  # The Conservation of Momentum in Collisions

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• Pages: 3 (1060 words)
• Published: June 23, 2018
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## Introduction:

Newton’s third law of motion states that force occur in pairs. It also states that every action has an equal and opposite reaction. This means that for every force applied on an object, there is an equal force in magnitude which is operating in an opposite direction. During the interaction of two objects which have masses, there is bound to be change in momentum, but as we know it, momentum is usually conserved in systems. The interaction may constitute either an elastic or inelastic collision. In a perfectly elastic collision there is no loss in kinetic energy of any kind during the interaction of the objects whereras an inelastic collisioncan be described as one in which some of the kinetic energy is lost or otherwise changed into some other form of energy during the interaction. Any slight interaction between the two or more objects will change some of the kinetic energy into a form of internal energy. This means that there are no large scale impacts which can be perfectly elastic. Momentum is conserved in inelastic collisions. However, the kinetic energy through the collisions cannot be accounted for since some of it is converted to other forms of energy. In ideal gases and conditions, c

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ollisions tend to perfectly elastic collisions.

In hard spheres, collisions may be nearly elastic, hence it is useful to note the extent of an elastic collision. The assumption of momentum conservationin addition to the conservation of kinetic energy enables the calculation of the final velocities in two-body collisions.

## Investigation 1:

In the investigation 1, the following equipments were used: air table, two paper sheets, spark timer, and ruler.

### Method

The first step was to put the large sheet of paper over the carbonized paper and the air was turned on After I set the spark timer to 30Hz the pucks were released and the remote button was pressed to set things in motion. Thereafter, the sheet of paper was removed and found the tracks of the motion of both pucks before and after they had collided with each other. Then the distance that the pucks had travelled before and after the collisions were measured in meters, (?x in meters), taking into account every two points using the ruler. The spark timer was set to 0.0333s between each pulse and the spark timer. The following equation wasused to find the velocity for each puck before and after the collision:

V = (Delta x) / (2 X Delta t)

All the velocities were calculated using the length of corresponding tracks and the time taken. I factored in the time interval between adjacent dots as 30ms.

As part of my investigation, I also found the magnitude and direction of the difference vector. The way I did this was by dividing the magnitude of the “difference” by the magnitude of the initial sum of velocity vectors (V1+V2).

Percentage difference in momentum:

The conservation of Momentum equation = (m1xv1) + (m2xv2) = (m1xv1’) + (m2 xv2’)

Since the masses of the pucks ar

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equal, the equation becomes:

V1 + V2 = V1’ + V2’

To calculate V1 + V2 and V1’ + V2’, I calculated the resultant vector which is the (hypo) in my table.

[{(V’1 + V’2) – (V1 + V2)} / (V1 + V2)] x 100 =15.38461538 %

## Percent change in Kinetic Energy:

From my measured speeds by setting the masses of both pucks equal. This was the case in an elastic collision. However, in this experiment, the collision is not very elastic; there is some loss in kinetic energy because of air resistance or friction”

[%u2206K = K (f) – K (i)]:

K (f) = ?m (v%u2081’? + v%u2082’?)

K (i) = ?m (v%u2081? + v%u2082?)

Since the masses of the pucks are equal, then;

K (f) = v%u2081’? + v%u2082’? K (i) = v%u2081? + v%u2082?

{ [ (v’1^2 + v’2^2)- ( v1^2 + v2^2 ) ] / ( v1^2 + v2^2 ) } X 100 = 31.08287881%

The Percentage change in kinetic energy is therefore= 31.08287881%

## Investigation 2:

The aim of this second experiment was to bring out an inelastic type of collision. For this experiment, I used the same apparatus that were used in Investigation 1, only with the addition of Velcro strips that were to help in the sticking together of the pucks. Both pucks were wrapped with the Velcro strip round them and trail practices were done just to get the rhythm of the experiment. The track was confirmed to be level and tthe distances between every two points were measured using the ruler. The spark timer was set to 0.0333s. The velocities of the two pucks before collision (V1 and V2) and the velocity of the resultant stuck pucks; the following equation was used;

## Percentage Difference in Momentum:

The conservation of Momentum equation = (m1*v1) + (m2*v2) = 2 (m1 + m2) (V’)

Since the masses of the pucks are equal, the equation becomes:

V1 + V2 = 2 V’

To calculate V1 + V2, I calculated the resultant vector using a ruler after drawing a line through the middle. Through that I found the results to be:

Percentage Difference = {|V1 + V2| – |2V’| / |V1 + V2|} X 100 = 65.49127273%

## Percent Change in Kinetic Energy:

The kinetic energy of the pucks before the collision should not be equal to the kinetic energy after the collision, since this is an inelastic collision.

%u2206K = K (f) – K (i): K (f) = ?m (v%u2081’? + v%u2082’?) K (i) = ?m (v%u2081? + v%u2082?)

Because the masses were equa, the equation used can be:

K (f) = v%u2081’? + v%u2082’? K (i) = v%u2081? + v%u2082?

Percentage Difference = {[2v’^2 – (v1^2 + v2^2)] / (v1^2 + v2^2)} X 100

= 54.48648649%

## Conclusion:

These experiments involved measuring the conservation of momentum by applying Newton’s Third law of Motion. In investigation 1, the collision of the masses showed that there was exerted an equal and opposite force to the initial force which resulted to an elastic collision. Here, the momentum and kinetic energy were conserved. In Investigation 2, there was observed an inelastic type of collision where there was some loss of momentum and kinetic energy too.

These result that were obtained showed that there was no perfect

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