it is going faster when you let it go. Having given all the pucks a similar shove, you expect there
to be something about the motion of each of the pucks that is the same as the corresponding
characteristic of the motion of all the other pucks. We have already established that the smaller
the mass of the puck, the greater the speed, and the greater the kinetic energy of the puck.
Experimentally, we find that all the pucks have one and the same value of the product mv ,
where v
m is equal to the
is the post-shove puck velocity. Further, we find that the value of
v
product of the constant force F and the time interval ∆t for which it was applied.
That is,
F∆t = mv
The product of the force and the time interval for which it is applied is such an important
quantity that we give it a name, impulse, and a symbol J .
J = F ∆t (26-1)
Also, as you probably recall from chapter 4, by definition, the product of the mass of an object,
and its velocity, is the momentum p of the object.
Thus, the results of the experiment described above can be expressed as
J = p
The experiment dealt with a special case, the case in which each object was initially at rest. If
we do a similar experiment in which, rather than being initially at rest, each object has some
known initial velocity, we find, experimentally, that the impulse is actually equal to the change
in momentum.
J = ∆p (26-2)
Of course if we start with zero momentum, then the change in momentum is the final
momentum.
Equation 26-2, J = ∆p , is referred to as the Impulse-Momentum Relation. It is a cause and
effect relationship. You apply some impulse (force times time) to an object, and the effect is a
change in the momentum of the object. The result, which we have presented as an experimental
result, can be derived from Newton’s second law of motion. Here we do so for the case in which
the force acting on the object is constant during the time interval under consideration. Note that
the force which appears in the definition of impulse is the net external force acting on the object.
Consider the case of a particle, of mass m, which has but one, constant force (which could
actually be the vector sum of all the forces) acting on it.
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