Solutions 259
determine the motion of the centre of mass of the
system. Since the masses of the balls are equal,
the initial position of the centre of mass is at a
height (h 1 h 2 )/2, and its initial velocity v is
horizontal. Then the centre of mass will move
along a parabola characterized by the following
equation:
h-Fh i g x a
.\i \h (^) (i)
2
2
k 2/k v
where x is the horizontal coordinate of the centre
of mass, and h is its vertical coordinate. At the
moment the first ball touches the ground at a
distance x = 1, the height H of the centre of mass,
according to expression (1), is
H h12h2 g \I 1 ‘ 2
k 2 v
Since the masses of the balls are equal, the second
ball must be at a height H2 = 2H at this instant.
Therefore,
1 - (^12) =k+ha—g(- 1 --) 2
3.21. Let the resistance of half the turn be R.
Then in the former case, we have fifteen resistors
of resistance R connected in parallel, the total
resistance being R/15.
In the latter case, we have the same fifteen re-
sistors connected in series, the total resistance
being 15R. Therefore, as a result of unwinding,
the resistance of the wire will increase by a factor
of 225.
3.22. It can easily be noted from symmetry con-
siderations that the potentials of points A and C
(Fig. 210) at any instant of time will be the same.
Therefore, the closure of the key K will not lead
to any change in the operation of the circuit, and
the coil AC will not be heated.
17*