| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time to ground |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT mechanics question requiring standard application of kinematic equations with constant acceleration. Parts (a) and (b) use basic equations with given values, part (c) requires finding times when the particle is at a specific height (standard two-solution problem), and part (d) is a routine sketch. All techniques are textbook exercises with no novel problem-solving required. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(A\) to \(B\): \(V^2 = 24^2 + 2(-g)(-2.5)\) | M1A1 | OR: \(0 = 24^2 - 2gh\) and \(V^2 = 2g(h+2.5)\) oe |
| \(V = 25\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Equation in \(t\): e.g. \(25 = -24 + gt\), or \(2.5 = \frac{(25+(-24))t}{2}\), or \(2.5 = -24t + \frac{1}{2}gt^2\), or \(2.5 = 25t - \frac{1}{2}gt^2\) | M1A1 | Or find \(t_{UP} = \frac{24}{g}\) and \(t_{DOWN} = \frac{25}{g}\) and add |
| \(t = 5\) (s) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| From \(A\) to \(C\): \(10 = 24t + \frac{1}{2}(-g)t^2\) | M1A1 | |
| Complete method to find required time e.g. solving quadratic, finding positive difference in roots | M1 | Allow if answer given as range \(t_1, t, t_2\) |
| \(4\), \(4.0\) or \(3.98\) (s) | A1 | |
| ALT 1: \(W^2 = 24^2 - 2\times10g\), then \(0 = W - g\left(\frac{1}{2}t\right)\) | M1A1, M1 | |
| \(4.0\) or \(3.98\) (s) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape: symmetrical with respect to gradients, end point higher than start point, vertex on horizontal axis | B1 | Shape must appear symmetrical regarding gradients |
| Correct labels (24, 25 and 5) following through on answers to (a) and (b), provided they are positive. Ignore incorrect time when \(v=0\) | B1ft | Neither mark available if using a velocity-time graph |
## Question 6:
**Part (a):**
| Working | Mark | Guidance |
|---------|------|----------|
| $A$ to $B$: $V^2 = 24^2 + 2(-g)(-2.5)$ | M1A1 | OR: $0 = 24^2 - 2gh$ and $V^2 = 2g(h+2.5)$ oe |
| $V = 25$ | A1 | |
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| Equation in $t$: e.g. $25 = -24 + gt$, or $2.5 = \frac{(25+(-24))t}{2}$, or $2.5 = -24t + \frac{1}{2}gt^2$, or $2.5 = 25t - \frac{1}{2}gt^2$ | M1A1 | Or find $t_{UP} = \frac{24}{g}$ and $t_{DOWN} = \frac{25}{g}$ and add |
| $t = 5$ (s) | A1 | |
**Part (c):**
| Working | Mark | Guidance |
|---------|------|----------|
| From $A$ to $C$: $10 = 24t + \frac{1}{2}(-g)t^2$ | M1A1 | |
| Complete method to find required time e.g. solving quadratic, finding positive difference in roots | M1 | Allow if answer given as range $t_1, t, t_2$ |
| $4$, $4.0$ or $3.98$ (s) | A1 | |
| **ALT 1:** $W^2 = 24^2 - 2\times10g$, then $0 = W - g\left(\frac{1}{2}t\right)$ | M1A1, M1 | |
| $4.0$ or $3.98$ (s) | A1 | |
# Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape: symmetrical with respect to gradients, end point higher than start point, vertex on horizontal axis | B1 | Shape must appear symmetrical regarding gradients |
| Correct labels (24, 25 and 5) following through on answers to (a) and (b), provided they are positive. Ignore incorrect time when $v=0$ | B1ft | Neither mark available if using a velocity-time graph |
---\begin{enumerate}
\item A particle is projected vertically upwards from a point $A$ with speed $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\end{enumerate}
The point $A$ is 2.5 m vertically above the point $B$.\\
Point $B$ lies on horizontal ground.\\
The particle moves freely under gravity until it hits the ground at $B$ with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ After hitting the ground the particle does not rebound.\\
(a) Find the value of $V$.\\
(b) Find the time taken for the particle to reach $B$.
The point $C$ is 10 m vertically above $A$.\\
(c) Find the length of time for which the particle is above $C$.\\
(d) Sketch a speed-time graph for the motion of the particle from projection to the instant that it reaches $B$. (No further calculations are required.)
\hfill \mbox{\textit{Edexcel M1 2024 Q6 [12]}}