One Dimensional Problems with F(t)

Newton’s Second Law with SF = F(t) and a = dv/dt becomes:   F(t) = ma = m dv(t)/dt .

Multiplying through by dt gives:   m dv = dp = F(t) dt  , where p = mv, so dp = m dv.

We now integrate both sides (with the appropriate limits of integration included):  poò pf dp = 0ò t F(t) dt .

The left side is easy:   pf - po = 0ò t F(t) dt = Dp = impulse .

Dividing through by m (since p = mv) gives:  v(t) = vo + (1/m) 0ò t F(t) dt .

To get x(t), we use the definition of velocity:  v = dx(t)/dt = vo + (1/m) 0ò t F(t) dt .

We can multiply both sides by dt to get:  dx = [vo + (1/m) 0ò t F(t) dt] dt .

We now integrate both sides (with appropriate limits of integration):  x(t) = xo + 0ò t [vo + (1/m) 0ò t F(t) dt] dt .

Integrating term by term gives:   x(t) = xo + vot + 0ò t [(1/m) 0ò 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 symbol m for the value of 5 kg, Fo for the value 10 Nt, and use the symbol w 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(q) to first order gives sin(q) » q, 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(q) = 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: 0òt sin²(w t) dt = 1/2 (t - [sin(w t) cos(w t)/w ]). 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 "s" from the path of the proton.

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) = ke2s / [s2 + vo2t2]3/2 where t = 0 when the proton passes closest to the electron.

b) Calculate the impulse delivered by this force.

d) Using the above results, calculate the approximate final momentum and final kinetic energy of the electron.

e) Show that the condition for the original assumption to be valid is: ke2/s << 1/2mvo2 .

Solution:

 t=0

a) Look at the diagram, and consider the x and y components of the force at some position:

Fx  =  (ke2/r2 )*(cos(q ))

Fy  =  (ke2/r2 )*(sin(q ))

where   r2  =  s2 + vo2t2  ,  cos(q )  =  vot / r  and  sin(q)  =  s/r .
Note that r is always positive, but (vo t) can be positive for positive time, but negative for negative time.

Therefore:  Fx  =  ke2 cos(q) /r2  =  ke2vot / r3  =  ke2vot / [s2 + vo2t2]3/2] ,  and

Fy  =  ke2 sin(q) /r2  =  ke2s / r3  =  ke2s / [s2 + vo2t2]3/2].

b) Note that the force components are now functions of time only (along with constants: k, e, s, and vo).  Thus we start with the definition of acceleration:

ay = dvy/dt,  or  dvy = ay dt,  or integrating both sides:

voò v dvy   =   toò t ay dt ,   or    vy(t) – vyo   =   toò t ay dt .

Also, there is just the one force, so  Fy = may ,  or   ay = Fy/m .

Since vyo = 0 (electron initially at rest), we have:

vy(t)   =   vyo + (1/m) toò t F(t) dt   =   0 + (1/m) toò t [ke2s / [s2 + vo2t2]3/2 ] dt ,

which we re-write as:  vy(t) = (ke2s/m) I ,  where  I  =  toò t [1/ [s2 + vo2t2]3/2 ] dt  and  to is negative infinity.

Let’s now try to do the integration of I. If we divide numerator and denominator in the integral by (1/s3), we get:

I  =  (1/s3) toò t [1/ [1 + (vot/s)2]3/2 ] dt .

If we let u = (vot/s)   [note that u is a dimensionless quantity], then  du = (vo/s)dt  so that dt = (s/vo)du;

then  I  =  (1/s3) uoò u [1/ [1 + u2]3/2 ] (s/vo)du   =   (1/s2vo) I'   where  I' = uoòu [1/ [1 + u2]3/2 ] du .

We can look up this integral, and we get:  I'  =  uoò 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/s) 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/s  so  [u/[1+u2]1/2 ]  =  [(vot/s) / [1+(vot/s)2]1/2] .

Therefore,  vy(t)  =  (ke2s/m)*(1/s2vo)*{ [(vot/s) / [1+(vot/s)2]1/2] - (-1) }.

To get the full impulse, we let t go to infinity, so  vyf  =  (ke2s/m(*(1/s2vo)*{1 - (-1)}  =  2ke2/msvo . Notice that the units do work out!

The full impulse delivered is  Dp  =  mvyf - mvyo   =   mvyf - 0   =   2ke2/svo .

d) The final momentum of the electron will be the same as the full impulse:  pyf = 2ke2/svo = mvyf.

The final kinetic energy of the electron will be: KEf  =  ½mvyf2  =  2k2e4 / s2mvo2 .

e) 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 Dt 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 Dt to be much less than s:  1/2 vyf Dt << s .

To get an idea of Dt , we can consider the impulse: F Dt = Dp . 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:

Dt  =  Dp / Fmax  =  (2ke2/svo) / (ke2/s2)  =  (2s) /vo .

Therefore the condition: 1/2 vyf Dt << s becomes

1/2 vyf (2s/vo) << s,   or  with vyf = 2ke2/msvo

1/2 (2ke2/msvo) (2s/vo) << s,   or

2ke2/mvo2 << s,   or

ke2/s << 1/2 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!