Recognizing that mv ′ is the final momentum and that v
m is the initial momentum we realize
that we have
F∆t = p′ − p
On the left, we have what is defined to be the impulse, and on the right we have the change in
momentum (equation 26-2):
J = ∆p
This completes our derivation of the impulse momentum relation from Newton’s 2nd Law.
Conservation of Momentum Revisited
Regarding the conservation of momentum, we first note that, for a particle, if the net external
force on the particle is zero, then the impulse, defined by J = F ∆t , delivered to that particle
during any time interval ∆t, is 0. If the impulse is zero then from J = ∆p , the change in
momentum must be 0. This means that the momentum p is a constant, and since p =
v
m , if the
momentum is constant, the velocity must be constant. This result simply confirms that, in the
absence of a force, our impulse momentum relation is consistent with Newton’s 1st Law of
Motion, the one that states that if there is no force on a particle, then the velocity of that particle
does not change.
Now consider the case of two particles in which no external forces are exerted on either of the
particles. (For a system of two particles, an internal force would be a force exerted by one
particle on the other. An external force is a force exerted by something outside the system on
something inside the system.) The total momentum of the pair of particles is the vector sum of
the momentum of one of the particles and the momentum of the other particle. Suppose that the
particles are indeed exerting forces on each other during a time interval ∆t. To keep things
simple we will assume that the force that either exerts on the other is constant during the time
interval. Let’s identify the two particles as particle #1 and particle #2 and designate the force
exerted by 1 on 2 as F . Because this force is exerted on particle #2, it will affect the motion of
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particle #2 and we can write the impulse momentum relation as
F ∆t = ∆p (26-3)
12
2
Now particle #1 can’t exert a force on particle 2 without particle #2 exerting an equal and
opposite force back on particle 1. That is, the force F exerted by particle #2 on particle #1 is
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the negative of F .
12
F = −F
21
12
Of course F (“eff of 2 on 1”) affects the motion of particle 1 only, and the impulse-momentum
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relation for particle 1 reads
1
83