| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question with straightforward application of formulas. Part (i) uses work-energy principle with PE and KE changes. Part (ii) applies F=ma at equilibrium then P=Fv. Part (iii) reverses the process. All steps are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors6.02a Work done: concept and definition6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{1}{2}.700.20^2\) or \(\frac{1}{2}.700.15^2\) | B1 | either K.E. |
| \(700 \times 9.8 \times 400\sin5°\) | B1 | correct P.E. |
| \(\frac{1}{2}.700.15^2 + 700.9.8.400\sin5° = \frac{1}{2}.700.20^2 + \text{W.D.}\) | M1 | for 4 terms with W.D. |
| W.D. \(= 178,000 \text{ J}\) | A1 | 4 |
| (ii) \(D = 200 + 700.9.8\sin5°\) | M1 | |
| \(D = 798 \text{ N}\) | A1 | may be implied |
| \(P = Dx15 = 12,000 = 12\text{kW}\) | A1 | 3 |
| (iii) \(D' = 11,968 \div 20 = 598\) | M1 | |
| \(D' - 700.9.8\sin5° - 200 = 700a\) | M1 | |
| \(a = 0.285 \text{ ms}^{-2}\) (\(\pm\)) | A1 | 3 |
| Alternative for false assumption | (D = 445, a = -0.21875) | |
| (i) \(D - 700 \times 9.8\sin5° = 700a\) and \(15^2 = 20^2 + 2a. 400\) | M1 | |
| W.D. \(= 400xD = 178,000\) | A1 | 2 marks (out of 4) maximum |
(i) $\frac{1}{2}.700.20^2$ or $\frac{1}{2}.700.15^2$ | B1 | either K.E.
$700 \times 9.8 \times 400\sin5°$ | B1 | correct P.E.
$\frac{1}{2}.700.15^2 + 700.9.8.400\sin5° = \frac{1}{2}.700.20^2 + \text{W.D.}$ | M1 | for 4 terms with W.D.
W.D. $= 178,000 \text{ J}$ | A1 | 4 | or 178 kJ
(ii) $D = 200 + 700.9.8\sin5°$ | M1 |
$D = 798 \text{ N}$ | A1 | may be implied
$P = Dx15 = 12,000 = 12\text{kW}$ | A1 | 3 | AG (11,968W)
(iii) $D' = 11,968 \div 20 = 598$ | M1 |
$D' - 700.9.8\sin5° - 200 = 700a$ | M1 |
$a = 0.285 \text{ ms}^{-2}$ ($\pm$) | A1 | 3 | allow 0.283 (from 12kW) of constant acceleration
**Alternative for false assumption** | | (D = 445, a = -0.21875)
(i) $D - 700 \times 9.8\sin5° = 700a$ and $15^2 = 20^2 + 2a. 400$ | M1 |
W.D. $= 400xD = 178,000$ | A1 | 2 marks (out of 4) maximum | 106 A car of mass 700 kg is travelling up a hill which is inclined at a constant angle of $5 ^ { \circ }$ to the horizontal. At a certain point $P$ on the hill the car's speed is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The point $Q$ is 400 m further up the hill from $P$, and at $Q$ the car's speed is $15 \mathrm {~ms} ^ { - 1 }$.\\
(i) Calculate the work done by the car's engine as the car moves from $P$ to $Q$, assuming that any resistances to the car's motion may be neglected.
Assume instead that the resistance to the car's motion between $P$ and $Q$ is a constant force of magnitude 200 N.\\
(ii) Given that the acceleration of the car at $Q$ is zero, show that the power of the engine as the car passes through $Q$ is 12.0 kW , correct to 3 significant figures.\\
(iii) Given that the power of the car's engine at $P$ is the same as at $Q$, calculate the car's retardation at $P$.
\hfill \mbox{\textit{OCR M2 2005 Q6 [10]}}