One Dimensional Problems with F(t)
Basics:
From Newton’s 2nd Law: F(t) = ma = m dv(t)/dt
m dv = dp = F(t) dt
pf
po
dp =
t
0
F(t) dt
pf - po = 0t F(t) dt = p = impulse
v(t) = vo + (1/m) 0t F(t) dt
v = dx(t)/dt = vo + (1/m) 0t F(t) dt
dx = [vo + (1/m) 0t F(t) dt] dt
x(t) = xo + 0t [vo + (1/m) 0t F(t) dt] dt
x(t) = xo + vot + 0t [(1/m) 0t F(t) dt] dt
In order to find the motion with forces that depend on time only, we need to be able to do
two integrations.
Collected Homework: Problem #5: Given that the force on a particle of mass 5 kg
initially at rest at the location xo = 0 m is F(t) = 10 Nt * sin2( [2 rad/sec]*t).
Suggestion: in this problem use the symbols: m for the value of 5 kg, Fo for the
value 10 Nt, and  for the value of 2 rad/sec. Only use the numerical values
when plotting the graphs for x(t) and v(t).
a) find v(t);
b) find x(t);
c) Since the Taylor series expansion for sin() to first order gives sin()  , find
x(t) and v(t) for a particle of mass, m, initially at rest at the location xo = 0 m given
F(t) = 10 Nt * ([2 rad/sec]*t)2 .
d) Plot v(t) from part a with the v(t) from part c over a range of times from t = 0
seconds to t = 1 second; then plot x(t) from part b with the x(t) from part c over a
range of times from t = 0 seconds to t = 1 second.
e) Since the average of sin2() = ½, find x(t) and v(t) when a constant force of
F(t) = 5 Nt is applied, that is, F(t) = ½ Fo .
f) Plot v(t) from part a with the v(t) from part e over a range of times from t = 0
seconds to t = 100 seconds; then plot x(t) from part b with the x(t) from part e
over a range of times from t = 0 seconds to t = 100 seconds.
HINT: 0t sin²( t) dt = 1/2 (t - [sin( t) cos( t)/ ]). Extra 10% if you can derive the
result for the above integral.
Example done in class: Problem 2-4 in text: A high speed proton of electric charge "e"
moves with constant speed vo in a straight line past an electron of mass "m" and charge "e" initially at rest. The electron is at a distance "a" from the path of the proton.
At first glance, this looks like the force depends on distance rather than on time.
However, under one assumption we can convert this to a force that depends on time.
Later in the course we will solve this scattering problem exactly, but for now we will do
it approximately.
a) Assume that the proton passes so quickly that the electron does not have time to move
appreciably from its initial position until the proton is far away. Show that the
component of force in a direction perpendicular to the line along which the proton moves
is:
F(t) = ke2a / [a2 + vo2t2]3/2 where t = 0 when the proton passes closest to the
electron.
b) Calculate the impulse delivered by this force.
c) Using the above results, calculate the approximate final momentum and final kinetic
energy of the electron.
d) Show that the condition for the original assumption to be valid is: ke2/a << ½mvo2 .
Solution:
t=0
vo t
a
r

e-
a) look at the diagram, and consider the x and y components of the force at some
position:
Fx = (ke2/r2 )*(cos())
Fy = (ke2/r2 )*(sin())
where r2 = a2 + vo2t2 , cos() = vot / r , and sin() = a/r . Note that r is always positive,
but (vo t) can be positive for positive time, but negative for negative time.
b) vy(t) = vyo + (1/m) tot F(t) dt = 0 + (1/m) tot [ke2a / [a2 + vo2t2]3/2 ] dt
since vyo = 0 (electron initially at rest at t = -∞), we have:
vy(t) = (ke2a/m) I , where I = tot [1/ [a2 + vo2t2]3/2 ] dt and to is negative infinity.
If we divide numerator and denominator in the integral by (1/a3), we get:
I = (1/a3) tot [1/ [1 + (vot/a)2]3/2 ] dt
If we let u = (vot/a), [note that u is a dimensionless quantity]
then I = (1/a3) uou [1/ [1 + u2]3/2 ] (a/vo)du
= (1/a2vo) I' where I' =
u
uo
[1/ [1 + u2]3/2 ] du
We can look up this integral, and we get: I' = uou [1/ [1 + u2]3/2 ] du = [u/[1+u2]1/2 ]uou
This can be verified by taking the derivative of [u/[1+u2]1/2 ] and showing that it equals
[1/ [1 + u2]3/2 ].
Lower limit: In our case, to = negative infinity, so uo = (voto/a) must also equal negative
infinity. The expression [u/[1+u2]1/2 ] for u=uo is equal to -1 since in the denominator the
1 can be neglected compared to infinity squared, and u/[u2]1/2 = -1. The minus sign
comes from the fact that uo is negative.
Upper limit: u = vot/a so [u/[1+u2]1/2 ] = [(vot/a) / [1+(vot/a)2]1/2] .
Therefore, vy(t) = (ke2a/m)*(1/a2vo)*{ [(vot/a) / [1+(vot/a)2]1/2] - (-1) }.
To get the full impulse, we let t go to infinity, so
vyf = (ke2a/m(*(1/a2vo)*{1 - (-1)}
= 2ke2/mavo . Notice that the units do work out!
The full impulse delivered is p = mvyf - mvyo = mvyf - 0 = 2ke2/avo .
c) The final momentum of the electron will be the same as the full impulse:
pyf = 2ke2/avo .
The final kinetic energy of the electron will be: KEf = ½mvyf2 = 2k2e4 / a2mvo2 .
d) When will the original assumption be valid? (The original assumption was that the
electron does not move appreciably as the proton passes.)
We need some t around t=0 (closest approach) to see how far the electron moves during
that time when the force is near its maximum. Due to the symmetry, the electron should
be moving at half its speed at the time of closest approach. Hence we need the distance
covered during the t to be much less than a: ½ vyf t << a .
To get an idea of t , we can consider the impulse: F t = p . If we had the maximum
force applied for a small time (instead of a changing force applied for infinite time) to
give the same change in momentum (impulse), then we could calculate that time to be:
t = p / Fmax = (2ke2/avo) / (ke2/a2) = (2a) /vo .
Therefore the condition: ½ vyf t << a becomes
½ vyf (2a/vo) << a, or with vyf = 2ke2/mavo
½ (2ke2/mavo) (2a/vo) << a, or
2ke2/mvo2 << a, or
ke2/a << ½ m vo2 .
Another way of looking at this condition is to consider the motion from the point of view
of the proton: In this case, it is the electron that is moving with the speed vo in the -x
direction. For the approximation to be valid, the electric potential energy of the electron
at its closest approach must be much less than the kinetic energy of the electron as it
passes!