| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.3 This is a straightforward application of Newton's second law in vector form (F=ma) requiring vector addition and scalar multiplication. Part (i) involves simple algebraic manipulation to find the unknown force, and part (ii) uses the constant acceleration equation v=u+at. While it requires competence with 3D vectors, the problem-solving is routine and mechanical with no conceptual challenges beyond standard M1 content. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either using suvat: Use of \(\mathbf{v}=\mathbf{u}+t\mathbf{a}\) | M1 | Substitution required. Must be vectors |
| \(\mathbf{v}=4\mathbf{i}-2t\mathbf{j}\) | A1 | |
| Use of \(\mathbf{r}=(\mathbf{r}_0+)\ t\mathbf{u}+\frac{1}{2}t^2\mathbf{a}\) | M1 | Substitution required. \(\mathbf{r}_0\) not required. Must be vectors |
| \(+3\mathbf{j}\) | B1 | May be seen on either side of a meaningful equation for \(\mathbf{r}\) |
| \(\mathbf{r}=4t\mathbf{i}+(3-t^2)\mathbf{j}\) | A1 | Accept \(\mathbf{r}=3\mathbf{j}+4t\mathbf{i}-\frac{1}{2}\times 2\times t^2\mathbf{j}\) in correct notation |
| Or using integration: \(\mathbf{v}=\int\mathbf{a}\,dt\) | M1 | Attempt at integration. Condone no \(+c\). Must be vectors |
| \(\mathbf{v}=4\mathbf{i}-2t\mathbf{j}\) | A1 | cao |
| \(\mathbf{r}=\int\mathbf{v}\,dt\) | M1 | Integrate their \(\mathbf{v}\), must contain 2 components. Must be vectors |
| \(+3\mathbf{j}\) | B1 | May be seen on either side of a meaningful equation for \(\mathbf{r}\) |
| \(\mathbf{r}=4t\mathbf{i}+(3-t^2)\mathbf{j}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v}(2.5)=4\mathbf{i}-5\mathbf{j}\) | B1 | FT their \(\mathbf{v}\) |
| Angle is \((90+)\arctan\dfrac{5}{4}\) | M1 | Award for arctan attempted. FT their values. Allow argument \(\pm(\text{i cpt})/(\text{j cpt})\) or \(\pm(\text{j cpt})/(\text{i cpt})\) |
| \(=141.34019\ldots\) so \(141°\) (3 s.f.) | A1 | cao |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either** using suvat: Use of $\mathbf{v}=\mathbf{u}+t\mathbf{a}$ | M1 | Substitution required. Must be vectors |
| $\mathbf{v}=4\mathbf{i}-2t\mathbf{j}$ | A1 | |
| Use of $\mathbf{r}=(\mathbf{r}_0+)\ t\mathbf{u}+\frac{1}{2}t^2\mathbf{a}$ | M1 | Substitution required. $\mathbf{r}_0$ not required. Must be vectors |
| $+3\mathbf{j}$ | B1 | May be seen on either side of a meaningful equation for $\mathbf{r}$ |
| $\mathbf{r}=4t\mathbf{i}+(3-t^2)\mathbf{j}$ | A1 | Accept $\mathbf{r}=3\mathbf{j}+4t\mathbf{i}-\frac{1}{2}\times 2\times t^2\mathbf{j}$ in correct notation |
| **Or** using integration: $\mathbf{v}=\int\mathbf{a}\,dt$ | M1 | Attempt at integration. Condone no $+c$. Must be vectors |
| $\mathbf{v}=4\mathbf{i}-2t\mathbf{j}$ | A1 | cao |
| $\mathbf{r}=\int\mathbf{v}\,dt$ | M1 | Integrate their $\mathbf{v}$, must contain 2 components. Must be vectors |
| $+3\mathbf{j}$ | B1 | May be seen on either side of a meaningful equation for $\mathbf{r}$ |
| $\mathbf{r}=4t\mathbf{i}+(3-t^2)\mathbf{j}$ | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v}(2.5)=4\mathbf{i}-5\mathbf{j}$ | B1 | FT their $\mathbf{v}$ |
| Angle is $(90+)\arctan\dfrac{5}{4}$ | M1 | Award for arctan attempted. FT their values. Allow argument $\pm(\text{i cpt})/(\text{j cpt})$ or $\pm(\text{j cpt})/(\text{i cpt})$ |
| $=141.34019\ldots$ so $141°$ (3 s.f.) | A1 | cao |
---4 The three forces $\left. \begin{array} { r } - 1 \\ 14 \\ - 8 \end{array} \right) \mathrm { N } , \left( \begin{array} { r } 3 \\ - 9 \\ 10 \end{array} \right) \mathrm { N }$ and $\mathbf { F } \mathrm { N }$ act on a body of mass 4 kg in deep space and give it an acceleration of $\left. \quad \begin{array} { r } - 1 \\ 2 \\ 4 \end{array} \right) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(i) Calculate $\mathbf { F }$.
At one instant the velocity of the body is $\left. \begin{array} { r } - 3 \\ 3 \\ 6 \end{array} \right) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(ii) Calculate the velocity and also the speed of the body 3 seconds later.
\hfill \mbox{\textit{OCR MEI M1 Q4 [8]}}