I've been analyzing "ball hits a rod in space" type collisions, where speeding ball transfers part of its kinetic energy during elastic collision to the motionless rod, making it gain linear and angular momentum. There are many videos explaining such a scenario, I think I understood basic concept.

I tried to write a motion equations of similar, yet bit more complicated situation, where motionless rod B is hit by rod A. Rod A has only linear momentum ($V_a$), is going to hit rod B at a distance r off the center of mass of rod B, at the angle of α. Masses and lengths (and therefore moments of inertia) of both rods are given. I'd expect, that - after the elastic collision - both rods are going to have nonzero linear momentums and angular momentums.

rod hits rod collision

By my understanding, to describe this collision, 4 equations of motion are needed:

  1. conservation of kinetic energy
  2. conservation of linear momentum
  3. conservation of angular momentum in relation to the center of mass of rod A
  4. conservation of angular momentum in relation to the center of mass of rod B

However I've some troubles with points 3 and 4.

  1. $\frac{1}{2}m_av_a^2 = \frac{1}{2}m_av_a'^2 + \frac{1}{2}I_aω_a'^2 + \frac{1}{2}m_bv_b'^2 + \frac{1}{2}I_bω_b'^2$
  2. $m_av_a = m_av_a' + m_bv_b'$
  3. ???
  4. ???

I'd appreciate some help :)

edit1. fixed 1 and 2 eqs.


edit2. okay, I've spent some time on this and here are my thoughts:

I can define an impulse $J_b$, that affects rod B, and is equal to the local linear momentum gained. And the same can be done for rod A. Sum of those two impulses is equal to 0.

$$J_a = \Delta p_a = m_a * (v_a' - v_a)$$ $$J_b = \Delta p_b = m_b * (v_b' - v_b)$$ $$J_a + J_b = 0$$

Those impulses are the source of angular momentum. Considering distribution of $J_a$ vector, we can finally define equation 3 and 4.

  1. $$I_a\omega_a = J_a * cos(\alpha)l = m_a (v_a' - v_a)*cos(\alpha)l$$
  2. $$I_b\omega_b = J_b * r = m_b v_b'*r$$

What do you think? I'd appreciate if anyone could evaluate the correctness of my reasoning. :)

  • $\begingroup$ You do not need #1, since not all contacts conserve energy. What you need is the contact law describing the relative separation speed after the contact in terms of the speed before the contact and the coefficient of restitution. $\endgroup$ Commented Jun 2, 2023 at 16:13
  • $\begingroup$ You have $\omega_A'$ drawn in the wrong sense. Body A is going to kind of rotate about the contact point in a clock-wise fashion. $\endgroup$ Commented Jun 2, 2023 at 17:27

2 Answers 2


Note: Your equation 2 has assumed $v_a'$ towards the right, and I will do the same for my equations. Also, I'm assuminng $d$ as the perpendicular distance between the centres of mass of the two rods (perpendicular to their lines of motion). $d=r+\frac{l}{2}cos(\alpha)$

  1. $m_b v_a d = m_b d (v_a' - v_b') + I_b \omega_b' + I_a \omega_a'$

  2. $m_a v_a d = m_a d (v_b' - v_a') + I_b \omega_b' + I_a \omega_a'$

  • $\begingroup$ I'm sorry, I don't know, where did you took those equations from, but these don't strike correct for me. The distance between two centers of mass is not $d = r + \frac{1}{2}cos(\alpha)$ and equations you proposed make the arrangement unsolvable... $\endgroup$
    – Hypasist
    Commented Jul 29, 2020 at 22:18
  • $\begingroup$ @Hypasist, I have made some edits to the answer to clarify my derivation. I've just applied conservation of angular momentum with respect to the two centres of masses, as you havewritten in the question. $\endgroup$
    – dnaik
    Commented Jul 30, 2020 at 2:06

At the moment of impact, this is the picture.


All vectors on body A are shown in a negative sense, and vectors on body B in a positive sense.

Conservation of momentum, linear and angular, is achieved by applying the same impulse $\color{magenta} J$ to both bodies in equal and opposite sense, on the same line of action.

