2 A train of mass 150000 kg ascends a straight slope inclined at \(\alpha ^ { \circ }\) to the horizontal with a constant driving force of 16000 N . At a point \(A\) on the slope the speed of the train is \(45 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Point \(B\) on the slope is 500 m beyond \(A\). At \(B\) the speed of the train is \(42 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). There is a resistance force acting on the train and the train does \(4 \times 10 ^ { 6 } \mathrm {~J}\) of work against this resistance force between \(A\) and \(B\). Find the value of \(\alpha\).

**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| PE gain $= 150000g \times 500\sin\alpha$ $(= 75000000g\sin\alpha)$ | B1 | Correct expression for PE gain |
| $\frac{1}{2} \times 150000 \times 45^2 - \frac{1}{2} \times 150000 \times 42^2$ $(= 19575000)$ | B1 | Correct expression for KE loss |
| | M1 | For 5 term work energy equation (or 4 terms if using loss in KE as 1 term) |
| $150000g \times 500\sin\alpha = 19575000 + 16000 \times 500 - 4 \times 10^6$ | A1 | |
| $\alpha = 1.8$ | A1 | |
| **Total: 5** | | |

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2 A train of mass 150000 kg ascends a straight slope inclined at $\alpha ^ { \circ }$ to the horizontal with a constant driving force of 16000 N . At a point $A$ on the slope the speed of the train is $45 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Point $B$ on the slope is 500 m beyond $A$. At $B$ the speed of the train is $42 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. There is a resistance force acting on the train and the train does $4 \times 10 ^ { 6 } \mathrm {~J}$ of work against this resistance force between $A$ and $B$. Find the value of $\alpha$.\\

\hfill \mbox{\textit{CAIE M1 2019 Q2 [5]}}