Collision Response without Rotation

\[\begin{aligned}v'_a=v_a+\Delta v_a\\ v'_b=v_b+\Delta v_b \end{aligned}\]

\[\begin{aligned}v'_a=v_a-\frac{J}{m_a}\\  v'_b=v_b+\frac{J}{m_b} \end{aligned}\]

\[\begin{aligned}v'_a=v_a-\frac{J}{m_a}=v_a-(j_n\cdot n+j_t\cdot t)\frac{1}{m_a} \\ v'_b=v_b+\frac{J}{m_b} =v_b+(j_n\cdot n+j_t\cdot t)\frac{1}{m_b} \end{aligned}\]

\[(v_a'-v_b') = (v_a-v_b)+J(\frac{1}{m_a}+\frac{1}{m_b})\]

which can be expressed as:

\[v_{ab}' = v_{ab}+J(\frac{1}{m_a}+\frac{1}{m_b})\]

Dcompose velocity

\[v=v\cdot n+v\cdot t\]

Decomposition J

\[J=j_n\cdot n+j_t\cdot t\]

 \[v'_a\cdot n=\left( v_a-\left[ \frac{j_n}{m_a}n+\frac{j_t}{m_a}t \right]  \right)\cdot n=v_a\cdot n-\frac{j_n}{m_a}\]

\[v'_b\cdot n=\left( v_b+\left[ \frac{j_n}{m_b}n+\frac{j_t}{m_b}t \right]  \right)\cdot n=v_b\cdot n+\frac{j_n}{m_b}\]

Subtracting the preceding two equations results in the following:

\[((v'_a -v'_b)\cdot n = (v_a-v_b)\cdot n - j_n(\frac{1}{m_a}+\frac{1}{m_b})\]

Given how we have defined relative velocity the above equation simplifies to:

\[v'_{ab}\cdot n = v_{ab}\cdot n - j_n\left(  \frac{1}{m_a}+\frac{1}{m_b}  \right)\]

Given the definnition of the coefficient of restitution,

\[v'_{ab}\cdot n=-e(v_{ab}\cdot n) \]

we can say that:

\[-e(v_{ab}\cdot n)=  v_{ab}\cdot n-j_n\left( \frac{1}{m_a} +\frac{1}{m_b} \right)\]

Collecting the terms and solving for j(n), the impulse in the normal direction result in the following:

\[j_n=\frac{(1+e)(v_{ab}\cdot n)}{\frac{1}{m_a}+\frac{1}{m_b}}\]

We can adopt a similar approach to calculating the tangential component of J.

First perform the dot product with the t vector on both sides of the impulse equation.

\[v'_a\cdot t=\left( v_a-\left[ \frac{j_n}{m_a}n+\frac{j_t}{m_a}t \right] \right)\cdot t=v_a\cdot t-\frac{j_t}{m_a}\]

\[v'_b\cdot t=\left( v_b+\left[ \frac{j_n}{m_b}n+\frac{j_t}{m_b}t \right] \right)\cdot t=v_b\cdot t+\frac{j_t}{m_b}\]

Following similar steps adopted for the normal component, we get:

\[v'_{ab}\cdot t = v_{ab}\cdot t - j_t\left( \frac{1}{m_a}+\frac{1}{m_a} \right)\]

Given that:

\[v'_{ab}\cdot t=f(v_{ab}\cdot t) \]

Now substitute the equation for the friction coefficient:

\[f(v_{ab}\cdot t) = v_{ab}\cdot t - j_t\left( \frac{1}{m_a}+\frac{1}{m_b} \right)\]

Collecting the terms and solving for j(n), the impulse in the normal direction result in the following:
\[j_t=\frac{(1-f)(v_{ab}\cdot t)}{\frac{1}{m_a}+\frac{1}{m_b}}\]