Momentum Conservation in an Explosion


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?


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.