Interesting Facts About The Principle of Conservation of Momentum

Collins is the only astronaut in a space station far away from any planet. One day, He was outside of the space station for certain missions. He was equipped with a full set of space suit and a pair of jet engines which enable him to move freely in the space. Suddenly, he realized that both of his jet engines were out of fuel due to leakage. He was at 10 miles away from the space station while remained stationary.Collins is facing a survival issue now. He needs to get to the space station before the oxygen supply in his suit runs out within 2 hours. Now, if you were Collins, what should you do to save yourself?

     This problem shouldn't be difficult for anyone who had studied high school physics. Here is the solution. Collins may dismantle his empty jet engines and push them to the direction opposite to the space station. According to Newton's third law, a force opposites to the his pushing force will be exerted on himself by the empty jet engines. Therefore, he will be pushed back by that force to his space station and get saved. It is equally valid to understand this by taking the jets and Collins as an mechanical system. Since the total momentum of the system is conserved, by pushing the jets away, Collins will attain a momentum equals in magnitude but opposites to the direction which he pushes the jets away. Therefore, by adjusting the direction of pushing, Collins may find his way back.

    In the end, Collins was saved by one of the important laws of nature, the principle of conservation of momentum. Being one of the core principles in the classical physics, we may have already learned it during our high school years, but the importance of this principle and how it is linked to other physics principles may not be clear to all of us. Here I will present some interesting facts about the principle of conservation of momentum.

1.) This principle of conservation of momentum implies the Newton's third Law of motion.
Before we going into details, lets us have a flashback on the definitons of the principle/law above:

Principle of Conservation of Momentum:If the net external force on a N-particle system is zero, the system's total momentum is constant.
Related Equation: 1.) for an elastic collision : \begin{equation}m_{A}u_A + m_{B}u_B=m_{A}v_A + m_{B}v_B... ...(1)\end{equation} Related Equation:2.) for an inelastic collision : \begin{equation}m_{A}u_A + m_{B}u_B=(m_A+ m_B)v... ...(2)\end{equation}
Newton's third law of motions: Whenever an object A exerts a force on object B (that is, FBA), the object B will exerts a force on object A(that is, FAB) simultaneously which is equal in magnitude but opposite in direction.
Related Equation: \begin{equation}F_{AB}=-F_{BA}... ...(3)\end{equation} Our Question now : How do we know that both of the law/principle above are related to each others?

Solution :
As what we have in the equations above, equation(1) and (2) basically tells us that the summation of the momentum of both object A and object B before they collide together is equal to their summation of the momentum after collision. For an elastic collision as described by equation(1), after the collision, what happens to object A and object B is that they are separated and rebound to their own velocity, while for an inelastic collision as described by equation(2), after the collision, object A and object B will stick together to move with a uniform velocity.

Let works out the equation(1) : \begin{equation}m_{A}u_A + m_{B}u_B=m_{A}v_A + m_{B}v_B... ...(1)\end{equation} rearrange the equation, we will have : \begin{equation}m_{A}(v_A -u_A)=-m_{B}(v_B - u_B)... ...(4)\end{equation} divide equation(4) by the time of collision, t, we will have \begin{equation}\frac{m_{A}(v_A -u_A)}{t}=-\frac{m_{B}(v_B - u_B)}{t}\end{equation} \begin{equation}F_{A}=-F_{B}... ...(5)\end{equation} As we should expected, equation(5) is just equation (3), since during a collision, all of the force exerted on A are from B and vice versa.

It is also worthwhile to mention here that equation(2) is also equivalent to equation(3), the proving methods are similar and will be left to the reader as an exercise.

The result shown in equation(5) is important, because it shows us that, evidently, the principle of conservation of momentum implies the Newton's Third Law of motion and vice versa,and hence they may serve as the foundation of each other. A more careful and concrete discussion will be presented in the coming article, but it is enough for us to stop here first.

