| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring vector addition, finding magnitudes using Pythagoras in 3D, applying F=ma, and using constant acceleration kinematics. All steps are routine applications of standard formulas with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| \( | 2i - 2j + k | = 3\) |
| \(\mathbf{F}_3 = (\pm)(2i - 2j + k)\) | A1 | |
| The possible resultants of the three forces are: \(15i + 6j\) and \(7i + 14j - 4k\) | A1 A1 | |
| Both have magnitude \(\sqrt{261}\) N | A1 AG | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| The vector equation of motion is: \( | \mathbf{F}_1 + \mathbf{F}_2 | = |
| Equation of motion: \(2 | a | = 15\) |
| \( | a | = 7.5\) ms\(^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| The particle moves in the straight line defined by the direction of the resultant force with the constant acceleration 7.5 ms\(^{-2}\). Applying the formula \(s = ut + 0.5at^2\), obtain \(t = 4\) seconds | M1 A1 | [2] |
### (i)
$|2i - 2j + k| = 3$ | M1 |
$\mathbf{F}_3 = (\pm)(2i - 2j + k)$ | A1 |
The possible resultants of the three forces are: $15i + 6j$ and $7i + 14j - 4k$ | A1 A1 |
Both have magnitude $\sqrt{261}$ N | A1 AG | [5]
### (ii)
The vector equation of motion is: $|\mathbf{F}_1 + \mathbf{F}_2| = |11i + 10j - 2k| = 15$ | B1 |
Equation of motion: $2|a| = 15$ | M1 |
$|a| = 7.5$ ms$^{-2}$ | A1 | [3]
### (iii)
The particle moves in the straight line defined by the direction of the resultant force with the constant acceleration 7.5 ms$^{-2}$. Applying the formula $s = ut + 0.5at^2$, obtain $t = 4$ seconds | M1 A1 | [2]Two forces $\mathbf{F}_1$ and $\mathbf{F}_2$ are given by
$$\mathbf{F}_1 = 13\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}, \quad \mathbf{F}_2 = -2\mathbf{i} + 6\mathbf{j} + \mathbf{k},$$
in which the units of the components are newtons. A third force, $\mathbf{F}_3$, of magnitude 6 N acts parallel to the vector $2\mathbf{i} - 2\mathbf{j} + \mathbf{k}$.
\begin{enumerate}[label=(\roman*)]
\item Find the two possible resultants of $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$, and show that they have the same magnitude. [5]
\end{enumerate}
A particle, $P$, of mass 2 kg is initially at rest at the origin. The only forces acting on $P$ are $\mathbf{F}_1$ and $\mathbf{F}_2$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the magnitude of the acceleration of $P$. [3]
\item Find the time taken for $P$ to travel 60 m. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q11 [10]}}