| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Collision with friction aftermath |
| Difficulty | Standard +0.3 This is a standard three-part mechanics question requiring routine application of SUVAT with friction, conservation of momentum for a coalescing collision, and energy/friction calculations. Each part follows directly from the previous with no novel insight required, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-0.4 \times 6g = 6a\) | \*B1 | Resolve horizontally using Newton's second law; 2 relevant terms; must be either \(-0.4 \times 6g = 6a\) or \(0.4 \times 6g = 6a\). |
| \(v^2 = 20^2 + 2 \times (-4) \times 12.5\) | DM1 | Use complete suvat method to get an equation in \(v\) or \(v^2\) – must be using \(u = 20\), \(s = 12.5\) and *their* \(a\). |
| \(v^2 = 300 \Rightarrow v = 10\sqrt{3}\) | A1 | AG. Condone correct expression for \(v\) or \(v^2\) followed by correct answer. |
| Alternative method: | ||
| \(RF = 0.4 \times 6g\) | \*B1 | Correct application of \(F = \mu R\) for \(P\). |
| \(0.5 \times 6 \times 20^2 - 0.5 \times 6 \times v^2 = 12.5 \times (0.4 \times 6g)\) | DM1 | 3 relevant terms; dimensionally correct; allow sign errors only. |
| \(v^2 = 300 \Rightarrow v = 10\sqrt{3}\) | A1 | AG. Condone correct expression for \(v\) or \(v^2\) followed by correct answer. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6 \times 10\sqrt{3} = (6+2)v'\) | M1 | For use of conservation of momentum, 3 non-zero terms, allow sign errors. Use of 20 is M0. |
| \(v' = 7.5\sqrt{3}\) | A1 | \(12.99038\ldots\) |
| Initial KE \(= \frac{1}{2} \times 6 \times \left(10\sqrt{3}\right)^2\ [=900]\) | B1 | Either initial kinetic energy or final kinetic energy correct. Allow unsimplified. |
| Final KE \(= \frac{1}{2} \times 8 \times \left(7.5\sqrt{3}\right)^2\ [=675]\) | ||
| Loss of KE \(= 225\) J | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = \left(their\ 7.5\sqrt{3}\right)^2 + 2 \times (their\ {-4}) \times s\) | M1 | Use complete suvat method to find distance. This must be using *their* \(v'\) from part (b), so it is dependent on scoring the first M mark in part (b) and either *their* \(a\) from part (a), or from \(\pm 0.4 \times 8g = 8a\). |
| [Distance \(=\)] \(21.1\) m | A1 | 21.1 or better (21.09375). |
| 2 |
## Question 4:
### Part 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-0.4 \times 6g = 6a$ | **\*B1** | Resolve horizontally using Newton's second law; 2 relevant terms; must be either $-0.4 \times 6g = 6a$ or $0.4 \times 6g = 6a$. |
| $v^2 = 20^2 + 2 \times (-4) \times 12.5$ | **DM1** | Use complete suvat method to get an equation in $v$ or $v^2$ – must be using $u = 20$, $s = 12.5$ and *their* $a$. |
| $v^2 = 300 \Rightarrow v = 10\sqrt{3}$ | **A1** | AG. Condone correct expression for $v$ or $v^2$ followed by correct answer. |
| **Alternative method:** | | |
| $RF = 0.4 \times 6g$ | **\*B1** | Correct application of $F = \mu R$ for $P$. |
| $0.5 \times 6 \times 20^2 - 0.5 \times 6 \times v^2 = 12.5 \times (0.4 \times 6g)$ | **DM1** | 3 relevant terms; dimensionally correct; allow sign errors only. |
| $v^2 = 300 \Rightarrow v = 10\sqrt{3}$ | **A1** | AG. Condone correct expression for $v$ or $v^2$ followed by correct answer. |
| | **3** | |
### Part 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6 \times 10\sqrt{3} = (6+2)v'$ | **M1** | For use of conservation of momentum, 3 non-zero terms, allow sign errors. Use of 20 is M0. |
| $v' = 7.5\sqrt{3}$ | **A1** | $12.99038\ldots$ |
| Initial KE $= \frac{1}{2} \times 6 \times \left(10\sqrt{3}\right)^2\ [=900]$ | **B1** | Either initial kinetic energy or final kinetic energy correct. Allow unsimplified. |
| Final KE $= \frac{1}{2} \times 8 \times \left(7.5\sqrt{3}\right)^2\ [=675]$ | | |
| Loss of KE $= 225$ J | **A1** | |
| | **4** | |
### Part 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = \left(their\ 7.5\sqrt{3}\right)^2 + 2 \times (their\ {-4}) \times s$ | **M1** | Use complete suvat method to find distance. This must be using *their* $v'$ from part (b), so it is dependent on scoring the first M mark in part (b) and either *their* $a$ from part (a), or from $\pm 0.4 \times 8g = 8a$. |
| [Distance $=$] $21.1$ m | **A1** | 21.1 or better (21.09375). |
| | **2** | |
---4 Two particles $P$ and $Q$, of masses 6 kg and 2 kg respectively, lie at rest 12.5 m apart on a rough horizontal plane. The coefficient of friction between each particle and the plane is 0.4 . Particle $P$ is projected towards $Q$ with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $P$ immediately before the collision with $Q$ is $10 \sqrt { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
In the collision $P$ and $Q$ coalesce to form particle $R$.
\item Find the loss of kinetic energy due to the collision.\\
The coefficient of friction between $R$ and the plane is 0.4 .
\item Find the distance travelled by particle $R$ before coming to rest.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q4 [9]}}