Okay, say I have the following problem and I want to calculate the velocity of the ball as soon as it is released and the maximum height it reaches.sophiecentaur said:Do you know the formulae for GPE and KE? They contain the quantities that you need to know about. What do you need to calculate? You haven't actually asked a practical question yet. To find how high a ball will go (or a number of other similar questions, you can rely on the fact that (ideally) no energy is lost so that the GPE at the maximum height will equal the initial KE + GPE in the system. This is referred to as Energy Conservation.
The 'Force' of the kick is not enough to predict how fast the ball will travel. You need to specify either the distance over which the kick is applied (giving the work done) or the time (the Impulse). This is a collision problem (which is what the kick is, basically). Either way, it is not easy to specify those initial conditions because a kick does not apply a constant force during the contact. Text book problems usually specify some very simplified conditions in their set questions but, to apply this to a practical problem is a bit of a nightmare - as are all mechanical calculations involving muscles and skeletons. In many ways it's easier to work backwards from the height reached and to work out either the KE or the Impulse imparted to the ball.UncertaintyAjay said:Calculating velocity isn't too hard. If your foot produces a force of 30 N, the net force on the ball is 30-4.9= 25.1 N
So the acceleration of the ball upwards is 25.1/0.5= 50.2 m/s^2. Use v^2 = u^2 + 2as to calculate the velocity at release. So v^2 = 0+2*50.2*0.05
Fabian901 said:Thanks a lot for the answers! Obviously in practical situations the force would not be constant across that small distance, but if we knew how the force varied and plotted it on a force displacement graph would we still be able to calculate KE (the area under the curve) and therefore the balls velocity when released?
Also, now that we know all the variables, would GPE and KE (at the time the ball is released) have the following values:
GPE = mgh = 0.5 x 9.81 x 0.3 = 1.47 J
KE = 0.5mv^2 = 0.5 x 0.5 x 5.02 = 1.255 J
Would all that KE be converted to GPE as the ball reaches maximum height?
One more question. Are we exerting a force through a distance every time we are giving motion to an object? Kicking a ball would be a clear example but if we are playing ping pong for instance, are we exerting a force through a distance every time we hit the ball without table tennis racket? Obviously that distance would be excessively small but it would still not equal 0 would it?
I see! Thanks a lot for the answer!sophiecentaur said:Air resistance will have an effect on a relative low mass / large area object. You can work out the drag factor for an spherical object in air. Incorporating that would give a better idea of final height. Low velocities will produce less effects from drag, of course.
Even a table tennis ball impact takes a finite time (so does the impact between the steel balls in a Newton's Cradle). In many calculations about collisions, they use a quantity called Coefficient of Restitution which is the ratio of the separation speed to the approach speed. It's 1 (100%) for a perfect elastic collision and can be anything down to Zero (for two lumps of putty). That is a simplified figure which assumes that the same proportion is lost for all speeds but is pretty successful in practice.
In many collision calculations, the actual time or distance during impact can be ignored and you can just work on the assumption that Momentum is Conserved and deal with the initial and final situations. COR comes in very handy for this, of course. I remember doing endless collision calculations in A Level Applied Maths.