One Dimensional Conservation of Momentum and Energy

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One Dimensional Conservation of Momentum and Energy

One Dimensional Conservation of Momentum and Energy

Purpose

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

Theory

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

i

1 1i

2 2i

(1)

2 1

Initial energe E = m1v

1i

+ m2v

2 2

1 2

1 2i

(2)

Final momenturm P = m v + m v

f

1 1 f 2 2 f

2 1

Final energe E

f

= m v1

f

+ m2v

2 2

1 2

1 2 f

(3)

(4)

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

no motion.)

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:

⎛ m1

− m2

⎞ ⎛ 2m2


v1

f

= ⎜ v1

i

v2i

m1

m

⎟ +


2

m1

m

⎟ (5)

⎝ + ⎠ ⎝ +

2 ⎠

⎛ 2m1

⎞ ⎛ m1

− m2


v2

f

= ⎜ v1

i

v2i

m1

m

⎟ −


2

m1

m

⎟ (6)

⎝ + ⎠ ⎝ +

2 ⎠

If m 1 = m 2 , and v 2i =0 then Equation (5) becomes


⎛ 2m2


v

1 ⎜


f

= v2i

= v2i

= 0 (7)

⎝ m1

+ m2


⎛ 2m1


v2

f

= ⎜ v1

i

= v1

i

m1

m

⎟ (8)

⎝ +

2 ⎠

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?

Apparatus

1. two carts with “ladders” and magnets

2. a track

3. a computer connected with sensors

4. several iron blocks

Procedures

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

moving.

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.

Note

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

before collision.

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