klwong said:
p=mv
since speed is slower, during
the backword process the final momenton of the black ball is smaller?
What you're actually wanting to look at is the momentum transfer between each pair of objects upon acceleration and collision.
Let's simplify things real quick. Let's assume we only have one ball, the black ball.
Upon firing the gun imparts X momentum to the black ball and -X momentum on itself and the spaceship to which it is attached. Then, when the ball is caught, it imparts X momentum to the spacecraft and the spacecraft imparts -X momentum to the ball. Since ##X + (-X) = 0## and ##-X+X = 0## the net momentum on both the ball and the spaceship at the end of this is zero.
However, the spaceship will have moved to the right in between the ball being fired and being caught. Let's say that distance is ##D##.
Now, the robot carries the ball back to the gun. In order to do so it has to apply a force on the spaceship that points left so that it can accelerate to the right. This force accelerates the spaceship to the left, imparting some amount of momentum to the spaceship. Let's say this is ##X/10##. The robot thus has momentum ##-X/10## by conservation of momentum.
The robot then decelerates when it gets to the gun, which requires a force applied to the spaceship that slows both the ship and the robot to zero momentum. The distance moved by the spacecraft is slightly more than before after the ball was fired, say ##-1.05D##, as we have to include the robot's mass into this.
Then the robot moves back to its starting position, the spacecraft moves ##0.05D##, and everything ends up back in its exact original position. ##D - 1.05D + 0.05D = 0## after all. If you don't believe that the spacecraft will move back to its exact position then you'll have to delve into a more complicated and in-depth analysis that I won't get into here.
Adding six more balls changes nothing. You just have more momentum transfers to take into account, each of which conserves momentum. Just think about it. If every momentum transfer conserves momentum, then there is no setup possible in which the sum of all the momentum transfers can lead to anything but whatever the original momentum was, which in this case was zero. So we don't even need to look at varying setups with many different moving objects or at setups that have complicated, curved paths. By virtue of conservation of momentum we know that there is NO setup that can result in a reactionless drive.