| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Block on horizontal plane motion |
| Difficulty | Moderate -0.3 This is a standard M1 friction problem requiring application of SUVAT equations, resolution of forces, and friction law F=μR. Part (a) is straightforward kinematics, part (b) involves resolving forces in two directions and using F=ma, and part (c) applies limiting equilibrium. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[2 = \frac{1}{2}\times a \times 25\right]\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) OE using \(u=0\), \(s=2\) and \(t=5\) |
| \(a = 0.16 \text{ ms}^{-2}\) | A1 | Allow \(a = \frac{4}{25}\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 5g - X\sin 30\) | B1 | |
| \(X\cos 30 - F = 5a\) | M1 | Apply Newton's 2nd law to the block, using their \(a\) |
| \(X\cos 30 - 0.4(5g - X\sin 30) = 5\times 0.16\) | M1 | Use \(F = 0.4R\) to obtain an equation in \(X\) only, using their \(R\) which must involve \(5g\) and a component of \(X\) only |
| \(X = 19.5\) (3sf) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = (5g - 25\sin 30)\) \([R = 37.5]\) | B1 | |
| \(F = 25\cos 30\) \(\left[F = \frac{25\sqrt{3}}{2}\right]\) | B1 | |
| \(\mu = \frac{F}{R} = 0.577\) (3sf) | B1 | Allow \(\mu = \frac{\sqrt{3}}{3}\) or \(\mu = \frac{1}{\sqrt{3}}\) |
| 3 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[2 = \frac{1}{2}\times a \times 25\right]$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ OE using $u=0$, $s=2$ and $t=5$ |
| $a = 0.16 \text{ ms}^{-2}$ | A1 | Allow $a = \frac{4}{25}$ |
| | **2** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 5g - X\sin 30$ | B1 | |
| $X\cos 30 - F = 5a$ | M1 | Apply Newton's 2nd law to the block, using their $a$ |
| $X\cos 30 - 0.4(5g - X\sin 30) = 5\times 0.16$ | M1 | Use $F = 0.4R$ to obtain an equation in $X$ only, using their $R$ which must involve $5g$ and a component of $X$ only |
| $X = 19.5$ (3sf) | A1 | |
| | **4** | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = (5g - 25\sin 30)$ $[R = 37.5]$ | B1 | |
| $F = 25\cos 30$ $\left[F = \frac{25\sqrt{3}}{2}\right]$ | B1 | |
| $\mu = \frac{F}{R} = 0.577$ (3sf) | B1 | Allow $\mu = \frac{\sqrt{3}}{3}$ or $\mu = \frac{1}{\sqrt{3}}$ |
| | **3** | |5\\
\includegraphics[max width=\textwidth, alt={}, center]{a96ca3b4-6d35-4512-a0a1-3f28443fd051-08_286_661_260_742}
A block of mass 5 kg is being pulled along a rough horizontal floor by a force of magnitude $X \mathrm {~N}$ acting at $30 ^ { \circ }$ above the horizontal (see diagram). The block starts from rest and travels 2 m in the first 5 s of its motion.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the block.
\item Given that the coefficient of friction between the block and the floor is 0.4 , find $X$.\\
The block is now placed on a part of the floor where the coefficient of friction between the block and the floor has a different value. The value of $X$ is changed to 25, and the block is now in limiting equilibrium.
\item Find the value of the coefficient of friction between the block and this part of the floor.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q5 [9]}}