinetic energy. Using the equation
1
2
2
𝑚𝑣1𝑖 +
1
2
2
1
2
𝑚𝑣2𝑖 =
determine if it is perfectly elastic or not.
m = 0.0055 kg
2
1
2
2
1
2
𝑚𝑣1𝑖 =
1
2
(0. 0055)(0. 8339)1𝑖 =
𝑚𝑣1𝑓 +
1
2
2
2
𝑚𝑣2𝑓 ← no kinetic energy in the target ball initially
1
2
2
(0. 0055)(0. 0545)1𝑓 +
0.001912 J ≠ 0.000541 J
The collision is not perfectly elastic.
1
2
2
(0. 0055)(0. 44)2𝑓
2
𝑚𝑣1𝑓 +
1
2
2
𝑚𝑣2𝑓 we can
7. Give at least 2 sources of error in this lab. [A 2]
1. When XT1 or XI1 is in projectile motion, there are other forces acting on the ball besides
gravity, Air resistance would affect the final displacement and lead to errors in calculations.
2. Human error when displacements were measured there could be error in using the measuring
tool slightly off the actual point (slightly to the left of the center a point/slightly to the right of
the center of a point).
3. Instrumental Error when the measuring tape doesn't have enough decimal places, to display the
measurement accurately enough.
Citations
Hirsch, A., & Martindale, D., (2003). Nelson physics 12. K. Martindale.