# Collision, Rotation and Kinetic Energy [closed]

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?

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):

$$v^2 = v_0^2 - \frac{3v_0²}{2}$$

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

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

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.

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

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• 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. – John Rennie Jun 19 '17 at 6:00
• @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. – Thiago Ururay Jun 19 '17 at 6:25
• 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. – Thiago Ururay Jun 19 '17 at 6:26