| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position from velocity and initial conditions |
| Difficulty | Standard +0.3 This is a straightforward M1 mechanics question requiring application of Newton's second law (F=ma) to find acceleration, then standard constant acceleration kinematics (s=ut+½at²) in vector form. All steps are routine calculations with no problem-solving insight needed, though the vector notation and multi-part structure make it slightly above average difficulty for basic M1 content. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.02e Two-dimensional constant acceleration: with vectors3.03d Newton's second law: 2D vectors |
| Answer | Marks |
|---|---|
| 1(i) | (−i+16j+72k)+(−80k)=8a |
| Answer | Marks |
|---|---|
| ⎝ 8 ⎠ | M1 |
| E1 | Use of N2L. All forces present. |
| Need at least the k term clearly derived | 2 |
| (ii) | ⎛ 1 ⎞ |
| Answer | Marks |
|---|---|
| = 3i + 4k | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | Use of appropriate uvast or integration (twice) | |
| Correct substitution (or limits if integrated) | 3 | |
| (iii) | 32+42 =5so 5 m | B1 |
| surd form | 1 | |
| (iv) | 4 |
| Answer | Marks |
|---|---|
| = 53.130… so 53.1° (3 s. f.) | M1 |
| A1 | 3 |
| Answer | Marks |
|---|---|
| cao | 2 |
Question 1:
--- 1(i) ---
1(i) | (−i+16j+72k)+(−80k)=8a
⎛ 1 ⎞
a=⎜− i+2j−k⎟m s –2
⎝ 8 ⎠ | M1
E1 | Use of N2L. All forces present.
Need at least the k term clearly derived | 2
(ii) | ⎛ 1 ⎞
r=4(i−4j+3k)+0.5×16⎜− i+2j−k⎟
⎝ 8 ⎠
= 3i + 4k | M1
A1
A1 | Use of appropriate uvast or integration (twice)
Correct substitution (or limits if integrated) | 3
(iii) | 32+42 =5so 5 m | B1 | FT their (ii) even if it not a displacement. Allow
surd form | 1
(iv) | 4
arctan
3
= 53.130… so 53.1° (3 s. f.) | M1
A1 | 3
Accept arctan . FT their (ii) even if not a
4
displacement. Condone sign errors.
(May use arcsin4/5 or equivalent. FT their (ii)
and (iii) even if not displacement. Condone sign
errors)
cao | 2
8
2 mark Sub
(i) either
Need j cpt 0 so 18t2−1=0 M1 Need not solve
1
⇒t2 = . Only one root as t > 0 E1 Must establish only one of the two roots is valid
18
or
Establish sign change in j cpt B1
Establish only one root B1
2
(ii) v = 3 i + 36t j M1 Differentiate. Allow i or j omitted
A1
Need i cpt 0 and this never happens E1 Clear explanation. Accept ‘i cpt always there’ or equiv
3
(iii) x=3t and y=18t2−1 B1 Award for these two expressions seen.
Eliminate t to give
⎛x⎞ 2 t properly eliminated. Accept any form and brackets
y=18⎜ ⎟ −1 M1
⎝3⎠ missing
so y=2x2 −1 AA11 ca
3
8
PhysicsAndMathsTutor.comA rock of mass 8 kg is acted on by just the two forces $-80$k N and $(-\mathbf{i} + 16\mathbf{j} + 72$k$)$ N, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors in a horizontal plane and k is a unit vector vertically upward.
\begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of the rock is $\left(\frac{1}{8}\mathbf{i} + 2\mathbf{j}\right)$ k$)$ ms$^{-2}$. [2]
\end{enumerate}
The rock passes through the origin of position vectors, O, with velocity $(\mathbf{i} - 4\mathbf{j} + 3$k$)$ m s$^{-1}$ and 4 seconds later passes through the point A.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the position vector of A. [3]
\item Find the distance OA. [1]
\item Find the angle that OA makes with the horizontal. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q1 [8]}}