Review question

# How fast do these connected particles move? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7982

## Solution

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.

Calculate

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.

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 $T=\quantity{9.6}{N}.$

Calculate

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.

Calculate

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 $2\times0.09+0.018=\quantity{0.198}{m}.$