# Question 6:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T = 2058x/5.25]$ | M1 | For using $T = \lambda x/L$ |
| $2058x/5.25 = 80 \times 9.8 \quad (x = 2)$ | A1 | |
| $OP = 7.25\text{m}$ | A1 | [3 marks total] AG From $5.25 + 2$ |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial $PE = (80 + 80)g(5) \quad (= 7840)$ or $(80+80)gX$ used in energy equation | B1 | |
| Initial $KE = \frac{1}{2}(80+80)3.5^2 \quad (= 980)$ | B1 | |
| [Initial $EE = 2058x^2/(2\times 5.25) \quad (= 784)$, Final $EE = 2058 \times 7^2/(2\times 5.25) \quad (= 9604)$, or $2058(X+2)^2/(2\times 5.25)$] | M1 | For using $EE = \lambda x^2/2L$ |
| [Initial energy $= 7840 + 980 + 784$, final energy $= 9604$] or $1568X + 980 + 784 = 196(X^2 + 4X + 4) \rightarrow 196X^2 - 784X - 980 = 0$ | M1 | For attempting to verify compatibility with the principle of conservation of energy, or using the principle and solving for X |
| Initial energy $=$ final energy or $X = 5 \rightarrow$ P&Q just reach the net | A1 | [5 marks total] AG |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[PE\text{ gain} = 80g(7.25 + 5)]$ | M1 | For finding PE gain from net level to O |
| $PE\text{ gain} = 9604$ | A1 | |
| $PE\text{ gain} = EE\text{ at net level} \rightarrow$ P just reaches O | A1 | [3 marks total] AG |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| For any one of 'light rope', 'no air resistance', 'no energy lost in rope' | B1 | |
| For any other of the above | B1 | [2 marks total] |

## First Alternative Method for Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[160g - 2058x/5.25 = 160v\,dv/dx]$ | M1 | For using Newton's second law with $a = v\,dv/dx$, separating the variables and attempting to integrate |
| $v^2/2 = gx - 1.225x^2 \quad (+C)$ | A1 | Any correct form |
| | M1 | For using $v(2) = 3.5$ |
| $C = -8.575$ | A1 | |
| $[v(7)^2]/2 = 68.6 - 60.025 - 8.575 = 0 \rightarrow$ P&Q just reach the net | A1 | [5 marks total] AG |

## Second Alternative Method for Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ddot{x} = g - 2.45x \quad (= -2.45(x-4))$ | B1 | |
| | M1 | For using $n^2 = 2.45$ and $v^2 = n^2(A^2 - (x-4)^2)$ |
| $3.5^2 = 2.45(A^2 - (-2)^2) \quad (A = 3)$ | A1 | |
| $[(4-2)+3]$ | M1 | For using 'distance travelled downwards by P and Q $=$ distance to new equilibrium position $+ A$' |
| distance travelled downwards by P and Q $= 5 \rightarrow$ P&Q just reach the net | A1 | [5 marks total] AG |

---