| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.3 This is a standard M1 SUVAT question with straightforward application of kinematic equations and Newton's second law. Part (a) is a 'show that' using v²=u²+2as at maximum height, part (b) applies s=ut+½at², and part (c) uses work-energy or forces during impact. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-stage nature and marks allocation. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02a Work done: concept and definition6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = u^2 + 2as\): \(0 = u^2 - 2 \times 9.8 \times 25.6\) | M1 A1 | |
| \(u^2 = 501.76 \Rightarrow u = 22.4\) (★) | A1 so | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1.5 = 22.4T - 4.9T^2\) | M1 A1 | |
| \(4.9T^2 - 22.4T - 1.5 = 0\) | ||
| \(T = \frac{22.4 \pm \sqrt{22.4^2 + 4 \times 1. \times 4.9}}{9.8}\) | M1 | |
| \(= 4.64\) s | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Speed at ground \(v = 22.4 - 9.8 \times 4.64\) | M1 | |
| \(v = -23.07\) | A1 | |
| (or \(v^2 = 22.4^2 + 2 \times 9.8 \times 1.5\), \(v = 23.05\)) | ||
| \(v^2 = u^2 + 2as\): \(0 = 23.07^2 + 2 \times a \times 0.025\) | M1 A1 ft | |
| (\(\rightarrow a = -10644.5\)) | ||
| \(F - 0.6g = 0.6a\) | M1 | |
| \(F = 6390\) N (3 sf) | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Air resistance; variable \(F\); | B1 | (1 mark) |
## Part (a)
$v^2 = u^2 + 2as$: $0 = u^2 - 2 \times 9.8 \times 25.6$ | M1 A1 |
$u^2 = 501.76 \Rightarrow u = 22.4$ (★) | A1 so | (3 marks)
## Part (b)
$-1.5 = 22.4T - 4.9T^2$ | M1 A1 |
$4.9T^2 - 22.4T - 1.5 = 0$ | |
$T = \frac{22.4 \pm \sqrt{22.4^2 + 4 \times 1. \times 4.9}}{9.8}$ | M1 |
$= 4.64$ s | A1 | (4 marks)
## Part (c)
Speed at ground $v = 22.4 - 9.8 \times 4.64$ | M1 |
$v = -23.07$ | A1 |
(or $v^2 = 22.4^2 + 2 \times 9.8 \times 1.5$, $v = 23.05$) | |
$v^2 = u^2 + 2as$: $0 = 23.07^2 + 2 \times a \times 0.025$ | M1 A1 ft |
($\rightarrow a = -10644.5$) | |
$F - 0.6g = 0.6a$ | M1 |
$F = 6390$ N (3 sf) | A1 | (6 marks)
## Part (d)
Air resistance; variable $F$; | B1 | (1 mark)
**Total: 14 marks**
---A ball is projected vertically upwards with a speed $u$ m s$^{-1}$ from a point $A$ which is 1.5 m above the ground. The ball moves freely under gravity until it reaches the ground. The greatest height attained by the ball is 25.6 m above $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that $u = 22.4$. [3]
\end{enumerate}
The ball reaches the ground 7 seconds after it has been projected from $A$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, to 2 decimal places, the value of $T$. [4]
\end{enumerate}
The ground is soft and the ball sinks 2.5 cm into the ground before coming to rest. The mass of the ball is 0.6 kg. The ground is assumed to exert a constant resistive force of magnitude $F$ newtons.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, to 3 significant figures, the value of $F$. [6]
\item State one physical factor which could be taken into account to make the model used in this question more realistic. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q7 [14]}}