物理難題求解答～～～～?
2015-11-09 9:58 am
A stone is dropped at t = 0. A second stone, with 4 times the mass of the first, is dropped from the same point at t = 170 ms. (a) How far below the release point is the center of mass of the two stones at t = 480 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?
回答 (1)
2015-11-09 12:55 pm
(a) How far below the release point is the center of mass of the two
stones at t = 480 ms? (Neither stone has yet reached the ground.)
L1 = 0.5*g*t^2 = 0.5*9.8*0.48^2 = 1.12896 m
L2 = 0.5*9.8*(0.48-0.17)^2 = 0.47089 m
ΔL = L1 - L2 = 0.65807 m
(b) How fast is the center of mass of the two-stone system moving at
that time?
V1 = √(2g*L1) = √(2*9.8*1.12896) = 4.704 m/s
V2 = √(2*9.8*0.47089) = 3.038 m/s
stones at t = 480 ms? (Neither stone has yet reached the ground.)
L1 = 0.5*g*t^2 = 0.5*9.8*0.48^2 = 1.12896 m
L2 = 0.5*9.8*(0.48-0.17)^2 = 0.47089 m
ΔL = L1 - L2 = 0.65807 m
(b) How fast is the center of mass of the two-stone system moving at
that time?
V1 = √(2g*L1) = √(2*9.8*1.12896) = 4.704 m/s
V2 = √(2*9.8*0.47089) = 3.038 m/s
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