# Linear Momentum Inelastic Collusion Biology Essay

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Momentum is a vector quantity found by multiplying the mass of an object by its velocity. Since mass is scalar and velocity is a vector the momentum acts directly in the same direction as the original velocity vector within it. Derived from the same logic and reasoning that lead to the formulation of Newton's third law, whereby the in any interaction between forces the force that one object exerts on another is equal in magnitude and opposite in direction to the force of which that object exerts on it, the law of conservation of momentum states, that as long as there are no external forces which do work on a given system the total momentum shall be conserved from a pre-collision state to a post-collision state. In other words the initial and final momentum magnitudes of any particle or set of particles in a given system shall be preserved; this is of course true regardless of the changes in direction that an object may go through.

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The law of conservation of momentum holds true for both elastic and inelastic collisions. However in both of these cases the formulas which prove these two concepts are identical for elastic and partially inelastic collisions for this formula when we have one body A and a second body B, the formula, mAvAi + mBvBi = mAvAf + mBvBf . The formula that provides us with a preservation of momentum for a collection of bodies in a system where the collision produced between the two causes the bodies to stick together or move as one unified object with the same velocity is , mAvAi + mBvBi = (mA+ mB) -vf. This collision is called a perfectly inelastic collision or a completely inelastic collision. When a collision is inelastic, the kinetic energy is not conserved. It turns out that the conservation of momentum is still valid, but the final kinetic energy may be less than the initial value. The difference is converted into other forms of energy such as heat, potential energy or physical deformation. If potential energy is released during the collision, the final kinetic energy may be larger than the initial value! Of course the total kinetic plus potential energy of the system is still conserved. In the inelastic collision analyzed in this lab the two bodies stick together after collision and have, therefore, a common final velocity For our experiment since the second body, body B will initially be at rest and therefore equal to zero we rewrite the formula as mAvAi = (mA+ mB) -vf or mAvA1 = (mA+ mB) v2, when we can change the subscripts from i to 1 and f to 2. Using this formula we shall test the law of conservation of momentum with an artificially created completely inelastic collision which was made by attaching Velcro to both of the colliding ends of the two gliders on the air track, this will cause the two gliders to move as one after the collision. However be sure to keep in mind that although these formulas for momentum are vectors meaning they have components in all three directions, that is in the x, y and z direction, since we are on a level plane and the angles between the two air tracks at collision is equal to zero, the velocity will only stay in the x direction since this is a directly head on collision. Otherwise if it were not a head on collision we would have to consider the velocities and therefore the momentum in all three directions in the calculation of the components of these quantities. However since there is only movement in one direction the velocities found in the x-direction are therefore the correct calculations for the total momentum, and hence there is no need for vector addition in this experiment.

## Objective

To test the law of conservation of momentum

## Apparatus

Air track

Two Gliders (glider A and glider B) with vectors

Two stop watches

## Experimental Procedure/ Flowchart-

Measure and record the masses of glider A, label it mA, and glider B, label it mB.

Place glider A at one end of the track and the other glider (glider B) 40 cm from the front edge of glider A.

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Push glider A towards glider B with a small enough force so as to make the two come into contact with 3 to 7 seconds. Whatever time you get record this amount as t1 on Table

Repeat the previous procedures 4 times to acquire a total of 5 trails.

When the two gliders come into contact with one another both of them will move as a result of the collision, from here measure the time it takes for both of them to travel 40 cm and record this time on Table I as t2.

From these values compute the velocity it takes for glider A to travel 40 cm to collide with glider B as v1 using the normal formula Velocity = (âˆ†x/âˆ†t) = (40 cm/t1)

For the v2 section of table I record the time it takes for both glider A and glider B to travel 40 cm after the collision produced in Section IV of the above experimental procedure. Here use the formula v2=(40cm,t2)

## Results

Table I Time measurements and velocities determination

Trail

t1 (s)

t2(s)

v1= 40 cm/t1

v2= 40 cm/t2

1

1.53 s

2.84 s

0.26 m/s

0.14 m/s

2

1.17 s

2.37 s

0.34 m/s

0.17 m/s

3

1.3 s

2.68 s

0.31 m/s

0.15 m/s

4

1.75 s

3.44 s

0.23 m/s

0.12 m/s

5

153 s

3.12 s

0.26 m/s

0.13 m/s

6

1.48 s

2.94 s

0.27 m/s

0.14 m/s

Conversion from 40cm to 0.4m

40cm- (1-10-2) = 0.4 m

1cm

Calculations for Velocity (m/s)

