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Investigating Force, Mass and Acceleration using a Trolley

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Investigating Force, Mass and Acceleration using a Trolley Aim of experiment: Using simple trolley experiments, take suitable measurements from them to find the Mass of the trolley, the fraction force acting in the system and to investigate any other effects found when using apparatus. Let M = the mass of the trolley (A), MA = Mass added onto the trolley MB = the mass of the hanging mass (B) t = Tension in the string g = Gravitational force = 9.8 N R = the resistive force acting in the system. i.e. Fraction between the track and the trolley, fraction force in the pulley and air resistance. The Diagram above shows the setup of apparatus for the particular experiments. The acceleration meter measures the acceleration of the system by measuring the time taken through each of the 4cm cardboard and the time interval (t) between the two cardboards. The speed (u, v) through each of them is calculated and the acceleration is calculated using the formulae. a = (v - u)/ t Two sets of experiments are carried out: First experiment: 0.09 kg of mass (MA) is placed on the trolley and the hanging mass (MB) is 0.01 kg at the start. The acceleration of the system is measured. ...read more.


N on the system. Using the equation obtained before: a = (MBg - R) / (M + MA + MB) Let k = factor of M 1 ? k ? 5 a = (MBg - R) / (kM + MB) a = (0.06g - R) / (kM + 0.06) The Table below shows the results obtained form this experiment. k Mass/(kM)kg Acceleration1 / ms-2 Acceleration2 / ms-2 Acceleration3 / ms-2 Average Accel / ms-2 1 1M 1.22 1.22 1.20 1.21 2 2M 0.61 0.61 0.61 0.61 3 3M 0.41 0.40 0.41 0.41 4 4M 0.30 0.29 0.30 0.30 5 5M 0.22 0.23 0.21 0.22 The Graph show that a is approximately equals to 1.24 k-1. The Acceleration is inversely proportional to k and as k doubles, the acceleration halves. In order to prove that a = (0.06g - R) / (kM + 0.06) supports the graph and the data from the second experiment. A graph with inverse acceleration (1/a) against k is plotted. Graph showing 1/a against k By making a into 1/a in the equation. The equation of the graph could be formed. a = (0.06g - R) / (kM + 0.06) 1/a = (kM + 0.06) / (0.06g - R) 1/a = kM/ (0.06g - R) ...read more.


1.53 0.03 1.74 1.72 1.7 1.72 0.04 1.97 1.94 1.93 1.95 0.04 A same graph for the first experiment is plotted but with error bars. The maximum and minimum of the mass of trolley and the resistive force is calculated. The line represent the maximum regression F = 0.4933A+ 0.0439 M + 0.1 = 0.493 R = 0.0439 N M = 0.393 kg The line represent the minimum regression F = 0.4739A + 0.0425 M + 0.1 = 0.474 R = 0.0425 N M = 0.374 kg 0.393 - 0.383 = 0.01 0.0439 - 0.0434 = 0.001 0.374 - 0.383 = -0.01 (2d.p.) 0.0425 - 0.0434 = -0.001 (3d.p.) Therefore the mass of the trolley is approximately 0.38 +/- 0.01 Resistive force is approximately 0.043 N +/- 0.001 The final result I obtain is only an approximate value due to the fact that there is uncertain errors caused by the light gate, the inclination of the track and the variation of the mass used. In the second experiment the resistive force of the system is proved to be inconstant as the mass of the trolley increase and this could cause errors on the results I obtained from the first experiment because the mass of the trolley is not constant and I assumed the resistive force to be constant in the calculation. ?? ?? ?? ?? JUSTIN WONG 4/23/2007 ...read more.

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