As a result, the response is a change in velocity, both linear and angular, for both bodies $\color{blue}{\Delta v_A}$, $\color{blue}{\Delta \omega_A}$, $\color{blue}{\Delta v_B}$, and $\color{blue}{\Delta \omega_B}$

The four equations (two bodies, linear and angular) that describe the response are summarized below

$$ \left.\begin{aligned}m_{A}\Delta v_{A} & =-J\\ m_{B}\Delta v_{B} & =+J\\ I_{A}\Delta\omega_{A} & =-(r_{A}\cos\alpha)J\\ I_{B}\Delta\omega_{B} & =-(r_{B})J \end{aligned} \right\} \begin{aligned}\Delta v_{A} & =-\tfrac{1}{m_{A}}J\\ \Delta v_{B} & =+\tfrac{1}{m_{B}}J\\ \Delta\omega_{A} & =-\tfrac{1}{I_{A}}(r_{A}\cos\alpha)J\\ \Delta\omega_{B} & =-\tfrac{1}{I_{B}}(r_{B})J \end{aligned} \tag{1,2,3,4}$$

Yet the above is not sufficient to solve the problem. You need to describe what type of contact this is (plastic, elastic, fully elastic) and describe the law of collision relating to the relative speeds before and after impact.

$$ (v_A^\star + \Delta v_A^\star) - (v_B^\star + \Delta v_B^\star) = - \epsilon \, (v_A^\star - v_B^\star) \tag{5}$$

where $\epsilon = 0 \ldots 1$ is the coefficient of restitution, and the star superscript denotes velocities resolved at the point of contact and along the contact normal direction for each body

$$ \begin{aligned} v_A^\star & = v_A + (r_A \cos \alpha) \omega_A \\ v_B^\star & = v_B - (r_B) \omega_B \end{aligned} \tag{6,7}$$


$$ \begin{aligned} \Delta v_A^\star & = \Delta v_A + (r_A \cos \alpha) \Delta \omega_A \\ \Delta v_B^\star & = \Delta v_B - (r_B) \Delta \omega_B \end{aligned} \tag{8,9}$$

which makes equation (5) after some re-arranging equal to

$$\left(\Delta v_{A}+(r_{A}\cos\alpha)\Delta\omega_{A}\right)-\left(\Delta v_{B}-(r_{B})\Delta\omega_{B}\right)=-\left(1+\epsilon\right)\,v_{{\rm imp}} \tag{10}$$

where $$v_{{\rm imp}}=\left(v_{A}+(r_{A}\cos\alpha)\omega_{A}\right)-\left(v_{B}-(r_{B})\omega_{B}\right) \tag{11}$$ is the relative impact speed.

Now use equations (1),(2),(3) and (4) above to find a single equation in terms of the unknown impulse $J$

$$\boxed{\vphantom{\begin{array}{c} \\ \\ \\ \end{array}}J=\frac{\left(1+\epsilon\right)\,v_{{\rm imp}}}{\tfrac{1}{m_{A}}+\tfrac{1}{I_{A}}(r_{A}\cos\alpha)^{2}+\tfrac{1}{m_{B}}+\tfrac{1}{I_{B}}(r_{B})^{2}}} \tag{12}$$

And once $J$ is calculated, then (1),(2),(3),(4) are used to find the response.

Note that the above is complicated and gets even worse with 2D contacts involving vectors. Worst yet are 3D contacts with inertia tensors and other complexities beyond this question's scope.

But there is a general shortcut that can be applied in most cases

Notice the solutions has the form of $$ J = \left(1+\epsilon\right)\,m_{\rm eff}\,v_{{\rm imp}} $$ where $m_{\rm eff}$ is the effective reduced mass of the system.

Given two bodies A, B and their perpendicular distances to the impulse line of action from the respective centers of mass as $d_A$ and $d_B$, then the effective mass is evaluated as

$$ m_{\rm eff} = \frac{1}{ \tfrac{1}{m_A} + \tfrac{d_A^2}{I_A} + \tfrac{1}{m_B} + \tfrac{d_B^2}{I_B}} $$

you can see with the special case of two spheres colliding that have $d_A=0$ and $d_B=0$ the above is $m_{\rm eff} = \frac{1}{\tfrac{1}{m_A} + \tfrac{1}{m_B}} = \frac{m_A m_B}{m_A + m_B}$ which is a familiar expression for reduced mass.

To evaluate the effective mass you would feel if you hit the rod at the impact point, you consider only their respective parts.

$$\begin{aligned} m_{\rm eff}^A & = \frac{1}{ \tfrac{1}{m_A} + \tfrac{d_A^2}{I_A} } \\ m_{\rm eff}^B & = \frac{1}{ \tfrac{1}{m_B} + \tfrac{d_B^2}{I_B} } \\ \end{aligned}$$

Notice that the values are maximum when the line of action goes through the center of mass and the perpendicular distances are zero. The further away the point of impact is, the less the effective mass is.


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