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ONE-DIMENSIONAL COLLISIONS
Prior to your scheduled laboratory session please read this entire guide
and do the required prior calculations. There will be a short quiz at the
end of the session. Before leaving the laboratory upload the worksheet
with all cells in blue and purple filled. After the laboratory session, you
have to prepare a report (filling only pages 7-11) and deliver it to your
laboratory teacher within one week.
Objectives
Correctly apply the conservation laws of kinetic energy and linear momentum.
Compute the restitution coefficient in elastic collisions.
Propagate the errors using a spreadsheet for making the calculations and doing the
data treatment or using software like Maple or Maxima (Open Source).
Introduction
The objective of this practice is to verify the conservation law of the linear momentum
in the absence of external forces, as well as to calculate coefficients of restitution in one
dimension to determine the type of collision that has happened. For this purpose, the
velocity of objects must be measured before and after a collision and apply the laws of
conservation of the linear momentum and kinetic energy.
Collisions
The collision laws describe the movement of two bodies that interact with each other.
Although we do not know the forces that describe the interaction, we know that in the
absence of external forces, during the collision, the linear and the angular momentum of
the system formed by the two bodies are conserved. If the interaction is elastic, the
kinetic energy is also conserved. In the case of the track that is used in the practice, the
problem is simplified for the following reasons:
(a) The friction between the track and the mobile (the wagon) is negligible (due to
sapphire bearings of the wagon). Therefore, energy is not dissipated due to
friction.
(b) All movement is one-dimensional.
(c) The track is horizontal and therefore there are no variations of gravitational
potential energy.
(d) The point of contact is in the line that unites the center of mass of the two
wagons and, therefore, the interaction does not generate rotations in the
wagons.
Thus, in the study of the interaction between two mobile that collide on a track like the
one in the practice, there are two important magnitudes that can be conserved: the linear
momentum and the kinetic energy of the system. If there are no external forces, then the
linear momentum is conserved:
F
ext
d ⃗ p
dt
p
1
p
2
p
1
'
p
2
'
where the subscripts 1 and 2 correspond to the wagon 1 and 2, respectively, and the
accent mean that the magnitudes are after the collision.
On the other hand, the kinetic energy is conserved in elastic collisions but not in the
inelastic ones where the lost energy during the collision (or nonconservative work),
W
NC
, can be calculated as follows:
W
NC
= E
c
= E
c 1
'
+ E
c 2
'
−( E
¿ c 1 + E
c 2
❑
If we consider a one-dimensional collision and that the wagon 2 is initially at rest in the
laboratory's reference system before the impact (
E
c 2
=0, p
2
), from equation (1) and
equation (2) it is obtained that:
W
NC
= E
c 1
'
+ E
c 2
'
− E
c 1
p
1
= p
1
'
2
'
In the case of a totally elastic collision without loss of energy (W
NC
=0), equation (3) can
be rewritten as:
v
2
'
− v
1
'
= v
1
In a general case, the loss of energy in the collision can be quantified by evaluating the
ratio between the relative velocities of the wagons before and after the collision. This
quotient is called the restitution coefficient ( e ) which in our particular case can be
written as :
e =
v
2
'
− v
1
'
v
1
Thus, the restitution coefficient gives us an idea of the percentage of energy that is not
dissipated in the interaction. This coefficient can range from 0 to 1 and we distinguish 3
cases:
(i) When
e =1, all the energy is conserved in the interaction (W
NC
=0) and,
therefore, it is completely elastic.
(ii) When e =0, the dissipation of kinetic energy is maximum and that is the case
of the totally inelastic collision. In this type of collision, the two wagons
come together and, therefore, at the same speed. In this case, equation (4)
can be written as:
m
1
v
1
=( m
1
2
) v ' (7)
For this part we will need the two wagons. The wagon that will depart with a certain
initial speed will be called m 1
and we will plug the plate (on the left) and the magnet (on
the right to attach it to the trigger). We will position the two photoelectric cells as
separated as possible from each other, but leaving about 20 cm free at each end because
at the initial moment acts a force on the wagon to propel it and if we measure the time
too close, the readings would not be correct. A good option is to place the first in the
mark of 50 cm of the rule of the track and the second to the mark of 100 cm. The wagon
that initially will be at rest will be m 2
and we will place it between the two photoelectric
cells. In this wagon we will plug the elastic rubber on the right and on the left the cork
to dampen the blow with the end holder of the track. In this wagon we will also screw
the metallic rod on the top that will serve us to keep the masses in place. We must
remember to place the black shutter plates in each of the wagons and make sure that
they continue in place after each collision. At the end of the track we will plug a cork
to reduce the impact of the wagons.
To measure the time of each wagon through the photoelectric cells, we will connect the
cell 1 (the closest to the trigger) in the connectors nº 3 of the timer and the cell 2 (the
farthest one of the trigger) in the connectors nº 1 of the timer. The yellow connectors of
the cell must go to the yellow of the timer, the red ones with the red ones and the blue of
the cell with the white of the timer. The timer must be in mode 6 (collisions) and we
select it by pressing the "Mode" button 6 times. We will know that we have selected the
right mode when the red light is lit to the left of the drawing with 4 arrows ( ⇆⇄ )
above numbers 1 and 3. In addition we must make sure that the switch located below of
the "Mode" button is to the right ( ↴ ). In this mode the time measured on the 4 screens,
from left to right, will be ( t ’ 2
/ - / t 1
/ t ’ 1
). The second screen starting from the left will not
be used. We must remember to press the Reset button before each measurement.
