Ignoring air resistance!

Quote: Wheres My Shirt "Which would cause a force on the ball, and as I said, the mass only effects the force, so I took that into account "Mass would have no effect on speed, direction, or the path of the ball" It's not as straightforward as you might think; if the two balls have the same shape and velocity at release, they don't necessarily have the same trajectory ... Let's ignore gravity and just look at the air resistance - the force from that air resistance is proportional to the velocity squared (roughly speaking): F = kv^2 (k is some friction factor) We're expecting the force due to air resistance to be the same on both balls, as it only depends on the shape of the ball and the velocity. As we all know, F = ma, and so; kv^2 = ma kv^2 = m d/dt v or, another way, k (dr/dt)^2 = m d^r/dt^2 So we see that how the velocity of the ball changes in the face of air resistance contains a term dependent on the mass of the ball. I suspect that the mass of the ball is therefore significant in the case of 2 otherwise identical balls thrown at the same speed in the presence of air resistance. I think the trajectories may well be different, but I've not bothered to plot any of them! Edit - rather than bothering to solve the equations of motion numerically, I used some Google-fu and came up with this: jersey.uoregon.edu/vlab/Cannon/ which I think demonstrates the problem. Click the drag on button, and fire the cannon. then only change the density of the projectile, and try again!

Quote: Wheres My Shirt "Which would cause a force on the ball, and as I said, the mass only effects the force, so I took that into account "Mass would have no effect on speed, direction, or the path of the ball" It's not as straightforward as you might think; if the two balls have the same shape and velocity at release, they don't necessarily have the same trajectory ... Let's ignore gravity and just look at the air resistance - the force from that air resistance is proportional to the velocity squared (roughly speaking): F = kv^2 (k is some friction factor) We're expecting the force due to air resistance to be the same on both balls, as it only depends on the shape of the ball and the velocity. As we all know, F = ma, and so; kv^2 = ma kv^2 = m d/dt v or, another way, k (dr/dt)^2 = m d^r/dt^2 So we see that how the velocity of the ball changes in the face of air resistance contains a term dependent on the mass of the ball. I suspect that the mass of the ball is therefore significant in the case of 2 otherwise identical balls thrown at the same speed in the presence of air resistance. I think the trajectories may well be different, but I've not bothered to plot any of them! Edit - rather than bothering to solve the equations of motion numerically, I used some Google-fu and came up with this: jersey.uoregon.edu/vlab/Cannon/ which I think demonstrates the problem. Click the drag on button, and fire the cannon. then only change the density of the projectile, and try again!

Right, so what we've learned here is that...

Momentum, i.e. mass x velocity, isn't relevant to why the ball travels forward.

Mass is only relevant in relation to the ball, and that only matters in relation to forces acting upon it (i.e. the player, air resistance and gravity).

Velocity is the only important factor in why the ball travels forward. That is, the velocity of the player in a given direction is transferred to the ball at the time of release, and it's path will be defined by that in conjunction with: the angle of release and the force applied to the ball at that angle of release (as well as the forces of gravity and air resistance).

So really, momentum appears to have absolutely nothing to do it. Stevo should rename it to a more accurate "transferable velocity" rule!

Seriously though, you all need to get laid.

Or have a damn good monkey spank.

What Wellsy said on the last page was actually correct

But you didn't think so

I just pointed out that your refutation was only valid in the case of no air resistance. Your references to the force required was purely i the context of getting the ball up to that velocity as you passed it, not what happened afterwards

Which is not true in the case of air resistance being present.

Absolutely, momentum would only really apply in a perfect elastic reaction - whereby the momentum of the player is transferred to the ball. Considering the player has about 500 kgm/s, that would mean a player would need to throw the ball at about mach 3 for momentum to have any application.

That's quite impressive! I get bored after about 22

Not true, I actually changed my wording so that that would not be the case - "mass would only effect the required force to get it to that speed and in that direction".

Just because I didn't make clear what was applying the force doesn't change the fact I didn't exclude air resistance.

Can you work out the path of the ball then (other forces acting on it like air resistance and gravity) using trigonometry then?

i.e. If a static player passes a ball backwards at a 45degree angle at 10m/s, then using trigonometry it's velocity in a directly backwards direction is 7m/s.

If then a second player makes the exact same pass but is running forwards at 10m/s, then it would travel forward at 3m/s? (10m/s forwards - 7m/s backwards = 3m/s forwards).

Using Newtonian physics you can, yes.

To your example, then yes, the ball would travel forward at approximately 3m/s. At such slow speeds air resistance wouldn't have too much of an effect anyway.