(^142) Aptitude Test Problems in Physics
stant velocity. They can be found from the fol-
lowing equations:
mat = Tl — mg, mat = T2 - mg,
= T2, al = —2a 2 ,
where m is the mass of each load, and T 1 and T2
are the tensions of the ropes acting on the left and
right loads. Solving the system of equations, we
obtain al = —(2/5)g and a 2 = (1/5)g. Thus, the
acceleration of the left load is directed down-
wards, while that of the right load upwards. The
time of fall of the left load can be found from the
equation
04gt 2
h vt
.
0,
2
whence2.5v V 6.25v 2 5ht = (^) g g2 i g •
During this time, the right load will move up-
wards. Consequently, the left load will be the
first to touch the floor.
1.42. Each time the block will move along the
inclined plane with a constant acceleration; the
magnitudes of the accelerations for the downward
and upward motion and the motion along the
horizontal guide will be respectively
a 1 = pg cos a — g sin a,
a 2 = pg cos a g sin a,
a = tig cos a
(Fig. 150). Here a is the slope of the inclined plane
and the horizontal, and p is the coefficient of
friction. Hence we obtain
a—
aid-a2
2 '