CollisionTwo blocks of masses, m1 = 2.00kg and m2 = 4.00kg, are each released from rest at a height of 5.00 meters on a frictionless track and...

To have an ahead on collision the blocks should traverse
in opposite directions. The speed gained by each of the blocks in opposite directions
under gravity through the frictionless smooth track  is given by
:

v^2  = 2gs = 2*9.81*5.
Or

(i) v = 9.904544412m/s is the speed in opposite
directions before impact. m2 has +9.90..m/s anticlocwise  and m1 has -9.90..m/s  clock
wise say.

ii) The velocities of the blocks after collision
is given by:

v1f (final velocity of m1 ) =
 [2m2b+(m1-m2)a]/(m1+m2) where a and b are the initial velocities of the blocks m1 and
m2, in this case -v and +v

=[2*4v+(2-4)(-v)]/(2+4) = 10v/6
= 5v/3 = (5*9.904544412m/s)/3 = +16.5076 m/s anti
clockwise.

v2f (final velocity v) =
 [2m1a+(m2-m1)a]/(m1+m2)=2*2(-v)+(4-2)V]/(2+4) = (-4+2)v/6 = -v/3 =
-9.904544412m/3 m/s =
3.3015 clockwise.

iii)

The
first block with speed 16.5076m/s could move a height h meter, given by 2gh =
16.5076..^2 Or h = 16.5076^2/(2g) = 13.8889 meter of height , where g is the
acceleration due to gravity and is assumed  9.81m/s^2

The
second block could go as high as  3.3015^2/(2g) = 0.5556 meter of
height.