One Dimensional Conservation of Momentum and Energy
1. Study the conservation of momentum and energy in one dimension by examining the
motion of colliding carts on a track.
2. Determine velocity for one or two carts before and after their collision
1. Momentum and energy
For bodies travelling in one dimension, position and velocity can be simply specified;
position will be a distance from the origin, and velocity will be either positive or negative
based on the direction of travel. (For instance, an object travelling to the right may be
considered to have a positive velocity, while one travelling to the left will have a negative
velocity.) When two objects moving in one dimension collide, the following equations hold:
Initial momenturm P = mv + m v
Initial energe E = m1v
Final momenturm P = m v + m v
1 1 f 2 2 f
Final energe E
= m v1
1 2 f
In above equations, m 1 and m 2 are masses for cart 1 and cart 2, and V 2 and V 1 are their
velocities. The subscripts i and f refer to initial and final states. Since velocities can be
positive or negative, this means that in the above equations, momentum can be positive or
negative, while energy must always be positive (unless it is zero, in which case there is
2. Elastic collision
For any collision, momentum should be conserved. Energy will also be conserved if a
collision is elastic. When two bodies collide, an elastic collision doesn’t lose energy
because the mechanic energy doesn’t convert into heat while an inelastic collision does
lose energy. To have an elastic collision, carts are equipped with magnets. Once they are
collided, their bodies don’t touch each other because magnets push them away. It enables
the collision elastic. Once the elastic collision holds, the following equation can be used
to calculate the final velocities for cart 1 and 2:
⎞ ⎛ 2m2
= ⎜ v1
⎝ + ⎠ ⎝ +
⎞ ⎛ m1
= ⎜ v1
⎝ + ⎠ ⎝ +
If m 1 = m 2 , and v 2i =0 then Equation (5) becomes
= 0 (7)
= ⎜ v1
However, if m 2 = 3 m 1 , and v 2i =0 then Equation (5) becomes
v = −v
/ 2 (9)
1 f 1i
v = v / 2 (10)
2 f 1i
3. Dependent measurement
This experiment will require calculations with repeated dependent measurements. A
“ladder” consists of thirteen alternative black and transparent lines, shown in Figure 3.
When a cart carrying a “ladder” moves along the track it will pass through two inferred
light beams collimating to two sensors. The light beams either are passed through or
blocked out by the “ladder”. If passed the inferred lights are then received by the sensors.
If blocked the sensors receive nothing at all. As the result, the computer counts the
traveling times of the 13 lines of the “ladder”. With measuring distances of the lines and
the traveling times, the velocity of the cart can be determined.
Figure 1 Two carts collide when m 1 =m 2
Figure 2 Two carts collide when m 2 >>m 1
Figure 3 A “ladder” on a glass plate
For an example, the following data is the traveling time of 13 lines of the “ladder”
displayed on the computer:
0.0123 (T 1 )
0.1234 (T 2 )
0.2345 (T 3 )
0.3456 (T 4 )
0.4567 (T 5 )
0.5678 (T 6 )
0.6789 (T 7 )
0.7890 (T 8 )
0.8911 (T 9 )
1.0911 (T 10 )
1.1911 (T 11 )
1.2980 (T 12 )
1.4134 (T 13 )
Obviously, using 6d/(T 7 - T 1 ), 6d/(T 8 - T 2 ), 6d/(T 9 - T 3 ), 6d/(T 10 - T 4 ), 6d/(T 11 - T 5 )
and 6d/(T 12 - T 6 ) to calculate the average velocity is much better than calculating
d/(T 2 - T 1 ) or 12d/(T 13 - T 1 ). Do you know why?
1. two carts with “ladders” and magnets
2. a track
3. a computer connected with sensors
4. several iron blocks
1. Make sure the track is level before you begin this experiment.
2. Without recording data, try a sample collision by having one cart (m 1 ) collide with the
other (m 2 =m 1 ) which is initially at rest, as shown in Figure 1. Then do the same thing the
other way around, i.e. the cart which was previously at rest will now be the one initially
3. As shown in Figure 2, adding two iron blocks on a cart (m 2 ) so that m 2 =3m 2 , do step 2
again. Compared with the first case: m 1 =m 2 , find out what are different for V 1 and V 2 .
4. Measure the black “ladder" spacing with its possible error (remember that an uncertainty
of measuring a distance is ½ the smallest division).
5. Let the Cart 1collide with the Cart 2 and record the data of times from the computer for
two cases respectively: m 1 =m 2 and m 2 =3m 1 . Weigh the masses of carts and iron blocks,
and record m 1 and m 2 for the two cases.
6. Calculate the velocities, distance divided by time, for both carts before and after the
collision using the method of differences mentioned in Dependent Measurement, and
attach the results with their possible errors. Do the equations (7), (8), (9) and (10) hold?
If yes, why? If not exactly, why? Discuss it in your lab report.
1. The names of the program for measuring times are “software” and “carts2” respectively.
2. Four regions of traveling times of two carts measured by the computer will be shown as
below. Left data is of Cart 1 and right is of Cart 2. Up region is the data of before
collision and down region is the data after collision. However, if Cart 2 is at rest before
collision Data 3 will be the data of after collision for Cart 2 because there is none of data