Section 1-7 - Impulse & Momentum

Yet Another Picture

Consider two objects moving in one dimension, each with its proper initial velocity:

We'll let the directions of the various vectors be represented by the signs of those quantities, just like we did back in Sections 2 & 3.  The figure not withstanding, the velocities could be in either direction.  Suppose that these objects collide; we certainly would expect the motions of each mass to change due to the force exerted on it by the other.

From the Third law, we know that these forces are equal in magnitude and opposite in direction:
F_1,2 = - F_2,1
Now, let's make use of the Second Law to see the results of each of these forces on the corresponding mass:
F_1,2 = m₂a₂ = - F_2,1 = - m₁a₁
m₂a₂ =  - m₁a₁
Remember the definition of acceleration?  Substitute it in:
m₂ (Dv₂/Dt₂) =  - m₁(Dv₁/Dt₁)
What can we say about the two time intervals?

Our derivation was for a one dimensional problem.  However, we could add a subscript x to each of the quantities above, then repeat the derivation for the y-components and again for the z-components, and we would see that momentum is conserved independently in each of the three dimensons.

Note that, as we said above, the whole notion of the conservation of momentum is nothing more than a re-statement of Newton's Second and Third Laws of Motion.

As an aside, we can now verify Newton's Third Law (remember that we skipped on that in Section 4) indirectly by verifying the conservation of total momentum, as you will do in Lab.

Impulse

Consider the collision above, except that we shall add in an external force, due to the action of an object outside of our system of masses:
Let's repeat the calculation from the first section with this force (on m₁, let's say) included:
F_ext + F_2,1 = m₁a₁         F_1,2 = m₂a₂
Remember the Third Law:
F_2,1 =  - F_1,2
so that
m₁a₁ - F_ext = - m₂a₂
Substitute as before:
m₁(Dv₁/Dt) - F_ext = - m₂(Dv₂/Dt)
m₁Dv₁ - F_extDt= - m₂Dv₂
m₁(v_1f - v_1o) - F_extDt= - m₂ (v_2f - v_2o)
F_extDt = m₁(v_1f - v_1o) + m₂ (v_2f - v_2o)
F_extDt = m₁v_1f - m₁v_1o + m₂ v_2f - m₂v_2o
F_extDt = Dp₁+Dp₂ = Dp_total.
Since it doesn't matter which was mass 1 and which was mass 2, this works for any situation.  The quantity on the left we will call the impulse, the force times the time interval over which it was applied:
J = FDt,
so that
J = Dp_total.
What is the force is not constant in time?  Analogously with the work, we must write an integral:  J = / F dt.

Let's review the three pictures:

• The velocity of an object will remain constant unless the object is acted on by a force, in which case F_tot = ma.
• The kinetic energy of an object will remain constant unless the object has work performed on it, in which case W_tot = DKE.
• The momentum of an object will remain constant unless the object is acted on by an impulse, in which case J_tot = Dp.

• The total mechanical energy of a system will remain constant unless the system has work performed on it by non-conservative forces, in which case W_NC = DE_tot.
• The total momentum of a system will remain constant unless the system is acted on by an external impulse, in which case J_ext = Dp_tot.

Incidentally, there is a fourth picture that is very useful when the path of an object must be determined, but we will ignore that picture in this class.

Completely Elastic and Completely Inelastic Collisions

Is kinetic energy conserved here?
(KE_tot)_o = ¹/₂ m₁ v_1o²
(KE_tot)_f  = ¹/₂ (m₁ + m₂) v_f² = ¹/₂ (m₁ + m₂) [[m₁/(m₁+ m₂)] v_1o]² = ¹/₂ m₁ v_1o² * [m₁/(m₁+ m₂)] < ¹/₂ m₁ v_1o²
No, some was lost, probably to sound, deformation of the objects, thermal energy, et c.

Now, let's consider a totally elastic collision, in which no kinetic energy is lost during the collision (although, it can be transferred from one object to the other).  We should write one equation representing conservation of momentum (in one dimension only) and another representing the fact that the kinetic energy is the same before and after.
N.B.: There is a shorter, but perhaps much less revealing, solution at the end of this section (see One Last Note):

m₁v_1o + m₂v₂₀ = m₁v_1f + m₂v_2f
¹/₂m₁v_1o² + ¹/₂m₂v₂₀² = ¹/₂m₁v_1f²+ ¹/₂m₂v_2f²
Once again, let's let one object start at rest, just to make the math easier:
m₁v_1o = m₁v_1f + m₂v_2f
¹/₂m₁v_1o² = ¹/₂m₁v_1f² + ¹/₂m₂v_2f²
Now, take a really deep breath.

