| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time to ground |
| Difficulty | Moderate -0.3 This is a straightforward SUVAT problem requiring standard application of kinematic equations with gravity. Part (a) involves finding initial velocity from the upward journey, then calculating total time using displacement to ground. Part (b) is a routine sketch of a V-shaped speed-time graph. The multi-step nature and two parts elevate it slightly above pure recall, but it remains a standard textbook exercise 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0^2 = u^2 - 2 \times g \times 19.6\) | M1 A1 | Attempt at relevant suvat using \(s = 19.6\); correct no. of terms, condone sign errors |
| \(-24.5 = uT - \frac{1}{2}gT^2\) | M1 A1 | Attempt at another relevant suvat using 24.5 or 44.1; correct no. of terms |
| Produce equation in \(T\) only and solve for \(T\) | DM1 | Dependent on both M marks; equation must have solutions |
| \(T = 5\) | A1 | N.B. If \(g = 9.8\) not used, A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| V-shape starting on speed axis, with point on \(t\)-axis | B1 | Shape only |
| Second line longer than first, approximately equal angles, \(T\) or their answer for \(T\) marked | DB1 | Dependent on first B1; B0 if clearly unequal angles; N.B. if graph reflected, B0 DB0 |
## Question 4:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0^2 = u^2 - 2 \times g \times 19.6$ | M1 A1 | Attempt at relevant suvat using $s = 19.6$; correct no. of terms, condone sign errors |
| $-24.5 = uT - \frac{1}{2}gT^2$ | M1 A1 | Attempt at another relevant suvat using 24.5 or 44.1; correct no. of terms |
| Produce equation in $T$ only and solve for $T$ | DM1 | Dependent on both M marks; equation must have solutions |
| $T = 5$ | A1 | N.B. If $g = 9.8$ not used, A0 |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| V-shape starting on speed axis, with point on $t$-axis | B1 | Shape only |
| Second line longer than first, approximately equal angles, $T$ or their answer for $T$ marked | DB1 | Dependent on first B1; B0 if clearly unequal angles; N.B. if graph reflected, B0 DB0 |
---4. At time $t = 0$, a small ball is projected vertically upwards from a point $A$ which is 24.5 m above the ground. The ball first comes to instantaneous rest at the point $B$, where $A B = 19.6 \mathrm {~m}$ and first hits the ground at time $t = T$ seconds.
The ball is modelled as a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $T$.
\item Sketch a speed-time graph for the motion of the ball from $t = 0$ to $t = T$ seconds.\\
(No further calculations are needed in order to draw this sketch.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q4 [8]}}