A particle $P$ of mass $0.5\,\text{kg}$ is fastened to one end of a light elastic string with natural length $0.6\,\text{m}$ and modulus of elasticity $12\,\text{N}$. The opposite end of the string is fixed at point $O$. Particle $P$ is fired vertically downwards at speed $2\,\text{m s}^{-1}$ from the position $0.5\,\text{m}$ vertically beneath $O$. At an instant when the acceleration of $P$ is $4\,\text{m s}^{-2}$ downwards,
(main)[6]
Determine the extension of the string and the speed of $P$.
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use Newton’s second law in the vertical direction: $0.5\times 4 = 0.5g - T$” …