| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time to ground |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem requiring standard application of kinematic equations with gravity. Part (a) uses s = ut + ½at² directly. Part (b) requires using v² = u² + 2as twice and solving simultaneous equations, but follows a predictable pattern for M1. Slightly above average due to the two-part structure and algebraic manipulation in part (b), but still routine mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(h = -20 \times 5 + \frac{1}{2} \times 9.8 \times 25\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) to find \(h\). Must quote correct formula using 20 & 5, condone slips in substitution. Accept complete alternative solutions via maximum height (max ht 20.4..., time to top 2.04...). Accept complete alternative methods using other suvat equations. |
| (correctly substituted equation) | A1 | Correctly substituted equation(s). Condone use of premature approximation. |
| \(h = 22.5\) | A1 (3) | Final answer. Accept 22.5 or 23. Maximum 3 s.f. \(-22.5\) is A0. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(V^2 = 20^2 + 2 \times 9.8 \times 22.5\) OR \(V = -20 + (5 \times 9.8)\) | M1 | First ball – use of suvat to find \(V\) or \(V^2\). Follow their \(h\). |
| \(V^2 = 841\) or \(V = 29\) | A1 | Correct only (condone \(-29\)) |
| \(\left(\frac{3}{4}V\right)^2 = w^2 + 2 \times 9.8 \times 22.5\) | M1 | Second ball – suvat equation in \(V\) (or their \(V\)) to find \(w\). Must be using the \(\frac{3}{4}\). |
| \(w^2 = \frac{9}{16} \times 841 - 2 \times 9.8 \times 22.5\) | A1ft | Correctly substituted equation with their \(V\) and their \(h\). |
| \(w = 5.66\) | A1 (5) | or 5.7. Answer correct to 2 s.f. or to 3 s.f. |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $h = -20 \times 5 + \frac{1}{2} \times 9.8 \times 25$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ to find $h$. Must quote correct formula using 20 & 5, condone slips in substitution. Accept complete alternative solutions via maximum height (max ht 20.4..., time to top 2.04...). Accept complete alternative methods using other suvat equations. |
| (correctly substituted equation) | A1 | Correctly substituted equation(s). Condone use of premature approximation. |
| $h = 22.5$ | A1 (3) | Final answer. Accept 22.5 or 23. Maximum 3 s.f. $-22.5$ is A0. |
> **Note:** Do not ignore subsequent working if they reach 22.5 and then move on to do further work.
## Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $V^2 = 20^2 + 2 \times 9.8 \times 22.5$ OR $V = -20 + (5 \times 9.8)$ | M1 | First ball – use of suvat to find $V$ or $V^2$. Follow their $h$. |
| $V^2 = 841$ or $V = 29$ | A1 | Correct only (condone $-29$) |
| $\left(\frac{3}{4}V\right)^2 = w^2 + 2 \times 9.8 \times 22.5$ | M1 | Second ball – suvat equation in $V$ (or their $V$) to find $w$. Must be using the $\frac{3}{4}$. |
| $w^2 = \frac{9}{16} \times 841 - 2 \times 9.8 \times 22.5$ | A1ft | Correctly substituted equation with their $V$ and their $h$. |
| $w = 5.66$ | A1 (5) | or 5.7. Answer correct to 2 s.f. or to 3 s.f. |
**Total: [8]**
---2. A ball is thrown vertically upwards with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A$, which is $h$ metres above the ground. The ball moves freely under gravity until it hits the ground 5 s later.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $h$.
A second ball is thrown vertically downwards with speed $w \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from $A$ and moves freely under gravity until it hits the ground.
The first ball hits the ground with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the second ball hits the ground with speed $\frac { 3 } { 4 } V \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the value of $w$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q2 [8]}}