The question I'm trying to answer:

"Two balls, 4kg and 6kg, are joined by a light inextensible string which is initially slack. The lighter ball is projected away from the heavier mass at $12ms^{-1}$. Find the final speed of both balls and the impulse in the string when it becomes taut."

From the description, I assumed that the ball which is not being projected remains in rest until the string becomes taut. I'm not sure if this assumption is right or not (I don't have solutions) but I think it is by simply imagining the situation. Now, to find the velocity of the balls when the string becomes taut we can use the conversation of momentum:

$$4(0) + 6(12)=(4+6)v$$ $$v=\frac{72}{10}=7.2ms^{-1}$$

So the speed of both balls is $7.2ms^{-1}$.

Now, finding the impulse:

$$I=(m_1 + m_2)v - mu=(4+6)7.2 - 6(12)=0kgms^{-1}$$.

Now, I know the impulse can't be zero so I'm wondering where my logic has failed me.


closed as off-topic by AccidentalFourierTransform, tpg2114, Jon Custer, John Rennie, Kyle Kanos Feb 5 '17 at 12:20

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The impulse equals the change in momentum of one of the balls. You have calculated the change in total momentum, ie final momentum of system minus initial momentum of system. This will always be zero if momentum is conserved.

However, even with the correct calculation of impulse your solution is wrong. You assume that the balls have the same speed after the string becomes taut. This is equivalent to a completely inelastic collision. Real strings do not behave like this. They behave like springs which are extremely stiff. Springs conserve energy; within their elastic limit strings are observed to do the same. The "collision" between the balls, mediated by the taut string, is elastic.

So in addition to conservation of momentum you also need to apply conservation of kinetic energy. The simplest way of doing the latter is to make the relative speed of separation equal the relative speed of approach.

See my answer to Force Transfer Between two bodies linked by a rope.

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    $\begingroup$ An easier way to analyze this system is from the frame of reference of the center of mass (inertial frame). The problem then reduces to simple oscillatory motion! But based on the wording of the question "slack string" I think it's supposed to be solved the way he did it. $\endgroup$ – ps95 Feb 5 '17 at 0:24
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    $\begingroup$ With a string rather than spring, motion in the CM frame will be an oscillation but not simple harmonic because the force from the string only acts at the endpoints. In between collisions (either directly or via the string) the velocity of each is constant. $\endgroup$ – sammy gerbil Feb 5 '17 at 0:31
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    $\begingroup$ Yes, the relative velocity of approach (after the string jerks taut) is also $12ms^{-1}$ but the velocities are not interchanged - ie $(0,+12) \to (+12,0) ms^{-1}$. This would only happen if the masses were equal. You must also apply conservation of momentum. When you do this you get $(+\frac{48}{5}, -\frac{12}{5}) ms^{-1}$. ... As prakharsingh95 says, the centre of momentum (=CM) moves with constant velocity of $+\frac{24}{5}ms^{-1}$ while the balls "oscillate" about the CM (not SHM). $\endgroup$ – sammy gerbil Feb 5 '17 at 20:19
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    $\begingroup$ @sammygerbil when the string is fully taught, all the energy is stored in the string cm frame.until the string becomes slack, the motion is simple oscillatory, with total energy known. When the string becomes slack, it's a simple collision problem with constant velocities. The overall motion is periodic and while it is not shm, it can be analyzed using its concepts $\endgroup$ – ps95 Feb 5 '17 at 21:09
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    $\begingroup$ @prakharsingh95 : The string is taut, and the balls collide, instantaneously. Effectively all of the motion is constant velocity. We agree that the motion is periodic and is not SHM. I do not understand what you mean by "while [motion] is not shm, it can be analyzed using its concepts." What analysis? What concepts? $\endgroup$ – sammy gerbil Feb 6 '17 at 16:32

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