This is a conservation of momentum question. Initial momentum of the system is the momentum of ball 1 plus the momentum of ball 2. The final momentum of the system should be 0 since the balls stand still after the collision.
mv + mv = 0
mv = -mv
(20)(20) = -(40)Vi
400 = -40Vi
Vi = -10
So ball 2 should travel at 10m/s to the left
The weight of a 76 kg astronaut on Earth is 750N.
The weight of a 76 kg astronaut on the Moon is 130N.
The weight of a 76 kg astronaut on Mars is 280N.
The mass is the same everywhere, the weight chages with change of location (gravity field). I rounded all answers to two significant figures. Hope this helps.<span>
</span>
Assuming the cannonball is fired horizontally, its horizontal velocity stays at a constant 76 m/s. At the point it hits the ground, it has a speed of 89 m/s, so if its vertical velocity at that moment is 
, we have
(taking the negative square root because we take the downward direction to be negative)
Recall that
where 
and 
are the initial and final velocities, respectively; 
is the acceleration; and 
is the change in position of a body. In the cannonball's case, it starts with 0 vertical velocity and is subject to a downward acceleration with magnitude 
. So we have
(which is negative because we take the cannonball's starting position at the top of the cliff to be the origin) so the cliff is about 109 m high.
We know, 
.
We know, kinetic energy is given by :
Hence, this is the required solution.
