Well, a light truck has a mass : m
a heavy truck has a mass : M
where : M > m
so if the have the same momentum :
m.V = M.V'
M > m, then V > V'
Let's set values ; m = 4 ; M = 8
4V = 8V' >>>> V = 2V'
Kinetic energy :
4*V^2 / 2 = 2V^2 >>>> light truck
8*V^2 / 8 = V^2 >>>> heavy truck
interesting, the light truck has more kinetic energy
2007-03-01 02:36:59
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answer #1
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answered by anakin_louix 6
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Given each truck has same momentum:
m1v1 = M2V2,
where m1, v1 are mass and velocity of LIGHT truck
where M2, V2 are mass and velocity of HEAVY truck.
V2 =(m1v1) / M2
So kinetic energy of heavy truck is (1/2) M2V2^2 =
(1/2) M2 [m1v1 / M2] ^2 simplified to [(1/2) (m1v1)^2] / M2
and you'll notice the bracketed numerator is the kinetic energy * of the light truck! So despite your intuition the light truck has the greater KE.
[Intuition might suggest a 10 lb weight would fall twice as fast as a 5 lb weight. Sometime you can't trust it.]
*Footnote/Amendment/SLIGHT CORRECTION: The bracketed quantity is m1 TIMES the KE of lighter truck; so the KE of heavier truck is
m1/M2 (which is less than 1)times the KE of the lighter truck. The original premise is OK; my supporting algebra could use closer perusal!
2007-03-01 03:27:52
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answer #2
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answered by answerING 6
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Well, the simple answer in this case is the best. The one with the most kinetic energy will require the most energy to stop (the heavy truck). Force = mass x acceleration, right?
2007-03-01 02:36:09
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answer #3
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answered by f1le_f0und 3
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The heavy truck will and here is why
kinetic energy
n. The energy possessed by a body because of its motion, equal to one half the mass of the body times the square of its speed.
2007-03-01 02:36:59
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answer #4
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answered by scrow_80 3
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2007-03-05 02:32:14
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answer #5
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answered by Anonymous
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