# Is the Principle of Conservation of Momentum applied incorrectly in M1?

I' was doing this specimen paper in Pearson Advance Subsidiary level Mechanics 1( M1). However, I do not agree with the fact that the mark scheme wants you to use the Principle Conservation of Momentum in this when Speed is cut both ways. How is it possible to use the Conservation of momentum when it is not consereved. Or is my idea of conservation of momentum incorrect?

I believe CLM means Conservation Law of Momentum?

• This question would be more appropriate for the Physics Stack Exchange – Cesareo Mar 25 '20 at 9:54
• Momentum of a closed system is always conserved. Momentum of a single particle is not necessarily conserved. Otherwise every particle would have to continue traveling in the same direction at the same speed forever regardless of interactions. Some momentum must be exchanged between your particles in order to change their velocities. – David K Mar 25 '20 at 12:19

When solving such problems, you have to identify what system of particles you're considering, analyze all forces etc involved in the problem, set up the equation appropriately and then solve the resulting math (which for such problems is the easiest part in my opinion).

Here, we consider our system to consist of the two particles $$P$$ and $$Q$$. The only forces present are the forces exerted by one particle on another; there is no "external" force (i.e no external gravity/ external electric/magnetic fields etc). Since the net external force is $$0$$, this implies that the total momentum of the two particles IS conserved (recall that $$\dfrac{d \vec{P}_{\text{total}}}{dt} = \sum \vec{F}_{\text{external}}$$; so if the RHS is 0, then the LHS is $$0$$, and derivative being zero implies the quantity is conserved).

Notice that I used vectors throughout. So, to apply the momentum conservation equation, we have to work with vectorial quantities. This is a one-dimensional problem; I shall use $$\hat{\xi}$$ to mean a unit vector along the axis. We have two particles; one has mass $$m$$, with initial VELOCITY (which is a vector) $$u \hat{\xi}$$, and final velocity $$\dfrac{u}{2} (-\hat{\xi})$$ (negative because you're told it moves in opposite direction).

The second particle has mass $$km$$, with initial velocity $$2u(-\hat{\xi})$$ (it has speed $$2u$$, but is moving in the opposite direction), and final velocity $$u \hat{\xi}$$. Thus, momentum conservation implies that \begin{align} \vec{P}_{\text{total, initial}} &= \vec{P}_{\text{total, final}} \\ m \cdot (u \hat{\xi}) + km \cdot \left(-2u\hat{\xi} \right) &= m \cdot \left( -\dfrac{u}{2} \hat{\xi}\right) + km \left( u \hat{\xi}\right). \end{align} This is a straight-forward linear equation, which upon solving, indeed yields $$k = \dfrac{1}{2}$$.

Side-Remark:

By the way, this is a one-dimensional problem, so it is not actually necessary to introduce vectors, but when I learnt this stuff, the $$\pm$$ stuff and directions was treated in a very ad-hoc manner, so I just presented the work in a manner which is more easily generalized to more complex/higher dimensional problems. (I know this wasn't the main point of your question, but I just felt compelled to add this remark)

This should be posted in physics stack exchange but anw...

Review again vector vs scalar, momentum and velocity are vectors, speed is scalar. In addition, momentum is velocity vector scaled by its related mass. Speed decrease does not mean momentum is not conserved, speed constant does not mean momentum is conserved (once again, vector vs scalar)

Review again how to define system and environment. If You define both particles as Your system, then total momentum of this system should be conserved as there is no interaction with environment.