2.)The principle of conservation of momentum itself implies the kinetic energy must not be conserved in an inelastic collision:
It is easy to understand this by taking an example. Imagine that you fire a bullet to a wall. The bullet will be embedded into the wall and both bullet and wall will have zero velocity after their collision. The kinetic energy of the system consisted by the bullet and the wall is clearly disappeared, in other words, not conserved. It is interesting to ask where did the kinetic energy go? Yes, we must expect that the kinetic energy is transformed to other forms of energy such as heat energy, sound energy and molecular potential energy. Thus, it is easy for us to conclude that as long as these transformations of energy occur, the kinetic energy will never be conserved after an inelastic collision. But wait! How do we know that other forms of energy wouldn't transform to kinetics energy concurrently to compensate its lost? If this happens, how can we rest assured that the kinetics energy must not conserved in ANY case of inelastic collision?

Here, Let us assume that the kinetic energy is conserved after an inelastic collision, let us see what will happen:

We start from equation(2) : \begin{equation}m_{A}u_A + m_{B}u_B=(m_A+ m_B)v... ...(2)\end{equation} rearrange it, we will get: \begin{equation}m_{A}(v-u_{A})= - m_{B}(v-u_{B})... ...(6)\end{equation} Since we assumed that kinetic energy is conserved such that \begin{equation}\frac{1}{2}m_{A}{u_A}^2+\frac{1}{2}m_{B}{u_B}^2=\frac{1}{2}(m_{A}+m_{B}){v_A}^2... ...(7)\end{equation} Multiply this equation by 2 and rearrange, we will have: \begin{equation}m_{A}(v^2-{u_A}^2)=-m_{B}(v^2-{u_B}^2)\end{equation} \begin{equation}m_{A}(v-u_A)(v+u_A)=-m_{B}(v-u_B)(v+u_B)... ...(8)\end{equation} Substitute equation(6)into(8), we will have: \begin{equation}v + u_A = v + u_B\end{equation} or \begin{equation}u_A = u_B\end{equation} This is an impossible result to for a collision to occur! This is because if object A and object B have the same initial velocity(remember that velocity is a vector where direction counts), they will ended up in an endless chasing race whereby the chaser will never win, and there will be no collision at all. Thus, we must conclude if an inelastic collision is going to happen, then equation (6) and (7) can't be true simultaneously. Since equation (2) has the support of the principle of conservation of momentum, we must hence conclude that equation(7) can't be true, whereby in an inelastic collision, the kinetic energy of the system can't be conserved, provided that the system is not acted upon by any external force.

3.) The principle of conservation of momentum implies the Newton's Law of Restitution in an elastic collision.

Ponder about this question: Two coins, named coin A and coin B moving on a frictionless level surface collide together and seperate after their collision. Before they collide, velocity of coin A is 3ms-1 to the North while velocity of coin B is 4ms-1 to the East. While their masses remain UNKNOWN, we know that the velocity of coin A after the collision is 2ms-1 to the direction of S45W. Can you find the direction of the motion of coin B after the collision?

We might think that we don't have enough information to solve it due to the lack of information regarding the mass of both coins, but actually we do. The key of solution is hidden beneath the truth that the collision of coins here is an elastic collision in which the Newton's Law of Restitution holds.

The Newton's law of restitution can be derived from the principle of conservation of momentum in an elastic collision. In a perfectly elastic collision, it is assumed that the kinetic energy of the objects will not be transformed to the other forms of energy, in other words, the kinetic energy of the system is conserved. Therefore, we have:
\begin{equation}\frac{1}{2}m_{A}{u_A}^2+\frac{1}{2}m_{B}{u_B}^2=\frac{1}{2}m_{A}{v_A}^2+\frac{1}{2}m_{B}{v_B}^2... ...(9)\end{equation} Multiply equation(9)by 2 and rearrange it, we will have: \begin{equation}m_{A}({v_A}^2-{u_A}^2)=m_{B}({u_B}^2-{v_B}^2)\end{equation} \begin{equation}m_{A}(v_A-u_A)(v_A+u_A)=m_{B}(u_B-v_B)(u_B+v_B)\end{equation} Substituting equation(4), we will have: \begin{equation}v_A+u_A=u_B+v_B\end{equation} Or \begin{equation}v_B-v_A=-(u_B+u_A)... ...(10)\end{equation} Equation (10) is called the Newton's Law of restitution where the relative velocity between two objects is equal in magnitude but opposite in direction after the collision. The coins question above is now solvable to us, since we can solve it easily by using equation(10).

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