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200 Puzzling Physics Problems P48

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Here is the question:

Two small identical smooth blocks $A$ and $B$ are free to slide on a frozen lake. They are joined together by a light elastic rope of length $L \sqrt(2)$ which has the property that it stretches very little when the rope becomes taut. At time t = 0, A is at rest at x = y = 0 and B is at x = L; y = 0 moving in the positive y-direction with speed V. Determine the positions and velocity of A and B at time
$t = 2L/V$

The hints and solutions are found here: https://dejanphysics.files.wordpress.com/2016/10/gnadig_1.pdf (under pdf pages 59 and 113)

I was confused because the hint seemed strange where it said that energy is conserved. Since the string becomes taut at some point, then doesn't some of the energy go into the spring potential energy of the rope, so not all of it is used on the blocks?

I assumed this, so I wrote out conservation of angular momentum about the CM of the system:

Before momentum = Momentum after:

$m(v)(L/2) = 2m(v_1)(\frac{L}{2\sqrt{2}})$ (*)

where v_1 is the velocity after they start both moving. Also, in the translational aspect, the CM moves with velocity $V/2$, so I was getting that right. But when I used formula (*) to find the angular velocity and positions of A,B and their velocities, I was getting very wrong answers. Can someone please explain where my mistake is?

asked Mar 27 in Physics Problems by andrey333 (130 points)

1 Answer

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Best answer

You have not explained how you applied your formula, nor what answers you got, so I cannot explain what mistake you made.

Yes, elastic potential energy is stored in the rope when it becomes taut. But this energy is released immediately and the rope becomes slack again.

The rope does not remain taut. Perhaps this is what is confusing you. It used to be common to assume that the rope was totally inelastic and remained taut. But it is more realistic to treat the rope as a very stiff elastic spring.

The rope provides a 'jerk' or impulse which changes the momentum of each block. It is equivalent to an elastic collision between the two blocks, except that in this collision the blocks are initially moving apart, and afterwards are moving closer together. Momentum and kinetic energy are both conserved. To conserve energy it is easiest to use the Law of Restitution : relative speed of separation = relative speed of approach, along the line joining the two blocks. The speeds perpendicular to the line joining the two blocks are not affected by the collision : this is equivalent to saying that angular momentum is conserved.

Whenever the distance between the blocks becomes 0 or L there is a collision which changes their momenta. Between collisions the blocks move as free particles.


Since there are no horizontal external forces, the CM of the 2 blocks moves with constant velocity $\frac12 V$ in the +y direction. The rope confines the blocks inside a circle of diameter $L\sqrt2$ centred on the CM. This circle moves along with the CM. When the blocks collide with this circle (which happens simultaneously) they are 'reflected' from it. Each is turned through 90 degrees. The blocks move symmetrically around a square of side L inside the circle, each with speed $\frac12 V$ in the CM frame of reference .


After time $t=2L/V$ the CM has travelled a distance of $(\frac12 V)t=L$ and lies at $(\frac12 L, L)$ in the ground frame of reference.

At this time each block has travelled the same distance $L$ around the square in the CM frame of reference. They started on a horizontal diameter of the square (red circles), at $(-\frac12 L, 0)$ and $(+\frac12 L, 0)$. At time $t$ they lie on a vertical diameter (blue circles), at $(0, -\frac12 L)$ and $(0, +\frac12 L)$ in the CM frame.

In the ground frame the positions of the blocks at time $t$ are :
$(0, -\frac12 L)+(\frac12 L, L)=(+\frac12 L, \frac12 L)$ and
$(0, +\frac12 L)+(\frac12 L, L)=(+\frac12 L, +\frac32 L)$.

answered Mar 28 by sammy gerbil (26,166 points)
selected Mar 28 by andrey333
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