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The gunpowder explosion in a gun results in an expansion of gases that causes a bullet to be propelled forward. The gun in turn 'kicks' or 'recoils' backwards. The recoil momentum (magnitude only) of a gun that kicks is ____ the momentum of the bullet that it fires.
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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
m1 • ∆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.
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The interaction between gun and bullet occurs due to the expanding gases. Before the gas expansion, both gun and bullet have a momentum of 0 units. Thus, the pre-explosion momentum of the system of two objects is 0 units. After the explosion, the total momentum must still be 0 units. But how can that be? If both objects are moving, how can the system have 0 units of total momentum after the explosion? The answer lies in the fact that momentum is a vector and has a magnitude and a direction (see Know the Law section). If the direction the bullet moves after the explosion is opposite the direction that the gun moves after the explosion, then one would have positive momentum and the other would have negative momentum. Now answer the question of how can the system of two objects have 0 units of total momentum after the explosion if both objects are moving? There is only way for this to happen and if you think about it you are likely to get at the answer.
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Momentum as a Vector:
Momentum is a vector and it has a direction. The direction of an object's momentum is in the same direction that the object is moving. An eastward moving object has a an eastward momentum.
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