Question

Regading my question yesterday about the ball being caught by the man on the scales (Sun. 3:47), I'm now wondering if, to get the final weight on the scales I need to add the actual ball's mass of .50 kg AS WELL AS the force of the ball on the man to his original weight in order to get my final answer.If i do must I add .50*9.8 or does the force of the ball on the man catching it include this ball weight.
Sorry, for the second question but you explain things better than my school teacher does.

Answer 1

If you just used the deacceleration of the ball to get the force as he caught it, yes you should add the weight.

Answer 2

It doesn't matter. What you are ultimately computing is the force exterted by the feet of the man on the scales. If you do the computation by including the ball as part of the system, then you must take into account the fact that the system's center of mass is accelerating downwards.

So, when you set up the force balance to compute the force exerted by the scales on the man's feet (which by Newton's third law is minus the force exerted by the feet on the scales), you have to deal with the fact that the total force F is:

F = m a

where m is the mass of the ball and a the accleration of the ball.

You can also define your system boundary such that the ball falls outside it. Then the total force is zero, but you now have to include the foce exerted by the ball on the man.

Answer 3

It is the net force on the ball that results in its acceleration upward (deacceleration)
Force down = m g
F = Force up = force from man on ball up (what we are looking for, equal and opposite down on man and scale)
Fnet = m a
F - m g = m a
F = m g + m a

Answer 4

Thanks!