| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.5 This is a straightforward variable acceleration question requiring direct substitution into given expressions, application of F=ma, and one integration with initial conditions. All steps are routine procedures with no problem-solving insight needed, making it slightly easier than average for A-level mechanics. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02a Kinematics language: position, displacement, velocity, acceleration3.02e Two-dimensional constant acceleration: with vectors3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(9\mathbf{i}\) m s\(^{-2}\); \((9\mathbf{i} - 12\mathbf{j})\) m s\(^{-2}\) | B1 | Award for either. Accept no units. isw e.g. finding magnitudes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{F} = 4(9\mathbf{i} - 12\mathbf{j}) = (36\mathbf{i} - 48\mathbf{j})\) N | B1 | Accept factored form. isw. FT a(3). Accept 60 N or their \(4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{v} = \int\binom{9}{-4t}dt = \binom{9t+C}{-2t^2+D}\) | M1 | Integration. At least one term correct |
| A1 | Neglect arbitrary constant(s) | |
| Using \(\mathbf{v} = 4\mathbf{i} + 2\mathbf{j}\) when \(t=1\): \(\binom{4}{2} = \binom{9+C}{-2+D}\) | M1 | Sub at \(t=1\) to find arbitrary constant(s) |
| \(C = -5,\ D = 4\) so \(\mathbf{v} = (9t-5)\mathbf{i} + (4-2t^2)\mathbf{j}\) | A1 | y form |
## Question 2:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $9\mathbf{i}$ m s$^{-2}$; $(9\mathbf{i} - 12\mathbf{j})$ m s$^{-2}$ | B1 | Award for either. Accept no units. isw e.g. finding magnitudes |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{F} = 4(9\mathbf{i} - 12\mathbf{j}) = (36\mathbf{i} - 48\mathbf{j})$ N | B1 | Accept factored form. isw. FT **a**(3). Accept 60 N or their $4|\mathbf{a}|$ |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{v} = \int\binom{9}{-4t}dt = \binom{9t+C}{-2t^2+D}$ | M1 | Integration. At least one term correct |
| | A1 | Neglect arbitrary constant(s) |
| Using $\mathbf{v} = 4\mathbf{i} + 2\mathbf{j}$ when $t=1$: $\binom{4}{2} = \binom{9+C}{-2+D}$ | M1 | Sub at $t=1$ to find arbitrary constant(s) |
| $C = -5,\ D = 4$ so $\mathbf{v} = (9t-5)\mathbf{i} + (4-2t^2)\mathbf{j}$ | A1 | y form |
---2 The acceleration of a particle of mass 4 kg is given by $\mathbf { a } = ( 9 \mathbf { i } - 4 t \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { 2 }$, where $\mathbf { i }$ and $\mathbf { j }$ are unit vectors and $t$ is the time in seconds.\\
(i) Find the acceleration of the particle when $t = 0$ and also when $t = 3$.\\
(ii) Calculate the force acting on the particle when $t = 3$.
The particle has velocity $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } { } ^ { 1 }$ when $t = 1$.\\
(iii) Find an expression for the velocity of the particle at time $t$.
\hfill \mbox{\textit{OCR MEI M1 Q2 [6]}}