Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 113

motion of the ball is the total time of the ascent
and descent of the body thrown upwards at a veloc-
ity 4 sin a in the gravitational field. Consequent-
ly, ti = 24 sin a/g. The motion of the ball along
the horizontal is the sum of two motions. Before
the collision with the wall, it moves at a velocity
4 cos a. After the collision, it traverses the same
distance backwards, but at a different velocity.
In order to calculate the velocity of the backward
motion of the ball, it should be noted that the ve-
locity at which the ball approaches the wall (along
the horizontal) is v 0 cos a + v. Since the impact
is perfectly elastic, the ball moves away from the
wall after the collision at a velocity v 0 cos a + v.
Therefore, the ball has the following horizontal
velocity relative to the ground:
(v 0 cos a ± v = vo cos a + 2v.

If the time of motion before the impact is t, by

equating the distances covered by the ball before

and after the collision, we obtain the following

equation:

v 0 cos a•t = (t^1 t) (v 0 cos a -I- 2v).

Since the total time of motion of the ball is t 1

24 sin a/g, we find that

t=

v 0 sin a (v 0 cos a+ 2v)

g (vo cos a+ v) •

11.9*. Figure 125 shows the top view of the trajecto-

ry of the ball. Since the collisions of the ball with

the wall and the bottom of the well are elastic, th e

magnitude of the horizontal component of the

ball velocity remains unchanged and equal to v.

The horizontal distances between points of two

successive collisions are A21 2 = A 244 2 = A 2 A 3 =

... = 2r cos a. The time between two successive
collisions of the ball with the wall of the well is

t = (^1) 2r cos a/v.
The vertical component of the ball velocity
does not change upon a collision with the wall and
reverses its sign upon a collision with the bottom.
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