0
$\begingroup$

enter image description here

Let's say we release a bob of mass $m$ attached the ceiling through a wire of length $l_0$. Now at the bottommost point another identical bob of mass $m$ gets gently attached to the bob. Now we are required to find the angle upto which the system rises.

First I tried to solve the question using Energy Conservation problem from the starting point to the last point (maximum $\theta$) (taking the bottommost point as the point of potential reference)

$$U_{initial} = mg(l_0)$$
$$K_{initial}=0$$ $$U_{final}= 2mgl_0(1-\cos\theta)$$ $$K_{final}=0$$

Using $$U_{initial}+K_{initial}=U_{final}+K_{final}$$ $$mg(l_0)=2mgl_0(1-\cos\theta)$$ $$\Rightarrow \theta= 60^0$$

Then I tried to solve the problem using the conservation of momentum

At the bottommost point velocity of the single mass is $\sqrt{2gl_0}$ Using conservation of momentum

$$m \cdot \sqrt{2gl_0}=2m \cdot v$$ (where $v$ is the velocity of the combined mass from the lowermost point ) $$v= {\sqrt{2gl_0} \over 2}$$

Now again using the energy conservation from the lowermost point to the last point

$$U_{initial} = 0$$
$$K_{initial}={1 \over 2}(2m)v^2$$ $$U_{final}= 2mgl_0(1-\cos\theta)$$ $$K_{final}=0$$

Using $$U_{initial}+K_{initial}=U_{final}+K_{final}$$ $${1 \over 2}(2m)v^2=2mgl_0(1-\cos\theta)$$ $${m(2gl_0) \over 4}= 2mgl_0(1-\cos\theta)$$ $$\Rightarrow \theta = cos^{-1}(3/4)$$

Why such contradiction exists between the answers? Can't the energy conservation law be applied here.

$\endgroup$
2
$\begingroup$

For the bobs to join together the collision has to be inelastic and so kinetic energy is not conserved during the joining of the two bobs.
Once you have found the speed after the joining of the the two masses by momentum conservation then you will see that there is a decrease in the kinetic energy.


Update which I have found great difficulty writing.

Whether the joining is gentle or not because there are no external horizontal forces during the collision momentum conservation in the horizontal direction must apply.

So the kinetic energy of the one bob before the joining is $\frac 1 2 \; m \; 2gl_o$ and the kinetic energy of the two bob after the joining is $\frac 1 2 \; 2m \;\dfrac{2gl_o}{4}$.

The translational kinetic energy of the centre of mass of the two bobs has decreased by a half.

If the joining is done by a pin attached to one bob going into the other bob then one can say that work needs to be done to drive the pin into the bob and the result is that some bonds between atoms are permanently broken and the bobs get hotter.

Suppose instead that the two bobs, $A$ and $B$ were so clean that when they collided together they were joined together by cohesive forces.

So the moving bob $A$ hits the stationary bob $B$, the bonds between the two bobs deform and the centre of mass of the two bobs moves off at half the speed of the initial speed of bob $A$ as determined by momentum conservation.

There are two extremes:

1 As a result of the collision the bonds between the two bobs are permanently deformed and the work needed to do this comes from the kinetic energy that bob $A$ had initially.
The collision between the bobs is inelastic.

2 The collision between the two bobs is elastic which means that the compressed bonds act like compressed springs and are storing elastic potential energy and also exert forces on the two masses.

The two masses separate, gaining kinetic energy and continue to do so until the bonds start to get stretched.
The two bobs then slow down losing kinetic energy and the bonds because they are being stretched gaining elastic potential energy.
Eventually the two bobs stop and the bonds now pull the two bobs together.

So in summary.
You have the centre of mass of the two bobs moving in an arc of a circle and the two bobs oscillating about their centre of mass with an energy associated with that oscillatory motion which is half the kinetic energy that bob $A$ had initially.


All this is rather abstract but the point I am trying to make is that the “missing” kinetic energy is not lost but it is no longer part of the translational kinetic energy of the two bob system.

