Momentum and Collisions - Mission MC6 Detailed Help

Two pop cans are at rest on a stand. A firecracker is placed between the cans and lit. The firecracker explodes and exerts equal and opposite forces on the two cans. Assuming the system of two cans to be isolated, the post-explosion momentum of the system ____.

The Law of Momentum Conservation:
If an interaction between object 1 and object 2 occurs in an isolated system, then the momentum change of object 1 is equal in magnitude and opposite in direction to the momentum change of object 2. In equation form
m• ∆v1 = - m2 • ∆v2

The total momentum of the system before the interaction (p1+ p2) is the same as the total momentum of the system of two objects after the interaction (p1' + p2'). That is,  
p1 + p2 = p1' + p2'

System momentum is conserved for interactions occurring in isolated systems.

A pop can's momentum can be calculated as the product of mass and velocity (see Formula Frenzy section). In this case, both cans are at rest before the explosion. Since their velocity is 0 m/s, their momentum is zero as well. The total momentum of the system of two cans is 0 units. Total system momentum is conserved during the explosion. A little extra reasoning on your part will get you the answer to this question very quickly.

The momentum (p) of an object can be calculated from knowledge of its mass (m) and velocity (v) using the formula: 

p = m • v


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