Momentum is a vector measure found by multiplying the mass of an object by its speed. Since mass is scalar and speed is a vector the impulse acts straight in the same way as the original speed vector within it. Derived from the same logic and concluding that lead to the preparation of Newton ‘s 3rd jurisprudence, whereby the in any interaction between forces the force that one object exerts on another is equal in magnitude and antonym in way to the force of which that object exerts on it, the jurisprudence of preservation of impulse provinces, that every bit long as there are no external forces which do work on a given system the entire impulse shall be conserved from a pre-collision province to a post-collision province. In other words the initial and concluding impulse magnitudes of any atom or set of atoms in a given system shall be preserved ; this is of class true regardless of the alterations in way that an object may travel through.

The jurisprudence of preservation of impulse holds true for both elastic and inelastic hits. However in both of these instances the expression which prove these two constructs are indistinguishable for elastic and partly inelastic hits for this expression when we have one organic structure A and a 2nd organic structure B, the expression, mAvAi + mBvBi = mAvAf + mBvBf. The expression that provides us with a saving of impulse for a aggregation of organic structures in a system where the hit produced between the two causes the organic structures to lodge together or travel as one incorporate object with the same speed is, mAvAi + mBvBi = ( mA+ megabit ) -vf. This hit is called a absolutely inelastic hit or a wholly inelastic hit. When a hit is inelastic, the kinetic energy is non conserved. It turns out that the preservation of impulse is still valid, but the concluding kinetic energy may be less than the initial value. The difference is converted into other signifiers of energy such as heat, possible energy or physical distortion. If possible energy is released during the hit, the concluding kinetic energy may be larger than the initial value! Of class the entire kinetic plus possible energy of the system is still conserved. In the inelastic hit analyzed in this lab the two organic structures stick together after hit and have, hence, a common concluding speed For our experiment since the 2nd organic structure, organic structure B will ab initio be at remainder and hence equal to zero we rewrite the expression as mAvAi = ( mA+ megabit ) -vf or mAvA1 = ( mA+ megabit ) v2, when we can alter the inferiors from I to 1 and f to 2. Using this expression we shall prove the jurisprudence of preservation of impulse with an unnaturally created wholly inelastic hit which was made by attaching Velcro to both of the clashing terminals of the two sailplanes on the air path, this will do the two sailplanes to travel as one after the hit. However be certain to maintain in head that although these expressions for impulse are vectors intending they have constituents in all three waies, that is in the ten, Y and z way, since we are on a degree plane and the angles between the two air paths at hit is equal to zero, the speed will merely remain in the ten way since this is a straight caput on hit. Otherwise if it were non a caput on hit we would hold to see the speeds and therefore the impulse in all three waies in the computation of the constituents of these measures. However since there is merely motion in one way the speeds found in the x-direction are therefore the right computations for the entire impulse, and therefore there is no demand for vector add-on in this experiment.

## Aim

To prove the jurisprudence of preservation of impulse

## Apparatus

Air path

Two Gliders ( glider A and sailplane B ) with vectors

Two halt tickers

## Experimental Procedure/ Flowchart-

Measure and record the multitudes of sailplane A, label it mA, and sailplane B, label it mB.

Topographic point sailplane A at one terminal of the path and the other sailplane ( sailplane B ) 40 centimeter from the forepart border of sailplane A.

Push sailplane A towards sailplane B with a little adequate force so as to do the two come into contact with 3 to 7 seconds. Whatever clip you get enter this sum as t1 on Table

Repeat the old processs 4 times to get a sum of 5 trails.

When the two sailplanes come into contact with one another both of them will travel as a consequence of the hit, from here mensurate the clip it takes for both of them to go 40 centimeter and record this clip on Table I as t2.

From these values compute the speed it takes for sailplane A to go 40 centimeter to clash with sailplane B as v1 utilizing the normal expression Velocity = ( a?†x/a?†t ) = ( 40 cm/t1 )

For the v2 subdivision of tabular array I record the clip it takes for both glider A and sailplane B to go 40 centimeter after the hit produced in Section IV of the above experimental process. Here use the expression v2= ( 40cm, t2 )

## Consequences

Table I Time measurings and speeds finding

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+ megabit ) v2

ma = 207.9g megabit = 206.9g ma + megabit = 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 jurisprudence for the preservation of mechanical energy is equal to E1 = E2, or K1 + U1 = K2 + U2. In order to happen the loss of mechanical energy we change this around to ( K2 + U2 ) – ( K1 + U1 ) . However because there is no alteration in the y-direction, as the experiment merely takes topographic point in the x-direction, the gravitative possible energy does non alter, like wise since there is non spring so possible energy U2 and U1 are non 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 entire mass of sailplane A and sailplane 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

## Decision

Due the information from the computations it is easy to see that we managed to hold a successful experiment, as it all indicates that the values of energy before the hit and after the hit were really near to one another bespeaking a saving of additive impulse. This I believe to be true despite the fact that the value of the incline obtained from the graph was non equal to the computation of mA/ ( mA+mB ) , which equal.50, while the incline of the graph indicates a value of 2.13. However when it comes to the informations obtained from the experiment in respects to the kinetic energy, as we expected in an inelastic hit merely the impulse is conserved while the kinetic energy is non as all the values indicate a loss in kinetic energy.