極難 力學題 (urgent)

1)

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Nov11/Crazymech2.jpg


6) Since the M >> 1, the change in the velocity of the man + boat can be neglected and hence their change in k.e. can be assumed to be zero.

So no matter whether throwing forward or backward, initial k.e. of the ball relative to the boat is 0 and final k.e. relative to the boat = 0.5 x 1 x 42 = 8 J which is the work done by the man on the ball.

7) The compression of the spring will be max. at the moment when both blocks have the common velocioty v, i.e.

(m1 + m2)v = m1vo

v = m1vo/(m1 + m2)

Then the loss of k.e. of the blocks is:

(1/2) [m1vo2 - (m1 + m2)v2]

= (vo2/2) [m1 - m12/(m1 + m2)]

= (vo2/2) [m1m2/(m1 + m2)]

which is also the e.p.e. stored in the spring, i.e.

(1/2)kx2 = (vo2/2) [m1m2/(m1 + m2)]

x2 = m1m2vo2/[k(m1 + m2)]

x = vo√{m1m2/[k(m1 + m2)]}

8) For the ball to be able to pass through O, the condition is:

e.p.e. stored in spring > work done against friction when the ball goes from A to O

kd2/2 > μmgd

μ < kd/(2mg)

Now, suppose that s is the overshoot of the ball after passing through O for the first time, then:

Work done against friction = μmg(d + s)

e.p.e. stored in the spring at overshoot s = ks2/2

So:

kd2/2 - ks2/2 = μmg(d + s)

(k/2)(d2 - s2) = μmg(d + s)

(k/2)(d - s) = μmg

d - s = 2μmg/k

s = d - 2μmg/k

Now, for the ball unable to pass through O from overshoot s:

e.p.e. stored in spring < work done against friction when the ball goes from overshoot s to O

ks2/2 < μmgs

μ > ks/(2mg)

= [k/(2mg)] (d - 2μmg/k)

= kd/(2mg) - μ

2μ > kd/(2mg)

μ > kd/(4mg)

8.
kd/4mg < μ那部分是該球反彈但不過O的情況，這部分的相關微分方程是x" = - kx/m + μg（但μ < kd/2mg那部分的相關微分方程是x" = - kx/m - μg）。