| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | Specimen |
| Marks | 6 |
| Topic | Forces, equilibrium and resultants |
| Type | Multiple particles with intermediate connections |
| Difficulty | Moderate -0.3 This is a standard connected particles problem requiring Newton's second law applied to the system and individual particles. The calculations are straightforward with clear given values, though it requires careful bookkeeping of forces and understanding of coupled motion. Slightly easier than average due to its routine nature and explicit setup. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
**(i)** $P - 1050 = 18000 \times 0.3$ **M1**
$P = 6450$ **A1**
**(ii)** New acceleration: $6450 - 2850 = 18000a$ **M1**
$A = 0.2$ **A1**
**(iii)** $6450 - 450 - T = 8000(0.2)$ **M1**
$T = 4400$ N **A1**8 Two trucks, $S$ and $T$, of masses 8000 kg and 10000 kg respectively, are pulled along a straight, horizontal track by a constant, horizontal force of $P$ N. A resistive force of 600 N acts on $S$ and a resistive force of 450 N acts on $T$. The coupling between the trucks is light and horizontal (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{b18b1bc5-bf26-4161-b5a5-764b00e97bea-5_215_1095_427_479}
The acceleration of the system is $0.3 \mathrm {~ms} ^ { - 2 }$ in the direction of the pulling force of magnitude $P$.\\
(i) Calculate the value of $P$.
Truck $S$ is now subjected to an extra resistive force of 1800 N . The pulling force, $P$, does not change.\\
(ii) Calculate the new acceleration of the trucks.\\
(iii) Calculate the force in the coupling between the trucks.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q8 [6]}}