A 1 meter long rod on the ice with mass $m_2=1$ kg is perpendicularly hit on one end by a point particle with mass $m_1=0.1$ kg. The collision is elastic and the point particle is bounced back in the same direction. After the collision the rod's frequency is $\nu =2$ Hz. What was the initial velocity of the point particle?

My attempt:

Since the collision is elastic, the kinetic energy of the system is the same before and after the collision: $$0.5m_1v_1^2=0.5J_2 \omega_2^2+0.5m_2v_2^2+0.5m_1v_3^2$$ Where $v_3$ is the velocity of the point particle after the collision.

Now, in the case of a rod: $$J=\frac{1}{12}L^2m$$ And, we also know: $$\omega_2=2 \pi \nu$$ And there are also no external forces, therefor the momentum of the system is the same before and after the collision: $$m_1\vec{v_1}=m_1 \vec{v_3}+m_2\vec{v_2}$$ Here $v_1$ is the quantity we're looking for, $v_3$ is the point particle's velocity after the collision and $v_2$ is the velocity of the rod's center of mass. It follows: $$\vec{v_2}=\frac{m_1 \vec{v_1}-m_1 \vec{v_3}}{m_2}$$ From this it follows: $$0.5m_1v_1^2=\frac{1}{24}L^2m_2 4 \pi^2 \nu^2+0.5m_2 \left|\frac{m_1 \vec{v_1}-m_1 \vec{v_3}}{m_2}\right|^2+m_1v_3^2$$ This is 1 equation with 2 unknowns, and this is where I get stuck. Any help is appreciated.

  • $\begingroup$ Have you considered conservation of momentum? $\endgroup$
    – Kyle Kanos
    Commented Jun 3, 2014 at 15:30
  • $\begingroup$ Yes, I used it in the 4th equation. $\endgroup$
    – gndz
    Commented Jun 3, 2014 at 15:32
  • $\begingroup$ I see velocities there, not momenta. $\endgroup$
    – Kyle Kanos
    Commented Jun 3, 2014 at 15:32
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    $\begingroup$ You know the angular momentum of the system must also be preserved during the collision in any frame of reference. Before, it's $m_1 v_1 L/2$ relative to the center of the stationary rod. After, it's an expression that will include $\omega$, $v_2$ and $v_3$ among others. That's your missing equation. $\endgroup$
    – Floris
    Commented Jun 3, 2014 at 17:12
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    $\begingroup$ no a point particle also has angular momentum.As $\vec L=\vec r\times m\vec v$, $|\vec I|=mvd$ where d is perpendicular distance from rod. $\endgroup$
    – RE60K
    Commented Jun 10, 2014 at 12:48

5 Answers 5


Since the collision is elastic, the kinetic energy of the system is the same before and after the collision: $$0.5m_1v_1^2=0.5J_2 \omega_2^2+0.5m_2v_2^2+0.5m_1v_3^2$$

This kind of problem has usually

3 equations: conservation of: 1. Ke, 2. p, 3. L, and

3 unknowns: $y= v_3, z =v_2, x = \omega$, when the initial velocity $v_1$ is known.

But in this case the unknown parameter is $v_1 = x$ and you know that $\omega (2\pi\nu) = 4\pi$, angular momentum $L (I\omega) =\pi/3$ and $Ke (L\omega) = 2\pi^2/3$. This simplifies the problem, because that means that also the linear velocity of the rod is known $v_2 (L/r[m_2])=\frac23 \pi$

Based on this, your KE equation becomes: $$x^2=y^2 + \frac{1}{m_1} \left[I\omega^2+\left(\frac{2\pi}{3}\right)^2\right]\rightarrow x^2=y^2+10\frac{(12+4)\pi^2}{9} \tag1$$

the second equation can regard p (or L): $$m_1x = m_1y + \frac{2\pi}{3} \rightarrow y= x-\frac{2\pi}{3m_1} \tag2$$

There are 2 unknowns and 2 equations: $$\left\{\begin{align}x^2&=y^2+\frac{160\pi^2}{9} \\ y&= x-\frac{20\pi}{3}\end{align}\right.$$ and you may solve that simple system for $x$. $$[\x^2]= \left[[\x^2] -x\frac{40\pi}{3} + \frac{400\pi^2}{9} \right]+ \frac{160\pi^2}{9}\rightarrow x = \frac{[3]}{[40 \pi]} * \frac{14\pi* [40\pi]}{3*[3]}$$

Knowing the rules of collisions, the solution can be found even more quickly, since the linear velocity of the rod: $v_2=2/3\pi$ summed to its rotational velocity: $v_\omega(\omega r)=2\pi$ is the velocity of the rod $v_{m'}= 8/3\pi$ considered as a point-mass $m'$ at the tip of the rod, and you know its value is $m'=m_2/4$ *

