| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: speed of projection |
| Difficulty | Moderate -0.8 This is a straightforward two-part SUVAT question requiring direct application of standard kinematic equations with gravity. Students need to use v² = u² + 2as to find u, then v = u + at to find T. The calculations are routine with no conceptual challenges or problem-solving insight required, making it easier than average but not trivial since it involves two connected parts and careful sign conventions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v^2 = u^2 + 2as \Rightarrow 17.5^2 = u^2 + 2 \times 9.8 \times 10\) leading to \(u = 10.5\) | M1 A1 | (3) |
| (b) \(v = u + at \Rightarrow 17.5 = -10.5 + 9.8T\) giving \(T = 2\frac{6}{7} \text{ (s)}\) | M1 A1 f.t. DM1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| - \(s = \left(\frac{u+v}{2}\right)T \Rightarrow 10 = \left(\frac{17.5 + (-10.5)}{2}\right)T\) giving \(\frac{20}{7} = T\) | M1A1 f.t. DM1A1 | (4) |
| - \(s = ut + \frac{1}{2}at^2 \Rightarrow -10 = 10.5t - 4.9t^2\) leading to \(T = 2\frac{6}{7} \cdot \frac{5}{7}\) (rejecting negative) | M1 A1 f.t. DM1 A1 | (4) |
| - \(s = vt - \frac{1}{2}at^2 \Rightarrow -10 = -17.5t + 4.9t^2\) leading to \(T = 2\frac{6}{7} \cdot \frac{5}{7}\) | M1 A1 DM1 A1 | (4) |
| (b) can be done independently of (a) | M1 A1 | A1 (4) [7] |
**(a)** $v^2 = u^2 + 2as \Rightarrow 17.5^2 = u^2 + 2 \times 9.8 \times 10$ leading to $u = 10.5$ | M1 A1 | (3)
**(b)** $v = u + at \Rightarrow 17.5 = -10.5 + 9.8T$ giving $T = 2\frac{6}{7} \text{ (s)}$ | M1 A1 f.t. DM1 A1 | (4)
**Alternatives for (b):**
- $s = \left(\frac{u+v}{2}\right)T \Rightarrow 10 = \left(\frac{17.5 + (-10.5)}{2}\right)T$ giving $\frac{20}{7} = T$ | M1A1 f.t. DM1A1 | (4)
- $s = ut + \frac{1}{2}at^2 \Rightarrow -10 = 10.5t - 4.9t^2$ leading to $T = 2\frac{6}{7} \cdot \frac{5}{7}$ (rejecting negative) | M1 A1 f.t. DM1 A1 | (4)
- $s = vt - \frac{1}{2}at^2 \Rightarrow -10 = -17.5t + 4.9t^2$ leading to $T = 2\frac{6}{7} \cdot \frac{5}{7}$ | M1 A1 DM1 A1 | (4)
**(b) can be done independently of (a)** | M1 A1 | A1 (4) [7]
---2. At time $t = 0$, a particle is projected vertically upwards with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 10 m above the ground. At time $T$ seconds, the particle hits the ground with speed $17.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find
\begin{enumerate}[label=(\alph*)]
\item the value of $u$,
\item the value of $T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q2 [7]}}