Abstract: The purpose of this lab was to analyze collisions between ...

Abstract:
The purpose of this lab was to analyze collisions between objects in one dimension. By
analyzing the motion of the carts, we were able to determine the total momentum. By
using the principal of conservation of momentum, we applied the equations for p in order
to solve for an unknown mass.
Procedure:
For this experiment, we used two carts on a track system with sonic rangers mounted at
each end. First we simulated an elastic collision. In order to do so, we set up the carts so
that the collision would occur between magnetic ends. With one cart in stationary
position, we pushed the other along the track and used a computer program to record to
motion of the carts. To observe an inelastic collision, we set up an attachment device to
connect the two carts when they collided. Then we set one car into motion and recorded
the collision with computer software. The momenta for the collisions was then calculated
because we knew the mass and velocity of each of the carts. We found the uncertainty by
using the equation:
M1/M2=(V2-U2)/(U1-V1),
where V and U are the initial and final velocities. Finally, we put an unknown weight on
top of one of the carts and recorded the velocity of the carts for an elastic collision. By
using conservation of momentum, as well as conservation of kinetic energy, for elastic
collisions, we were able to calculate the mass of the weight.

Data:
m(cart)=.556g unless otherwise specified.
Data for elastic Collisions
Test # Vi1(m/s) Vf1(m/s) Vi2(m/s) Vf2(m/s) V2- Total
U2/U1-V1 Momentum
Before the
collision
Error Total
Momentum
After the
collision
Error
1 0.245 0 0 0.243 0.991837 0.13622 0.001172 0.135108 0.001162
2 0.259 0 0 0.247 0.953668 0.144004 0.001239 0.137332 0.001181
3 0.193 0 0 0.184 0.953368 0.107308 0.000923 0.102304 0.00088
4 0.288 0 0 0.278 0.965278 0.160128 0.001377 0.154568 0.00133
5 0.21 0 0 0.202 0.961905 0.11676 0.001004 0.112312 0.000966
Data for inelastic ones
Test # Vi1(m/s) Vf1(m/s) Vi2(m/s) Vf2(m/s) V2- Total
U2/U1-V1 Momentum
Before the
collision
Error Total
Momentum
After the
collision
Error
1 0.238 0.119 0 0.115 0.966387 0.132328 0.017511 0.130104 0.004036
2 0.188 0.086 0 0.085 0.833333 0.104528 0.003243 0.095076 0.002949
3 0.225 0.107 0 0.113 0.957627 0.1251 0.003881 0.12232 0.003795
4 0.355 0.176 0 0.174 0.972067 0.19738 0.006123 0.1946 0.006037
5 0.287 0.14 0 0.14 0.952381 0.159572 0.00495 0.15568 0.004829
Data for Unknown
For this section, m1=.556 g+unknown; m2=.556 g
Test # Vi1 Vf1 Vi2 Vf2 Mass of
unknown
1 0.217 0.047 0 0.272 0.3336
2 0.214 0.042 0 0.267 0.307093
3 0.236 0.047 0 0.29 0.297122
Average Unknown Mass = .32 g ± .02
Force data

Test
#
Analysis:
Vi1 Vf1 Vi2 Vf2 Mass of
unknown
Acceleration
for cart 1
Acceleration
for cart 2
Force of
2 on 1
Force of
1 on 2
1 0.181 0.053 0.237 0.473469 -0.67 0.939 0.689744 0.522084
There calculations were all performed on Excel by entering the formulas for the columns
(V2-U2)/(U1-V1) and momentum (m1v1+m2v2). m was imputed as the mass of the cart as
the cart had no weights on it (.556 g) and v was velocity of the cart. Error was calculated
in a two step process. First, we halved the range of the (V2-U2)/(U1-V1) values and
divided it by the square root of the number of data values for each set. This % error value
was then multiplied by the momentum valued to obtain error. The mass of the unknown
was calculated using
m v m v m v m v
m
1 1o 2 2o 1 1 f 2 2 f
1
1
m ( v v )
2 2 f 2o
v v
1o 1f
m m m
cart unknown
m ( v v )
m m
2 2 f 2o
unknown
v1o v1f
cart
Summary and Answers:
Although our data seems to show that momentum isn’t perfectly conserved, it is
theoretically conserved in both elastic and inelastic collisions for a closed system. The
data reflects what we would expect once you take into account the forces that act on the
non-closed track system. One reason our data does not show perfect conservation of
momentum is because there is friction in the tracks which causes the carts to slow down.
For the elastic collision case, the difference between the velocities before and after should
be the same. This is because of the following:
Note: u denoted before, v denotes after, subscripts denote different masses
From conversation of kinetic energy:
1 2
m1u1 2
1 2
m2u2 2
1 2
m1v1 2
1 2
m2v2 2
2
m u
2
m u
2
m v
2
m v
1 1 2 2 1 1 2 2
m ( u v ) m ( v u )
2 2 2 2
1 1 1 2 2 2
m ( u v )( u v ) m ( v u )( v u
)
1 1 1 1 1 2 2 2 2 2

From conversation of momentum:
m u m u m v m v
1 1 2 2 1 1 2 2
m u m v m v m u
1 1 1 1 2 2 2 2
m ( u v ) m ( v u )
1 1 1 2 2 2
Dividing both sets of equations:
m1( u1 v1)( u1 v1) m2( v2 u2)( v2 u2)
m ( u v ) m ( v u )
1 1 1 2 2 2
u v v u
1 1 2 2
u u v v
1 2 2 1
Our data shows minor discrepancies (compared v10 to v2f since other values are 0). This
is once again because of friction in the track.
While our experiment shows one way to find mass without weighing the object, there are
countless other methods that can be utilized. You could find the period of the mass of a
spring by using . You could also find it’s mass vis-à-vis another mass using
gravitational pull ( ). Also calculations can be done with conversation of
energy to determine mass. Finally, Millikan and Thompson determined the mass of an
electron by using the charge to mass ratio and then finding the charge of that electron.
Similar methods can be used to find mass if one object is charged and then projected
through an electric field. The deflection due to the field can be used to determine mass.
Conclusion:
In this lab, we explored the law of the conversation of linear momentum. Using that law,
we determined the mass of an unknown block to be .32 g ± .02.