Linear Momentum Lab - Lab report PDF

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Lab 5 - Linear Momentum Introduction: The goal of the lab is to experimentally confirm the conservation of linear momentum. In the lab, to test the principle of conservation of linear momentum, my lab partner and I slid a steel ball down a track, which, at the end of the track, collided into a hole in a wooden block and become embedded in the block. The block swung as a pendulum as a result of the impact. Finally, we were able to calculate and compare the linear momentum of the steel ball before and after the collision, to accomplish the goal of this lab. A real world application for the

conservation of linear momentum is the launchof a rocket. Upon ignition, a rocket sends exhaust gases downward (downward momentum). To balance this downward momentum, the rocket moves upward and linear momentum is conserved.

Procedure: In the beginning of the lab, we try to determine the velocity of the steel ball before the impact with the block. First, we made sure that nothing interfered with the projection of the steel ball by moving the block onto the platform. Second, we placed a strip of white, waxed paper to the floor of the box. Third, we released the steel ball from its highest point and let it roll down the track, until it falls into the box and leaves a mark on the waxed paper. Fourth, we determined the height, b, through which the ball falls. Moreover, we performed ten trials of releasing the ball on the track, and from these trials we found the average value of r. Also, now that we have the

values of r and b, we were able to calculate the velocity of the steel ball as it leaves the track by using the formula V = √(1/2g(r^2/b)). In the second half of the lab, we looked for the velocity of the steel ball after the impact with the block. First, we determined the mass of the steel ball and recorded the mass of the wooden block. Second, we suspended the block at the appropriate level. Third, we measured the distance h, using the equation √(h^2-x^2). Fourth, we measured the horizontal distance x, the distance the block swings after impact, where after a couple of tests the block’s swing sometimes just barely flicks the slider or just fails to touch it. Fifth, we calculated the vertical distance y using the equation y = h-√(h^2-x^2). Finally, using the conservation of energy, we used the change in height to figure out the velocity of the system using the formula Vb = √(2gy).

Data/ Calculations:  Ball Drop Distance b (m) = 0.145  Uncertainty in b (m) = 0.001  Average r value (m) = 0.38  Uncertainty in r (m) = 0.01  h (m) = 0.42  Uncertainty h (m) = 0.001  Y (m) = 0.041  V (before impact) = 2.209 m/s  V(after impact) = 0.896 m/s  Relative difference = 6.08% Questions:

1) How do multiple measurements of r change the uncertainty? The more repetitions you make of a measurement, the better the estimate will be. So, if you take the average of these measurements, you will have a more precise estimate. Thus, the uncertainty of r will decrease with multiple measurements. 2) Within the limits of your experimental accuracy, is momentum conserved during the collision? I believe, within the limits of my experimental accuracy, that momentum is conserved during the collision. I calculated the linear momentum of the system before and after impact, and got only a relative difference of 6.08%. 3) Derive equation (1), starting from general physics principles. a) mgh = 1/2 mv^2 b) 2mgh=mv^2 (multiply both sides by 2) c) 2gh=v^2 (divide both sides by m) d) √2gh = v (take the square root of both sides) 4) From your results, compute the fractional loss of kinetic energy of translation during impact. Disregard rotational energy of the sphere. 

Ball (kg) = 0.0672, Block (kg)= 0.0878, Vi= 2.209 m/s, Vf= 0.896 m/s



KEi=1/2mv^2= 0.164 J



KEf=1/2(m +M)v^2= 0.062 J

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Fractional change in KE: (KEi - KEf)/KEi= |-0.214 J| =0.214 J

5) Derive an expression for the fractional loss of kinetic energy of translation in terms only of m and M, and compare with the value calculated in the preceding question. Consider the collision as a totally inelastic one.

a) Initial momentum: p = m*vi, Final momentum: p = (M + m)*vf b) These are equal because momentum is conserved: m*vi = (M + m)*vf c) solve for vf: vf = vi*m/(M + m) d) KEi = 1/2*m*vi^2, KEf = 1/2*(M + m)*vf^2 e) Substitute: KEf = 1/2*(M + m)*(vi*m/(M + m))^2 f) Simplify: KEf = 1/2 * m^2*vi^2/(M+m) g) Fractional change in KE: (KEi - KEf)/KEi h) Substitute: ((1/2*m*vi^2) - (1/2 * m^2*vi^2/(M+m)) )/(1/2*m*vi^2) i) Cancel vi: ((1/2*m) - (1/2 * m^2/(M+m)))/(1/2*m) j) Cancel 1/2: (m - m^2/(M+m) )/m k) Divide through by m: 1 - m/(M+m) l)

Simplify the fraction: M/(m + M)  Same KE lost as question 4

Conclusion: Concluding the lab, according to my data from the experiment, I believe there is a conservation of linear momentum. My lab partner and I, calculated that the there is only a 6.08% relative difference in linear momentum of the ball/block system before and after impact. Also,

we found the velocity of the ball/block system before impact to be 2.209 m/s and the velocity after impact to be 0.896 m/s. Moreover, there could be some sources of error in our experiment, where we could have released the steel ball at different positions and miscalculated the value of r. Nevertheless, I believe that I have gained an understanding of the conservation of linear momentum and proved it at the conclusion of this laboratory experiment....