B Conservation of Momentum for system of particles

[mark2142]

Can I do it like this to conserve total P of say 3 body system in y-direction?
$$F_{aby}=-F_{bay}$$
$$F_{cby}=-F_{bcy}$$
$$F_{cay}=-F_{acy}$$
$$F_{aby}+F_{cby}+F_{cay}=-F_{bay}-F_{bcy}-F_{acy}$$
$$\frac d{dt}(p_{ay}+p_{cy}+p_{cy})=-\frac d{dt}(p_{by}+p_{by}+p_{ay})$$
$$\frac d{dt}(p_{ay}+p_{cy}+p_{cy}+p_{by}+p_{by}+p_{ay})=0$$
$$\frac d{dt}(p_{ay}+p_{by}+p_{cy})=0$$
$$p_{ay}+p_{by}+p_{cy}=constant$$

 

[jbriggs444]

Can I do it like this to conserve total P of say 3 body system in y-direction?
$$F_{aby}+F_{cby}+F_{cay}=-F_{bay}-F_{bcy}-F_{acy}$$
$$\frac d{dt}(p_{ay}+p_{cy}+p_{cy})=-\frac d{dt}(p_{by}+p_{by}+p_{ay})$$

Please explain the above step. It looks like hideous garbling of Newton's second law.

As I decrypt it, the idea is that (for instance) the derivative of the momentum of c is equal to the force of a on c and is also equal to the force of b on c. But Newton's second actually says that the derivative of the momentum of c is equal to the sum of the force of a on c and the force of b on c.

It's $\sum F = ma$, not $F_1 = ma = F_2$.

 

[mark2142]

Newton's second actually says that the derivative of the momentum of c is equal to the sum of the force of a on c and the force of b on c.

Oh yes! It should be
$$(F_{aby}+F_{acy})+(F_{cby}+F_{cay})+(F_{bcy}+F_{bay})=0$$
$$dp_a/dt+dp_b/dt+dp_c/dt=0$$
$$p_a+p_b+p_c=constant$$
Thank you.
Also I understand that in any system if there are only internal transfers of momentum then the net P of the system can not change. Its just that the interaction is so complex between particles of different masses (atoms) that it seems impossible.