The Conservation of Momentum in Collisions Essay Example

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 are 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%

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 elastic collision as loss in both momentum and kinetic energy were recorded. This is because the set-up was perfect and the presence of friction in the system made it impossible to conserve either kinetic energy or momentum. In addition, errors in the set up and calculations were some causes of the discrepancies in results.