With what speed does the mass hit the ground at the bottom of the cliff?The following takes place at a colony on the Moon some time in the distant...

Let m1 = mass cannon+platform
=46kg.

 m2 = mass  ball = 10kg
 

 u1 = u2 = initial horizontal velocity of m1+m2 =
3ms^-1

 v1 = final velocity of cannon and platform =
-2ms^-1

Use m1u1 + m2u2 = m1v1 +
m2v2 to find m2v2 the horizontal momentum of the ball after firing.
Rearranging:

m2v2 = m1u1 +  m2u2 -
m1v1

= (46 x 3) + (10 x 3) - (46 x (-2)) =
260kgms^-1

Now let m1u1 be the same horizontal momentum
when it hits the block.

m1u1 + m2u2 = m1v1 + m2v2 note that
v1 = v2 which is the final horizontal velocity of ball and block combined. Rearrange for
v1:

v1 = (m1u1 + m2u2)/(m1 +
m2)

= 260 + 0(stationary block)/10 + (32/1.62 (mass
block))

= 8.74ms^-1 this will be the horizontal component
of velocity at impact at the base of the cliff.

Find
vertical component of velocity at impact. Use v^2 = u^2
+2as.

v = (u^2 + 2as)^0.5 = (0 + 2(1.62)(50))^0.5 =
12.73ms^-1

The impact speed will be the magnitude of the
horizontal and vertical velocity vectors added using Pythagorus'
theorem:

Impact speed = ( 8.74^2 +
12.73^2)^0.5

= 15.44ms^-1