Solutions 1171.11. Let us suppose that hail falls along the
vertical at a velocity v. In the reference frame
fixed to the motorcar, the angle of incidence of
hailstones on the windscreen is equal to the angle
of reflection. The velocity of a hailstone before it
strikes the windscreen is v — v^1 (Fig. 128). Since
Fig. 128
hailstones are bounced vertically upwards (from
the viewpoint of the driver) after the reflection,
the angle of reflection, and hence the angle of
incidence, is equal to (Ai is the slope of the
windscreen of the motorcar). Consequently, a -f-
2 131 = n/2, and tan a = v/v^1. Hence tan a =
tan (3t/2 — 213 1 ) = cot 213 1 , and v/v 1 = cot 213 1.
Therefore, we obtain the following ratio of the
velocities of the two motorcars:
v 1 cot 213 2 3.v (^2) cot 213 1
1.12. Let us go over to a reference frame moving
with points A and B. In this system, the veloci-
ties of points A and B are zero. Since the distances
AC and BC are constant, point C, on the one
hand, can move in a circle of radius AC with the
centre at point A, and on the other hand, in a
circle of radius BC with the centre at point B.
Therefore, the direction of velocity of point C
must be perpendicular both to the straight line
AC and the straight line BC. Since points A, B,
and C do not lie on the same straight line, the
direction of velocity at point C would be perpen-