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I'm new to this community and I think you can help me! I studying Rolling at college and I'm struggling with this question:

Figure shows a cube of mass m sliding without friction at speed $v_0$. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity - both speed and direction - after the collision?

Questions Figure

I tried this:

From kinetic energy conservation:

$$ \frac{mv_0^2}{2} = \frac{mv^2}{2} + \frac{I\omega^2}{2} $$

$$ I = \frac{Md^2}{12} = \frac{2md^2}{12} = \frac{md^2}{6} $$

$$ \frac{mv_0^2}{2} = \frac{mv^2}{2} + \frac{md^2\omega^2}{12} $$

$$ v_0^2 = v^2 + \frac{d^2\omega^2}{6} \implies (1) $$

From angular momentum conservation:

$$ \frac{mvd}{2} = I\omega $$

$$ \frac{mv_0d}{2} = \frac{md^2\omega}{6} $$

$$ v_0 = \frac{d\omega}{3} $$

$$ d^2\omega^2 = 9v_0^2 \implies (2) $$

Substituting (2) into (1):

$$ v^2 = v_0^2 - \frac{3v_0²}{2} $$

$$ v^2 = -\frac{v_0^2}{2} $$

But this is a wrong answer, the correct answer is:

$$ v = \frac{v_0}{5} $$

Could you help me? Thanks in advance!

PS: this is question 86 is from

UPDATE

Especial thanks to @ssj3892414 to point the right direction! I'll put here the correct equation.

From angular momentum:

$$ \frac{mv_0d}{2} = \frac{mvd}{2} + I\omega $$

$$ \frac{mv_0d}{2} = \frac{mvd}{2} + \frac{md^2\omega}{6} $$

$$ v_0 = v + \frac{d\omega}{3} $$

$$ d\omega = 3(v_0 - v) $$

$$ d^2\omega^2 = 9(v_0^2 - 2vv_0 + v^2) $$

$$ \frac{d^2\omega^2}{6} = \frac{3(v_0^2 - 2vv_0 + v^2)}{2} \implies (2) $$

Substituting (2) into (1):

$$ v_0^2 = v^2 + \frac{3}{2}v_0^2 - 3vv_0 + \frac{3}{2}v^2 $$

manipulating this equation gives:

$$ 5v^2 - 6vv_0 + v_0² = 0 \implies (v - v_0)(5v - v_0) = 0 $$

So, answers $v = v_0$ and $v_0 = 5v$, which the second one is the right answer!

Knight, Randall Dewey. Physics for scientists and engineers: a strategic approach. 3rd ed. Vol. I. Boston: Pearson, 2013, p. 353. Print.

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closed as off-topic by sammy gerbil, Yashas, John Rennie, Kyle Kanos, Jon Custer Jun 19 '17 at 13:21

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Jun 19 '17 at 6:00
  • $\begingroup$ @JohnRennie, I'm glad for your feedback! I'm a member of others communities (at SE or not) and I see the problems in just 'answer homework'. What I'd like to discuss is: I checked each point on the list How should I ask a homework question on this website? and my question matches. I saw an example of a good question and mine is similar: physics.stackexchange.com/q/16182/159371. $\endgroup$ – Thiago Ururay Jun 19 '17 at 6:25
  • $\begingroup$ So, (1) I couldn't get help in other questions; (2) my question is about collision and conservations, especific topics; (3) I referenced the source; (4) I used homework-and-exercises tag; (5) my question provides the problem text, I've showed my work done (this points to what I wasn't understand and doesn't seeing); (6) the answer provided a direction so I could look for a way to solve my problem without a resolution. So why is my question marked as looking for homework answer? It is a genuine doubt so I prevent future errors. $\endgroup$ – Thiago Ururay Jun 19 '17 at 6:26
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The angular momentum conservation equation is wrong. You haven't taken into account the angular momentum of the block after collision.

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  • $\begingroup$ I can't figure out how to express block's angular momentum formula after the collision... actually, after the collision does it rotates? I don't see it... $\endgroup$ – Thiago Ururay Jun 18 '17 at 6:27
  • $\begingroup$ Oh, I search a little and found the expression! Thanks for your guide, I'll update the question with updated expression! :) $\endgroup$ – Thiago Ururay Jun 18 '17 at 7:15
  • $\begingroup$ A little late on this, but as you wrote the expression for blocks final angular momentum is the same as its initial one, but initial velocity v0 replaced by final velocity v. Also note that the other solutions v=v0, means that the rod didn't move and the block went on its merry way. So its a valid solution as well $\endgroup$ – ssj3892414 Jun 18 '17 at 9:27

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