# Question 7:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Total mass of train $= 800\,000$ kg | B1 | Allow 800 (tonnes) |
| Total resistance $= 5R + 17R = 22R$ | B1 | |
| Newton's 2nd Law in direction of motion: $121\,000 - 22R = 800\,000 \times 0.11$ | M1 | Right elements must be present, consistent with candidate's answers above for total resistance and mass. No extra forces. |
| $22R = 121\,000 - 88\,000 \quad R = 1500$ | E1 | Perfect answer required |

## Part (ii)(A):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Resultant force on last truck $= 40\,000 \times 0.11$ | B1 | Award for $40\,000 \times 0.11\ (=4400)$ or $40 \times 0.11$ seen |
| Use of Newton's 2nd Law | M1 | Right elements must be present and consistent with answer above; no extra forces |
| $T - 1500 = 40\,000 \times 0.11$ | A1 | Fully correct equation or equivalent working |
| $T = 5900$ — tension is 5900 N | A1 | Cao. **Special case:** Award SC2 for perfect argument that tension in penultimate coupling is 11 800 N |

**Or** (Rest of train):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Resultant force on rest of train $= 760\,000 \times 0.11$ | B1 | Award for $760\,000 \times 0.11\ (=83\,600)$ or $760 \times 0.11$ seen |
| Use of Newton's 2nd Law | M1 | Right elements must be present consistent with answer above; no extra forces |
| $121\,000 - 31\,500 - T = 760\,000 \times 0.11$ | A1 | Fully correct equation or equivalent working |
| $T = 5900$ — tension is 5900 N | A1 | Cao |

## Part (ii)(B):

**Either** (Rest of train):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Newton's 2nd Law applied to the trucks: $S - 25\,500 = 680\,000 \times 0.11$ | M1, A1 | Right elements must be present; no extra forces |
| $S = 100\,300$ — tension is 100 300 N | A1 | Cao |

**Or** (Locomotive):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Newton's 2nd Law applied to locomotive: $121\,000 - S - 5 \times 1500 = 120\,000 \times 0.11$ | M1, A1 | Right elements must be present; no extra forces |
| $S = 100\,300$ — tension is 100 300 N | A1 | Cao |

**Or** (By argument):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Each of the 17 trucks has the same mass, resistance and acceleration | M1 | |
| So tension in first coupling is 17 times that in last coupling | A1 | |
| $T = 17 \times 5900 = 100\,300$ | A1 | Cao. For this statement alone with no supporting argument allow SC2 |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolved component of weight down slope $= 800\,000 \times 9.8 \times \dfrac{1}{80} = 98\,000$ N | B1 | $m \times 9.8 \times \dfrac{1}{80}$ where $m$ is mass of object candidate is considering. Do not award if $g$ missing. Evaluation need not be seen |
| Newton's 2nd Law to whole train: $121\,000 - 33\,000 - 98\,000 = 800\,000a$ | M1, A1 | Right elements must be present consistent with candidate's component of weight down slope. No extra forces allowed |
| $a = -0.0125$; magnitude $0.0125\ \text{m s}^{-2}$, down the slope | A1 | Cao but allow rounding to $-0.012$ or $-0.013$ following earlier premature rounding. Negative sign must be interpreted as "Down the slope" or "decelerating" |

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking train as whole, force down slope $=$ resistance force | M1 | Equilibrium of whole train required. Evidence may come from correct force diagram. Allow missing $g$ for this mark only |
| $800\,000 \times 9.8 \times \sin\beta = 33\,000$ | A1 | |
| $\beta = 0.24°$ | A1 | |

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