The references:

/ Conservation of Momentum During the 17th century, Newton and others before him had measured the momentum of colliding objects before and after a collision and discovered a strange phenomenon. They discovered that the total momentum of the colliding objects was the same after the collision as it was before. For example, suppose two masses approach each other as shown below. Before the collision, the total momentum of the two spheres is Protal initial = M1V4i + MaVo; Where the "i" refers to the initial conditions m, Ow V2 4@ After the collision, the momentum of the two spheres is Protal final = M1V1 + MaVor Where the "f" refers to the final conditions m; vie0) Obvz Since the total initial momentum pyota) initial @nd the final momentum pigia) fingl are equal MqVai + MaVai = MyVqr + MoVoy The two masses in this example can be referred to as a "system."\" A system can simply mean a set of objects interacting with each other. An isolated system is one in which the only forces present are those between the objects of the system. For example, for the two masses above, the only forces acting on them are the equal and opposite forces that they exert on each other when they collide. There are no external net forces. If external forces do act, that is forces outside the system, then momentum may not be conserved. For example, if we take as a system a falling rock, momentum is not conserved since an external force, the force of gravity exerted by the earth, is acting on it and its momentum changes. Thus we can state the Law of Conservation of Momentum as follows. The Law of Conservation of Momentum states that if the net external force acting on an object, or system of objects, is zero, then the total momentum of the object or system of objects remains constant (is conserved). Law of Conservation of Momentum and Newton's Third Law As we have seen, the Law of Conservation of Momentum was discovered experimentally in the 17th century. It was this discovery that Newton used as a basis of his Third Law. We saw in the previous section that m1V1i + m2V2i = M1V1f+ m2v2f By grouping terms, we see that m1V1i - miV1f = m2v2f - m2V2i -m1 ( V1f - V1i ) = m2 ( V24 - V2i) We now divide both sides of the equation by At, the time for which the colliding objects are in contact. Vif li = mz V2f - V2i -m At At But If li = and 21 2i = a At At Therefore -m1a1 = m2a2 In other words -F1 = F2 where F1 and F2 are the forces on m, and m2 respectively. That is, the colliding forces exert forces on each other during the collision that are equal in magnitude but opposite in direction. This is what Newton's Third Law says.Conservation of Momentum, and Explosions One of the special cases where the law of conservation of momentum applies is one where an object explodes. In such a case, the initial momentum is the momentum of the object before the explosion. If the object is initially at rest, then the momentum of the object is zero. If the object is moving, then the momentum of the object is its mass times its velocity. After the object explodes, the sum of the momenta of the individual pieces is equal to the momentum of the original piece. For example, if a mass of 40.0 kg is initially at rest, then its momentum is Pinitial = mv = (40.0 kg)(0 m/s) = 0 kg.m/s Let us suppose that this mass now explodes into two pieces, so that one piece of mass 10.0 kg moves to the right at 4.00 m/s. We can determine the velocity of the other mass. The sum of the momenta must equal 0 kg.m/s. Therefore 0 kg.m/s = m,V1 + m2V2 0 kg.m/s = (10.0 kg)(4.00 m/s) + (30.0 kg) v2 0kg . m/s -(10.0kg )(4.00m/s) V = = -1.33 m/s 30.0kg Suppose the original 40.0 kg mass was moving to the right at 2.00 m/s instead of being at rest. The initial momentum is Pinitial = mv = (40.0 kg)(2.00 m/s) = 80.0 kg.m/s This mass explodes into two pieces so that the 10.0 kg mass moves to the right at 4.00 m/s. We use the conservation of momentum to determine the velocity of the second piece. 80.0 kg.m/s = m,V, + m2V2 80.0 kg.m/s = (10.0 kg)(4.00 m/s) + (30.0 kg) v2 80.0kg . m/s -(10.0kg )(4.00m/s) V2 = = 1.33 m/s 30.0kg Applications of the principles of conservation of momentum could be recoil velocities of cannons while firing projectiles, or throwing objects off of masses such as boats which can recoil. These same principles can be applied to more than two pieces after an explosion. They can also be applied to situations where an object explodes so that pieces move in two or three dimensions. Summary In this lesson, we studied the law of conservation of momentum. This important law in physics has many applications. It allowed Newton to formulate his action- reaction law, the third of his laws. We applied knowledge of the conservation principle to the special situation of an explosion in one dimension. A system can simply mean a set of objects interacting with each other. An isolated system is one in which the only forces present are those between the objects of the system. The Law of Conservation of Momentum states that if the net external force acting on an object, or system of objects, is zero, then the total momentum of the object or system of objects remains constant (is conserved). From the Law of Conservation of Momentum, it is possible to derive Newton's Third Law of motion, the action-reaction law, which states that when two objects are in contact, they exert forces on each other that are equal in magnitude and opposite in direction. The principles involved in the Law of Conservation of momentum can be applied to determine the momentum and the velocity of masses as a result of an explosion, a sudden break up of a mass into smaller masses. Other examples of where these principles could be applied in explosion-like problems such as two ice skaters or two astronauts pushing each other apart. Question(s): 1. The physics of a person jumping onto a moving sled. A 75.0 kg person runs at 3.0 m/s and jumps onto a sled of mass 10.0 kg already moving in the same direction as the person at 2.0 m/s. a) What is the initial momentum of the sled-person system? b) What will the final momentum of the sled and the person be? c) What is the velocity of the sled and the person after the person jumps on the sled? 2. The physics of a rifle and a bullet. An unconstrained rifle of mass 5.0 kg fires a 50.0 g bullet at a speed of 300.0 m/s with respect to the earth. a) What is the initial momentum of the rifle-bullet system? b) What is the recoil velocity of the rifle