| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | March |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Collision on slope |
| Difficulty | Challenging +1.2 This is a two-part mechanics problem requiring (a) application of energy methods or equations of motion with friction on a slope (standard M1 content), and (b) analysis of collision followed by motion of two beads with different friction coefficients. Part (a) is routine 'show that' work. Part (b) requires careful tracking of both beads' motion post-collision until they meet again, involving multiple phases and solving simultaneous equations. While multi-step, the techniques are standard M1 fare with no novel insight required, placing it moderately above average difficulty. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.6^2 = 0 + 2a \times 0.45\) | M1 | Use of constant acceleration equations to find \(a\) |
| \(a = 0.4\) | A1 | |
| \(R = 0.1g \times \cos\alpha = 0.1g \times \frac{24}{25} = 0.1g \times \cos 16.3°\left[R = \frac{24}{25} = 0.96\right]\) | B1 | Must use a value for \(\cos\alpha\) |
| \(0.1g \times \frac{7}{25} - F = 0.1 \times 0.4\ [0.28 - F = 0.04 \rightarrow F = 0.24]\) | M1 | Newton's second law, 3 terms |
| \(F = \mu \times 0.1g \times \frac{24}{25}\left[F = \frac{24\mu}{25} = 0.96\mu\right]\) | M1 | Use of \(F = \mu R\), where \(R\) is a component of \(0.1g\) |
| \(\mu = 0.25\) | A1 | AG Must be from exact working \(\mu = 0.25\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt PE loss or KE gain | M1 | Use of either \(\text{PE} = mgh\) or \(\text{KE} = \frac{1}{2}mv^2\) |
| PE loss \(= 0.1 \times g \times 0.45\sin 16.3° = 0.1 \times g \times 0.45 \times \frac{7}{25}\left[=\frac{63}{500} = 0.126\right]\), KE gain \(= \frac{1}{2} \times 0.1 \times 0.6^2\left[=\frac{9}{500} = 0.018\right]\) | A1 | Both correct |
| \(R = 0.1g \times \cos\alpha = 0.1g \times \frac{24}{25} = 0.1g \times \cos 16.3°\left[R = \frac{24}{25} = 0.96\right]\) | B1 | Must use a value for \(\cos\alpha\) |
| \(0.1 \times g \times 0.45 \times \frac{7}{25} = \frac{1}{2} \times 0.1 \times 0.6^2 + F \times 0.45\) \(\left[\frac{63}{500} = \frac{9}{500} + \mu \times \frac{54}{125}\right]\) or \([0.126 = 0.018 + \mu \times 0.432]\) | M1 | Use of work-energy equation: PE loss \(=\) KE gain \(+\) WD against friction |
| \(F = \mu \times 0.1g \times \frac{24}{25}\left[F = \frac{24\mu}{25} = 0.96\mu\right]\) | M1 | Use of \(F = \mu R\), where \(R\) is a component of \(0.1g\) |
| \(\mu = 0.25\) | A1 | AG Must be from exact working \(\mu = 0.25\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.1 \times 0.6 = 0.5v\) | M1 | Use of conservation of momentum, 2 terms |
| \(v = 0.12\) | A1 | |
| For \(B\): \(0.5g \times \frac{7}{25} - 0.275 \times 0.5g \times \frac{24}{25} = 0.5a\) [leading to \(a = 0.16\)] | B1 | Apply Newton's second law for particle \(B\), 3 terms. Allow correct unsimplified expression in \(a\) only |
| \(s_A = 0 + \frac{1}{2} \times 0.4t^2 \quad s_B = 0.12t + \frac{1}{2} \times 0.16t^2\) | \*M1 | Attempt expression for either \(s_A\) or \(s_B\). Must see \(u_A = 0\) and \(u_B \neq 0\) but \(u_B\) must have been found from a momentum equation |
| For both \(s_A\) and \(s_B\) and attempt to solve \(s_A = s_B\) to find \(t\) | DM1 | Must be from 3 terms leading to a 2-term quadratic. If energy used in 7(a) then must find \(a = 0.4\) for \(A\). *Their* working must lead to a positive \(t\) value |
