particles A and B connected by a string over a peg, each 0.09 metres above the floor

The diagram shows two particles \(A\) and \(B\), connected by a light inextensible string which passes over a smooth fixed peg. The system is held with the string taut and with \(A\) and \(B\) each at a height of \(\quantity{0.09}{m}\) above a fixed horizontal plane; it is then released from rest. When \(B\) reaches the plane is becomes stationary.


  1. the tension in the string while both particles are in motion,

[Take \(g\) to be \(\quantity{10}{m\,s^{-2}}\).]

Let the acceleration of the particles be \(a\), measured upwards on \(A\) and downwards on \(B\), and let the tension in the string be \(T\). The following diagram shows the forces acting on the two particles.

forces T and 0.8g acting on A and T and 1.2g acting on B
Applying Newton’s second law, \(F=ma\), to each of the particles and using \(g=10\) we have \[\begin{align*} T-8 &= 0.8a \\ 12-T &= 1.2a. \end{align*}\]

Adding gives us \(4=2a\), \(a=2\). Substituting this into one of the equations gives \[T=\quantity{9.6}{N}.\]


  1. the speed of the particles when \(B\) reaches the plane,
We know that \(B\) starts at rest and travels \(\quantity{0.09}{m}\). We know its acceleration and want to find its final velocity, so we use the equation of motion \[\begin{align*} v^2 &= u^2+2as. \\ v^2 &= 0+2\times2\times 0.09 = 0.36 \\ v &= 0.6 \end{align*}\]

The particles are moving at \(\quantity{0.6}{m\,s^{-1}}\) when \(B\) reaches the plane.


  1. the maximum height above the plane attained by \(A\), assuming that \(A\) does not reach the height of the fixed peg.
Assuming that neither the peg nor the string interfere with its motion, \(A\) will now be under the influence only of gravity. We know its initial velocity is \(\quantity{0.6}{m\,s^{-1}}\) upwards and its acceleration is \(g\) downwards. We want to know its upwards displacement, so we’ll use the same equation of motion as before. \[\begin{align*} v^2 &= u^2+2as \\ 0 &= 0.6^2+2(-10)s \\ s &= \frac{0.36}{20} = 0.018 \end{align*}\]

This is how far it rises after the string becomes slack but this happens when \(A\) is \(2\times\quantity{0.09}{m}\) above the plane, so the maximum height above the plane is \[2\times0.09+0.018=\quantity{0.198}{m}.\]