(^158) Aptitude Test Problems in Physics
ities of the washer and the block become equal.
Then the washer will slide down the block, the
block being accelerated until the washer passes
through the lowest position. Thus, the block will
have the maximum velocity at the instants at
which the washer passes through the lowest posi-
tion during its backward motion relative to the
block.
In order to calculate the maximum velocity of
the block, we shall write the momentum conserva-
tion law for the instant at which the hlock is sepa-
rated from the wall:
m2 1/ 2 gr = m2v2,
and the energy conservation law for the instants at
which the washer passes through the lowest posi-
tion:
m 2 gr
MIA m 2 v 2
2 '^2 '
This system of equations has two solutions:
(1) vi = 0, v 2 = ii2gr,
2m 2 ms—m l
(2) v 1 , y 2gr, v 2 =
mi+ m2 mi+ m2
Solution (1) corresponds to the instants at which
the washer moves and the block is at rest. We are
interested in solution (2) corresponding to the in-
stants when the block has the maximum velocity:
2m 2 112gr
vimax — m1+ m2 •
1.56. Let us go over to a reference frame fixed to
the box. Since the impacts of the washer against
the box are perfectly elastic, the velocity of the
washer relative to the box will periodically reverse
its direction, its magnitude remaining equal to v.
It can easily be seen that the motion of the washer
will be repeated with period 2At, where At =
— 2r)/v is the time of flight of the washer be-
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