| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a standard M1 projectile motion question with routine calculations using SUVAT equations. All parts follow textbook methods: time of flight from vertical motion, range from horizontal motion, conceptual understanding of mass independence, maximum height formula, and symmetry of velocity. No problem-solving insight required, just systematic 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Vertical component: \(u_y = 20\sin 50°\) | M1 | Using vertical component of velocity |
| \(s = u_y t + \frac{1}{2}at^2\), with \(s=0\): \(0 = 20\sin 50° \cdot t - \frac{1}{2}(9.8)t^2\) | M1 | Forming equation with \(s=0\) |
| \(t(20\sin 50° - 4.9t) = 0\) | A1 | Correct equation |
| \(t = \frac{20\sin 50°}{4.9} = 3.13\) s | A1 | Correct answer shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u_x = 20\cos 50°\) | M1 | Horizontal component used |
| \(PQ = 20\cos 50° \times 3.13 = 40.2\) m | A1 | Accept 40.2 m |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| The distance would be the same | B1 | Correct conclusion |
| Mass does not affect trajectory under gravity alone (only gravity acts) | B1 | Valid reason |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Maximum height when \(v_y = 0\): \(0 = 20\sin 50° - 9.8t\) | M1 | Setting vertical velocity to zero |
| \(t = \frac{20\sin 50°}{9.8} = 1.565\) s | A1 | Correct time |
| \(h = 20\sin 50°(1.565) - \frac{1}{2}(9.8)(1.565)^2\) | M1 | Substituting into \(s = ut + \frac{1}{2}at^2\) |
| \(h = 12.0\) m | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Magnitude = \(20\) m s\(^{-1}\) | B1 | By symmetry |
| Direction: \(50°\) below horizontal | B1 | Angle stated below horizontal |
# Question 6:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical component: $u_y = 20\sin 50°$ | M1 | Using vertical component of velocity |
| $s = u_y t + \frac{1}{2}at^2$, with $s=0$: $0 = 20\sin 50° \cdot t - \frac{1}{2}(9.8)t^2$ | M1 | Forming equation with $s=0$ |
| $t(20\sin 50° - 4.9t) = 0$ | A1 | Correct equation |
| $t = \frac{20\sin 50°}{4.9} = 3.13$ s | A1 | Correct answer shown |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u_x = 20\cos 50°$ | M1 | Horizontal component used |
| $PQ = 20\cos 50° \times 3.13 = 40.2$ m | A1 | Accept 40.2 m |
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| The distance would be the same | B1 | Correct conclusion |
| Mass does not affect trajectory under gravity alone (only gravity acts) | B1 | Valid reason |
## Part (d)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Maximum height when $v_y = 0$: $0 = 20\sin 50° - 9.8t$ | M1 | Setting vertical velocity to zero |
| $t = \frac{20\sin 50°}{9.8} = 1.565$ s | A1 | Correct time |
| $h = 20\sin 50°(1.565) - \frac{1}{2}(9.8)(1.565)^2$ | M1 | Substituting into $s = ut + \frac{1}{2}at^2$ |
| $h = 12.0$ m | A1 | Correct answer |
## Part (e)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Magnitude = $20$ m s$^{-1}$ | B1 | By symmetry |
| Direction: $50°$ below horizontal | B1 | Angle stated below horizontal |
---6 A ball is kicked from the point $P$ on a horizontal surface. It leaves the surface with a velocity of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $50 ^ { \circ }$ above the horizontal and hits the surface for the first time at the point $Q$. Assume that the ball is a particle that moves only under the influence of gravity.\\
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-12_317_1118_513_461}
\begin{enumerate}[label=(\alph*)]
\item Show that the time that it takes the ball to travel from $P$ to $Q$ is 3.13 s , correct to three significant figures.
\item Find the distance between the points $P$ and $Q$.
\item If a heavier ball were projected from $P$ with the same velocity, how would the distance between $P$ and $Q$, calculated using the same modelling assumptions, compare with your answer to part (b)? Give a reason for your answer.
\item Find the maximum height of the ball above the horizontal surface.
\item State the magnitude and direction of the velocity of the ball as it hits the surface.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c022c936-72bc-4cf9-8f98-285f12c1d479-13_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q6 [13]}}