| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Forces as vectors |
| Difficulty | Moderate -0.8 Part (i) requires recognizing that parallel forces means G is a scalar multiple of F, leading to straightforward simultaneous equations. Part (ii) is direct application of F=ma with vector addition. Both parts are routine mechanics exercises requiring only standard recall and basic calculation, making this easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(-6 = -2 \times 3\) | M1 | May be implied |
| so \(y = 3 \times 3 = 9\) and \(z = -4 \times 3 = -12\) | A1 | Both correct [Award 2 for both correct answers seen WW] |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} -2 \\ 3 \\ -4 \end{pmatrix} \begin{pmatrix} 3 \\ -5 \\ -1 \end{pmatrix} = 5a\) | M1 | Use of Newton's 2nd Law in vector form for all 3 cpts of attempted resultant. Treat use of wrong vectors as MR. |
| \(a = \begin{pmatrix} 0.2 \\ -0.4 \\ -1 \end{pmatrix}\) so accn is \(\begin{pmatrix} 0.2 \\ -0.4 \\ -1 \end{pmatrix} \text{ m s}^{-2}\) | B1 | Correct LHS |
| A1 | The acceleration may be written as a magnitude in a given direction. | |
| Magnitude is \(\sqrt{0.2^2 + (-0.4)^2 + (-1)^2} = 1.09544\ldots\) so \(1.10 \text{ m s}^{-2}\) (3 s.f.) | M1 | FT their values. Condone missing brackets. Condone no – signs. |
| F1 | Accept 1.1. Accept surd form. Must come from a vector with 3 non-zero components for a | |
| 5 | ||
| 7 |
**(i)**
$-6 = -2 \times 3$ | M1 | May be implied
so $y = 3 \times 3 = 9$ and $z = -4 \times 3 = -12$ | A1 | Both correct [Award 2 for both correct answers seen WW]
| | 2 |
**(ii)**
$\begin{pmatrix} -2 \\ 3 \\ -4 \end{pmatrix} \begin{pmatrix} 3 \\ -5 \\ -1 \end{pmatrix} = 5a$ | M1 | Use of Newton's 2nd Law in vector form for all 3 cpts of attempted resultant. Treat use of wrong vectors as MR.
$a = \begin{pmatrix} 0.2 \\ -0.4 \\ -1 \end{pmatrix}$ so accn is $\begin{pmatrix} 0.2 \\ -0.4 \\ -1 \end{pmatrix} \text{ m s}^{-2}$ | B1 | Correct LHS
| A1 | The acceleration may be written as a magnitude in a given direction.
Magnitude is $\sqrt{0.2^2 + (-0.4)^2 + (-1)^2} = 1.09544\ldots$ so $1.10 \text{ m s}^{-2}$ (3 s.f.) | M1 | FT their values. Condone missing brackets. Condone no – signs.
| F1 | Accept 1.1. Accept surd form. Must come from a vector with 3 non-zero components for a
| | 5 |
| | 7 |
---3 Force $\mathbf { F }$ is $\left( \begin{array} { r } - 2 \\ 3 \\ - 4 \end{array} \right) \mathrm { N }$, force $\mathbf { G }$ is $\left( \begin{array} { r } - 6 \\ y \\ z \end{array} \right) \mathrm { N }$ and force $\mathbf { H }$ is $\left( \begin{array} { r } 3 \\ - 5 \\ - 1 \end{array} \right) \mathrm { N }$.\\
(i) Given that $\mathbf { F }$ and $\mathbf { G }$ act in parallel lines, find $y$ and $z$.
Forces $\mathbf { F }$ and $\mathbf { H }$ are the only forces acting on an object of mass 5 kg .\\
(ii) Calculate the acceleration of the object. Calculate also the magnitude of this acceleration.
\hfill \mbox{\textit{OCR MEI M1 2011 Q3 [7]}}