In the second equation, we can multiply through by 2 to get:
m₁v_1o² = m₁v_1f²+ m₂v_2f²
and in both equations, we can divide through by m₁ to get:
v_1o = v_1f + (m₂/m₁)v_2f
v_1o² =v_1f² + (m₂/m₁)v_2f²
To lessen the amount of writing we have to do, let's let g represent the ratio of the masses: g = m₂/m₁, so that
v_1o = v_1f + g v_2f
v_1o² =v_1f² + g v_2f²
Now, we're ready to start.  Solve the first equation for v_1f
v_1f = v_1o - g v_2f
and substitute into the second equation:
v_1o² = [v_1o - g v_2f]²+ g v_2f²
Write more explicitly the square:
v_1o² = v_1o² - 2g v_1ov_2f + g²v_2f²+ g v_2f²
Subtract v_1o² from each side:
0 =  - 2g v_1ov_2f + g² v_2f² + g v_2f²
Factor out a gamma
0 =  - 2 v_1ov_2f + g v_2f² + v_2f²
and a v_2f :
0 =  [- 2v_1o + (g + 1) v_2f] v_2f
This quadratic equation has one obvious solution: v_2f = 0.  Let's look for the other solution after dividing out the (now assumed non-zero) v_2f:
0 = - 2v_1o + (g + 1) v_2f
v_2f  = 2v_1o/(g + 1)
Remember that g = m₂/m₁, so
v_2f  = 2v_1o/[(m₂/m₁) + 1] = [2m₁/(m₁ + m₂)]v_1o

Once again, remember that these solutions are only valid if one mass had been initially at rest, the collision was totally elastic, and motion was restricted to one dimension.  To be clear, you should label whichever mass was not initially moving mass two.

What if neither mass had been at rest?  Well, we could go back and re-do the derivation with the two extra terms, but here is a neat trick: we can always pick a new frame of reference (indicated below by a prime) in which v_2o' is zero, then calculate the final velocities in that frame, then convert back to the original frame, all using relative velocities.

Example:
Find the final velocities of the masses given for this initial condition.
In one dimension, letting the signs of the velocities represent the directions:
m₁ = 5 kg     m₂ = 2 kg     v_1o = 5 m/s     v_2o = - 6 m/s
Let's make a chart:
 

	vo	convert to new frame in which v2o' = 0	vo'	vf'	convert back to original frame by reversing the previous  transformation	vf
m1	+5 m/s	add 6	+ 11 m/s	v1f' =[(m1 - m2)/(m1 + m2)]v1o' =           =  [(5 - 2)/(5 + 2)]11 = 4.71 m/s	subtract 6	- 1.29 m/s
m2	- 6 m/s	add 6	0 m/s	v2f ' = [2m1/(m1 + m2)]v1o' =     = [2*5/(5 + 2)]11= 15.7 m/s	subtract 6	9.71 m/s

Just FYI, here are the actual solutions (asserted without proof) which would be obtained by solving the two equations without assuming v_2o = 0:
v_1f = [(m₁ - m₂)/(m₁ + m₂)]v_1o + [2m₂/(m₁ + m₂)]v_2o
v_2f = [2m₁/(m₁ + m₂)]v_1o + [(m₂ - m₁)/(m₁ + m₂)]v_2o
Someday, when I have more interest, I'll show how this answer is equivalent to the method we used.

Here is an additional interesting derivation for a totally elastic collision.  Here, we do not need to assume that m₂ is intially at rest.

m₁v_1o + m₂v₂₀ = m₁v_1f + m₂v_2f
¹/₂m₁v_1o² + ¹/₂m₂v₂₀² = ¹/₂m₁v_1f²+ ¹/₂m₂v_2f^{2
Rearrange each:}
m₁v_1o - m₁v_1f = m₂v_2f - m₂v_{20   =>}   m₁(v_1o - v_1f) = m₂(v_2f - v₂₀)
m₁(v_1o² - v_1f ²) = m₂(v_2f ² - v_2o²)  =>  m₁(v_1o - v_1f )(v_1o + v_1f ) = m₂(v_2f  - v_2o)(v_2f  + v_2o)  
Now, divide the second equation by the first:
v_1o + v_1f  = v_2f  + v_2o
and rearrange again to get:
v_1o - v_2o  = -(v_1f  - v_2f).
That is, the realtive velocity of the masses before the collision is the negative of the relative velocity after the collision.  Let's check our example from earlier.  Before the collision, the realtive velocity was
5 - (-6) = 11 m/s. 
After, it was -1.29 - 9.71 =  -11 m/s.

