A light elastic string has natural length $3\,\text{m}$ and modulus of elasticity $45\,\text{N}$. A particle $P$ with weight $6\,\text{N}$ is fastened at the midpoint of the string. The two ends are attached to fixed points $A$ and $B$ on the same vertical line, with $A$ above $B$, and $AB = 4\,\text{m}$. The particle $P$ is released from rest from a position $1.5\,\text{m}$ vertically beneath $A$.
(i)[4]
Calculate the distance $P$ travels after release before it first comes to rest instantaneously at a point vertically above $B$. (You may assume that, at this point, the part of the string joining $P$ to $B$ is slack.)
(ii)[5]
Show that the greatest speed of $P$ is reached when it is $2.1\,\text{m}$ below $A$, and determine this greatest speed.
(iii)[3]
Calculate the greatest magnitude of the acceleration of $P$.
Worked solution & mark scheme
This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: “$45 \times 1^2/(2 \times 1.5) + 0.6g h = 45 h^2/(2 \times 1.5)$” …