# Conservation of Momentum

# In Single-Particle Systems

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The simplest possible physical system consists of a single, isolated
particle, and in some respects application of the Law
of Conservation of Momentum to such systems is trivial - but not
quite.
Suppose that a single, isolated object has no net
force acting on it (either because there is nothing pushing or
pulling on it, or because all of the forces that act on it cancel or
balance exactly). Newton's First
Law insists that the object will keep whatever velocity
it currently has. Since its mass
and velocity don't change, its momentum
won't change either. This is true whatever the object happens to be -
momentum must be conserved in any single-object system, as long as no
net force acts on the system.

Of course, it a net force *does* act on the object, Newton's
Second Law says that the object will accelerate
- its velocity will change. How much? The impulse-momentum
equation says that the object's change in momentum (mass times
change in velocity) equals the impulse
exerted on it (force times time). Does this violate the Law of
Conservation of Momentum? No. The Law of Conservation of Momentum
says that momentum will only be conserved** if** no net force (no
impulse) acts on the system.

##

Example: A Rock In Space - A Numerical Example

Suppose a rock is floating in space, far enough from any other
matter that all forces on it can be ignored.

momentum = (mass)(velocity) = (2 kg)(4 m/s) = 8 kg m/s

If no forces act on it, the net force on the rock is zero, so
Newton's First Law guarantees that the rock's velocity will not
change. If its velocity and its mass don't change, its momentum must
also remain the same - momentum is conserved.

Suppose,
on the other hand, that a force of 1 Newton acts on the rock for 6
seconds. This means that there is an impulse on the rock.

impulse = (force)(time) = (1 N)(6 s) = 6 Ns

From the impulse-momentum equation, we get:

impulse = change in momentum = (mass)(change in velocity)

At the end of the 5 second time interval, the velocity of the rock
will have increased by 3 m/s. The momentum of the rock will be:

momentum = mv = (2 kg)(7 m/s) = 14 kg m/s

last update April 23, 2000 by JL
Stanbrough