| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Standard +0.8 This is a multi-part vector projectile question requiring: (a) finding initial velocity components using constant acceleration equations, (b) solving a quadratic inequality for when speed ≤10, and (c) using perpendicular velocity vectors (dot product = 0) to find height difference. Part (b) requires careful algebraic manipulation and part (c) needs geometric insight about perpendicular velocities, making this moderately challenging but still within standard M2 scope. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal component: \(p = 8\) | B1 | |
| Vertical component: \(-12 = q - 3g\) | M1 | Complete method to find q using suvat. Condone sign errors. |
| \(q = 17.4\) | A1 | 17 or better |
| Speed \(= \sqrt{8^2 + 17.4^2}\) | M1 | Use of Pythagoras to find speed using their velocity. Independent M mark |
| \(= 19.2\) (19) (ms\(^{-1}\)) | A1 | 3 sf or 2 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Use of Pythagoras to find vertical component | M1 | |
| vertical component \(= \pm 6\) | A1 | Seen or implied. Accept without \(\pm\) |
| \(-6 = 6 - 9.8T\) | DM1 | Complete method using suvat to find required time. Dependent on the previous M1 |
| \(T = 1.22\) (1.2) | A1 | 3 sf or 2 sf. Not \(\frac{60}{49}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use suvat and Pythagoras to form an equation in t | M1 | Or an inequality |
| \(8^2 + (17.4 - gt)^2 = 100\) | A1 | Correct unsimplified equation for t. Accept inequality |
| Solve for T | DM1 | Complete method to obtain T. Dependent on the previous M1 |
| \(T = 1.22\) (1.2) | A1 | 3 sf or 2 sf. Not \(\frac{60}{49}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Velocity perpendicular \(\Rightarrow\) vertical component \(= \frac{2}{3} \times 8\) | M1 | Complete method to find vertical component of velocity at B |
| \(= \frac{16}{3}\) | A1 | |
| \(\left(\frac{16}{3}\right)^2 - (-12)^2 = -2g(-h)\) | DM1 | Complete method to find the required vertical distance using their vertical component of the velocity. Dependent on the previous M1 |
| \(h = 5.90\) (5.9) (m) | A1 | Max 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{8}{17.4 - gt} \cdot \frac{8}{-12} = 0\) and time \(= 3 - t\) | M1 | Complete method to find the time from B to A |
| Time \(= 3 - 1.23\ldots = 1.768\ldots\) | A1 | |
| \(s = vt - \frac{1}{2}gt^2 = 12t - 4.9t^2\) | DM1 | Complete method to find the required vertical distance using their time. Dependent on the previous M1 |
| \(s = 5.9\) (m) | A1 | Max 3 sf |
## 8a
Horizontal component: $p = 8$ | B1 |
Vertical component: $-12 = q - 3g$ | M1 | Complete method to find q using suvat. Condone sign errors.
$q = 17.4$ | A1 | 17 or better
Speed $= \sqrt{8^2 + 17.4^2}$ | M1 | Use of Pythagoras to find speed using their velocity. Independent M mark
$= 19.2$ (19) (ms$^{-1}$) | A1 | 3 sf or 2 sf
(5)
## 8b
Use of Pythagoras to find vertical component | M1 |
vertical component $= \pm 6$ | A1 | Seen or implied. Accept without $\pm$
$-6 = 6 - 9.8T$ | DM1 | Complete method using suvat to find required time. Dependent on the previous M1
$T = 1.22$ (1.2) | A1 | 3 sf or 2 sf. Not $\frac{60}{49}$
(4)
## 8b alt
Use suvat and Pythagoras to form an equation in t | M1 | Or an inequality
$8^2 + (17.4 - gt)^2 = 100$ | A1 | Correct unsimplified equation for t. Accept inequality
Solve for T | DM1 | Complete method to obtain T. Dependent on the previous M1
$T = 1.22$ (1.2) | A1 | 3 sf or 2 sf. Not $\frac{60}{49}$
(4)
## 8c
Velocity perpendicular $\Rightarrow$ vertical component $= \frac{2}{3} \times 8$ | M1 | Complete method to find vertical component of velocity at B
$= \frac{16}{3}$ | A1 |
$\left(\frac{16}{3}\right)^2 - (-12)^2 = -2g(-h)$ | DM1 | Complete method to find the required vertical distance using their vertical component of the velocity. Dependent on the previous M1
$h = 5.90$ (5.9) (m) | A1 | Max 3 sf
(4)
## 8c alt
$\frac{8}{17.4 - gt} \cdot \frac{8}{-12} = 0$ and time $= 3 - t$ | M1 | Complete method to find the time from B to A
Time $= 3 - 1.23\ldots = 1.768\ldots$ | A1 |
$s = vt - \frac{1}{2}gt^2 = 12t - 4.9t^2$ | DM1 | Complete method to find the required vertical distance using their time. Dependent on the previous M1
$s = 5.9$ (m) | A1 | Max 3 sf
[13]8. [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in a vertical plane, with $\mathbf { i }$ being horizontal and $\mathbf { j }$ being vertically upwards.]
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1dea68fe-7916-41ed-894e-6b48f8d989bb-28_426_1145_347_338}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
At time $t = 0$, a small ball is projected from a fixed point $O$ on horizontal ground. The ball is projected from $O$ with velocity ( $p \mathbf { i } + q \mathbf { j }$ ) $\mathrm { ms } ^ { - 1 }$, where $p$ and $q$ are positive constants. The ball moves freely under gravity.
At time $t = 3$ seconds, the ball passes through the point $A$ with velocity ( $8 \mathbf { i } - 12 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of the ball at the instant it is projected from $O$.
For an interval of $T$ seconds the speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of the ball is such that $v \leqslant 10$
\item Find the value of $T$.
At the point $B$ on the path of the ball, the direction of motion of the ball is perpendicular to the direction of motion of the ball at $A$.
\item Find the vertical height of $B$ above $A$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2020 Q8 [13]}}