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Newton's law of restitution. Could someone tell me what the easiest form of this law to use is? I usually try to use e=(v(2)-v(1))/(u(1)-u(2)).

Does the law also work if I don't know the direction of v(1) and v(2)?

So for example, if a particle moving with u(1) hits a steady object and I don't know if the particle's speed vector will change its direction, can I still use the law in the form e=(v(2)-v(1))/(u(1)-u(2)) with u(2)=0 (since the object was steady originally)?

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The law states this:

  • Two objects masses and with velocities $m_1$, $m_2$, $\vec{v}_1$ and $\vec{v}_2$ collide with contact normal $\vec{n}$.
  • The final velocities are $\vec{v}_1^\star$ and $\vec{v}_2^\star$ such that the coefficient of restitution $\epsilon$ is defined by $$\vec{n}\cdot \left( \vec{v}_2^\star - \vec{v}_1^\star \right) = -\epsilon \;\left( \vec{n}\cdot \left( \vec{v}_2 - \vec{v}_1 \right) \right)$$
  • If you consider the impact speed along the contact normal $v_{impact} =\vec{n}\cdot \left( \vec{v}_2 - \vec{v}_1 \right)$ and the rebound speed along the contact normal $v_{rebound} =\vec{n}\cdot \left( \vec{v}_2^\star - \vec{v}_1^\star \right)$
  • The collision law can now be stated as $$\boxed{ v_{rebound} = -\epsilon \;v_{impact} }$$
  • The last part of the impact mechanics states that there is an exchange in momentum along the contact normal with equal and opposite parts on the two masses $$m_1 \vec{v}_1\star = m_1 \vec{v}_1 - J\, \vec{n}\\m_2 \vec{v}_2\star = m_2 \vec{v}_2 + J\, \vec{n}$$
  • Combining the above gives the law $$ J =(1+\epsilon) \frac{m_1 m_2}{m_1+m_2} v_{impact} $$

NOTE: This is the easiest way for me to understand cotacts, $\cdot$ is the vector dot product and $\vec{n}\cdot\vec{n}=1$.

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