Body A Trail 1 0.4m/1.53s = 0.26143 m/s 0.26 m/s = v1

Body A Trail 2 0.4m/1.17s = 0.34188 m/s 0.34 m/s = v1

Body A Trail 3 0.4m/1.30s = 0.307692 m/s 0.31 m/s = v1

Body A Trail 4 0.4m/1.75s = 0.22857m/s 0.23 m/s = v1

Body A Trail 5 0.4m/1.53s = 0.26143 m/s 0.26 m/s = v1

Body A Trail 6 0.4m/1.48s = 0.27027 m/s 0.27 m/s = v1

Body A and B combined Trial 1 0.4m/2.84s = 0.14089 0.14 m/s = v2

Body A and B combined Trial 2 0.4m/2.37s = 0.16877 0.17 m/s = v2

Body A and B combined Trial 3 0.4m/2.68s = 0.14925 0.15 m/s = v2

Body A and B combined Trial 4 0.4m/3.44s = 0.11627 0.12 m/s = v2

Body A and B combined Trial 5 0.4m/3.12s = 0.12820 0.13 m/s = v2

Body A and B combined Trial 6 0.4m/2.94s = 0.13605 0.14 m/s = v2

Calculations for Momentum, mAvA1 = (mA+ mB) v2

mA = 207.9g mB = 206.9g mA + mB = 414.6g

Trail 1 0.2079kg-0.26m/s = 0.4146g-0.14 m/s 0.054054 kgm/s = 0.058044 kgm/s

Trail 2 0.2079kg-0.34m/s = 0.4146g-0.17 m/s 0.070686 kgm/s = 0.069972 kgm/s

Trail 3 0.2079kg-0.31m/s = 0.4146g-0.15 m/s 0.064449 kgm/s = 0.062190 kgm/s

Trail 4 0.2079kg-0.23m/s = 0.4146g-0.12 m/s 0.047817 kgm/s = 0.049752 kgm/s

Trail 5 0.2079kg-0.26m/s = 0.4146g-0.13 m/s 0.054054 kgm/s = 0.053898 kgm/s

Trail 6 0.2079kg-0.27m/s = 0.4146g-0.14 m/s 0.056133 kgm/s = 0.058044 kgm/s

## Discussion

The equation of mechanical energy is E = K + U. And since the law for the conservation of mechanical energy is equal to E1 = E2, or K1 + U1 = K2 + U2. In order to find the loss of mechanical energy we change this around to (K2 + U2) - (K1 + U1). However because there is no change in the y-direction, as the experiment only takes place in the x-direction, the gravitational potential energy does not change, like wise since there is not spring then potential energy U2 and U1 are not considered. Therefore the equation for the loss of Mechanical Energy is the same as the equation for the loss of kinetic energy which is K2- K1. Finally K is the kinetic energy which is equal to (1/2) Mv2, where M represents the combined or total mass of glider A and glider B, MA + MB.

Calculations for Kinetic Energy Loss and Mechanical Energy Loss, (1/2) Mv22- (1/2) Mv12

Trial 1 (1/2) (0.4146) (0.14)2 - (1/2) (0.4146) (0.26)2 = (0.0040630J-0.014013J) = -0.00995J

Trial 2 (1/2) (0.4146) (0.17)2 - (1/2) (0.4146) (0.34)2 = (0.0059909J-0.023963J) = - 0.01797J

Trial 3 (1/2) (0.4146) (0.15)2 - (1/2) (0.4146) (0.31)2 = (0.0046642J-0.019921J) = -0.01525J

Trial 4 [(1/2) (0.4146)] - ((0.12)2-(0.23)2) = -0.00798J

Trial 5 [(1/2) (0.4146)] - ((0.13)2-(0.26)2) = -0.01051J

Trial 6 [(1/2) (0.4146)] - ((0.14)2-(0.27)2) = -0.01104J

## Conclusion

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Examples of our workDue the data from the calculations it is easy to see that we managed to have a successful experiment, as it all indicates that the values of energy before the collision and after the collision were very close to one another indicating a preservation of linear momentum. This I believe to be true despite the fact that the value of the slope obtained from the graph was not equal to the calculation of mA/(mA+mB), which equal .50, while the slope of the graph indicates a value of 2.13. However when it comes to the data obtained from the experiment in regards to the kinetic energy, as we expected in an inelastic collision only the momentum is conserved while the kinetic energy is not as all the values indicate a loss in kinetic energy.