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System (rod+ball) collides elastically with a block

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The following link directs to the picture of the problem:

https://imgur.com/gallery/1a8cX?s=wa

Questions:

A) Are momentum, angular momentum and energy conserved? Under which circumstances?
B) Angular velocity of the ball immediately before the collision?
C) M/m rate so as to make the system (rod+ball) stay at rest after collision.
D) Velocity of the block after the collision.
E) Obtain an equation for D taking into account there is friction (Mu coefficient) ground-block along distance d and height h.

These are my answers:

A) - Linear momentum is conserved before/after the collision, due to the lack of external forces. However, there is a force exerted by the pivot on the system during the collision. Therefore, linear momentum of the system (ball+rod) IS NOT CONSERVED.

 - Angular momentum is conserved due to the fact that torque exerted on the rod by the ball is equal and opposite to the torque exerted on the ball by the rod. (Meaning that AM of the rod is equal and opposite to AM of the ball; therefore AM of the system IS CONSERVED).
 - Energy IS CONSERVED IF FRICTION of any kind IS NEGLECTED. If we considered friction ground-block along distance d, energy would not be conserved after the elastic collision.

B) First, I obtain system's center of mass: I consider the rod a puntual mass and remembering both rod and ball masses are equal to m: Xcm= (m(l/2)+ml)/2m= 3/4l
As energy is conserved before the collision and system start and ends at rest, we can assure gravitational potential energy transforms to kinetic energy(note omega equals to angular velocity):
2mg(3/4l)=1/2(2m(3/4l)^2)omega^2 Therefore: omega= sqroot(8g/3l)

C) First we obtain CoM's velocity before the collision:
V=(omega)radius; therefore Vcm:sqroot(3gl/2)
Now to obtain the velocity of the block after the collision we use momentum conservation principle:
2mVcm=MVblock; therefore: Vblock=2m/Msqroot(3gl/2)
Finally, to relate both M and m we use the principle of conservation of kinetic energy at an elastic collision:
1/22mVcm=1/2MVblock; therefore: M/m=2

D) Vblock= sqroot(3gl/2)

E) Energy is not conserved along d, however we now deltaE=muMgd. Therefore, Vblock= (mu)gd.
We have a projectile trajectory. Block has an initial velocity just at its horizontal component. Therefore:
D= d + Vblock(t) and y= h - 1/2g(t^2) Solving for t at second equation: t=sqroot(2h/g)
Finally, D= d(1 + mu(g)sqroot(2h/g))

Please if you could check my answers I would be greatful. If there's anything wrong or that can be improved let me know.
Thanks

asked Apr 2 in Physics Problems by JD_PM (490 points)
edited Apr 6 by JD_PM
Is there some reason you think your answers might be wrong? Is there something you are unsure about?
See the Meta question http://physics.qandaexchange.com/?qa=253/how-to-upload-an-image for instructions about uploading images.

Is there some text to go with the image?
It is just I want to be sure I am right, as I am a physics students who is preparing himself. This is an exam-type problem (first year of physics degree)
I tried to post the pic as you said but did not work.
The text says the following: there is a system (ball+rod) which collides elastically against a block. The system starts at rest, besides, after the collision the system stays at rest too. The block is at rest until is hit by the system. The block goes along a distance d before falling from a height h, reaching a distance D. There's friction between block and ground. To sum up, I want to point out that even if I'm right with the answers if you wanted to add something in order to improve them, it would be great. Thanks
The instructions for posting images via PasteBoard do work, if you follow them correctly.
Well, I can try it again. I posted the image via PasteBoard and copied the link as instructed, however I will cheak it

1 Answer

2 votes
 
Best answer

A) Your answer is not clear about whether linear momentum is conserved. As you state, there is an external force acting during the collision between the ball $m$ and the block $M$. This external force is the reaction at the hinge (pivot), which is attached to the outside world. This means that linear momentum might not be conserved. However, angular momentum about the hinge is conserved.

B) The question states that "the system" is at rest after the collision. Doesn't this mean the ball + rod? You have calculated the angular velocity immediately before the collision.

I can see that you have equated PE lost to KE gained, using $\frac12 I \omega^2$ for KE. But how have you calculated moment of inertia $I$?

The moment of inertia of the rod about one end is $\frac13 mL^2$, not $m(\frac12 L)^2=\frac14 mL^2$. The moment of the (point) mass $m$ is $mL^2$. So the total moment of inertia of ball+rod is $\frac43 mL^2$ not $\frac98 mL^2$.

Your calculation method for moment of inertia is wrong. The moment inertia of an extended object is not the moment of inertia of its CM. For example, if the rod were pivoted at its CM, your method gives MI=0 instead of $\frac{1}{12}mL^2$. However, the method does work for calculating the change in PE (provided the gravitational field is uniform).

C) Here you need to apply conservation of angular momentum not linear momentum.

The KE of the ball+rod immediately before the collision equals the KE of the block immediately after the collision :
$\frac12 I\omega^2=\frac12 Mu^2$
Also, the angular momentum of the ball+rod equals that of the block :
$I\omega=MuL$
so
$I^2\omega^2=M^2u^2L^2$
Dividing the AM equation by the KE equation we get
$I=ML^2$
However, $I=\frac43 mL^2$ therefore $M=\frac43 m$.

D) The velocity $u$ of the block immediately after the collision is given by $\frac12 Mu^2=32mgL$ where $M=\frac43m$ therefore $u=\frac32\sqrt{gL}$.

E) Your calculation of the velocity of the block as it leaves the table does not seem to be correct.

The initial KE of the block before it slides across the rough part of the table is $\frac12 Mu^2=\frac12 M \frac94 gL=\frac98 MgL$ from above. The work done on the block by friction is $\mu Mgd$. So the velocity $v$ of the block as it leaves the table is given by
$\frac12 Mv^2=\frac98 MgL-\mu Mgd$
$v=\sqrt{(\frac94L-2\mu d)g}$.
It is not possible to simplify this expression unless you are given a value for $\mu d$ in terms of $L$. EG if $\mu d=\frac58 L$ then $v=\sqrt{gL}$.

Judging by the diagram $D$ is measured from the edge of the table, not from the initial position of the block. Then $D=vt$ where $h=\frac12 gt^2$ which you have got right.

answered Apr 2 by sammy gerbil (26,166 points)
selected Apr 9 by JD_PM
Okey, as momentum is not conserved we can't use Vcm=Vblock; got it Thanks

Then now I obtained for D:
D= sqroot(2*(9/4*Lh - 2*mu*d*h)) do you agree?
But now, if d went to infinity we would get a complex number as our D. How would you explain that case?
Thanks a lot
The block loses kinetic energy because of friction. If $d$ were infinite the the block would have to start with an infinite amount of kinetic energy in order to get to the edge of the table. If the block starts with a finite amount of kinetic energy, it can only go a finite distance before losing all that energy and stopping.

In this case the block stops (ie $v \to 0$) when $\frac94L = 2\mu d$. This gives you the maximum distance which the block can travel. $d$ cannot increase any further than this, so $v$ does not become a complex number.
Oh, You're totally right! V is not able to become a complex number, I was wrong.

Just to finish; do you agree with D=sqroot(2h(9/4L-2(mu)d))?
Yes that is correct.
Thanks A LOT Sammy Gerbil, this is the kind of site I 've been looking for a long time. As you could see, at the end everything was wrong. Thanks to you I improved.
I'll be back soon
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