| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion with resistance force |
| Difficulty | Standard +0.3 This is a straightforward two-part mechanics question requiring standard SUVAT application in part (a) and basic Newton's second law with constant resistance in part (b). Both parts involve routine procedures with clear setups and no conceptual surprises, making it slightly easier than average for M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.5 = 0 + \frac{1}{2}gt^2\) | M1 | Complete method to find the time taken using \(a = g\) |
| Correct unsimplified equation in \(t\) only | A1 | Correct unsimplified equation in \(t\) only |
| \(t = 0.55\) or \(0.553\) (s) | A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.5 = 0 + \frac{1}{2}a(0.6)^2\) | M1 | Complete method to form an equation in \(a\) only, \(a \neq g\), using \(t = 0.6\) |
| Correct unsimplified equation in \(a\) only | A1 | Correct unsimplified equation in \(a\) only |
| \(0.2g - R = 0.2a\) | M1 | Use \(F = ma\) to form an equation of motion with correct terms, condone sign errors, \(a \neq g\) |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| \(R = 0.293\), \(0.29\) | A1 | Cao. N.B. Penalise use of \(g = 9.81\) once for whole question. Also penalise answers as fractions once (penalise first if both are fractions) |
# Question 3:
## Part 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.5 = 0 + \frac{1}{2}gt^2$ | M1 | Complete method to find the time taken using $a = g$ |
| Correct unsimplified equation in $t$ only | A1 | Correct unsimplified equation in $t$ only |
| $t = 0.55$ or $0.553$ (s) | A1 | Cao |
## Part 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.5 = 0 + \frac{1}{2}a(0.6)^2$ | M1 | Complete method to form an equation in $a$ only, $a \neq g$, using $t = 0.6$ |
| Correct unsimplified equation in $a$ only | A1 | Correct unsimplified equation in $a$ only |
| $0.2g - R = 0.2a$ | M1 | Use $F = ma$ to form an equation of motion with correct terms, condone sign errors, $a \neq g$ |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| $R = 0.293$, $0.29$ | A1 | Cao. **N.B.** Penalise use of $g = 9.81$ once for whole question. Also penalise answers as fractions once (penalise first if both are fractions) |
---\begin{enumerate}
\item Two students observe a book of mass 0.2 kg fall vertically from rest from a shelf that is 1.5 m above the floor.
\end{enumerate}
Student $A$ suggests that the book is modelled as a particle falling freely under gravity.\\
(a) Use student $A$ 's model to find the time taken for the book to reach the floor.
Student $B$ suggests an improved model where the book is modelled as a particle experiencing a constant resistance to motion of magnitude $R$ newtons.
Given that the time taken for the book to reach the floor is 0.6 seconds,\\
(b) use student $B$ 's model to find the value of $R$
\hfill \mbox{\textit{Edexcel M1 2023 Q3 [8]}}