Momentum and Collisions - Mission MC6 Detailed Help

Students of varying mass are placed on large carts and deliver impulses to each other's carts, thus changing their momenta. In some cases, the carts are loaded with equal mass; in other cases they are unequal. In some cases, the students push off each other; in other cases, only one team does the pushing. For each situation, list the letter of the team which ends up with the greatest speed.

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.

Both teams start at rest prior to the explosive push-off. The momentum of each individual team is 0 units and the total momentum of the system before the explosion is 0 units. Since momentum is conserved, the total system momentum after the explosion is 0 units as well. For the system to have 0 units of momentum after the explosion, the two teams must be moving in opposite directions with an equal amount of momentum. So if the masses of the two interacting teams are equal, their post-explosion velocity will be equal as well. But if the teams have unequal mass, then the team with less mass has a greater velocity.

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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