(^138) Aptitude Test Problems in Physics
whence sin a = v 1 /v 2 = 10/20 = 1/2, a = 30°.
Then the time of motion of the balls before colli-
sion is t = s/(v 2 cos a) 0.6 s.
Since the balls are heavy, the role of the air
drag can easily be estimated. The nature of motion
of the first ball will not change significantly since
the acceleration due to the air drag is amax =
1 m/s^2 even if the mass of each ball is 10 g, and
the maximum velocity of the first ball is v 1 =
10 m/s. This acceleration does not change the total
time of motion of the first ball by more than 1%.
Since the air drag is directed against the velocity
of the ball, we can make the balls collide by impart-
ing the same vertical velocity component to the
second ball as that of the first ball provided that in
subsequent instants the vertical projections of the
accelerations of the balls are identical at any in-
stant of time. For this purpose, the angle a formed
by the velocity vector of the second ball with
the horizontal at the moment it is shot off must
be equal to 30°.
1.37. Let us write the equation of motion for the
ball at the moment when the spring is com-
pressed by Ax:
ma = mg — k Ax.
As long as the acceleration of the ball is positive,
its velocity increases. At the moment when the
acceleration vanishes, the velocity of the ball at-
tains the maximum value. The spring is com-
pressed thereby by Al such that
mg — k Al = 0,
whence
Al =
mg
k •
Thus, when the velocity of the ball attains the
maximum value, the ball is at a height
mg
from the surface of the table.
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