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I am currently studying for the GRE Physics subject test by working through published past tests. My question is about problem 44 from the test GR8677:

Top-down diagram of point mass and stick before collision

$44.$ A uniform stick of length $L$ and mass $M$ lies on a frictionless horizontal surface. A point particle of mass $m$ approaches the stick with speed $v$ on a straight line perpendicular to the stick that intersects the stick at one end, as shown above. After the collision, which is elastic, the particle is at rest. The speed $V$ of the center of mass of the stick after the collision is
(A) $\frac{m}{M}v$
(B) $\frac{m}{M+m}v$
(C) $\sqrt{\frac{m}{M}}v$
(D) $\sqrt{\frac{m}{M+m}}v$
(E) $\frac{3m}{M}v$

My approach was to first write down expressions for the conservation of energy, linear momentum, and angular momentum. The problem states that the particle is at rest after the collision, which simplifies the equations:

$$\frac{1}{2} m v^2 = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2$$

$$m v = M V$$

$$\frac{1}{2} L m v = I \omega$$

where $I=ML^2/12$ is the moment of inertia of the stick about its CM and $\omega$ is the stick's angular velocity. The most natural next step is to solve the linear momentum equation, giving us the correct answer (A). This is the solution used here and here.

However, adhering to my penchant for valuing understanding above efficiency, I attempted to verify this answer by combining the other two conservation equations. I solved the angular momentum equation for $\omega$ to obtain $$\omega = \frac{L m v}{2 I}.$$ I then solved the energy equation for $V$ and substituted in this result: $$V^2 = \frac{1}{M}(m v^2 - I \omega^2)$$ $$= \frac{1}{M}\left( m v^2 - I \left( \frac{L m v}{2 I} \right)^2 \right)$$ $$= \frac{1}{M}\left( m v^2 - \frac{(L m v)^2}{4 I} \right)$$ $$= \frac{m v^2}{M}\left( 1 - \frac{L^2 m}{4 (M L^2 / 12)} \right)$$ $$= \frac{m v^2}{M} \left( 1 - 3\frac{m}{M} \right)$$ $$\Longrightarrow V = v \sqrt{ \frac{m}{M} \left( 1 - 3\frac{m}{M} \right) }$$ $$= v \frac{m}{M} \sqrt{ \frac{M}{m} - 3 }$$

I see several problems with this result. First, it does not immediately resemble any of the answers, though it can be made to match either (A) or (E) with a choice of $M/m=4$ or $M/m=12$, respectively. Second, if $M/m < 3$, the velocity of the stick becomes imaginary which (to me) does not have an obvious physical interpretation. I also do not see why there should be any restriction on the mass ratio in the first place.

Is this approach illegitimate or does it contain an error? If so, why/where? If not, why does the result not coincide with the first approach?

In the first solution link above, there are many comments going back-and-forth on the reasons for the discrepancy but, frustratingly, no conclusion. Any insights would be greatly appreciated.

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3 Answers 3

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The GRE question seems to be erroneous. You see, if one specifies that the collision is elastic then they cannot also specify what would be the speed of the ball (the mass $m$) after the collision. That over-constraints the problem. That is the reason you are finding inconsistencies in your two ways to approach the problem. Put another way, there are two variables $V$ and $ω$ but there are three equations. So, it is expected that one would find inconsistent solutions.

The correct way to pose the question would be to either delete the word elastic (then the kinetic energy equation goes away and we have two equations for two variables) or to keep the word elastic but do not specify what happens to the ball after the collision (then we will have three variables to solve out of three equations).

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  • $\begingroup$ It would be interesting to the see the GRE solution and even the reasoning behind it. $\endgroup$ Aug 8, 2019 at 0:20
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    $\begingroup$ @AaronStevens I think they would have gone with Rob's reasoning that since there is one specific fine-tuning of parameters which can allow everything that is described in the question to happen, one should conclude that it is the case, i.e., $M=4m$. $\endgroup$
    – youpilat13
    Aug 8, 2019 at 1:28
  • $\begingroup$ @AaronStevens Why would there not be a unique solution in that case? I didn't get it. $\endgroup$
    – youpilat13
    Aug 8, 2019 at 1:29
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    $\begingroup$ The GRE was probably just trying to get students to realize that you don't need to think too complicated. Linear momentum is conserved, so $V=\frac mMv$. The other information is irrelevant. Unfortunately it also made things too constrained. $\endgroup$ Aug 8, 2019 at 2:17
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What you've done is to find the condition under which an elastic collision can end with the incoming particle at rest. Conservation of momentum with that condition gives $MV=mv$, and conservation of angular momentum fixes $\omega$. If you require energy conservation as well, you find an inconsistency unless $\sqrt{M/m-3}=1$.

This constraint means that, if you did this experiment with $M\neq 4m$ and observed the incoming ball to stop, the collision should have either been inelastic (energy lost to heat, noise, etc.) or super-elastic (energy added from some internal source that isn't specified here).

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I think they just want you to ignore irrelevant information. Linear momentum is conserved, so we get $$V=\frac mMv$$

They threw in the extra information about an elastic collision I guess to throw students off. Unfortunately it leads to an actual contradictory situation.

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