homework and exercises - Find the speed of two balls joined by a string and the impulse in the string when the string goes taut

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.

Answer

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.