| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, horizontal |
| Difficulty | Standard +0.3 This is a standard two-part connected particles question requiring Newton's second law applied to a system and individual bodies. Part (a) is routine calculation finding tension given driving force. Part (b) requires the insight that slack rope means zero tension and equal accelerations, but this is a common textbook scenario. Slightly above average due to the two-part structure and the conceptual element in part (b), but still well within standard M1 territory. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For van: \(2500 - 700 - T = 3600a\); For trailer: \(T - 300 = 1200a\); For system: \(2500 - 700 - 300 = (3600 + 1200)a\) | M1 | Apply Newton's 2nd law to the van or to the trailer or to the system of van and trailer. Correct number of terms. |
| For any two correct | A1 | |
| Obtain an equation in \(T\) only \(\left[a = \frac{5}{16} = 0.3125\right]\) | M1 | |
| Tension in the rope \(= T = 675\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For van: \(-F - 700 = 3600a\); For trailer: \(-300 = 1200a\); System: \(-F - 700 - 300 = (3600+1200)a\) | M1 | Apply Newton's 2nd law to any two of the van, the trailer and the system with braking force \(F\) and with \(T = 0\). |
| Least possible value of braking force \(= F = 200\) N | A1 | Allow \(F = -200\) |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For van: $2500 - 700 - T = 3600a$; For trailer: $T - 300 = 1200a$; For system: $2500 - 700 - 300 = (3600 + 1200)a$ | M1 | Apply Newton's 2nd law to the van or to the trailer or to the system of van and trailer. Correct number of terms. |
| For any two correct | A1 | |
| Obtain an equation in $T$ only $\left[a = \frac{5}{16} = 0.3125\right]$ | M1 | |
| Tension in the rope $= T = 675$ N | A1 | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For van: $-F - 700 = 3600a$; For trailer: $-300 = 1200a$; System: $-F - 700 - 300 = (3600+1200)a$ | M1 | Apply Newton's 2nd law to any two of the van, the trailer and the system with braking force $F$ and with $T = 0$. |
| Least possible value of braking force $= F = 200$ N | A1 | Allow $F = -200$ |
---2 A van of mass 3600 kg is towing a trailer of mass 1200 kg along a straight horizontal road using a light horizontal rope. There are resistance forces of 700 N on the van and 300 N on the trailer.
\begin{enumerate}[label=(\alph*)]
\item The driving force exerted by the van is 2500 N .
Find the tension in the rope.\\
The driving force is now removed and the van driver applies a braking force which acts only on the van. The resistance forces remain unchanged.
\item Find the least possible value of the braking force which will cause the rope to become slack.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q2 [6]}}