| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Inclined road towing |
| Difficulty | Standard +0.3 This is a standard M1 pulley/connected particles question requiring Newton's second law applied to two connected bodies on inclined planes. Part (a) involves straightforward force resolution with given tension. Part (b) adds a kinematic calculation using suvat equations. While multi-step with several parts, it follows a completely standard template with no novel problem-solving required—slightly easier than the average A-level question due to its routine nature. |
| Spec | 3.03d Newton's second law: 2D vectors3.03q Dynamics: motion under force in plane6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of Newton's second law for van or trailer or system | \(M1^*\) | Must have correct number of relevant terms. Allow sign errors. Allow sin/cos mix. Allow \(g\) missing. Masses must be correct for their equation(s). Forces must have components (or not) as required. Must have either 0.05 or \(\sin 2.86\ldots\) or \(\sin 2.9\), not just \(\sin\theta\) |
| Trailer: \(450 + 750g \times 0.05 - 300 = 750a\), \([525 = 750a]\) | \(A1\) | For any two correct equations |
| Van: \(D + 4500g \times 0.05 - 2500 - 450 = 4500a\), \([D - 700 = 4500a]\) | ||
| System: \(D + 4500g \times 0.05 + 750g \times 0.05 - 2500 - 300 = (4500 + 750)a\), \([D - 175 = 5250a]\) | ||
| For attempt to solve for \(a\) or \(D\) | \(DM1\) | Must get to '\(a =\)' or '\(D =\)'. Must have correct number of relevant terms. \(g\) must be present. Allow sign errors. Allow sin/cos mix. If no working shown, answers should be correct for their equations |
| \(a = 0.7\ \text{ms}^{-2}\) and \(D = 3850\ \text{N}\) | \(A1\) | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of Newton's second law for van or trailer or system | \(M1^*\) | Must have correct number of relevant terms. Allow sign errors. Allow sin/cos mix. Allow \(g\) missing. Masses must be appropriate. Forces must have components (or not) as required. Must have either 0.09 or \(\sin 5.16\ldots\) or \(\sin 5.2\), not just \(\sin\theta\) |
| Trailer: \(T - 300 - 750g \times 0.09 = 750a\), \([T - 975 = 750a]\) | \(A1A1\) | A1 for one correct equation, second A1 for another correct equation. If using Van and Trailer equations, must be using the same \(T\) for both to get the second A1 |
| Van: \(9100 - 2500 - 4500g \times 0.09 - T = 4500a\), \([2550 - T = 4500a]\) | ||
| System: \(9100 - 2500 - 300 - (4500 + 750)g \times 0.09 = (4500 + 750)a\), \([1575 = 5250a]\) | ||
| For attempt to solve for \(a\) or \(T\) | \(DM1\) | Must get to '\(a =\)' or '\(T =\)'. Must have correct number of relevant terms. \(g\) must be present. Allow sign errors. Allow sin/cos mix. If no working shown, answers should be correct for their equations |
| \(T = 1200\ \text{N}\) and \(a = 0.3\ \text{ms}^{-2}\) | \(A1\) | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v^2 = 20^2 + 2 \times their\ 0.3 \times 375\) | \(M1\) | For use of \(v^2 = 20^2 + 2a \times 375\) or other complete method to find \(v^2\) or \(v\). For info time taken \(t = \frac{50}{3}\) |
| \(v = 25\ \text{ms}^{-1}\) | \(A1FT\) | FT their value of \(a\), i.e. \(v = \sqrt{400 + 750 \times their\ a}\). Provided it does not lead to root of negative value |
| Alternative Method: Using energy | ||
| System: \(\frac{1}{2} \times (4500+750)(v^2 - 20^2) + (4500+750)g \times 375 \times 0.09 = (9100-2500-300) \times 375\) | \((M1)\) | Must include all appropriate terms. Allow sign errors. \(g\) must be present. Allow their value of \(T\) in place of 1200 |
| Van: \(\frac{1}{2} \times 4500(v^2 - 20^2) + 4500g \times 375 \times 0.09 = (9100 - 2500 - their\ 1200) \times 375\) | ||
| Trailer: \(\frac{1}{2} \times 750(v^2 - 20^2) + 750g \times 375 \times 0.09 = (their\ 1200 - 300) \times 375\) | ||
| \(v = 25\ \text{ms}^{-1}\) | \((A1FT)\) | FT their value of \(T\) if using Van or Trailer |
| 2 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of Newton's second law for van or trailer or system | $M1^*$ | Must have correct number of relevant terms. Allow sign errors. Allow sin/cos mix. Allow $g$ missing. Masses must be correct for their equation(s). Forces must have components (or not) as required. Must have either 0.05 or $\sin 2.86\ldots$ or $\sin 2.9$, not just $\sin\theta$ |
