132 Aptitude Test Problems in Physics
where
2cooi/TV2g)
In the second case (after the jerk), the motion
can be regarded as if the sledge stuck in the sand
had a velocity v 1 > vo and having traversed the
distance xo were decelerated to the velocity vo
(starting from this moment, the second case is
observed). The motion of the sledge after the jerk
can be represented as a part of the total vibratory
motion according to the law
x (t) = x 1 sin coot, v (t) = v 1 cos coot
starting from the instant t 2 when the velocity of
the sledge becomes equal to vo. As before, x 1 =
v 1 /coo. Besides,
mg 2 MV? MV:
2/ Px° 2 2
whence
The
jerk is
v1 = xocoo^ l/2.
distance covered by the sledge after the
x1—xo =
coo coo coo
Consequently, the ratio of the braking lengths is
Xi — Xo =
y2-1.
xo
In order to determine the time of motion of the
sledge after the jerk, we must find the time of mo-
tion of the sledge from point xo to point x 1 by
xo , 17-- m rt.
V0 k^ g
coo
It can easily be seen that the time before the sledge
comes to rest is equal to quarter the "period of vi-
brations". Therefore,
31/2
V1
vo
1
(v1— v0) = X0 (17-2 -.