The initial velocity $x$ can be found in a very simple way with the trivial 1-D formula (using the velocity of CoM) : $x=v_{m'}*1.75$:

$$v_i =v_{m'} \frac{m_1+m'}{2(m_1 )}=\pi\frac{8}{3}\left[\frac{.35}{.2}\right]$$

$x = 14.66076... =\pi14/3$


* linear momentum is of course the same: $m_2*v_2=m'*v_{m'} \rightarrow m' =( v_2/v_{m'}= 2/3*3/8) = 0.25$, but It is not even necessary to calculate it, since its value at CM, CoP, tip varies linearly (1, 3/4, 1/4), and therefore at the tip it is always $m_2/4$

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    $\begingroup$ that's a really brilliant solution, linus! $\endgroup$
    – user59485
    Commented Mar 14, 2015 at 11:03
  • 1
    $\begingroup$ @Floris, what lesson? what is not clear in the answer? there is a post explaining all aspects. :) $\endgroup$
    – user59485
    Commented Mar 31, 2015 at 5:48

enter image description here Using conservation of energy:

$$\left[\frac 12m_1v_i^2\right]_{particle}=\left[\frac 12m_1v_f^2\right]_{particle}+\left[\frac 12I\omega^2+\frac 12m_2v_2^2\right]_{rod}\\\text{where }I=\frac{ml^2}{12}=\frac1{12}$$

$$\frac {v_i^2}{20}=\frac {v_f^2}{20}+2\frac {\pi^2}{3}+\frac {v_2^2}2$$


(Alternative):Using coefficient of restitution=$1$

$$v_{i}=\frac{lw}{2} + v_2 +v_f$$

Using conservation of Momentum:


$$ \frac{v_i}{10}=\frac{-v_f}{10}+v_2$$

$$ v_i+v_f=10v_2$$

Using conservation of angular Momentum:

$$m_1v_i\left(\frac l2\right)=m_1\left(-v_f\right)\left(\frac l2\right)+I\omega$$



So, $$10v_2=\frac{20\pi}3\implies v_2=\frac{2\pi}3$$ $$(v_i-v_f).\frac{20\pi}3=\frac{40\pi^2}3+10\left(\frac{2\pi}3\right)^2=\frac{160\pi^2}9\implies v_i-v_f = \frac{8\pi}3$$ Also, $$v_i+v_f=\frac{20\pi}3$$ So: $$v_i=\frac{14\pi}3,v_f=2\pi$$

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    $\begingroup$ ω is 2πν=4π so energy is not 1/6 but 2/3 * π^2 $\endgroup$
    – bobie
    Commented Oct 21, 2014 at 13:47
  • $\begingroup$ @bobie see update $\endgroup$
    – RE60K
    Commented Mar 22, 2015 at 3:05

The collision of a rod with a point mass is the similar to the collision of two masses but with the effective mass of the rod being

$$ m' = m_{rod} \frac{I_{rod}}{I_{rod}+m_{rod} r^2} $$ where $r$ is the distance between the point of impact and the center of mass, and $I_{rod}$ is the mass moment of inertia about the center of mass. If the rod is slender with length $\ell$ then $$ I_{rod} = \frac{m}{12} \ell^2 \\ r = \frac{\ell}{2} \\ m' = m_{rod} \frac{1}{4} $$

So the momentum exchanged is $$ J = \frac{(1+\epsilon)\, v}{\frac{1}{m_{point}} + \frac{1}{m'}} $$ where $v$ is the impact speed and $\epsilon$ the coefficient of restitution.

The final velocity of the point mass is $$v_{point} = v - \frac{J}{m_{point}}$$

The final velocity of the rod center of mass is $$v_{rod} = \frac{J}{m_{rod}} \\ \omega_{rod} = \frac{r J}{I_{rod}} $$

NOTE: that the rod will rotate about a point with distance $c$ from the center of mass (in the opposite side from the impact point) located at $c =\frac{I_{rod}}{m_{rod} r} = \frac{\ell}{6}$. This is known as the instant center of rotation, and the point of impact is the center of percussion of the rotation center C.

