| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2015 |
| Session | June |
| Marks | 4 |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward two-part mechanics question requiring basic resolution of forces and Newton's second law. Part (i) involves simple perpendicular component calculation (F = 240sin25°), and part (ii) applies F=ma with parallel components. Both are standard textbook exercises with no problem-solving insight required, making this easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions |
**Question 6**
**(i)** $240\sin 25$
$= 101.428\ldots \approx 101\ \text{N}$
— M1: Resolve perpendicular to direction of travel. Allow sin/cos error. A1: c.a.o. **[2]**
**(ii)** $1100a = 240\cos 25 - 100$
$\therefore a = 0.1068\ldots \approx 0.107\ \text{ms}^{-2}$
— B1: Resolve 240 in direction of travel. Allow consistent sin/cos error. M1: N2L in direction of travel. Allow 1 error, omission or extraneous term. A1: All terms correct. A1: c.a.o. **[4]**6\\
\includegraphics[max width=\textwidth, alt={}, center]{9ddae838-2639-4952-bbc0-3944a81e5762-3_401_1224_1315_456}
The diagram shows a barge being towed along a canal by a force of 240 N at an angle of $25 ^ { \circ }$ to its direction of motion. A force, $F \mathrm {~N}$, perpendicular to the direction of motion, is applied to the barge to keep it moving in the direction shown.\\
(i) Find the magnitude of $F$.\\
(ii) The mass of the barge is 1100 kg and there is a resistance force of 100 N parallel to the direction of motion. Find the acceleration of the barge.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2015 Q6 [4]}}