| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Particle on rough inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part variable force mechanics problem requiring resolution of forces on an incline, application of F=ma with position-dependent force, and use of the chain rule v(dv/dx) = a. While it involves several steps and the variable force adds complexity beyond basic inclined plane problems, the techniques are standard M2 content with clear signposting through the 'show that' part (i), making it moderately above average difficulty but not requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2v\,dv/dx = 0.2g\sin60 - 0.3 \times 0.2g\cos60 - 0.6x\) | M1A1 | Uses Newton's Second Law parallel to the plane. Correct equation |
| \(v\,dv/dx = 5\sqrt{3} - 1.5 - 3x\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = (5\sqrt{3} - 1.5)/3 \ (= 2.39)\) | B1 | Uses \(a = 0\) |
| \(\int v\, dv = \int(5\sqrt{3} - 1.5 - 3x)\, dx\) | M1 | Separates the variables and attempts to integrate |
| \(v^2/2 = 5\sqrt{3}\,x - 1.5x - 3x^2/2 \ (+c)\) | A1 | Allow \(c = 0\) without calculation seen |
| \(v = 4.13\) | A1 | Substitutes \(x = 2.39\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = 5\sqrt{3}\, x - 1.5x - 3x^2/2\) | M1 | Puts \(v = 0\) and attempts to solve a quadratic equation. |
| \(x = 4.77(35...)\) | A1 | |
| \(a = 5\sqrt{3} - 1.5 - 3 \times 4.77(35...)\) | M1 | |
| Magnitude of \(a = 7.16 \text{ m s}^{-2}\) | A1 | |
| 4 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2v\,dv/dx = 0.2g\sin60 - 0.3 \times 0.2g\cos60 - 0.6x$ | M1A1 | Uses Newton's Second Law parallel to the plane. Correct equation |
| $v\,dv/dx = 5\sqrt{3} - 1.5 - 3x$ | A1 | AG |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = (5\sqrt{3} - 1.5)/3 \ (= 2.39)$ | B1 | Uses $a = 0$ |
| $\int v\, dv = \int(5\sqrt{3} - 1.5 - 3x)\, dx$ | M1 | Separates the variables and attempts to integrate |
| $v^2/2 = 5\sqrt{3}\,x - 1.5x - 3x^2/2 \ (+c)$ | A1 | Allow $c = 0$ without calculation seen |
| $v = 4.13$ | A1 | Substitutes $x = 2.39$ |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = 5\sqrt{3}\, x - 1.5x - 3x^2/2$ | **M1** | Puts $v = 0$ and attempts to solve a quadratic equation. |
| $x = 4.77(35...)$ | **A1** | |
| $a = 5\sqrt{3} - 1.5 - 3 \times 4.77(35...)$ | **M1** | |
| Magnitude of $a = 7.16 \text{ m s}^{-2}$ | **A1** | |
| | **4** | |7 A particle $P$ of mass 0.2 kg is released from rest at a point $O$ on a rough plane inclined at $60 ^ { \circ }$ to the horizontal, and travels down a line of greatest slope. The coefficient of friction between $P$ and the plane is 0.3 . A force of magnitude $0.6 x \mathrm {~N}$ acts on $P$ in the direction $P O$, where $x \mathrm {~m}$ is the displacement of $P$ from $O$.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 \sqrt { } 3 - 1.5 - 3 x$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of $P$ at a displacement $x \mathrm {~m}$ from $O$.\\
(ii) Find the value of $x$ for which $P$ reaches its maximum velocity, and calculate this maximum velocity.\\
(iii) Calculate the magnitude of the acceleration of $P$ immediately after it has first come to instantaneous rest.\\
\hfill \mbox{\textit{CAIE M2 2017 Q7 [11]}}