| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward projectile motion question requiring standard SUVAT equations with vertical/horizontal components. Part (a) is recall, parts (b) and (c) involve routine calculations (time to max height using v=u+at, then using s=ut+½at², and finally horizontal distance). The 'show that' in (b)(ii) provides the answer, reducing difficulty. No novel problem-solving or geometric insight required—purely algorithmic application of memorized formulas. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Ball is a particle | B1 | One appropriate assumption |
| No air resistance | B1 (2) | Second appropriate assumption |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = 12\sin40° - 9.8t\) | M1, A1 | Equation to find time at maximum height; Correct equation |
| \(t = \frac{12\sin40°}{9.8} = 0.787 \text{ s}\) | M1, A1 (4) | Solving for \(t\); Correct time |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 12\sin40°\times0.7871 - 4.9\times0.7871^2\) | M1 | Substituting time into expression for height |
| \(= 3.04 \text{ m}\) | A1, A1 (3) | Correct expression; AG correct height from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.44 = 12\sin40°t - 4.9t^2\) | M1 | Equation for time based on height being 2.44 |
| \(4.9t^2 - 12\sin40°t + 2.44 = 0\) | A1, A1 | Correct LHS; Correct RHS |
| \(t = 0.4385 \text{ or } 1.136\) | m1, A1 | Solving quadratic; Correct time(s) |
| \(s = 12\cos40°\times1.136 = 10.4 \text{ m}\) | M1, A1 (7) | Substituting larger time into horizontal displacement expression; Correct distance |
## Question 8:
### Part (a)
Ball is a particle | B1 | One appropriate assumption
No air resistance | B1 (2) | Second appropriate assumption
### Part (b)(i)
$0 = 12\sin40° - 9.8t$ | M1, A1 | Equation to find time at maximum height; Correct equation
$t = \frac{12\sin40°}{9.8} = 0.787 \text{ s}$ | M1, A1 (4) | Solving for $t$; Correct time
### Part (b)(ii)
$h = 12\sin40°\times0.7871 - 4.9\times0.7871^2$ | M1 | Substituting time into expression for height
$= 3.04 \text{ m}$ | A1, A1 (3) | Correct expression; AG correct height from correct working
### Part (c)
$2.44 = 12\sin40°t - 4.9t^2$ | M1 | Equation for time based on height being 2.44
$4.9t^2 - 12\sin40°t + 2.44 = 0$ | A1, A1 | Correct LHS; Correct RHS
$t = 0.4385 \text{ or } 1.136$ | m1, A1 | Solving quadratic; Correct time(s)
$s = 12\cos40°\times1.136 = 10.4 \text{ m}$ | M1, A1 (7) | Substituting larger time into horizontal displacement expression; Correct distance8 A football is placed on a horizontal surface. It is then kicked, so that it has an initial velocity of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ above the horizontal.
\begin{enumerate}[label=(\alph*)]
\item State two modelling assumptions that it would be appropriate to make when considering the motion of the football.
\item \begin{enumerate}[label=(\roman*)]
\item Find the time that it takes for the ball to reach its maximum height.
\item Hence show that the maximum height of the ball is 3.04 metres, correct to three significant figures.
\end{enumerate}\item After the ball has reached its maximum height, it hits the bar of a goal at a height of 2.44 metres. Find the horizontal distance of the goal from the point where the ball was kicked.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q8 [16]}}