Now, generally, I have not found this relationship to be much use.  It can however easily be used as a quick check on the accuracy of the first method we worked out.

Two Dimensional Collisions

• Try to find a frame of reference (see above) for which one mass is initially at rest; we've already seen the benefit of such a transformation.
• Try to pick a co-ordinate system in which the initial momentum of the other mass is along one of the axes (say the x-axis), so that no components will be necessary, i.e., v_1ox = v_1o and v_1oy = 0.

Here are a couple of interesting special cases (which both assume no rotation of the objects):

Totally inelastic collision:
Since the objects stick together, v_f1 = v_f2 (call them just v_f ) and q ₁= q₂ (call the angle just q).
x: m₁v_1o = m₁v_f cosq + m₂v_f cosq = (m₁ + m₂) v_f cosq
y: 0 = m₁v_f sinq + m₂v_f sinq = (m₁ + m₂) v_f  sinq.
O.K., so the second relationship implies that q = 0, which simplifies the first to
m₁v_1o = (m₁ + m₂) v_f
so that
v_f  = m₁v_1o/(m₁ + m₂)
No surprise there, it's become an essentially one dimensional problem again.

Now try a totally elastic collision:
x: m₁v_1o = m₁v_1f cosq₁+ m₂v_2f cosq₂
y: 0 = m₁v_1f sinq₁ - m₂v_2f sinq₂.
In addition, we have conservation of kinetic energy:
¹/₂m₁v_1o² = ¹/₂m₁v_1f² + ¹/₂m₂v_2f ²
You might imagine that this will not be easy to solve (you'd be right!).

Let's look at a very specific case, that of the masses being equal (mass two is still initially at rest):
v_1o = v_1f + v_2f  (masses cancel)
v_1o² = v_1f² + v_2f ² (halves and masses cancel)
where I've written the momentum equations in vector form.  The first equation says we can make a figure like this:

and the second, which looks a lot like the Pythagorean theorm, is only going to be true if the triangle is a right triangle so that v_if and v_2f are at right angles to one another, a nice result.

Three Dimensional Collisions

A 0.5 kg block is released from rest at the top of a frictionless, curves track 2.5 metres above the top of a 2 metre high table.  At the bottom of the track, this mass collides elastically with a 1 kg mass that is initially at rest.

a) Find the velocities of the two masses immediately after the collision.
b) How high back up the track does the 0.5 kg mass rise?
c) How far from the base of the table does the 1 kg mass land?
d) How far from the base of the table does the 0.5 kg mass land?

This is a good problem, in that it requires you to choose which of the three 'pictures' we have developed to use in each section.  Keep in mind that the three pictures are essentially identical, but that one may be much more convenient to use that the other two in a given situation.
Mass one starts from rest (label situation A) and slides down the incline until just prior to collision with mass two (call this B).  We choose to use conservation of mechanical energy in this case, since there are no non-conservative forces doing work (The normal force from the incline is perpendicular to the direction of motion, and the weight can be converted to a PE term.  We don't want to use impulse-momentum, since both of those forces are enternal and provide an impulse, the normal in a complicated way, and we would still need to find the time to travel down the incline).
PE_A + KE_A = PE_B + KE_B
m₁gh₁ + 0 = 0 + ¹/₂m₁v_1B²
where I have set PE=0 at the table top.
v_1B = 7 m/s  (I will omit numerical calculation steps in this example)
For the collision (B refers to just before, C refers to just after), we will use a combination of conservation of momentum (no external forces acting horizontally, no net force vertically) and, since the collision is elastic, conservation of kinetic energy:
m₁v_1B + 0 = m₁v_1C + m₂v_2C
¹/₂m₁v_1B² + 0 = ¹/₂m₁v_1C² + ¹/₂m₂v_2C²
We already worked out the solution to these equations (for m₂ initially at rest):
v_1C = [(m₁-m₂)/(m₁+m₂)]v_1B = - 2.3 m/s
v_2C = [2m₁/(m₁+m₂)]v_1B = + 4.67 m/s.
Let's look at what happens with mass two, which is now launched horizontally off the table.  The quickest way to approach this is with the kinematic equations:
Dy = v_oyt + ¹/₂a_yt² = 0 - ¹/₂a_gt²  =>  -h₂ = -¹/₂a_gt²  =>  t = 0.64 sec
Dx = v_oxt + ¹/₂a_xt² = v_2Ct + 0 = 2.99 m from the base of the table.
I've skipped on some of the background justification here too.
To find the altitude to which the smaller mass rises again on the incline (label this event D), use once again conservation of mechanical energy:
¹/₂m₁v_1C² + 0 = m₁gh_D + 0
h_D = 0.27 m
Mass one now slides back down the plane to the bottom (label this E); find this speed once again with conservation of mechanical energy:
m₁gh_D + 0 = 0 + ¹/₂m₁v_1E²
v_1E = +2.3 m/s (we have to add the sign ourselves after we take the square root to indicate that mass one moves to the right).
Then, mass one is launched horizontally, so use the same relationships as above:
Dy = v_oyt + ¹/₂a_yt² = 0 - ¹/₂a_gt²  =>  -h₂ = -¹/₂a_gt²  =>  t = 0.64 sec
Dx = v_oxt + ¹/₂a_xt² = v_1Et + 0 = 1.49 m from the base of the table.