| Trailer: $450 + 750g \times 0.05 - 300 = 750a$, $[525 = 750a]$ | $A1$ | For any two correct equations |
| Van: $D + 4500g \times 0.05 - 2500 - 450 = 4500a$, $[D - 700 = 4500a]$ | | |
| System: $D + 4500g \times 0.05 + 750g \times 0.05 - 2500 - 300 = (4500 + 750)a$, $[D - 175 = 5250a]$ | | |
| For attempt to solve for $a$ or $D$ | $DM1$ | Must get to '$a =$' or '$D =$'. Must have correct number of relevant terms. $g$ must be present. Allow sign errors. Allow sin/cos mix. If no working shown, answers should be correct for their equations |
| $a = 0.7\ \text{ms}^{-2}$ and $D = 3850\ \text{N}$ | $A1$ | |
| | **4** | |
---
## Question 5(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of Newton's second law for van or trailer or system | $M1^*$ | Must have correct number of relevant terms. Allow sign errors. Allow sin/cos mix. Allow $g$ missing. Masses must be appropriate. Forces must have components (or not) as required. Must have either 0.09 or $\sin 5.16\ldots$ or $\sin 5.2$, not just $\sin\theta$ |
| Trailer: $T - 300 - 750g \times 0.09 = 750a$, $[T - 975 = 750a]$ | $A1A1$ | A1 for one correct equation, second A1 for another correct equation. If using Van and Trailer equations, must be using the same $T$ for both to get the second A1 |
| Van: $9100 - 2500 - 4500g \times 0.09 - T = 4500a$, $[2550 - T = 4500a]$ | | |
| System: $9100 - 2500 - 300 - (4500 + 750)g \times 0.09 = (4500 + 750)a$, $[1575 = 5250a]$ | | |
| For attempt to solve for $a$ or $T$ | $DM1$ | Must get to '$a =$' or '$T =$'. Must have correct number of relevant terms. $g$ must be present. Allow sign errors. Allow sin/cos mix. If no working shown, answers should be correct for their equations |
| $T = 1200\ \text{N}$ and $a = 0.3\ \text{ms}^{-2}$ | $A1$ | |
| | **5** | |
---
## Question 5(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v^2 = 20^2 + 2 \times their\ 0.3 \times 375$ | $M1$ | For use of $v^2 = 20^2 + 2a \times 375$ or other complete method to find $v^2$ or $v$. For info time taken $t = \frac{50}{3}$ |
| $v = 25\ \text{ms}^{-1}$ | $A1FT$ | FT their value of $a$, i.e. $v = \sqrt{400 + 750 \times their\ a}$. Provided it does not lead to root of negative value |
| **Alternative Method: Using energy** | | |
| System: $\frac{1}{2} \times (4500+750)(v^2 - 20^2) + (4500+750)g \times 375 \times 0.09 = (9100-2500-300) \times 375$ | $(M1)$ | Must include all appropriate terms. Allow sign errors. $g$ must be present. Allow their value of $T$ in place of 1200 |
| Van: $\frac{1}{2} \times 4500(v^2 - 20^2) + 4500g \times 375 \times 0.09 = (9100 - 2500 - their\ 1200) \times 375$ | | |
| Trailer: $\frac{1}{2} \times 750(v^2 - 20^2) + 750g \times 375 \times 0.09 = (their\ 1200 - 300) \times 375$ | | |
| $v = 25\ \text{ms}^{-1}$ | $(A1FT)$ | FT their value of $T$ if using Van or Trailer |
| | **2** | |
---5 A van of mass 4500 kg is towing a trailer of mass 750 kg down a straight hill inclined at an angle of $\theta$ to the horizontal where $\sin \theta = 0.05$. The van and the trailer are connected by a light rigid tow-bar which is parallel to the road. There are constant resistance forces of 2500 N on the van and 300 N on the trailer.
\begin{enumerate}[label=(\alph*)]
\item It is given that the tension in the tow-bar is 450 N .
Find the acceleration of the trailer and the driving force of the van's engine.\\
On another occasion, the van and trailer ascend a straight hill inclined at an angle of $\alpha$ to the horizontal where $\sin \alpha = 0.09$. The driving force of the van's engine is now 9100 N , and the speed of the van at the bottom of the hill is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The resistances to motion are unchanged.
\item \begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the van and the tension in the tow-bar.
\item Find the speed of the van when it has travelled a distance of 375 m up the hill.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q5 [11]}}