In the real world the collision might well send (shock) waves within the bobs.
As time went on those waves would die down and as a result the bobs would have a higher temperature.

So by whatever mechanism you chose to join the two bobs together half of the initial kinetic energy that bob $A$ had would end up as heat, sound and work done permanently deforming the bobs.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So that means that I can't use the first method since at the time of collision some kinetic energy disburses into potential one? $\endgroup$ – user118752 Sep 19 '16 at 9:13
  • $\begingroup$ OP says "gently", that suggests no collision at all , like mild magnetism etc $\endgroup$ – user104372 Sep 19 '16 at 9:15
  • 1
    $\begingroup$ I was thinking the same also! $\endgroup$ – user118752 Sep 19 '16 at 9:17
  • 2
    $\begingroup$ Whether it be gently or not the two bobs stick together in the end and so in some way kinetic energy decreases. If is was by a pin in one bob going into the other then work is done permanently breaking bonds and heat is produced. $\endgroup$ – Farcher Sep 19 '16 at 9:23
  • 1
    $\begingroup$ In this problem, the trick lies in recognising first of all that this is an inelastic collision. Obviously the lost energy is converted into heat, spring energy or some other form of energy but that is immaterial for reaching the correct answer. This situation of collision of a moving mass with a stationary mass and then both the masses sticking together and moving together in the same direction can not satisfy both energy conservation and momentum conservation simultaneously. It has to be treated as an inelastic collision where momentum conservation holds. $\endgroup$ – user103515 Sep 19 '16 at 13:17
0
$\begingroup$

The velocity (v1) of the mass m just before sticking to the 2nd mass m is sqrt(2gl0)

Now use conservation of momentum to calculate the velocity(v2) just after the 2 masses stick together. It comes out to be sqrt(2gl0)/2.

Calculate the KE just before sticking. it come out to be (mgl0)

Calculate the KE just after sticking. it comes out be be (mgl0)/2.

So, you can see clearly that the total energy is not conserved in this problem. When masses stick together energy is not conserved. Its an inelastic interaction. You can not apply both energy and momentum conservation simultaneously to this problem. You can check by writing the energy and momentum conservation equations to this problem.

But momentum is always conserved since you have neglected friction(with air). So your second method in which you first calculated the velocity using momentum conservation and then applied conservation of energy after the masses sticked together is correct.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "g l0" should be "g l0(1- cos theta)" $\endgroup$ – user104372 Sep 20 '16 at 7:34
0
$\begingroup$

You do not need complex calcs, if there is an inelastic collision speed becomes 1/2 and height 1/4,(else, the speed of the bob[s] becomes v/sqrt2), height ($l_0*(1-cos\theta)$ is just half, then find the corresponding angle. If the original angle was indeed 60°, in this case the new angle will be = 41.41°, because $\theta'= arccos (1-\frac{cos\theta}{2})$, (1-1/4 = 3/4)

What is the given answer? From what they consider the correct answer we can deduce what conditions they imagined. Did you draw the picture yourself, since it is not accurate. If the second bob is attached below, that lowers CoM and that has to be taken into account

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If the final velocities of the two bobs is $\dfrac{v}{\sqrt 2}$ then momentum in the horizontal direction is not conserved. The collision between the two bobs must be inelastic. $\endgroup$ – Farcher Sep 19 '16 at 10:33
  • $\begingroup$ @Farcher, let's wait for the given answer, and we'll know what they had in mind. From OP's proceedings this seems to be most likely, If it is inelastic then v becomes half. $\endgroup$ – user104372 Sep 19 '16 at 10:36
  • 1
    $\begingroup$ The given answer is $ \theta = cos^{-1}3/4$ $\endgroup$ – user118752 Sep 19 '16 at 12:55
  • 1
    $\begingroup$ I truly appreciate the way you have made it simple for me to do, and not clicking a tick on your answer doesn't mean I didn't accept it. Its just that I wanted to know te reason by KE doesn't remain conserved and hence I chose the first one. $\endgroup$ – user118752 Sep 20 '16 at 4:07