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Please consider the following situation (ignoring air friction, and assuming all strings have no mass):

Ball A with mass m1 is hanging from the ceiling by the blue string.

Ball B with mass m2 is hanging by another string, from ball B. So currently, the system is at rest.

Now, somebody cuts the blue string.

Which of the following scenarios would happen, and exactly why? Also, does the answer depend on the values of m1 and m2 at all?

  1. Both ball A and ball B will start to accelerate to the ground (with acceleration g)?

  2. Ball A will start to accelerate downwards (with acceleration g). Ball B, however, will stay exactly in its place, until ball A collides with it from above.

  3. Something else?

My intuition seems to be around option 1. In order to stay at rest, ball B needs some force to cancel out the weight m2g. But once ball A is accelerating downwards, the string connecting the two balls is no longer fully stretched. So I can't see why ball B won't start to accelerate downwards at this point, at the same rate as ball A (a = g m/s^2).

Please explain the correct intuition and show the relevant equations (in terms of Newton's laws).

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  • $\begingroup$ Since ball B has inertia and is acted upon by a tension force due to A and the ceiling, once this tension force goes to zero, Ball B will wait a little bit and then start falling. So, I think the distance between A and B while falling will be a little less than the string's length. $\endgroup$
    – Anky Physics
    Commented Jan 17 at 14:56
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    $\begingroup$ @AnkyPhysics Why does B wait a bit before starting to fall? $\endgroup$
    – Aviv Cohn
    Commented Jan 17 at 15:10
  • $\begingroup$ @AnkyPhysics would be correct if we account for the propagation time of the force down the string. In reality, when you cut the top of the string, the bottom doesn't "know" instantaneously, as that would require the information to travel down the string faster than the speed of light. But assuming a massless idealized string, the force does propagate instantaneously, and both balls start falling the instant the string is cut. $\endgroup$
    – Nuclear Hoagie
    Commented Jan 17 at 15:39
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    $\begingroup$ @AvivCohn In a real string, tension does not go to 0 throughout the whole string instantaneously - the relaxation propagates from where you cut it. Gravity is indeed always pulling down, but tension still pulls up after the string is cut. Imagine a string 1 light-year in length - when you cut the top, it's simply impossible for the bottom to be affected in less than 1 year. Whether you pull harder on the top of the string, cut it, or do anything else to affect the tension, it won't affect the bottom immediately. It's easier to visualize imagining the string as a slinky or a loose spring. $\endgroup$
    – Nuclear Hoagie
    Commented Jan 17 at 16:12
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    $\begingroup$ Voting to reopen. Yet another abuse of the catch-all "homework-like" closing reason. This is clearly a conceptual question, not asking for calculations. $\endgroup$
    – gandalf61
    Commented Jan 18 at 8:27

4 Answers 4

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Before the blue string is cut there is tension $T=m_2g$ in the black string which supports the weight of the lower ball.

Once the blue string is cut, the upper ball starts to accelerate downwards, the black string becomes slack, and the tension in that string falls to zero. Once the tension in the black string has reached zero, the only forces acting on the two balls are their respective weights and they will both accelerate downwards with acceleration $g$.

In real life the tension in the string cannot fall to zero instantaneously, so there will be some (very short) interval between when the blue string is cut and when the lower ball reaches an acceleration of $g$. However, since we are instructed to ignore air friction and the mass of the black string, we can also probably assume that the black string is an ideal string that does not stretch. With an ideal string then the tension will fall to zero instantaneously and both balls will immediately accelerate downwards with acceleration $g$. So the answer that you are probably expected to give is option $1$.

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As BioPhysicist says, the real world is complicated. It depends on properties of the string. Strings are a little bit like springs. This is because they are made of atoms. Atoms are connected by bonds, and bonds always have at least a little stretch. Sound propagates through bonds stretching.

The problem is supposed to be simpler than that so you can focus on how forces work, and not on distractions. The string is idealized as massless. There is no friction.

In this kind of problem, a string has no mass, air friction, stretch, stiffness, or any other complication.

The important properties of the string are if you pull the ends apart, you set up tension in the string. The force on one end is the same as the force on the other. The length is constant so long as there is tension. It is flexible and wraps around pulleys without changing the force.

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In the case of dropping a slinky, it is option 2.

For the masses on the string, a good approximation is option 1 if the mass of the string is negligible, but it does take some time for the "signal" that the blue string was cut to propagate through the black string, so there will be a delay in the time from when the string is cut to when mass B moves. The delay won't be long enough for ball A to collide with ball B though unless the string is sufficiently short.

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  • $\begingroup$ Can you elaborate on this delay? Why won't B start to fall immediately? What happens in the equations in each moment? $\endgroup$
    – Aviv Cohn
    Commented Jan 17 at 15:11
  • $\begingroup$ @AvivCohn I am not sure what you mean by "what happens in the equations". The string is not rigid, so it takes time for signals to propagate. Think of a wave traveling on a string; it has a finite speed. $\endgroup$
    – BioPhysicist
    Commented Jan 17 at 15:27
  • $\begingroup$ "Option 2" means that the lower ball does not move at all until the upper ball collides with it. I have not done experiments to determine how fast waves propagate along stretched slinkys. I'm guessing that the wave would propagate at constant speed. Meanwhile, the upper ball starts with zero velocity, and constantly accelerates. So, the wave gets a head start, because for some interval of time, its constant velocity must be greater than the velocity of the falling ball; but as the ball accelerates it must eventually overtake the wave. The question is, how long does that take? $\endgroup$
    – Solomon Slow
    Commented Jan 17 at 16:32
  • $\begingroup$ Note that Option 2 is wrong because it says that the top ball will drop with initial acceleration $g$, but it is actually more than $g$. $\endgroup$
    – naturallyInconsistent
    Commented Jan 18 at 0:51
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The correct answer is 3). Let's ignore transient effects in the strings (both!). At t=0 two forces act on A, m$_1$g due to gravity and m$_2$g due to the black string. it will accelerate by ( m$_1$g + m$_2$g)/m$_1$. After a short time, the acceleration is just g. On B the string acceleration works in the opposite direction and the initial acceleration is zero. At any time A will have a larger speed than B and in the end catch up with B. The details depend on the spring constant and the masses.

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