Procedure and Discussion Case
Procedure and Discussion
It is essential for the two balls to strike the floor at the same time so that they will cancel out in the equation:
viA = vfA + vfB
∆xiA/∆t = ∆xfA/∆t + ∆xfB/∆t
∆xiA = ∆xfA + ∆xfB
By the time being the same at initial and final, in other words constant, we can further prove that the value of the distance is the same as the value of the momentum. If the two balls were to not strike the floor at the same time, it indicates that there is a change in momentum since they are indirectly proportionate. This is not possible because momentum is always conserved in collisions, thus if the time is different it is not a true collision, thus further resulting in false data in the experiment.

As seen in the equation v= ∆d/∆t, v is directly proportional to ∆d, therefore the measured value of ∆diA (from point X to where it hits the tracing paper) can be said the same for the value of viA. As velocity increases, distance increases. As velocity decreases, distance also decreases.

As seen in the equation p=mv, p is directly proportional to v, therefore the value of viA (that is derived from ∆diA) is the same for the value of piA. As momentum increases, velocity increases. As momentum decreases, velocity also decreases.

If v= ∆d/∆t if substituted as v in p=mv, then p=m∆d/∆t. As seen in the equation, p is directly proportional to ∆d, therefore to find out piA, it is possible by measuring the distance from the collision of the incident ball on the target ball (point X) to where the target ball lands. Since the mass of both the balls are the same, they are not an effect on the situation. Also since the time of flight collision is constant throughout the system, it can be ignored for finding out the momentum as well. The momentum and distance in this case are the same in measured value although different in units. As momentum increases, distance increases. As momentum decreases, distance also decreases.

According to the results on the tracing paper, 1-pfb is on an angle in comparison to piA which means that the ramp was not perfectly aligned, causing a very slight two dimensional collision. This is in fact a source of error during experimentation, therefore aside from the slight error, it is ideally a one dimensional, head on collision.

Since, in theory, it is a one dimensional collision, it is expected that the incident ball will travel at a velocity of v down the ramp, and at collision its final velocity will transfer to the target ball which had been stationary on a screw at the end of the ramp. Since the incident balls momentum is transferred at collision, ideally, it is suppose to come to rest in the position of where the target had been stationary. With that

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Value Of The Momentum And Collision Of The Incident Ball. (April 4, 2021). Retrieved from https://www.freeessays.education/value-of-the-momentum-and-collision-of-the-incident-ball-essay/