Momentum Conservation in an Explosion

## Question:

In the case of an explosion, when a mass M with initial velocity
u exploseds, it splits into two massses m1 & m2. After the
explosion, m1 has zero horizontal velocity while m2 has a
horizontal velocity v. Is it possible to express the velocity v
by the given symbols above? What is the external force, internal
force and total momentum in this case? Is the momentum conserved?
## Answer:

The momentum is conserved. That is always a given. We must be
careful though how we state the conservation of momentum. It is
only conserved as long as there is no addition or loss of energy
into the system. If we restrict ourselves to a short enough time
before and after the explosion then the energy added by the force
of gravity may be neglected and we may consider momentum to be
conserved for the duration of that observation.
Under the restrictions stated above the initial momentum was
Mu so the final momentum then must also be Mu. The explosion
causes the particle m1 to come to rest, to fall straight down and
the particle m2 to carry away all the initial momentum with it.
That means that m2v=Mu or v=Mu/m2

Neglecting air resistance, a real stretch here, the external
force before and after the explosion is only gravity. The
internal force is the force of the explosion, impulsive in
nature. Unless we know or estimate the duration of the explosion
we can not even get the average internal force on the the initial
large mass. We may assume that the internal force on the initial
mass is the external force on each of the two masses ejected from
the site of the explosion, if the initial object failed in a
brittle manner, without distortion.

Regards,

JDJ