0 = m dv_R,O + (-dm)v_F,R. 

Now, let's change our notation to match most textbooks.  The velocity of the rocket v_R,O we'll make just plain v, and the absolute value of the (negative) velocity of the spent fuel relative to the rocket v_F,R we'll call v_EXH.

m dv =  - v_EXH  dm.

Now, we can go two ways with this to find out two separate quantities. 
1) m dv =  - v_EXH  dm
    Divide both sides by dt
    m dv/dt =  - v_EXH  dm/dt

    But, dv/dt is the acceleration of the rocket:

    ma =  - v_EXH  dm/dt
    and, since the only force acting on the rocket is from the escaping exhaust, the left hand side is equal to that force (called the thrust):
    F_THRUST = - v_EXH  dm/dt, where we remember that v_EXH is a positive number and that dm is a negative number.

2) m dv =  - v_EXH  dm
    dv =  - v_EXH  dm/m
    Intergrate each side:
    _vo^v dv =  - v_{EXH  mo}^mdm/m
    v_f - v_o = - v_EXH  ln m/m_o = v_EXH  ln m_o/m
    v_f = v_o + v_EXH  ln m_o/m
    where we remember that v_EXH is a positive number.

Note:

As an aid to visualizing these three cases, consider the equations of two lines in the x-y plane:
ax + by = c
dx + ey = f

A solution of (x,y) for the pair corresponds to a point of intersection of the lines.  If the lines are parallel and separated, there is no solution.  If the lines are parallel and lie atop one another, there is an infinite number of solutions.  If the lines are not parallel, there is one solution.

Another Note:

In 1-d; let the sign of each quantity indicate the vector's direction:

Consider a solitary force F acting over a displacement Dx.  The work done will be
W = FDx = ¹/₂mv_f² - ¹/₂mv_o².

Now, let's write the same expression in a new frame of reference moving with velocity v_R relative to the first frame:
W' = FDx' = ¹/₂mv_f'² - ¹/₂mv_o'².

Keep in mind that:
v' = v + v_R   and Dx' = Dx + v_RDt, where Dt is the time interval (the same in both frames, at least in our Galilean transformation) over which the force is applied.

So, immediately, we can see that the amounts of work done by the same force as viewed in the two frames of reference are different:
W = FDx
W' = FDx' = F[Dx + v_RDt] = FDx + Fv_RDt = W + Fv_RDt

This then means that the change in kinetic energy is different in each of the frames, as we can see explicitly below:
DKE = ¹/₂mv_f² - ¹/₂mv_o².
DKE' = ¹/₂mv_f'² - ¹/₂mv_o'² = ¹/₂m[v_f + v_R]² - ¹/₂m[v_o+ v_R]² = ¹/₂m[v_f²+ 2v_fv_R + v_R²] - ¹/₂m[v_o² + 2v_ov_R + v_R²] =
           = ¹/₂mv_f² - ¹/₂mv_o² + ¹/₂m[2v_fv_R -  2v_ov_R ] = DKE + m[v_f  - v_o]v_R.

Does this make sense?  Subtract the two relationships:

W' - W = Fv_RDt = DKE' - DKE = m[v_f  - v_o]v_R.
Fv_RDt = m[v_f  - v_o]v_R.
Divide by v_R, assuming it's not zero:
FDt = m[v_f  - v_o] = Dp.
This is our impulse relationship.  The impulse is the same in each of the two reference frames:
Dp' = mv_f' - mv_o' = m[v_f + v_R] - m[v_o + v_R] = mv_f  - mv_o =Dp.

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