# Block collides elastically with a rod

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One block of mass m moves at velocity $v_{0}$ over a horizontal table of height H without friction. When it reaches the edge of the table, the block collides elastically against a rod of mass M and length L which turns about an axis (perpendicular to the paper) located at the opposed extreme of the collision (look at the image). Mass ratio: M/m=3/2.
Calculate :
a) Velocity of the block and angular velocity of the rod after the collision
b) Maximum height that the CM of the rod reaches after the collision (initially located as shown at the image)
c) Horizontal distance that the block reaches after the fall. What I tried: Note the energy immediately before the collision is the right side of the equation and immediately after the collision the left side (in the image is written the other way around and is wrong).

At a) I obtained an equation of the velocity of the block after the collision in function of the angular velocity of the rod after the collision. I need another equation so as to obtain both values. I think we have to use conservation of angular momentum here. I thought about this equation:

$$\omega_{f} = \frac{mL^2\omega_{o}}{\frac{ML^2}{3}+mL^2}$$

$$\omega_{o}=\frac{v_{o}}{L}$$

$$v_{f}^2=v_{o}-\frac{L^2\omega^2_{f}}{2}$$

Then I could obtain equations for final velocity and angular velocity in function of $v_{o}$ and $L$ At b) the solution is said to be $z=\frac{8v_{0}^2}{27g}$ But I do not get it.

edited Jul 20, 2018
Parts a) and b) are similar to your earlier question http://physics.qandaexchange.com/?qa=2575/rod-collides-in-a-perfect-inelastic-way-with-a-ball .

a) Yes you need to use conservation of angular momentum. You also need to use conservation of kinetic energy because the collision is perfectly elastic. The easiest way to do the latter is through Coefficient of Restitution $e=1$ : relative velocity of separation = relative velocity of approach.

b) KE of rod immediately after collision = gravitational PE of rod at maximum swing.
a) Are energy and AM approaches correct now?
b) Immediately after the collision there is also GPE right?  I took it into account as it can be seen in the image

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a) During the collision there is conservation of angular momentum, so
$$mLv_0=mLv_1+I\omega \text{............(1)}$$ where $v_0, v_1$ are the initial and final speed of the block, both measured to the right.

There is also conservation of kinetic energy. The speed of the end of the rod immediately after the collision is $L\omega$ and the collision is elastic so $$\text{ relative speed of separation = relative speed of approach}$$ $$L\omega-v_1=v_0 \text{............(2)}$$

The moment of inertia of the rod about one end is $I=\frac13 ML^2=\frac12 mL^2$ because $\frac{M}{m}=\frac32$. Insert in (1) and combine with (2) to get $$v_1=\frac13 v_0$$ $$\omega=\frac{4 v_0}{3L}$$

b) KE of rod immediately after the collision is $\frac12 I\omega^2$. Increase in gravitational PE of rod at maximum swing is $Mgh$ where $h$ is the maximum distance which the CM of the rod moves vertically upwards. Therefore $$Mgh=\frac12 I\omega^2=\frac12 (\frac 13 ML^2)(\frac{4v_0}{3L})^2$$ $$h=\frac{8v_0^2}{27g}$$

answered Jul 20, 2018 by (28,466 points)
edited Jul 21, 2018
Thanks Sr but why $I=\frac13 ML^2=\frac12 mL^2$? We agree with $I=\frac13 ML^2$  but I do not understand why you stated that is equal to what to me is the moment of inertia of a solid disk
At b) I understand the kinetic energy of the rod converts to gravitational potential energy. Then it is not necessary to take into consideration the GPE of the rod immediately after the collision isn't it?
a) The question states that $M/m=3/2$. I have used that relation.

b) Gravitational potential energy is always measured relative to some reference point. Whether the rod has GPE immediately after the collision depends on what reference point you use. But you only need to consider the **change** in GPE, not the absolute values before and after. So it does not matter what reference point you choose, nor whether the rod has GPE immediately after the collision - or indeed at any other instant!
Got it thanks :)
I have been thinking about the equation $mLv_0=mLv_1+I\omega$ Should not be       $-mLv_0$? Let me explain you why I say so. Vectorially speaking I obtained: $(x \hat i + y_{o} \hat j)(mv_{o} \hat i)=-y_omv_o \hat k$  as $\hat j \hat i = -\hat k$
I thought about your consideration: right direction as positive. But if that is the case, why $mLv_1$ is positive? The block goes backwards immediately after the collision
If $mLv_0$ is -ve then $mLv_1$ and $I\omega$ will also be -ve.

If you wish to define $v_1$ as +ve to the left you can do so. Then you will get $v_1=-\frac13 v_0$ showing that the direction is to the right.