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.

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

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.

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

Thanks!

To determine the final weight on the scales after the man catches the ball, you need to consider the weight of the man, the weight of the ball, and the force exerted by the ball on the man. Here's how you can calculate it:

1. Start by calculating the weight of the man alone. Let's say his weight is W1.

2. Then, calculate the weight of the ball. You mentioned the mass of the ball is 0.50 kg. The weight of an object is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2). So, the weight of the ball (Wb) would be 0.50 kg * 9.8 m/s^2.

3. Now, consider the force exerted by the ball on the man when the man catches it. This force can be determined using Newton's second law, which states that force (F) equals mass (m) multiplied by acceleration (a). In this scenario, the ball is decelerating to a stop in the man's hand, so the acceleration is negative. Let's say the negative acceleration is -2 m/s^2. So, the force exerted by the ball on the man (Fb) would be 0.50 kg * (-2 m/s^2).

4. To calculate the final weight on the scales, add the weight of the man, the weight of the ball, and the force exerted by the ball on the man to the man's original weight. The final weight on the scales (Ws) would be W1 + Wb + Fb.

So, to answer your question, you need to add the weight of the ball (0.50 kg * 9.8 m/s^2) to the original weight of the man, as well as the force exerted by the ball on the man (0.50 kg * -2 m/s^2).