<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/cad46445-8f22-46c9-af27-d381bea10139/Definition.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/cad46445-8f22-46c9-af27-d381bea10139/Definition.png" width="40px" /> Definition: $\vec{p}=m\vec{v}$ Defined against Force:
$$ \vec{F}=\frac{\text{d}\vec{p}}{\text{d}t} \text{ for a single particle: } \vec{F}=m\frac{\text{d}\vec{v}}{\text{d}t}=m\vec{a} $$
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The total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass:
$$ \vec P=M\vec{v}_\text{cm}=\sum^n_i m_i\vec{v_i} $$
Using what we know:
$$ M \vec{a}{cm}=\sum{i=1}^n \vec{F}i=\vec{F}\text{ext} $$
$$ \text{We get: }\quad \frac{\text{d}\vec{P}}{\text{d}t}=\vec{F}{\text{ext}} \text{ thus if: }\sum \vec{F}{\text{ext}}=0 \text{ then: } \quad \frac{\text{d}\vec{P}}{\text{d}t}=0 $$
$$ \vec{p}_f-\vec{p}i=\int{\vec{p}_i}^{\vec{p}f}\text{d} \vec p=\int{t_i}^{t_f} \vec F\,\text{d} t=\vec J $$
The area under the force-time curve is equal to the change in momentum of a body in a collusion which is equal to the magnetude of the impulse
<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/4afb84c8-07b8-4788-b951-cd1f188cac8f/impulse.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/4afb84c8-07b8-4788-b951-cd1f188cac8f/impulse.png" width="40px" /> Definition: The impulse $\vec{J}$ is defined as the time intergal of the force
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<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/2f5fbf0f-f122-4c50-a712-12662effa013/Principal_of_conservation_of_momentum.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/2f5fbf0f-f122-4c50-a712-12662effa013/Principal_of_conservation_of_momentum.png" width="40px" /> Principal of conservation of momentum: If no external forces act on a system of interacting bodies, then the total momentum of the system is conserved
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💷 Which means the velocity of the centre of mass is constant, if the mass is constant
<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/91e2c22c-a4df-415f-9de7-2151389c6c41/Elastic_collision.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/91e2c22c-a4df-415f-9de7-2151389c6c41/Elastic_collision.png" width="40px" /> An elastic collision is one where the kinetic energy is conserved, i.e., is the same before and after the collision
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$$ \begin{aligned} m_1v_{1i}+m_2v_{2i}&=m_1v_{1f}+m_2v_{2f} \\ \quad \qquad \qquad \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i^2}&=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2 \;\ E_k \text{ is conserved} \\ v_{1i}+v_{1f}=v_{2i}+v_{2f} \;\ &; \;\ v_{1i}-v_{2i}=-(v_{1f}-v_{2f}) \end{aligned} $$
💷 For an elastic collision the relative velocity of the two bodies before the collision equals the negative of the relative velocity of the two bodies after the collision
$$ \begin{aligned} v_{1f}&=\frac{(m_1-m_2)v_{1i}+2m_2 v_{2i}}{m_1+m_2} \\ v_{2f}&=\frac{(m_2-m_1)v_{2i}+2m_1 v_{1i}}{m_1+m_2} \end{aligned} $$
Special cases:
