| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Velocity direction at specific time/point |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with multiple parts requiring routine application of SUVAT equations and projectile motion formulas. Part (a) uses v²=u²+2as at maximum height, part (b) requires finding horizontal component from range equation, and part (c) involves finding when velocity direction matches displacement direction. All techniques are textbook standard with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(0^2 = u_v^2 - 2 \times 9.8 \times 10\) | M1 | Complete method using \(suvat\) to form an equation in \(u_v\) |
| \(u_v = 14\) * | A1 | Correct equation e.g. \(0 = u^2 - 20g\) |
| A1 | *Answer given* requires equation and working, including \(196\), seen | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Conservation of energy: \(\frac{1}{2}m(u_v^2 + u_v^2) = mg \times 10 + \frac{1}{2}mu_v^2 - \frac{1}{2}u_v^2 = 98\) | M1 | Initial KE = gain in GPE + final KE |
| \(u_v = 14\) * | A1 | Correct equation |
| A1 | *Answer given* | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\uparrow), -52.5 = 14t - \frac{1}{2}gt^2\) | M1 | Use the vertical distance travelled to find the total time taken. At most one error |
| A1 | Correct equation | |
| \(49t^2 - 140t - 525 = 0\) | ||
| \((t-5)(49t + 105) = 0 \quad t = 5\) | DM1 | Solve for \(t\). Dependent on the preceding M mark only |
| A1 | ||
| \((\rightarrow), 50 = 5u_H\) | M1 | Use their time of flight to form an equation in \(u_H\) |
| \(u_H = 10\) | A1 | only |
| \(u = \sqrt{10^2 + 14^2}\) | M1 | Use of Pythagoras with two non-zero components, or solution of a pair of simultaneous equations in \(u\) and \(a\). \(17.2\) or \(17\) (method involves use of \(g = 9.8\) so an exact surd answer is not acceptable) |
| \(= \sqrt{296} ; 17.2 \text{ m s}^{-1}\) | A1 | |
| (9) |
| Answer | Marks | Guidance |
|---|---|---|
| \(50 = u\cos\alpha t\) or \(50 = u_H t\) | M1 | First 3 marks for the quadratic as above. Used in their quadratic |
| \(49\left(\frac{50}{u_H}\right)^2 - 140\left(\frac{50}{u_H}\right) - 525 = 0\) | A1 | Correct quadratic in \(u_H\) |
| \(525(u_H)^2 + 140(u_H) - 1225000 = 0\) | ||
| Solve for \(u_H\) | DM1 | Dependent on the M mark for setting up the initial quadratic equation in \(t\) |
| \(u_H = 10\) | A1 | only |
| etc. | Complete as above | |
| (5) | ||
| 17 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan OBA = \frac{52.5}{50} = 1.05\) | B1 | Correct direction o.e. (accept reciprocal) |
| \(v_v = 1.05 \times 10 = 10.5\) | M1 | Use trig with their \(u_H\) and correct interpretation of direction to find the vertical component of speed. Working with distances is M0. (condone \(10 \div 1.05\)) |
| DM1 | Use \(suvat\) to form an equation in \(t\). Dependent on the preceding M. | |
| \((\uparrow), -10.5 = 14 - gt\) | A1 | Correct equation for their \(u_H\). For incorrect direction give A0 here. only |
| \(t = 2.5\) | A1 | only |
| (5) | ||
| 17 |
**(a)**
| $0^2 = u_v^2 - 2 \times 9.8 \times 10$ | M1 | Complete method using $suvat$ to form an equation in $u_v$ |
| $u_v = 14$ * | A1 | Correct equation e.g. $0 = u^2 - 20g$ |
| | A1 | *Answer given* requires equation and working, including $196$, seen |
| | (3) | |
**OR**
| Conservation of energy: $\frac{1}{2}m(u_v^2 + u_v^2) = mg \times 10 + \frac{1}{2}mu_v^2 - \frac{1}{2}u_v^2 = 98$ | M1 | Initial KE = gain in GPE + final KE |
| $u_v = 14$ * | A1 | Correct equation |
| | A1 | *Answer given* |
| | (3) | |
**(b)**
| $(\uparrow), -52.5 = 14t - \frac{1}{2}gt^2$ | M1 | Use the vertical distance travelled to find the total time taken. At most one error |
| | A1 | Correct equation |
| $49t^2 - 140t - 525 = 0$ | | |
| $(t-5)(49t + 105) = 0 \quad t = 5$ | DM1 | Solve for $t$. Dependent on the preceding M mark only |
| | A1 | |
| $(\rightarrow), 50 = 5u_H$ | M1 | Use their time of flight to form an equation in $u_H$ |
| $u_H = 10$ | A1 | only |
| $u = \sqrt{10^2 + 14^2}$ | M1 | Use of Pythagoras with two non-zero components, or solution of a pair of simultaneous equations in $u$ and $a$. $17.2$ or $17$ (method involves use of $g = 9.8$ so an exact surd answer is not acceptable) |
| $= \sqrt{296} ; 17.2 \text{ m s}^{-1}$ | A1 | |
| | (9) | |
**OR**
| $50 = u\cos\alpha t$ or $50 = u_H t$ | M1 | First 3 marks for the quadratic as above. Used in their quadratic |
| $49\left(\frac{50}{u_H}\right)^2 - 140\left(\frac{50}{u_H}\right) - 525 = 0$ | A1 | Correct quadratic in $u_H$ |
| $525(u_H)^2 + 140(u_H) - 1225000 = 0$ | | |
| Solve for $u_H$ | DM1 | Dependent on the M mark for setting up the initial quadratic equation in $t$ |
| $u_H = 10$ | A1 | only |
| etc. | | Complete as above |
| | (5) |
| | 17 | |
**(c)**
| $\tan OBA = \frac{52.5}{50} = 1.05$ | B1 | Correct direction o.e. (accept reciprocal) |
| $v_v = 1.05 \times 10 = 10.5$ | M1 | Use trig with their $u_H$ and correct interpretation of direction to find the vertical component of speed. Working with distances is M0. (condone $10 \div 1.05$) |
| | DM1 | Use $suvat$ to form an equation in $t$. Dependent on the preceding M. |
| $(\uparrow), -10.5 = 14 - gt$ | A1 | Correct equation for their $u_H$. For incorrect direction give A0 here. only |
| $t = 2.5$ | A1 | only |
| | (5) |
| | 17 | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12cd7355-f632-4a84-825f-a269851c6ec4-12_602_1175_237_386}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A small stone is projected from a point $O$ at the top of a vertical cliff $O A$. The point $O$ is 52.5 m above the sea. The stone rises to a maximum height of 10 m above the level of $O$ before hitting the sea at the point $B$, where $A B = 50 \mathrm {~m}$, as shown in Figure 4. The stone is modelled as a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that the vertical component of the velocity of projection of the stone is $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the speed of projection.
\item Find the time after projection when the stone is moving parallel to $O B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q7 [17]}}