| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Connected particles via tow-bar on horizontal surface |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question involving Newton's second law applied to connected particles. Part (a) requires simple F=ma for the system (1200-600=1000a), part (b) isolates the trailer (T-200=200a), and part (c) requires recognizing thrust means compression and applying F=ma to the trailer then car. While multi-part with 13 marks total, each step follows routine procedures taught in M1 with no novel insight required, making it slightly easier than average. |
| Spec | 3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks |
|---|---|
| (a) For whole system: \(1200 - 400 - 200 = 1000a\) | M1 A1 |
| \(a = 0.6 \text{ m s}^{-2}\) | A1 |
| (3) | |
| (b) For trailer: \(T - 200 = 200 \times 0.6\) | M1 A1 ft |
| \(T = 320 \text{ N}\) | A1 |
| OR: For car: \(1200 - 400 - T = 800 \times 0.6\) | OR: M1 A1 ft |
| \(T = 320 \text{ N}\) | A1 |
| (3) | |
| (c) For trailer: \(200 + 100 = 200f\) or \(-200f\) | M1 A1 |
| \(f = 1.5 \text{ m s}^{-2}\) (-1.5) | A1 |
| For car: \(400 + F - 100 = 800f\) or \(-800f\) | M1 A2 |
| \(F = 900\) | A1 |
| (N.B. For both: \(400 + 200 + F = 1000f\)) | |
| Total: [13] |
**(a)** For whole system: $1200 - 400 - 200 = 1000a$ | M1 A1 |
$a = 0.6 \text{ m s}^{-2}$ | A1 |
| **(3)** |
**(b)** For trailer: $T - 200 = 200 \times 0.6$ | M1 A1 ft |
$T = 320 \text{ N}$ | A1 |
**OR:** For car: $1200 - 400 - T = 800 \times 0.6$ | OR: M1 A1 ft |
$T = 320 \text{ N}$ | A1 |
| **(3)** |
**(c)** For trailer: $200 + 100 = 200f$ or $-200f$ | M1 A1 |
$f = 1.5 \text{ m s}^{-2}$ (-1.5) | A1 |
For car: $400 + F - 100 = 800f$ or $-800f$ | M1 A2 |
$F = 900$ | A1 |
(N.B. For both: $400 + 200 + F = 1000f$) | |
| **Total: [13]** |A car of mass 800 kg pulls a trailer of mass 200 kg along a straight horizontal road using a light towbar which is parallel to the road. The horizontal resistances to motion of the car and the trailer have magnitudes 400 N and 200 N respectively. The engine of the car produces a constant horizontal driving force on the car of magnitude 1200 N. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the car and trailer, [3]
\item the magnitude of the tension in the towbar. [3]
\end{enumerate}
The car is moving along the road when the driver sees a hazard ahead. He reduces the force produced by the engine to zero and applies the brakes. The brakes produce a force on the car of magnitude $F$ newtons and the car and trailer decelerate. Given that the resistances to motion are unchanged and the magnitude of the thrust in the towbar is 100 N,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the value of $F$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2009 Q6 [13]}}