  • 1
    $\begingroup$ Where does that initial formula come from? I am not familiar with it. Is it a well known result? Does it have a name? $\endgroup$
    – Floris
    Commented Mar 17, 2015 at 2:58
  • 1
    $\begingroup$ A body will never rotate about its center of mass unless a pure torque (no net force) is applied. $\endgroup$ Commented Mar 17, 2015 at 13:52
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    $\begingroup$ @Floris you can derive the first formula from the equations of motion with impact. See physics.stackexchange.com/a/155431/392 for hints on how to do it. $\endgroup$ Commented Mar 17, 2015 at 13:54
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    $\begingroup$ Being a homework type problem I wanted to just hint that the direction to go with this. My contribution is how to take a complicated problem (rod+mass) and turn in into a simpler one (two masses). $\endgroup$ Commented Mar 18, 2015 at 12:31
  • 2
    $\begingroup$ @linus there are easier ways to tackle this problem (using energy and momentum), and so my goal was to shed some light into what happens really (momentum exchange). Physics is not competitive, but collaborative. I am ok with not being awarded a question, or not getting a lot of votes. As you benefited from this post other will in the future too and that is enough. $\endgroup$ Commented Mar 19, 2015 at 13:29

One thing you can do is use conservation of angular momentum instead of linear - the point mass has angular velocity relative to the center of rotation of the rod. (If you are confused about this, imagine watching a car pass you on a road. The car moves in a straight line, but you rotate your head to follow it, giving it a [constantly changing] angular velocity.) Find the angular velocity of the particle the instant before the collision, apply conservation of angular momentum, and solve the problem from there.


The stroke of genius

The laws of Nature are wonderful: extremely simple - make a tremendously complex world, a few symbols can be extremely powerful and beautiful

all roads lead to Rome, you may go from London to Rome following many different routes, only a fool would get there via Moscow. But an eagle can fly over the Alps in a straight-line.

I'll try to explain what many readers do not understand:

....if you know shortcuts and are confident about applying them, by all means do. In my case, I could not understand exactly what he was doing so I chose to write a "pedestrian" answer for the benefit of other visitors

Visitors have not benefitted.

The standard solution

was fully and clearly explained last year, if there had not been a banal, trivial typo we would not be here now.

If the initial velocity is known, the unknown is $\omega$

enter image description hereenter image description here

and we need 3 equations:

$$ \left[\frac 12m_1v_i^2\right]_{m}=\left[\frac 12m_1v_f^2\right]_{m}+\left[\frac 12I\omega^2+\frac 12m_2v_2^2\right]_{rod}\tag1$$ $$ m_1 v_1 = m_1 v_3 + m_2 v_2 \tag2$$ $$ m_1 v_1 *r(=L/2) = m_1 v_3*r (=L/2) + I\omega \tag3$$

because there are 3 unknowns: $v_3$, $v_2$ and $\omega$

The solution of this problem

If $\omega$ is hnown ( = $4\pi$)

enter image description hereenter image description here

we do not need 3 equations because there are only 2 unknowns, 1. : $v$ and 2.: $v_3$, and the equations 2, 3 become obviously identical:

$$ m_1 v_1 = m_1 v_3 + m_2 v_2 \rightarrow .1x= .1y + \pi2/3\tag2$$ $$ r*m_1 v_1 = r*m_1 v_3 + L \rightarrow .1x = .1y +(L/r)\pi2/3\tag3$$

since the linear momentum of the rod is equal to its angular momentum divided by the radius (and mass which is = 1Kg) $$m_2v_2 = I\omega /r$$ This was reminded in a few comments, but no correction was made to the post

Using 3 equations instead of two is an unhealthy practice, and doing a monumental useless work is not only pedestrian but ... ehm... let's say: non-rational, and encouraging inexperienced or credulous visitors to do so is misleading

The simplest solution

The correct standard solution has been offered as a first choice, but was apparently rejected by the vast majority of the 'visitors'.

The ingenious solution has probably been overlooked, not-understood or considered not-worth-examining-at-all.

But it so easy to understand: any schoolboy knows that the velocity of a body at rest after an impact is $v = 2p/ M$

enter image description hereenter image description hereenter image description here

Considering the effective point-mass at the tip of the rod ( $m' =m_2/4$ ) we can determine the initial velocity of the point-particle just reversing the simple formula of the head-on collision

  • 2
    $\begingroup$ how is this different from mine $\endgroup$
    – RE60K
    Commented Mar 22, 2015 at 3:23
  • 1
    $\begingroup$ It is a shame that you use your otherwise good answer to attack mine - it detracts from what could have made a nice crisp summary of the physics (and yes I completely agree there can be better answers than the one I gave). I am baffled by your need to knock me for using three conservation laws from which I derive (in one line) my equation (4) which is essentially the one you claim "every schoolboy knows". Maybe it's been too long since I was a schoolboy? $\endgroup$
    – Floris
    Commented Mar 22, 2015 at 20:43
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    $\begingroup$ I have deleted my answer since it seems to have caused a lot of bad feeling; that was never my intention. $\endgroup$
    – Floris
    Commented Mar 30, 2015 at 13:30
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    $\begingroup$ Please, do not vote this answer: it only explained linus' answer. It has no merit of its own, and I will soon delete it. I posted only to explain the points some readers could not understand. $\endgroup$
    – user75366
    Commented Mar 31, 2015 at 8:18

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