phy難題急求解 - Ans.fyi 參考答案

phy難題急求解

2012-12-05 5:01 am

以下題目以弟百思不得其解,還望有人能夠解答萬分感激

題目一:
A rocket standing on its laughing platform points straight upward.Its engine are activated and eject gas
at a rate of 500kg/s.The molecules are expelled with a speed of 5km/s.How much mass can the rocket initially have if it is slowly to rise because of the engines?
題目二:
A light inelastic string passes around two small and smooth pulleys A and B in the same horizontal line at a distance 2a apart,and carries masses m1 at each end and a mass m2 (<2m1)at its middle point.The system is then released from rest with m2 at the middle point of AB.The system comes to instantaneous rest when m2 has fallen a distance.By applying the principle of conservation of mechanical energy,find the distance that m2 travlled?

回答 (1)

用戶25975

2012-12-05 5:47 pm

✔ 最佳答案

1) Expelling force = 500 x 5000 = 2500000 N

So the rocket can have, at max., a mass of 2500000/9.81 = 254842 kg initially for being able to rise.

2) Suppose that at the momentarily at rest instant, each of m1 has travelled a vertical distance h upwards, then:

Total length of the string segments joining m2 = 2(a + h)

Then vertical distance fallen by m2 = √[(a + h)2 - a2] = √(h2 + 2ah)

Using conservation of mech. energy, gpe gain by m1 = gpe loss by m2, so:

m2g√(h2 + 2ah) = 2m1gh

√(1 + 2a/h) = 2m1/m2

1 + 2a/h = 4(m1/m2)2

2a/h = (4m12 - m22)/m22

h = 2am22/(4m12 - m22)

h2 + 2ah = h(h + 2a)

= [2am22/(4m12 - m22)] [2am22/(4m12 - m22) + 2a]

= 4a2 [m22/(4m12 - m22)] [m22/(4m12 - m22) + 1]

= 4a2 [m22/(4m12 - m22)] [4m12/(4m12 - m22)]

= 4a2 m12m22/(4m12 - m22)2

Hence distance travelled by m2 is:

√(h2 + 2ah) = 2am1m2/(4m12 - m22)

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