mechanics questions (urgent)

2011-09-01 10:16 pm

1. A particle is accelerating uniformly along a straight line ABCD. If it travels the successive distances AB=x, BC=y, CD=z in equal intervals of time, prove that
a) z=2y-x
b) the ratio of the speeds at D and A is (5y-3x)/(3x-y)

2. A steel sphere falls from the rest through a viscous liquid. Which the acceleration falls exponentially but not linearly?

3. A heavy rope is held stationary on a smooth table with a small portion hanging outside the table. The total length of the rope is less than the height of the table. Why the acceleration of the rope increases with time but not constant?

4. If the question says 'two particles P and Q are set to travel in a circular paths of the same radius', does it imply that P and Q have the same mass if no further info is given?

5.

圖片參考：http://farm7.static.flickr.com/6088/6101888305_a0f4e0939a_b.jpg

The ans is D. Why? I get T=m l w^2 (w is the angular velocity). (the solution also states this)
So why (2) and (3) is incorrect? Which parts did I get wrong?

thanks

更新1:

for 2 http://farm7.static.flickr.com/6181/6102136087_9c8f143177_b.jpg why the a-t graph of the steel sphere is the 1 on the left but not on the right? the book says that equation can be proved. How? for 5, so (2) and (3) are incorrect if m, l, and w are kept unchanged?

更新2:

Sorry, would u mind showing the steps of getting v = (mg/k)[ 1 - exp(-kt/m)] and a = dv/dt = g.exp(-kt/m)? I've learnt calculus but I'm still not too familiar with it.

回答 (1)

天同

2011-09-02 12:16 am

✔ 最佳答案

1. Use equation: s = ut + (1/2)at^2
we get the floowing three equations,
x = ut + (1/2)at^2 ----- (1)
(x+y) = u(2t) + (1/2)a(2t)^2 = 2ut + 2at^2 ----- (2)
(x+y+z) = u(3t) + (1/2)a(3t)^2 = 3ut + (9/2)at^2 ----- (3)
solving the x, y and z from the above equations,
y = ut + (3/2)at^2 ----- (4)
z = ut + (5/2)at^2 ----- (5)

(a) 2y - x = 2[ ut + (3/2)at^2] - [ ut + (1/2)at^2] = ut + (5/2)at^2 = z
(b) 5y - 3x = 5[y = ut + (3/2)at^2] - 3[ ut + (1/2)at^2] = 2ut + 6at^2
3x-y = 3[ut + (1/2)at^2] - [ ut + (3/2)at^2 ] = 2ut
hence, (5y-3x)/(3x-y) = (2ut+6at^2)/2ut = (u+3at)/u = [u+a(3t)]/u = velocity at D/vel at A

2. Your question is not clear.

3. Consider an instant of time when the length of the rope hanging on the table is l. If the whole length of rope is L, then use: force = mass x acceleration
(l/L)Mg = Ma, where M is the mass of the rope and a is the acceleration
i.e. a = (l/L)g
But as the rope moves, the hanging portion increases with time, thus a increases with time.

4. Not necessarily. The centripetal force acting on P and Q may not be the same.

5. I don't think option D is the answer.
By resolving the tention T vertically and horizontally,
T.cos (theta) = mg
i.e. T = mg/cos(theta)
if other parameters remain unchanged, increasing of (theta) gives an increase of T. Hence, statement (1) is correct.

2011-09-01 20:17:38 補充：
2. The equation is: mg - kv = ma, where k is a constant
i.e. mg - kv = m(dv/dt)
solve this equation, will get v = (mg/k)[ 1 - exp(-kt/m)]
hence, a = dv/dt = g.exp(-kt/m)

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