| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem requiring students to apply the work-energy principle with friction on an inclined plane. The steps are standard: calculate GPE change using mgh (with h = x sin α), equate initial KE to GPE gained plus work done against friction, then solve for x. While it involves multiple energy components, the method is routine for M1 and requires no novel insight—slightly easier than average due to clear structure and given values. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([80x\sin 22.6\) or \(80x(5/13)]\) | M1 | For using PE change \(= mgh\); PE change \(= 8 \times g \times x \times \sin\alpha\) |
| \(= \frac{400}{13}x = 30.8x\) | A1 | Allow \(\alpha = 22.6\) used. Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| WD against friction \(= 15 \times x\) | B1 | |
| \(\frac{1}{2} \times 8 \times 5^2\) | B1 | |
| \(\frac{1}{2} \times 8 \times 5^2 = \frac{400}{13}x + 15x\) | M1 | For using KE loss = PE gain + WD against friction |
| \(x = \frac{260}{119} = 2.18\) | A1 | Total: 4 |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[80x\sin 22.6$ or $80x(5/13)]$ | M1 | For using PE change $= mgh$; PE change $= 8 \times g \times x \times \sin\alpha$ |
| $= \frac{400}{13}x = 30.8x$ | A1 | Allow $\alpha = 22.6$ used. **Total: 2** |
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| WD against friction $= 15 \times x$ | B1 | |
| $\frac{1}{2} \times 8 \times 5^2$ | B1 | |
| $\frac{1}{2} \times 8 \times 5^2 = \frac{400}{13}x + 15x$ | M1 | For using KE loss = PE gain + WD against friction |
| $x = \frac{260}{119} = 2.18$ | A1 | **Total: 4** |
---3 A particle of mass 8 kg is projected with a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a line of greatest slope of a rough plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 5 } { 13 }$. The motion of the particle is resisted by a constant frictional force of magnitude 15 N . The particle comes to instantaneous rest after travelling a distance $x \mathrm {~m}$ up the plane.\\
(i) Express the change in gravitational potential energy of the particle in terms of $x$.\\
(ii) Use an energy method to find $x$.
\hfill \mbox{\textit{CAIE M1 2016 Q3 [6]}}