| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question requiring systematic application of SUVAT equations in 2D. Parts (a)-(c) involve routine substitution into projectile motion formulas with given constraints, while part (d) requires finding the angle from velocity components. The multi-part structure and algebraic manipulation are typical for M2, but no novel insight is needed—just methodical application of standard techniques. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 = -2u\sin\theta + \frac{1}{2}g \times 4\) | M1 | Vertical distance. Condone sign errors. Must have used \(t=2\), but could be using \(u_y = u\sin\theta\) |
| \(-2 = u\sin\theta t - \frac{1}{2}gt^2\), \(u\sin\theta = g - 1\) | A1 | All correct |
| \(2u\cos\theta = 8\) (\(u\cos\theta = 4\)) | B1 | Horizontal distance. Accept \(u_x = 4\) o.e. |
| \(\tan\theta = \frac{g-1}{4} = 2.2\) | M1 | Divide to obtain expression for \(\tan\theta\) |
| A1 | Given answer - acceptable to quote and use equation for projectile path. Incorrect equation is 0/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u\cos\theta = 4\) | M1 | Use the horizontal distance and \(\theta\) to find \(u\) |
| \(u = \frac{4}{\cos\theta} = 9.66... = 9.7\) | A1 | NB \(\theta = 65.6°\) leading to 9.68 is an accuracy penalty |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6 = (1-g)T + \frac{1}{2} \times 9.87^2\) | M1 | Equation for vertical distance \(= \pm 6\) to give a quadratic in \(T\). Allow their \(u_y\) |
| \(4.9T^2 - 8.8T - 6 = 0\) | ||
| \(T = \frac{8.8 \pm \sqrt{[(-8.8)^2 + 24 \times 4.9}]}{9.8}\) | DM1 | Solve a 3 term quadratic |
| \(T = 2.323... = 2.32\) or \(2.3\) | A1 | 2.3 or 2.32 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 = 8.8^2 + 2g \times 6\) or \(v = -8.8 + gT\) | M1 | Use suvat to find vertical speed |
| A1 | Correct equation their \(u_y\), \(T\) | |
| \(v = 13.96...\), Horizontal speed \(= 4\) | ||
| \(\tan\alpha = \frac{v}{4}\) | DM1 | Correct trig. with their vertical speed to find required angle |
| A1 | Correct equation | |
| \(\alpha = 74.01... = 74°\) | A1 | \(74°\) or \(74.0°\). Allow 106 |
| Alternative: \(\frac{1}{2}m(9.6664)^2 + 6mg = \frac{1}{2}mv^2\) | M1 | Conservation of energy to find speed |
| \(v = 14.52719...\), \(\cos\alpha = \frac{4}{14.5}\) | DM1 | Correct method for \(\alpha\) |
| \(\alpha = 74.01... = 74°\) | A1 | Allow 106 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 = -2u\sin\theta + \frac{1}{2}g \times 4$ | M1 | Vertical distance. Condone sign errors. Must have used $t=2$, but could be using $u_y = u\sin\theta$ |
| $-2 = u\sin\theta t - \frac{1}{2}gt^2$, $u\sin\theta = g - 1$ | A1 | All correct |
| $2u\cos\theta = 8$ ($u\cos\theta = 4$) | B1 | Horizontal distance. Accept $u_x = 4$ o.e. |
| $\tan\theta = \frac{g-1}{4} = 2.2$ | M1 | Divide to obtain expression for $\tan\theta$ |
| | A1 | **Given answer** - acceptable to quote and use equation for projectile path. Incorrect equation is 0/5 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u\cos\theta = 4$ | M1 | Use the horizontal distance and $\theta$ to find $u$ |
| $u = \frac{4}{\cos\theta} = 9.66... = 9.7$ | A1 | NB $\theta = 65.6°$ leading to 9.68 is an accuracy penalty |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 = (1-g)T + \frac{1}{2} \times 9.87^2$ | M1 | Equation for vertical distance $= \pm 6$ to give a quadratic in $T$. Allow their $u_y$ |
| $4.9T^2 - 8.8T - 6 = 0$ | | |
| $T = \frac{8.8 \pm \sqrt{[(-8.8)^2 + 24 \times 4.9}]}{9.8}$ | DM1 | Solve a 3 term quadratic |
| $T = 2.323... = 2.32$ or $2.3$ | A1 | 2.3 or 2.32 only |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = 8.8^2 + 2g \times 6$ or $v = -8.8 + gT$ | M1 | Use suvat to find vertical speed |
| | A1 | Correct equation their $u_y$, $T$ |
| $v = 13.96...$, Horizontal speed $= 4$ | | |
| $\tan\alpha = \frac{v}{4}$ | DM1 | Correct trig. with their vertical speed to find required angle |
| | A1 | Correct equation |
| $\alpha = 74.01... = 74°$ | A1 | $74°$ or $74.0°$. Allow 106 |
| **Alternative:** $\frac{1}{2}m(9.6664)^2 + 6mg = \frac{1}{2}mv^2$ | M1 | Conservation of energy to find speed |
| $v = 14.52719...$, $\cos\alpha = \frac{4}{14.5}$ | DM1 | Correct method for $\alpha$ |
| $\alpha = 74.01... = 74°$ | A1 | Allow 106 |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ad18c22c-2fc5-4844-99b8-492f758bb24e-11_531_931_230_520}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A ball is thrown from a point $O$, which is 6 m above horizontal ground. The ball is projected with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal. There is a thin vertical post which is 4 m high and 8 m horizontally away from the vertical through $O$, as shown in Figure 2. The ball passes just above the top of the post 2 s after projection. The ball is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \theta = 2.2$
\item Find the value of $u$.
The ball hits the ground $T$ seconds after projection.
\item Find the value of $T$.
Immediately before the ball hits the ground the direction of motion of the ball makes an angle $\alpha$ with the horizontal.
\item Find $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q6 [15]}}