| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Horizontal force on slope |
| Difficulty | Standard +0.3 This is a standard A-level mechanics problem involving forces on an inclined plane with friction. Part (a) requires resolving forces in two directions and using limiting friction (F=μR), which is routine. Parts (b) and (c) involve applying F=ma and SUVAT equations in a two-phase motion problem. While multi-step, all techniques are standard M1 content with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 0.3g\cos\theta + 4\sin\theta = 3 \times \dfrac{24}{25} + 4 \times \dfrac{7}{25}\) \([=4]\) | M1 | Resolving forces perpendicular to the plane or parallel to the plane. Allow use of \(\theta = 16.3°\) |
| \(F = 4\cos\theta - 0.3g\sin\theta = 4 \times \dfrac{24}{25} - 3 \times \dfrac{7}{25}\) \([=3]\) | A1 | Two correct equations |
| \(3 = \mu \times 4\) | M1 | For use of \(F = \mu R\) |
| \(\mu = \dfrac{3}{4}\) | A1 | AG. Must be from correct and exact working, not using 16.3 |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F = \mu \times 0.3g\cos\theta = \dfrac{3}{4} \times 3 \times \dfrac{24}{25}\) \(\left[= \dfrac{54}{25} = 2.16\right]\) | B1 | |
| \(4 - \dfrac{3}{4} \times 0.3g \times \dfrac{24}{25} - 0.3g \times \dfrac{7}{25} = 0.3a\) | M1 | Use of Newton's second law |
| \(a = \dfrac{10}{3}\ \text{m s}^{-2}\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s_1 = \frac{1}{2} \times \frac{10}{3} \times 3^2 = 15\) and \(v = \frac{10}{3} \times 3 = 10\) | B1 FT | Distance \(s_1\) in 3s and \(v\) after 3s; FT \(a\) from (b) |
| \(-0.3g \times \sin\theta - \mu \times 0.3g\cos\theta = 0.3a\) leading to \(a = -10\) \(0 = 10^2 + 2 \times (-10) \times s_2\) | M1 | Apply Newton's 2nd law after 4 N removed, find \(a\) and use \(v^2 = u^2 + 2as\) to find extra distance \(s_2\) |
| \([s_2 = 5\) leading to total distance \(= s_1 + s_2 = 15 + 5 =]\ 20\) m | A1 | |
| Alternative method for Question 7(c) | ||
| Work done \(= 4 \times 0.5 \times \frac{10}{3} \times 3^2\ [= 60\ \text{J}]\) | B1 FT | \(WD = Fs\) and \(s = \frac{1}{2}at^2\) for 4 N force; FT \(a\) from (b) |
| \(60 = \mu \times 0.3g\cos\theta \times d + 0.3g \times d\sin\theta\) | M1 | WD by 4 N force \(=\) WD against \(F\) + PE gain |
| \(d = 20\) m | A1 | |
| 3 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 0.3g\cos\theta + 4\sin\theta = 3 \times \dfrac{24}{25} + 4 \times \dfrac{7}{25}$ $[=4]$ | M1 | Resolving forces perpendicular to the plane or parallel to the plane. Allow use of $\theta = 16.3°$ |
| $F = 4\cos\theta - 0.3g\sin\theta = 4 \times \dfrac{24}{25} - 3 \times \dfrac{7}{25}$ $[=3]$ | A1 | Two correct equations |
| $3 = \mu \times 4$ | M1 | For use of $F = \mu R$ |
| $\mu = \dfrac{3}{4}$ | A1 | AG. Must be from correct and exact working, not using 16.3 |
| | **4** | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = \mu \times 0.3g\cos\theta = \dfrac{3}{4} \times 3 \times \dfrac{24}{25}$ $\left[= \dfrac{54}{25} = 2.16\right]$ | B1 | |
| $4 - \dfrac{3}{4} \times 0.3g \times \dfrac{24}{25} - 0.3g \times \dfrac{7}{25} = 0.3a$ | M1 | Use of Newton's second law |
| $a = \dfrac{10}{3}\ \text{m s}^{-2}$ | A1 | |
| | **3** | |
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s_1 = \frac{1}{2} \times \frac{10}{3} \times 3^2 = 15$ and $v = \frac{10}{3} \times 3 = 10$ | **B1 FT** | Distance $s_1$ in 3s and $v$ after 3s; FT $a$ from **(b)** |
| $-0.3g \times \sin\theta - \mu \times 0.3g\cos\theta = 0.3a$ leading to $a = -10$ $0 = 10^2 + 2 \times (-10) \times s_2$ | **M1** | Apply Newton's 2nd law after 4 N removed, find $a$ and use $v^2 = u^2 + 2as$ to find extra distance $s_2$ |
| $[s_2 = 5$ leading to total distance $= s_1 + s_2 = 15 + 5 =]\ 20$ m | **A1** | |
| **Alternative method for Question 7(c)** | | |
| Work done $= 4 \times 0.5 \times \frac{10}{3} \times 3^2\ [= 60\ \text{J}]$ | **B1 FT** | $WD = Fs$ and $s = \frac{1}{2}at^2$ for 4 N force; FT $a$ from **(b)** |
| $60 = \mu \times 0.3g\cos\theta \times d + 0.3g \times d\sin\theta$ | **M1** | WD by 4 N force $=$ WD against $F$ + PE gain |
| $d = 20$ m | **A1** | |
| | **3** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{ba29ddb2-3558-4be1-a8a8-134e27a70149-10_220_609_260_769}
A particle $P$ of mass 0.3 kg rests on a rough plane inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 7 } { 25 }$. A horizontal force of magnitude 4 N , acting in the vertical plane containing a line of greatest slope of the plane, is applied to $P$ (see diagram). The particle is on the point of sliding up the plane.
\begin{enumerate}[label=(\alph*)]
\item Show that the coefficient of friction between the particle and the plane is $\frac { 3 } { 4 }$.\\
The force acting horizontally is replaced by a force of magnitude 4 N acting up the plane parallel to a line of greatest slope.
\item Find the acceleration of $P$.
\item Starting with $P$ at rest, the force of 4 N parallel to the plane acts for 3 seconds and is then removed. Find the total distance travelled until $P$ comes to instantaneous rest.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q7 [10]}}