| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Moderate -0.8 This is a straightforward vector projectile question requiring only direct reading of coefficients from the given equation and basic substitution. Parts (i) and (ii) involve minimal calculation, while part (iii) is a standard elimination of parameter exercise. All techniques are routine for M1 level with no problem-solving insight required. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.10d Vector operations: addition and scalar multiplication3.02e Two-dimensional constant acceleration: with vectors3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| A) Height \(5\) m | B1 | No units required; apply ISW if incorrect units given |
| B) \(g\) has been taken to be \(10\) m s\(^{-2}\) | B1 | Allow \(+10\) or \(-10\). No units required; apply ISW if incorrect units given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Displacement is \(\begin{pmatrix}150\\80\end{pmatrix} - \begin{pmatrix}90\\80\end{pmatrix}\) | M1 | Displacement must be given as a vector. Allow a description of a vector in words. Attempts at substitution for \(t\) and subtraction of vectors must be seen |
| \(= \begin{pmatrix}60\\0\end{pmatrix}\) | A1 | Cao. If the candidate then goes on to give a non-vector answer of "60 m", apply ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 30t\) | B1 | |
| \(y = 5 + 40t - 5t^2\) | B1 | |
| \(y = 5 + 40\times\left(\dfrac{x}{30}\right) - 5\times\left(\dfrac{x}{30}\right)^2\) | M1 | Attempt to eliminate \(t\) |
| \(y = 5 + \dfrac{4}{3}x - \dfrac{x^2}{180}\) | A1 | No errors |
## Question 7:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| A) Height $5$ m | B1 | No units required; apply ISW if incorrect units given |
| B) $g$ has been taken to be $10$ m s$^{-2}$ | B1 | Allow $+10$ or $-10$. No units required; apply ISW if incorrect units given |
**[2]**
---
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Displacement is $\begin{pmatrix}150\\80\end{pmatrix} - \begin{pmatrix}90\\80\end{pmatrix}$ | M1 | Displacement must be given as a vector. Allow a description of a vector in words. Attempts at substitution for $t$ and subtraction of vectors must be seen |
| $= \begin{pmatrix}60\\0\end{pmatrix}$ | A1 | Cao. If the candidate then goes on to give a non-vector answer of "60 m", apply ISW |
**[2]**
---
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 30t$ | B1 | |
| $y = 5 + 40t - 5t^2$ | B1 | |
| $y = 5 + 40\times\left(\dfrac{x}{30}\right) - 5\times\left(\dfrac{x}{30}\right)^2$ | M1 | Attempt to eliminate $t$ |
| $y = 5 + \dfrac{4}{3}x - \dfrac{x^2}{180}$ | A1 | No errors |
**[4]**7 A projectile P travels in a vertical plane over level ground. Its position vector $\mathbf { r }$ at time $t$ seconds after projection is modelled by
$$\mathbf { r } = \binom { x } { y } = \binom { 0 } { 5 } + \binom { 30 } { 40 } t - \binom { 0 } { 5 } t ^ { 2 }$$
where distances are in metres and the origin is a point on the level ground.
\begin{enumerate}[label=(\roman*)]
\item Write down\\
(A) the height from which P is projected,\\
(B) the value of $g$ in this model.
\item Find the displacement of P from $t = 3$ to $t = 5$.
\item Show that the equation of the trajectory is
$$y = 5 + \frac { 4 } { 3 } x - \frac { x ^ { 2 } } { 180 }$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q7 [8]}}