# How do masses on a spring accelerate when the connection between them is cut?

I'm learning about springs in Physics and have come across something I do not fully understand.

As in the picture, there are three masses (A, B and C), with mass ma, mb and mc, respectively. In the configuration above, masses A and B are connected by a light inelastic string. The masses of the string and the two springs are negligible and the whole system is stationary (in equilibrium).

What I want to work out is the acceleration of the three masses when the string between A and B is suddenly cut.

I currently understand from slinkies that C would stay stationary while S2 contracts. So B is accelerating downwards. Therefore, C would have no acceleration.

Also, S1 is carrying all three masses but after the string is cut, it only carries ma, so it would contract upwards with a force equivalent to g(mb+mc), because they are the forces taken away from the spring. But when I calculate the acceleration of A using F=ma, I end up with a=g and I am not sure if that is correct, because I also get a=g for the acceleration of B.

I am trying to express each acceleration in terms of ma, mb, mc and g.

I am not looking for answers, as I know this is not a homework site. All I ask is for pointers in the right direction and perhaps correct me if my understanding of the spring system is flawed.

Thank you.

• I think your reasoning is sound so far. I'm a bit confused how you wound up with the acceleration for A as g. The force on A should have been g(mb + mc), which would make A's acceleration g(mb + mc)/ma (so if mb + mc = ma then it will be just g).
– JMac
Jan 24, 2017 at 19:31
• Why is A's acceleration the force divided by ma? Would it not be the force divided by (mb + mc), which is how I got to g? Jan 24, 2017 at 19:35
• We know the string had a tension of g(mb + mc), so we know that is the force that was removed from ma. We then use the relationship F = maa (confusing wording, ma is mass A while the second a is acceleration of mass A), we know the force is g(mb + mc) so we equate that to get maa = g(mb + mc), when you solve for a you get g(mb + mc)/ma.
– JMac
Jan 24, 2017 at 19:40
• Oh I see now. So A accelerates upwards with a force equal to ma times a. But then why is this force equal to the previous force g(mb + mc)? Jan 24, 2017 at 19:47
• Yes, because the spring was countering that force but no longer has to, causing the spring to contract and the mass to accelerate.
– JMac
Jan 24, 2017 at 20:31