Please help~~

用戶85377

2011-02-26 8:28 pm

1. For no spilling of water, the minimum speed of a bucket of water at the highest point when whirled in a vertical circle is 2 ms^-1. What is the radius of the circle?

2. A light spring is fixed to the bottom of a vertical tube. A ball is released from rest at a height h above the upper end of the spring. After rebounding several times the ball eventually comes to rest and stays on top of the spring. Assume all contact surfaces are smooth and the spring obeys hooke's law throughout. Which of the following statements is/are correct?
(1) The compression of the spring is proportional to the mass of the ball.
(2) The compression of the spring is independent of the height h.
(3) The strain energy stored in the spring is equal to mgh.

3. A mass m slides on a frictless horizontal surface at speed v. It strikes a light spring of force constant k attached to a rigid wall. After a completely elastic encounter with the spring the mass moves in opposite direction. In terms of k, m and v, how long is the mass m in contact with the spring.

4. A satellite skimming over the surface of the Earth follows a circular orbit. By what factor must the speed of the satellite be increased so that it may escape from the earth permanently?

5. A man stands near a railway track when a train approaches and then passes him. The train sounds its whistle all the time. The highest note and the lowest note the man hears are of frequency 110Hz AND 90Hz. What is the actual frequency of the whistle?

回答 (1)

天同

2011-02-26 11:02 pm

✔ 最佳答案

1. For no spilling of water, centripetal acceleation = ghence, 2^2/R = g = 10where R is the radius of the circle, and take g = 10 m/s2
R = 0.4 m

2. Statements (1) and (2) are correctHooke's law applies, mg = kxi.e. x = mg/k
3. You could treat this impact motion as part of a simple harmonic motion. The contact time is clearly (1/2) of a period T.
since T = 2.pi.square-root[m/k]hence, contact time = T/2 = pi.square-root[m/k][ pi = 3.14159...]

4. Escape velocity Ve = square-root[2GM/R]where G is the Gravitational constant, M is the mass of earth and R is radius of earth.[you may refer to any physics textbook for proof of the equation]When satellite is skimming over earth surface,mv^2/R = GmM/R^2, where m is the mass of satellite and v is its speedv = square-root[GM/R]hence, Ve/v = square-root[(2GM/R)/(GM/R)] = 1.414
5. Use equation: change of frequency df = (fo).(c/u)
where fo is the actual frequency of the whistle, c is the speed of sound in air, u is the speed of the train
When the train is approaching,
110 - fo = (fo).(c/u)
When the train is going away,
fo - 90 = (fo).(c/u)

hence, 110 -fo = fo - 90
ie. fo = (110+90)/2 Hz = 100 Hz

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