Simple and Unquestionable Mathematical Proof Kinetic Energy is not ½mv² but mv By Gary Novak The proof is that only mv is conserved under all conditions, while ½mv² is not. The test is to use a rocket to replace the energy of falling objects and calculate the amount of fuel used for the added mass only. The rocket replaces or duplicates gravity to create the same velocities as dropping does. Then various conditions are compared. (Rockets are used to calculate force, which is independent of the definition of energy.) Here's a summary of the test:      A 4kg object dropped 1m (meter) has the same amount of ½mv² as a 1kg object dropped 4m, because force times distance equals ½mv² for an accelerating mass. But a rocket accelerating the masses to those velocities requires twice as much energy as fuel for the large mass as for the small one. (Also, the mv of the first is twice that of the second.) Therefore, both masses do not have the same energy; the rocket does not transform energy in proportion to ½mv²; ½mv² is not kinetic energy; and a gallon of fuel does not produce a consistent amount of ½mv². ----------------------------- A 4kg object dropped for 1s (second) has the same amount of mv (momentum) as a 1kg object dropped for 4s, because force times time equals mv for an accelerating mass. A rocket accelerating the masses to those velocities uses the same amount of energy as fuel for both masses. Therefore, both masses have the same amount of energy; the rocket transforms energy in proportion to mv; mv is kinetic energy; and a gallon of fuel produces a consistent amount of mv. The proof that there are no errors in the math is that the ratio for the first two tests is 2.000000, while the ratio of the second two tests is 1.000000. Any error would replace the zeros with other numbers. To determine the amount of energy in the falling objects, a mathematical model of a rocket can be used to add the energy. The total energetics can be calculated without doing an experiment. The rocket is constant powered, as rockets usually are. This means fuel is used at a constant rate. For this reason, burn time can replace actual fuel use. In other words, a two second burn uses twice as much fuel as a one second burn. Therefore, the method is to determine the time of burn required to give each of the objects their final velocities. The burn time is then adjusted for the payload only by subtracting the proportion of the mass due to the rocket. The rocket mass changes as fuel is burned; so an average mass is determined through calculus techniques. The Rocket Equations: The rocket is arbitrarily given three properties, and the rest is calculated. Its mass is 20kg; its rate of mass loss is 0.01kg/s; and the separation velocity of the exhaust is 1000m/s. x' = rate of change for x = dx/dt (' is rate of change) mo = mass of rocket at start = 20kg plus payload (21 or 24 kg) m' = rate of mass loss = -0.01kg/sec mt = mass of rocket at time T = mo + m't ve = separation velocity of exhaust = 10³m/sec (pos) F = force = -m've = -(-0.01)(10³) = 10 newtons v't = acceleration at time T = F/mt v = velocity = òv'dt = ò10/(mo - 0.01t)dt =      ò10³/(100mo - t)dt = -10³ln(100mo - t) + c      (-c is quantity at T = 0)      c = 10³ln(100mo) burn time = t = mo/0.01 - [(c-v)/10³]ex      ex is inverse of natural log for the preceding quantity. mass ratio averaged for loss of mass =      òmp/(mo - 0.01t) dt =      -100mpln(100mo - t) + c      -c is quantity at t = 0      c = 100mpln(100mo) Example: four kilograms dropped one meter. g = acceleration of gravity = 9.81 m/sec (rounded throughout) s = distance down = 1m t = time dropping = (2s/g)½ = 0.451523641 sec v = velocity after dropping = gt = 4.429446918 m/sec Constant C in burn time: c = 10³ln(100mo) = 10³ln[100(24)] = 7,783.224016 Total Time of Burn: t = mo/0.01 - [(c-v)/10³]ex =      24kg/0.01 - [(7,783.224016-4.429446918)/103]ex =      10.6071633 sec Fraction for Payload: mp = mass of payload = 4 kg tp = time for payload including mass ratio integrated with time =      òmp/(mo - 0.01t) dt =      -100mpln(100mo - t) + c c (- quantity at t = 0) = 100(4kg)ln[100(24kg)] = 3,113.289607 tp = -100(4kg)ln[100(24kg) - 10.6071633] + 3,113.289607 =      1.7717788 sec Explanation: This last step might look a little too magical even to an expert. Actually it is. By logical analysis, this last number would be divided by burn time to determine average mass fraction (Calculus averaging divides by the interval.). It would then be multiplied times burn time to determine the amount of time attributed to the payload. Since dividing and then multiplying by the same number is unnecessary, the integration quantity is in itself the exact quantity desired. It represents the amount of time attributed to the payload. Numbers for each object: A - 4 kg dropped 1 m B - 1 kg dropped 4 m C - 4 kg dropped 1 sec D - 1 kg dropped 4 sec Time Dropping: A - 0.45152364098573 sec B - 0.90304728197146 sec C - 1 sec D - 4 sec Velocity Dropping: A - 4.42944691807002 m/sec B - 8.85889383614004 m/sec C - 9.81 m/sec D - 39.24 m/sec Constant C in burn time: A - 7783.22401633603 B - 7649.69262371151 C - 7783.22401633603 D - 7649.69262371151 Total Time of Burn: A - 10.6071633272070 sec B - 18.5215158540212 sec C - 23.4288933861318 sec D - 80.8081749880098 sec Constant C in Mass Fraction: A - 3113.28960653441 B - 764.969262371151 C - 3113.28960653441 D - 764.969262371151 Time of Burn for Payload Only: A - 1.77177876722800 sec B - 0.885889383614004 sec C - 3.92400000000000 sec D - 3.92400000000000 sec Ratios: A/B – 2.00000000000000 C/D – 1.00000000000000 Details of Proof Energy Home This Page In Rich Text For Printing