Consider the following problem:

Two masses of 5 kg and 1 kg hang from a smooth pulley at the ends of a light inextensible string. The system is released from rest. After 2 seconds, the 5 kg mass hits a horizontal table:

i. How much further will the 1 kg mass rise?

ii. The 1 kg mass then falls and the 5 kg mass is jolted off the table. With what speed will the 5 kg mass begin to rise?

[6D Q. 4. Fundamental Applied Maths, 2nd Edition, Oliver Murphy]

It isn’t difficult to answer part i.: the answer is \displaystyle \frac89 g m.

However how to treat part ii.? First of all a picture to help us understand this problem:


The 1 kg mass has dropped under gravity through a distance of \displaystyle \frac89 g m. We can find the speed of the 1 kg mass using u=0,a=g,s=\frac89 g. Alternatively, we can use Conservation of (Mechanical) Energy.

Taking the final position as h=0, at its maximum height, the 1 kg mass has potential energy and no kinetic energy:

\text{PE}_0=mgh=1g\frac89 g=\frac{8}{9}g^2.

When it reaches the point where the string is once again taut, it has not potential energy but the potential energy it had has been transferred into kinetic energy:

\text{KE}_1=\frac12 mv^2=\frac12 v^2=\frac12 v^2,

and this must equal the potential energy \text{PE}_0:

\frac{8}{9}g^2=\frac12 v^2\Rightarrow v=\frac43g.

Now this is where things get trickier. My idea was to use conservation of momentum on the two particles separately. As this clever answer to this question shows, you can treat the 5 kg mass, string, and 1 kg mass as a single particle.

So the prior momentum is the mass of the 1 kg mass by \frac43 g:

p_0=m_0u=1\cdot \frac43 g=\frac43g.

The ‘after’ momentum is the mass of the 1 kg and 5 kg masses times the new velocity:

p_1=m_1v=6\cdot v.

By Conservation of Momentum, these are equal:

\frac43 g=6v\Rightarrow v=\frac{2}{9} g\text{ m s}^{-1}.