| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, inclined road |
| Difficulty | Standard +0.3 This is a standard M1 connected particles problem requiring Newton's second law applied to a two-body system on an incline. Part (i) involves routine resolution of forces and simultaneous equations to find acceleration and tension. Part (ii) requires finding deceleration when the driving force is removed and using kinematics. While multi-step, it follows a well-practiced template with no novel insight required, making it slightly easier than average. |
| Spec | 3.03g Gravitational acceleration3.03o Advanced connected particles: and pulleys3.03q Dynamics: motion under force in plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([2500 - 2000g \times 0.1 - 250 = 2000a]\) | M1 | For using Newton's 2nd law for the system or applying N2L to car and trailer and solving for \(a\); allow use of \(\alpha = 5.7°\) throughout |
| \(a = 1/8 = 0.125\ \text{ms}^{-2}\) | A1 | |
| \(2500 - T - 100 - 1200g \times 0.1 = 1200 \times 0.125\) or \(T - 150 - 800g \times 0.1 = 800 \times 0.125\) | M1 | For applying Newton's 2nd law either to car or trailer to set up equation for \(T\) |
| \(T = 1050\) N | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-2000g \times 0.1 - 250 = 2000a \Rightarrow [a = -1.125]\) | M1 | For applying Newton's 2nd law to system with no driving force to set up equation for \(a\) |
| \(0 = 30 - 1.125t\) | M1 | For using \(v = u + at\) |
| \(t = 26.7\) s | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}(2000)30^2 = 250s + 2000 \times g \times 0.1s \Rightarrow s = 400\) | M1 | Apply work/energy equation to find \(s\); KE loss \(=\) WD against \(F\) + PE gain (3 terms) |
| \([400 = \frac{1}{2}(30 + 0)t]\) | M1 | For using \(x = \frac{1}{2}(u+v)t\) |
| \(t = 26.7\) s | A1 | 3 |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[2500 - 2000g \times 0.1 - 250 = 2000a]$ | M1 | For using Newton's 2nd law for the system or applying N2L to car and trailer and solving for $a$; allow use of $\alpha = 5.7°$ throughout |
| $a = 1/8 = 0.125\ \text{ms}^{-2}$ | A1 | |
| $2500 - T - 100 - 1200g \times 0.1 = 1200 \times 0.125$ **or** $T - 150 - 800g \times 0.1 = 800 \times 0.125$ | M1 | For applying Newton's 2nd law either to car or trailer to set up equation for $T$ |
| $T = 1050$ N | A1 | 4 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2000g \times 0.1 - 250 = 2000a \Rightarrow [a = -1.125]$ | M1 | For applying Newton's 2nd law to system with no driving force to set up equation for $a$ |
| $0 = 30 - 1.125t$ | M1 | For using $v = u + at$ |
| $t = 26.7$ s | A1 | 3 | Allow $t = 80/3$ s |
**Alternative method for 5(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}(2000)30^2 = 250s + 2000 \times g \times 0.1s \Rightarrow s = 400$ | M1 | Apply work/energy equation to find $s$; KE loss $=$ WD against $F$ + PE gain (3 terms) |
| $[400 = \frac{1}{2}(30 + 0)t]$ | M1 | For using $x = \frac{1}{2}(u+v)t$ |
| $t = 26.7$ s | A1 | 3 | Allow $t = 80/3$ s |
---5 A car of mass 1200 kg is pulling a trailer of mass 800 kg up a hill inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.1$. The system of the car and the trailer is modelled as two particles connected by a light inextensible cable. The driving force of the car's engine is 2500 N and the resistances to the car and trailer are 100 N and 150 N respectively.\\
(i) Find the acceleration of the system and the tension in the cable.\\
(ii) When the car and trailer are travelling at a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the driving force becomes zero. The cable remains taut. Find the time, in seconds, before the system comes to rest.
\hfill \mbox{\textit{CAIE M1 2016 Q5 [7]}}