A 0.145 kg baseball pitched at 35.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 58.0 m/s. If the contact time between bat and ball is 2.00 10-3 s, calculate the force (assumed to be constant) between the ball and bat.
a) whats the change in momentum of the molecule?
let P be the change in momentum and positive velocity is in the direction of the wall
P = M*V(initial) + M*V(final)
V(initial) = -V(Final) = V
P = 2MV
P represents the change in momentum of the molecule, so by conservation of momentum the change in momentum of the wall is -P (as P-P = 0)
Let dt = (triangle)t
F = -change in momentum over change in time = -(-P)/(dt)= P/dt
which is your average force
F (average) = force per molecule * how many molecules hit in a second
you have 'force per molecule'
how many molecules hit is inversely dependent on the interval each between each molecule hitting the wall. What happens if you 1/2 the interval? you double the number of molecules hitting, what about *2 the interval? 1/2 the number of molecules hit.
thus: how many molecules hit per second = C * 1/t
where t is your interval
and C is some constant
To work out the constant, how many molecules hit it t = 1second? 1! so C = 1
then...
F = (P/dt) * (1/ t)
let P be the change in momentum and positive velocity is in the direction of the wall
P = M*V(initial) + M*V(final)
V(initial) = -V(Final) = V
P = 2MV
P represents the change in momentum of the molecule, so by conservation of momentum the change in momentum of the wall is -P (as P-P = 0)
Let dt = (triangle)t
F = -change in momentum over change in time = -(-P)/(dt)= P/dt
which is your average force
F (average) = force per molecule * how many molecules hit in a secondyou have 'force per molecule'
how many molecules hit is inversely dependent on the interval each between each molecule hitting the wall. What happens if you 1/2 the interval? you double the number of molecules hitting, what about *2 the interval? 1/2 the number of molecules hit.
thus: how many molecules hit per second = C * 1/t
where t is your interval
and C is some constant
To work out the constant, how many molecules hit it t = 1second? 1! so C = 1
then...
F = (P/dt) * (1/ t)
a)lol at the triangle(it is called delta) and also the long explanation by Nesus lol . how about this?
F*Δt = m*ΔV
F = m*ΔV/Δt = m (v2-v1)/Δt
v1 = + v
v2 = - v
Δv = -v - v = - 2v
F = m (-2v) / Δt = - 2vm/Δt
tipe? what is that? :o
F*Δt = m*ΔV
F = m*ΔV/Δt = m (v2-v1)/Δt
v1 = + v
v2 = - v
Δv = -v - v = - 2v
F = m (-2v) / Δt = - 2vm/Δt
QUOTE
(b)If molecules, all of this tipe, strike the wall at intervals a time t apart (on the average), what is the average force on the wall averaged over a long time?
tipe? what is that? :o
delta is an important topic in maths/physics. It is, as [B]Drude[B/] points out, the change in a variable. These changes will not affect constants like mass until we reach relatavistic velocities and normally they are concerned with velocity, distance and time. The change in one variable with respect to another can be represented graphically and at any point on the graph the gradient can be determined to evalute this rate of change.
Calculus has been extracted from this notion and it is one which you should familiarise yourself with fully.
And always try V=IR and if that doesn't work, F=ma, although in a year or two you may be considering both equations similar.
Calculus has been extracted from this notion and it is one which you should familiarise yourself with fully.
And always try V=IR and if that doesn't work, F=ma, although in a year or two you may be considering both equations similar.
you will also need to have the amount of time the particle is in contact with the wall. the way you explained the problem shows that it is only an inelastic collision which is about momentum. force deals with acceleration. so you will need how long the particle is in contact with the wall and the initial and final velocities. this'll give you the change in velocity; hence, the acceleration.