| Required time is \(t = 1\) s | A1 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.6^2 = 0 + 2a \times 0.45$ | M1 | Use of constant acceleration equations to find $a$ |
| $a = 0.4$ | A1 | |
| $R = 0.1g \times \cos\alpha = 0.1g \times \frac{24}{25} = 0.1g \times \cos 16.3°\left[R = \frac{24}{25} = 0.96\right]$ | B1 | Must use a value for $\cos\alpha$ |
| $0.1g \times \frac{7}{25} - F = 0.1 \times 0.4\ [0.28 - F = 0.04 \rightarrow F = 0.24]$ | M1 | Newton's second law, 3 terms |
| $F = \mu \times 0.1g \times \frac{24}{25}\left[F = \frac{24\mu}{25} = 0.96\mu\right]$ | M1 | Use of $F = \mu R$, where $R$ is a component of $0.1g$ |
| $\mu = 0.25$ | A1 | **AG** Must be from exact working $\mu = 0.25$ only |
**Alternative scheme for 7(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt PE loss or KE gain | M1 | Use of either $\text{PE} = mgh$ or $\text{KE} = \frac{1}{2}mv^2$ |
| PE loss $= 0.1 \times g \times 0.45\sin 16.3° = 0.1 \times g \times 0.45 \times \frac{7}{25}\left[=\frac{63}{500} = 0.126\right]$, KE gain $= \frac{1}{2} \times 0.1 \times 0.6^2\left[=\frac{9}{500} = 0.018\right]$ | A1 | Both correct |
| $R = 0.1g \times \cos\alpha = 0.1g \times \frac{24}{25} = 0.1g \times \cos 16.3°\left[R = \frac{24}{25} = 0.96\right]$ | B1 | Must use a value for $\cos\alpha$ |
| $0.1 \times g \times 0.45 \times \frac{7}{25} = \frac{1}{2} \times 0.1 \times 0.6^2 + F \times 0.45$ $\left[\frac{63}{500} = \frac{9}{500} + \mu \times \frac{54}{125}\right]$ or $[0.126 = 0.018 + \mu \times 0.432]$ | M1 | Use of work-energy equation: PE loss $=$ KE gain $+$ WD against friction |
| $F = \mu \times 0.1g \times \frac{24}{25}\left[F = \frac{24\mu}{25} = 0.96\mu\right]$ | M1 | Use of $F = \mu R$, where $R$ is a component of $0.1g$ |
| $\mu = 0.25$ | A1 | **AG** Must be from exact working $\mu = 0.25$ only |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.1 \times 0.6 = 0.5v$ | M1 | Use of conservation of momentum, 2 terms |
| $v = 0.12$ | A1 | |
| For $B$: $0.5g \times \frac{7}{25} - 0.275 \times 0.5g \times \frac{24}{25} = 0.5a$ [leading to $a = 0.16$] | B1 | Apply Newton's second law for particle $B$, 3 terms. Allow correct unsimplified expression in $a$ only |
| $s_A = 0 + \frac{1}{2} \times 0.4t^2 \quad s_B = 0.12t + \frac{1}{2} \times 0.16t^2$ | \*M1 | Attempt expression for either $s_A$ or $s_B$. Must see $u_A = 0$ and $u_B \neq 0$ but $u_B$ must have been found from a momentum equation |
| For both $s_A$ and $s_B$ and attempt to solve $s_A = s_B$ to find $t$ | DM1 | Must be from 3 terms leading to a 2-term quadratic. If energy used in **7(a)** then must find $a = 0.4$ for $A$. *Their* working must lead to a positive $t$ value |
| Required time is $t = 1$ s | A1 | |7 A bead, $A$, of mass 0.1 kg is threaded on a long straight rigid wire which is inclined at $\sin ^ { - 1 } \left( \frac { 7 } { 25 } \right)$ to the horizontal. $A$ is released from rest and moves down the wire. The coefficient of friction between $A$ and the wire is $\mu$. When $A$ has travelled 0.45 m down the wire, its speed is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mu = 0.25$.\\
Another bead, $B$, of mass 0.5 kg is also threaded on the wire. At the point where $A$ has travelled 0.45 m down the wire, it hits $B$ which is instantaneously at rest on the wire. $A$ is brought to instantaneous rest in the collision. The coefficient of friction between $B$ and the wire is 0.275 .
\item Find the time from when the collision occurs until $A$ collides with $B$ again.\\
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 2022 Q7 [12]}}