| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a standard two-part kinematics question using constant acceleration equations (SUVAT). Part (a) requires finding maximum height using v²=u²+2as, while part (b) involves finding times when the ball passes 7.5m height using s=ut+½at². Both parts are routine applications of well-practiced techniques with no conceptual challenges, making it slightly easier than average but still requiring careful calculation across multiple steps. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(u = 21, v = 0 \text{ (at max. ht.)}, a = -g \text{ use } v^2 = u^2 + 2as\) | M1 | |
| \(0 = 21^2 - 2gs \therefore s = 22.5 \text{ m}\) | M1 A1 | |
| \(\text{ball starts from } 1.9 \text{ m, so it reaches } 24.4 \text{ m above ground level}\) | A1 | |
| (b) \(s = 7.5 - 1.9 = 5.6, u = 21, a = -g, \text{ use } s = ut + \frac{1}{2}at^2\) | M1 | |
| \(5.6 = 21t - 4.9t^2 \text{ i.e. } 7t^2 - 30t + 8 = 0\) | M1 A1 | |
| \((7t - 2)(t - 4) = 0 \text{ giving } t = \frac{2}{7}, t = 4\) | M1 A1 | |
| \(\therefore \text{Barbara waits for } 3\frac{5}{7} (\approx 3.71) \text{ seconds}\) | A1 | (10 marks) |
(a) $u = 21, v = 0 \text{ (at max. ht.)}, a = -g \text{ use } v^2 = u^2 + 2as$ | M1 |
$0 = 21^2 - 2gs \therefore s = 22.5 \text{ m}$ | M1 A1 |
$\text{ball starts from } 1.9 \text{ m, so it reaches } 24.4 \text{ m above ground level}$ | A1 |
(b) $s = 7.5 - 1.9 = 5.6, u = 21, a = -g, \text{ use } s = ut + \frac{1}{2}at^2$ | M1 |
$5.6 = 21t - 4.9t^2 \text{ i.e. } 7t^2 - 30t + 8 = 0$ | M1 A1 |
$(7t - 2)(t - 4) = 0 \text{ giving } t = \frac{2}{7}, t = 4$ | M1 A1 |
$\therefore \text{Barbara waits for } 3\frac{5}{7} (\approx 3.71) \text{ seconds}$ | A1 | (10 marks)Andrew hits a tennis ball vertically upwards towards his sister Barbara who is leaning out of a window 7.5 m above the ground to try to catch it. When the ball leaves Andrew's racket, it is 1.9 m above the ground and travelling at $21 \text{ m s}^{-1}$. Barbara fails to catch the ball on its way up but succeeds as the ball comes back down.
Modelling the ball as a particle and assuming that air resistance can be neglected,
\begin{enumerate}[label=(\alph*)]
\item find the maximum height above the ground which the ball reaches. [4 marks]
\item find how long Barbara has to wait from the moment that the ball first passes her until she catches it. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q4 [10]}}