7. RECALL: In the previous lab about momentum, we had to do the percent difference twice. (Check the video at the bottom of the lab to remind yourself of what you watched/measured.) We had to complete the percent difference twice because momentum is a vector. If we were to analyze the same video from before but in terms of energy, the error percentage will be higher than than normally acceptable. Which factors of the experiment explain why there is a greater degree of error when calculating energy versus calculating momentum? Note that some of the factors listed will introduce considerable error in either calculation. Select all that apply: Formula for energy involves velocity, which is found from incorrect formula of kinematics. Unable to get the ruler to a straight line for the path of each puck. Units for the mass of each puck is in grams and not kilograms. Energy of the pucks has a greater error because this was a top-down view instead of side view. Sound lost to the environment when the pucks hit. Using the incorrect decimals for the mass. Rounding error using percent difference equation.
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Adi S.
Question 4 The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 seconds, whereas the standard deviation is found to be 0.24 seconds. The standard error (or standard deviation of the mean) therefore is 0.24 seconds divided by the square root of 16, yielding 0.06 seconds. Assuming that the uncertainty of this period arises solely from random error for these 16 trials, about how many additional trials would need to be completed to reduce the standard error to 0.04 seconds? Question 5 For a certain experiment, you must provide the kinetic energy of some object with its uncertainty. (The equation for kinetic energy K = 1/2mv^2, where m is the mass of the object and v is the object's speed.) To do this, you first measure the mass of the object to be 0.015 kg. Due to the limits on precision for mass using a scale, you can only measure the mass of an object within an uncertainty of 2.7 grams (so Ļm=2.7x10^-3 kg). In this same experiment, you measure the speed as 10 m/s, but you can only measure the speed with a precision Ļv=1.1 m/s. Hence, the mass and speed you report are m = 15Ā±2.7 g and v = 10Ā±1.1 m/s, respectively. How would you report the kinetic energy using the error propagation formula given in the introduction (assuming the speed and mass measurements are independent of each other)? Hint, the derivative of kinetic energy with respect to mass dK/dm = 1/2v^2 and the derivative of kinetic energy with respect to speed dK/dv = mv. (fill in the blank) K = 0.75 Ā± ______ J
Sri K.
Part C (5 pts): Susan measured the momentum of two cars before and after an elastic collision. She found the total initial momentum to be 1985 Ā± 9 kgĀ·m/s and the total final momentum to be 2003 Ā± 8 kgĀ·m/s . Was momentum conserved in this experiment? Explain using mathematical evidence. Part D (5 pts): In lab, you measured the speed of carts using LoggerPro. LoggerPro calculates the carts speed by dividing the flag length by the amount of time it takes the flag to go through the photogate (v = d/t). In LoggerPro, you accidentally entered the flag length to be 0.15m. It took the flag 0.25s to go through the photogate. LoggerPro reported the cart speed to be v = 0.15m/0.25s = 0.60 m/s . Is this speed greater than, less than or equal to the speed LoggerPro would calculate with the correct flag length of 0.10m entered? Explain using mathematical evidence.
Suman K.
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