1.a Determination of the mass of the wagons:
We will place the wagons with their plugs and the black shutter plates on the scale and
measure their mass m 1
and m 2
. Write the measures down on Table 1a
1.b Choosing the most suitable position of the trigger:
With m 1
without any extra mass and adding 400 g to the wagon 2 we will determine
which of the three possible positions of the trigger provides a better approximation to an
elastic collision, calculating the coefficient of restitution.
We will place the trigger in the first possible position and we will measure the velocities
before and after the collision. Release the trigger and measure the passage time of m 1
before the collision ( t 1
) and after the collision ( t ’ 1
) ( alert with the sign! ) and of mass m 2
after the collision ( t ’ 2
). Repeat the measurements 5 times and calculate the mean, the
random error t
a
and the total absolute error
t
. Write everything down on Table 1b.
Repeat the measures for the other two positions of the trigger and write down all the
times on Table 1b.
From these times we can calculate the velocities of the wagons knowing that the shutter
plates have a length of Δ x = 10 cm (which we assume without error):
v =
∆ x
t
And knowing the velocities we can calculate the coefficient of restitution using equation
(6) and the corresponding error. Write down the values on Table 1 b.
1c. Elastic collision against a wall
Once we have determined which is the best position of the trigger, we will always
use it in the rest of the practice.
To study an elastic collision against a wall we will keep wagon 1 without any extra
mass and with the same plugs of the previous section. We will take wagon 2 from the
track and in the end holder of the track we will place the plug with the elastic rubber.
We will disconnect the cables of the photoelectric cell 1 since now we will only
measure the time of one of the wagons just before and right after hitting against the end
of the track. We will keep the same mode of operation of the timer as in the previous
section so that now the time measured on the 4 screens, from left to right, will be ( t ’/ t /
- / - ) where t is the time before the collision and t ' is the time after.
We will place the trigger in the appropriate position and measure the velocities before
and after the collision. Release the wagon from the trigger and measure the time it takes
before the collision ( t ) and after the collision ( t ') ( alert with the sign!). Repeat the
measurements 5 times and calculate the mean, the random error t
a
and the total absolute
error t
. Write down everything on Table 1c.
Using equation (6) in the particular case in which v’ 2
equals 0 (because there is no
trolley 2) you can calculate the restitution coefficient. Write down the values on Table
1d.
1d. Elastic collisions between bodies of different masses
We will now measure elastic collisions between bodies of different masses and we will
verify the laws of conservation of the linear momentum and kinetic energy. To do this,
we will reassemble the track and the wagons as in section 1b and we will use the trigger
at the position determined in the same section.
We will start with the wagon 2 without any extra mass and we will push the wagon 1.
Measure the time of this wagon before the collision ( t 1
) and after the collision ( t ’ 1
( alert with the sign! ) and of mass m 2
after the collision ( t ’ 2
). We will only make 1
measure.
Repeat the procedure by adding to m 2
400 g, 600 g and 800 g. Write down all values on
Table 1d.
Calculate the linear momentum of each wagon before ( p 1
) and after the collision ( p’ 1
p’ 2
), as well as the linear momentum variation ( Δp = p’p = p’ 2
+ p’ 1
- p 1
) and the kinetic
energy variation ( Δp = p’E c
= E’
c,
+ E’
c,
– E
c,
). To facilitate the calculation of the kinetic
energy variation you can take advantage of the following equality:
As in the previous case, we simplify the calculation of errors and to validate the laws of
conservation of momentum and kinetic energy we just calculate the percentage change:
Δ p / p 1
and ΔE c
/E
c
. Write down all values on Table 2b.
Attach all the calculations in extra sheets
Part 1. Elastic collisions
Choosing the most suitable position of the trigger
Describe the algebraic expressions you use to calculate the absolute error of time and
the error of the restitution coefficient:
ε t
=
ε e
=
Paying attention to the value obtained by the coefficient of restitution in each case,
which is the most appropriate position of the trigger to perform an elastic collision in
this practice? Justify the answer:
Experiment: One-dimensional collisions.
Group:Date:
Professor (Laboratory):
Name:
Name:
Name:
Nom i cognoms:
Report
numerical values:
p’ 2
[algebraic expression]=
p’ 2
[m 2
] =
p’ 2
[m 2
p’ 2
[m 2
p’ 2
[m 2
Are the theoretical values equal to the values measured experimentally? Make a
hypothesis about the possible causes of difference.
Calculate the relative variation (in %) of the linear momentum and the kinetic energy.
Based on the calculated values, what can be stated about the conservation of the linear
momentum and kinetic energy in this experimental device? Write down possible causes
that explain the results.
Δp/p 1
[m 2
] = ΔE
c
/E c
[m 2
] =
Δp/p 1
[m 2
/E c
[m 2
Δp/p 1
[m 2
/E c
[m 2
Δp/p 1
[m 2
/E c
[m 2
Part 2. Inelastic collisions.
Conservation of the linear momentum and kinetic energy
Calculate the theoretical value of p’ for all the cases. Write down the algebraic
expression you use to calculate it and the corresponding numerical values:
p’ [algebraic expression]=
p’ [m 1
, m 2
] =
p’ [m 1
] =
p’ [m 1
, m 2
Calculate the relative variation (in %) of the linear momentum and the kinetic energy.
Based on the calculated values, what can be stated about the conservation of the linear
momentum and kinetic energy in this experimental device? Write down possible causes
that explain the results.
Δp/p 1
[m 1
, m 2
] = ΔE
c
/E c
[m 1
, m 2
] =
Δp/p 1
[m 1
] = ΔE
c
/E c
[m 1
] =
Δp/p 1
[m 1
, m 2
/E c
[m 1
, m 2
