| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - finding speed or height |
| Difficulty | Standard +0.8 This M2 question requires integrating energy conservation with projectile motion concepts across three parts. Part (a) is straightforward energy application, but parts (b) and (c) require recognizing that minimum speed occurs at the highest point (horizontal component only) and then using this insight with projectile equations. The multi-step reasoning connecting energy, velocity components, and trajectory geometry elevates this above routine M2 questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of conservation of energy (correct no. of terms) | M1 | Allow if all \(m\)'s cancelled |
| PE term correct | A1 | |
| KE terms correct | A1 | |
| \(h = 1.7\) (m) or \(1.68\) (m) (must be positive) | A1 | \((165/98)\); allow \(h = -1.68\) obtained then changed to \((+)1.68\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.5 = 4\cos\alpha\) | M1 | With usual rules; also allow \(2.5 = 4\sin\alpha\) |
| Correct equation | A1 | |
| \(51°\), \(51.3°\) or better | A1 | Since no \(g\) involved |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_B (\uparrow) = \sqrt{7^2 - 2.5^2}\) | M1 | |
| \(\frac{1}{2}\sqrt{171} = 6.54\) | A1 | |
| Use of \(v = u + at\) | M2 | 4 must be resolved and \(v\) must be a vertical component; \(v = 7\) or \(v = 2.5\) is M0 |
| \(-6.54 = 4\sin\alpha - gT\) | A1 | |
| \(T = 0.986..\) | A1 | Seen or implied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(s = ut + \frac{1}{2}at^2\) (4 must be resolved) | M1 | |
| \(-1.68.. = 3.122..T - 4.9T^2\) | A1 | |
| Solving a THREE TERM quadratic | M1 | This M mark and next A mark can be implied by a correct value for \(t\), but if \(t\) is wrong, must see clear attempt to solve quadratic using formula |
| Correct expression | A1 | |
| \(T = 0.986..\) | A1 | Seen or implied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.5 \times \text{their } T\) or \(4\cos(\text{their } \alpha) \times \text{their } T\) | M1 | Must have found a value for \(T\) |
| \(2.5\) (m) or \(2.46\) (m) | A1 |
## Question 7:
**Deduct only 1 mark in whole of question 7 for not giving an answer to either 2 sf or 3 sf, following use of g = 9.8, or g = 9.81. Deduct the final A mark in whichever part it first occurs.**
---
### Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of conservation of energy (correct no. of terms) | M1 | Allow if all $m$'s cancelled |
| PE term correct | A1 | |
| KE terms correct | A1 | |
| $h = 1.7$ (m) or $1.68$ (m) (must be positive) | A1 | $(165/98)$; allow $h = -1.68$ obtained then changed to $(+)1.68$ |
**N.B. Clear use of $v^2 = u^2 + 2as$ instead of energy is M0**
---
### Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.5 = 4\cos\alpha$ | M1 | With usual rules; also allow $2.5 = 4\sin\alpha$ |
| Correct equation | A1 | |
| $51°$, $51.3°$ or better | A1 | Since no $g$ involved |
---
### Question 7(c):
**Either method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_B (\uparrow) = \sqrt{7^2 - 2.5^2}$ | M1 | |
| $\frac{1}{2}\sqrt{171} = 6.54$ | A1 | |
| Use of $v = u + at$ | M2 | 4 must be resolved and $v$ must be a vertical component; $v = 7$ or $v = 2.5$ is M0 |
| $-6.54 = 4\sin\alpha - gT$ | A1 | |
| $T = 0.986..$ | A1 | Seen or implied |
**Or method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $s = ut + \frac{1}{2}at^2$ (4 must be resolved) | M1 | |
| $-1.68.. = 3.122..T - 4.9T^2$ | A1 | |
| Solving a **THREE TERM** quadratic | M1 | This M mark and next A mark can be implied by a correct value for $t$, but if $t$ is wrong, must see clear attempt to solve quadratic using formula |
| Correct expression | A1 | |
| $T = 0.986..$ | A1 | Seen or implied |
**Then (both methods):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.5 \times \text{their } T$ or $4\cos(\text{their } \alpha) \times \text{their } T$ | M1 | Must have found a value for $T$ |
| $2.5$ (m) or $2.46$ (m) | A1 | |7. A particle $P$ is projected from a fixed point $A$ with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ above the horizontal and moves freely under gravity. When $P$ passes through the point $B$ on its path, it has speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item By considering energy, find the vertical distance between $A$ and $B$.
The minimum speed of $P$ on its path from $A$ to $B$ is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the size of angle $\alpha$.
\item Find the horizontal distance between $A$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q7 [14]}}