| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against friction/resistance - inclined plane or slope |
| Difficulty | Moderate -0.3 This is a straightforward M2 energy method question requiring calculation of kinetic and potential energy changes, then using the work-energy principle. The setup is clear, all values are given, and it follows a standard two-part structure with no conceptual surprises—slightly easier than average due to its routine nature, though the multi-force work-energy application prevents it from being trivial. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| i | \(PE = 70 \times 3g\) | B1 |
| \(KE \text{ change} = 70(2.1^2 - 1.4^2)/2\) | B1 | |
| \(PE \text{ change} + KE \text{ change} = 2143.75 \text{ J}\) | M1 | Must include evaluation |
| A1 | Accept 2140. Allow all values to be negative. | |
| [4] | ||
| ii | \(20(90 + T) = 2143.75\) | M1 |
| \(T = 17.1875 \text{ N}\) | A1ft | \(ft(cv(2143.75))\) accept 17.2 |
| OR | ||
| \(70g \times 0.15 - 90 - T = 70(-0.06125)\) | M1 | Use of \(v^2 = u^2 + 2as\) to find \(a\) AND use of N2 law (4 terms) |
| \(T = 17.1875 \text{ N}\) | A1 | accept 17.2 |
| A1 | ||
| [3] |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| i | $PE = 70 \times 3g$ | B1 | |
| | $KE \text{ change} = 70(2.1^2 - 1.4^2)/2$ | B1 | |
| | $PE \text{ change} + KE \text{ change} = 2143.75 \text{ J}$ | M1 | Must include evaluation |
| | | A1 | Accept 2140. Allow all values to be negative. |
| | | [4] | |
| ii | $20(90 + T) = 2143.75$ | M1 | Work done = Energy change used |
| | $T = 17.1875 \text{ N}$ | A1ft | $ft(cv(2143.75))$ accept 17.2 |
| | **OR** | | |
| | $70g \times 0.15 - 90 - T = 70(-0.06125)$ | M1 | Use of $v^2 = u^2 + 2as$ to find $a$ AND use of N2 law (4 terms) |
| | $T = 17.1875 \text{ N}$ | A1 | accept 17.2 |
| | | A1 | |
| | | [3] | |1\\
\includegraphics[max width=\textwidth, alt={}, center]{65c47bd2-eace-4fec-b1e6-a0c904c4ec3f-2_314_931_242_607}
A sledge with its load has mass 70 kg . It moves down a slope and the resistance to the motion of the sledge is 90 N . The speed of the sledge is controlled by the constant tension in a light rope, which is attached to the sledge and parallel to the slope (see diagram). While travelling 20 m down the slope, the speed of the sledge decreases from $2.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $1.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and it descends a vertical distance of 3 m .\\
(i) Calculate the change in energy of the sledge and its load.\\
(ii) Calculate the tension in the rope.
\hfill \mbox{\textit{OCR M2 2011 Q1 [7]}}