| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Connected particles via tow-bar on horizontal surface |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question involving Newton's second law applied to connected particles. Part (a) requires finding acceleration then applying F=ma to the system. Part (b) involves separating the system and using ratio conditions. Part (c) uses equations of motion with constant deceleration. Part (d) asks for a qualitative comment. All techniques are routine for M1 with clear multi-step structure, making it slightly easier than average overall. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | acc° \(= \frac{25 - 0}{20} = \frac{5}{4}\) ms\(^{-2}\) | M1 A1 |
| for car and van, eqn. of motion is \(D - 1200 = 2000 \times \frac{5}{4}\) | M1 | |
| \(D = 3700\) N | A1 | |
| (b) | 1200 divided in ratio 1.25 : 0.75 i.e. 5 : 3 | M1 |
| car resistance \(= \frac{5}{8} \times 1200 = 750\) N, so caravan resistance is 450 N | A1 | |
| for car, eqn. of motion is \(3700 - 750 - T = 1250 \times \frac{3}{4}\) | M1 | |
| \(T = 1387.5\) N | A1 | |
| (c) | for van, \(450 = 750a \therefore a = \frac{3}{5}\) ms\(^{-2}\) | M1A1 |
| \(u = 25\), \(v = 0\), \(a = \frac{3}{5}\) use \(v^2 = u^2 + 2as\) | M1 | |
| \(0 = 625 - \frac{6}{5}s \therefore s = 520.8\) m | M1 A1 | |
| (d) | e.g. caravan may nose down at front, may not stay in a straight line so dist. likely to be less than that calculated in (c) | B1 B1 |
(a) | acc° $= \frac{25 - 0}{20} = \frac{5}{4}$ ms$^{-2}$ | M1 A1 |
| for car and van, eqn. of motion is $D - 1200 = 2000 \times \frac{5}{4}$ | M1 |
| $D = 3700$ N | A1 |
(b) | 1200 divided in ratio 1.25 : 0.75 i.e. 5 : 3 | M1 |
| car resistance $= \frac{5}{8} \times 1200 = 750$ N, so caravan resistance is 450 N | A1 |
| for car, eqn. of motion is $3700 - 750 - T = 1250 \times \frac{3}{4}$ | M1 |
| $T = 1387.5$ N | A1 |
(c) | for van, $450 = 750a \therefore a = \frac{3}{5}$ ms$^{-2}$ | M1A1 |
| $u = 25$, $v = 0$, $a = \frac{3}{5}$ use $v^2 = u^2 + 2as$ | M1 |
| $0 = 625 - \frac{6}{5}s \therefore s = 520.8$ m | M1 A1 |
(d) | e.g. caravan may nose down at front, may not stay in a straight line so dist. likely to be less than that calculated in (c) | B1 B1 | (15) |A car of mass 1.25 tonnes tows a caravan of mass 0.75 tonnes along a straight, level road. The total resistance to motion experienced by the car and the caravan is 1200 N. The car and caravan accelerate uniformly from rest to 25 m s$^{-1}$ in 20 seconds.
\begin{enumerate}[label=(\alph*)]
\item Calculate the driving force produced by the car's engine. [4 marks]
\end{enumerate}
Given that the resistance to motion experienced by the car and by the caravan are in the same ratio as their masses,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find these resistances and the tension in the towbar. [4 marks]
\end{enumerate}
When the car and caravan are travelling at a steady speed of 25 m s$^{-1}$, the towbar snaps. Assuming that the caravan experiences the same resistive force as before,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item calculate the distance travelled by the caravan before it comes to rest, [5 marks]
\item give a reason why your answer to $(c)$ may be unrealistic. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [15]}}