lOMoARcPSD|17922553
Lab Report of Linear Momentum
by Michael Sang
Physics A-DA-2021. 5225
lOMoARcPSD|17922553
Lab Report - Linear Momentum
Introduction:
The purpose of the lab is to experimentally confirm momentum conservation. In the lab, I
slid a steel ball onto a rail with my lab partner to test the principle of conservation of
momentum. The ball bounced into the hole in the wooden block at the end of the rail and
got buried in the block. The impact shook the block like a pendulum. Finally, to achieve the
goals of this lab, we were able to calculate and compare the linear momentum of the steel
ball before and after collision. A practical application of linear momentum conservation is
rocket launch. When ignited, the rocket sends exhaust gases downward (downward
momentum). To balance this downward momentum, the rocket moves upward, maintaining
linear momentum.
Procedure:
At the beginning of the lab, we will try to determine the velocity of the steel ball before it
hits the block. First, the block was moved to the platform so that it did not impede the throw of
the steel he ball. I then lined the bottom of the box with white wax paper. Third, release the steel
ball from the highest point and roll the track until it falls into the box, leaving a mark on the wax
paper. Fourth, we determined the height b from which the ball falls. Additionally, we performed
10 trials to release the ball on the lane and found the average value of r from these trials. Even
now that we got the
lOMoARcPSD|17922553
From the r and b values, we were able to use the velocity = √(1/2g(r^2/b)) to calculate the
velocity of the steel ball as it left the track.
The second half of the experiment examined the velocity of the steel ball after it hit the
block. First, we measured the mass of the steel ball and recorded the mass of the wooden block.
Then I stopped blocking at the appropriate level. Third, we measured the distance h using the
formula √(h^2-x^2). Fourth, we measured the horizontal distance x. This is the distance the block
swings after a collision, but after some testing, the block swing barely moves or sometimes
doesn't even touch the slider. Fifth, I calculated the vertical distance y using the formula y = h√(h^2-x^2). Finally, using conservation of energy, we used the change in altitude to calculate the
velocity of the system using the formula Vb = √(2gy).
Data/ Calculations:
 Relative difference = 6.08%
 Uncertainty h (m) = 0.001
 Average r value (m) = 0.38
 Uncertainty in r (m) = 0.01
 h (m) = 0.42
 Uncertainty in b (m) = 0.001

 Y (m) = 0.041
 V (before impact) = 2.209 m/s
 V(after impact) = 0.896 m/s
 Ball Drop Distance b (m) = 0.145
Questions:
lOMoARcPSD|17922553
1) How do multiple measurements of r change the uncertainty?
Repeating the measurements more often will give you a more accurate estimate.
Therefore, taking the average of these measurements gives a more accurate estimate. Therefore,
the uncertainty of r decreases with multiple measurements.
2) Within the limits of your experimental accuracy, is momentum conserved during the collision?
To the best of my experimental accuracy, I believe that momentum is conserved during
collisions. We calculated the linear momentum of the system before and after the collision and
found a relative difference of only 6.08%.
3) Derive equation (1), starting from general physics principles.
a) mgh = 1/2 mv^2
b) 2mgh=mv^2 (multiply both of the sides by 2)
c) 2gh=v^2 (divide both of the sides by m)
d) √2gh = v (get the square root of both sides)
4) From your results, compute the fractional loss of kinetic energy of translation during
impact.
Ignore the rotational energy of the sphere.

Ball (kg) = 0.0672, Block (kg)= 0.0878, Vi= 2.209 m/s, Vf= 0.896 m/s

KEi=1/2mv^2= 0.164 J

KEf=1/2(m +M)v^2= 0.062 J

Fractional/minor change in KE: (KEi - KEf)/KEi= |-0.214 J| =0.214 J
lOMoARcPSD|17922553
5) Derive an expression for the fractional loss of kinetic energy of translation in terms only
of m and M, and compare with the value calculated in the preceding question. Consider the
collision as a totally inelastic one.
a) Initial momentum: p = m*vi, Final momentum: p = (M + m)*vf
b) They're equal because momentum is conserved: m*vi = (M + m)*vf
c) solve for vf: vf = vi*m/(M + m)
d) KEi = 1/2*m*vi^2, KEf = 1/2*(M + m)*vf^2
e) Substitute: KEf = 1/2*(M + m)*(vi*m/(M + m))^2
f) Simplify: KEf = 1/2 * m^2*vi^2/(M+m)
g) Fractional change in KE: (KEi - KEf)/KEi
h) Substitute: ((1/2*m*vi^2) - (1/2 * m^2*vi^2/(M+m)) )/(1/2*m*vi^2)
i) Cancel vi: ((1/2*m) - (1/2 * m^2/(M+m)))/(1/2*m)
j) Cancel 1/2: (m - m^2/(M+m) )/m
k) Divide through by m: 1 - m/(M+m)
l) Simplify the fraction: M/(m + M)
 Is the same KE lost as question 4
Conclusion:
As a lab conclusion, I believe that linear momentum is conserved according to my
experimental data. My lab partner and I have calculated that the relative difference in linear
momentum of the ball/block system before and after collision is only 6.08%. We also found that
the ball/block system had a pre-collision velocity of 2.209 m/s and a post-collision velocity of
0.896 m/s. Additionally, in our experiments,
lOMoARcPSD|17922553
we may have miscalculated the value of r by dropping the steel ball at different positions, so there
could be several sources of error. However, at the end of this lab experiment, I think I understood
and proved momentum conservation.