| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: different start times, same height |
| Difficulty | Standard +0.3 This is a standard M1 SUVAT problem requiring systematic application of kinematic equations. Part (a) uses v²=u²+2as directly, part (b) requires setting up simultaneous equations for two particles' positions, and part (c) substitutes back to find height. While it involves multiple steps and careful bookkeeping of two particles, it follows a well-practiced template with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0^2 = U^2 - 2gH\) | M1 | Complete method to find equation in \(H\), \(U\) and \(g\) only. Condone sign errors |
| \(H = \frac{U^2}{2g}\) | A1 (2) | Correct expression for \(H\) in terms of \(U\) and \(g\). (A0 if they use \(h\) or \(s\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(s_P = \frac{1}{2}gt^2\) OR \(s_P = Ut - \frac{1}{2}gt^2\) | M1A1 | Complete method to find \(s_P\) in terms of \(t\), where \(t=0\) when \(Q\) is projected upwards |
| \(s_Q = \frac{1}{2}Ut - \frac{1}{2}gt^2\) OR \(s_Q = \frac{1}{2}U\!\left(t-\frac{U}{g}\right) - \frac{1}{2}g\!\left(t-\frac{U}{g}\right)^2\) | M1A1 | Complete method to find \(s_Q\) in terms of \(t\). Condone sign errors |
| \(s_P + s_Q = H \Rightarrow \frac{1}{2}Ut = \frac{U^2}{2g}\) OR \(s_P = s_Q \Rightarrow Ut - \frac{1}{2}gt^2 = \frac{1}{2}U\!\left(t-\frac{U}{g}\right) - \frac{1}{2}g\!\left(t-\frac{U}{g}\right)^2\) | M1 | Use of \(s_P + s_Q = H\) oe OR \(s_P = s_Q\) oe to obtain correct equation in \(t\), \(U\) and \(g\) only |
| \(t = \frac{U}{g}\) | A1 (6) | Correct expression for \(t\) in terms of \(U\) and \(g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(s_P = \frac{1}{2}g\!\left(\frac{U}{g}\right)^2\) OR \(s_P = U\!\left(\frac{2U}{g}\right) - \frac{1}{2}g\!\left(\frac{2U}{g}\right)^2\) and \(s_Q = \frac{1}{2}U\!\left(\frac{U}{g}\right) - \frac{1}{2}g\!\left(\frac{U}{g}\right)^2\) | M1 A1 | M1 sub their \(t\) value, provided it's POSITIVE, into their equation for \(s_P\) or \(s_Q\). A1 correct unsimplified expression |
| Collide at the point \(O\) or at the point of projection (at the same level as \(O\)) | A1 (3) | Correct conclusion. (At the same level as \(O\) is A0) |
## Question 3:
### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0^2 = U^2 - 2gH$ | M1 | Complete method to find equation in $H$, $U$ and $g$ only. Condone sign errors |
| $H = \frac{U^2}{2g}$ | A1 (2) | Correct expression for $H$ in terms of $U$ and $g$. (A0 if they use $h$ or $s$) |
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $s_P = \frac{1}{2}gt^2$ **OR** $s_P = Ut - \frac{1}{2}gt^2$ | M1A1 | Complete method to find $s_P$ in terms of $t$, where $t=0$ when $Q$ is projected upwards |
| $s_Q = \frac{1}{2}Ut - \frac{1}{2}gt^2$ **OR** $s_Q = \frac{1}{2}U\!\left(t-\frac{U}{g}\right) - \frac{1}{2}g\!\left(t-\frac{U}{g}\right)^2$ | M1A1 | Complete method to find $s_Q$ in terms of $t$. Condone sign errors |
| $s_P + s_Q = H \Rightarrow \frac{1}{2}Ut = \frac{U^2}{2g}$ **OR** $s_P = s_Q \Rightarrow Ut - \frac{1}{2}gt^2 = \frac{1}{2}U\!\left(t-\frac{U}{g}\right) - \frac{1}{2}g\!\left(t-\frac{U}{g}\right)^2$ | M1 | Use of $s_P + s_Q = H$ oe **OR** $s_P = s_Q$ oe to obtain correct equation in $t$, $U$ and $g$ only |
| $t = \frac{U}{g}$ | A1 (6) | Correct expression for $t$ in terms of $U$ and $g$ |
**N.B.** Must use EITHER the LH column OR the RH column, not a mixture.
### Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $s_P = \frac{1}{2}g\!\left(\frac{U}{g}\right)^2$ **OR** $s_P = U\!\left(\frac{2U}{g}\right) - \frac{1}{2}g\!\left(\frac{2U}{g}\right)^2$ and $s_Q = \frac{1}{2}U\!\left(\frac{U}{g}\right) - \frac{1}{2}g\!\left(\frac{U}{g}\right)^2$ | M1 A1 | M1 sub their $t$ value, provided it's POSITIVE, into their equation for $s_P$ or $s_Q$. A1 correct unsimplified expression |
| Collide at the point $O$ or at the point of projection (at the same level as $O$) | A1 (3) | Correct conclusion. (At the same level as $O$ is A0) |
---3. A particle, $P$, is projected vertically upwards with speed $U$ from a fixed point $O$. At the instant when $P$ reaches its greatest height $H$ above $O$, a second particle, $Q$, is projected with speed $\frac { 1 } { 2 } U$ vertically upwards from $O$.
\begin{enumerate}[label=(\alph*)]
\item Find $H$ in terms of $U$ and $g$.
\item Find, in terms of $U$ and $g$, the time between the instant when $Q$ is projected and the instant when the two particles collide.
\item Find where the two particles collide.
DO NOT WRITEIN THIS AREA\\
\includegraphics[max width=\textwidth, alt={}, center]{916543cb-14f7-486c-ba3c-eda9be134045-08_2666_99_107_1957}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q3 [11]}}