If you throw a ball horizontally while standing on roller skates, you roll backward with a momentum that matches that of the ball. will you roll backward if you go through the motions of throwing the ball, but instead hold on to it? explain. assuming you've answered them in terms of momentum conservation, answer them also in terms of newton's third law (or vice versa, if you answered already in terms of newton's third law).

man does not release the ball he will not back down

Explanation:

When you throw a ball we must define a system for the analysis, if we form a system formed by the ball and the pitcher, the forces that develop are internal, so the total amount of movement does not change,

        p₀ = pₙ

This implies that before the launch none had movement po is zero, so

       pₙ = 0

       m1v1 + m2 v2 = 0

       m1 v1 = m2 v2

Where index 1 is used by the pitcher and two for the ball, from this expression we see that the man receives a quantity of movement of equal magnitude than the ball that in the opposite direction.

If in the case that the man does not release the ball he will not back down because in the end the two are subject and the ball will stop

 To perform an analysis of the same problem using the third ce Newton law, when the thrower exerts a force on the ball (action), it exerts a force of equal magnitude on the man, but in the opposite direction, notice that each force is exerted on a different body so they cannot be added. This law also explains the throwback of the pitcher.

Now if the pitcher does not release the ball the two forces exist, but when the ball stops two other forces will be formed in the opposite direction that cancel the first ones, so neither of them will move

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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an energy transformation flow diagram is shown

where is it ?