| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question combining work-energy principle with projectile motion. Part (a) requires routine friction calculation, (b) applies work-energy principle directly, (c) and (d) are standard projectile motion. The tan α = 7/24 setup simplifies to a 7-24-25 triangle, making calculations straightforward. All techniques are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model3.03v Motion on rough surface: including inclined planes6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Normal reaction between \(P\) and ramp: \((R) = 0.3g\cos\alpha \left(0.3g \times \frac{24}{25} = 2.82...\right)\) | M1 | Seen or implied. Condone sin/cos confusion (implied by use of \(\frac{7}{25}\)) |
| Work done against friction \(= \frac{1}{5}R \times 15\) | M1 | Use of \(WD = \mu R \times \text{distance}\) with their \(R\) |
| \(= 8.47\ (8.5)\) (J) | A1 | 3 sf or 2 sf |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work-energy equation | M1 | All terms required. Dimensionally correct. Condone sign errors |
| \(\frac{1}{2} \times 0.3U^2 = \frac{1}{2} \times 0.3 \times 25^2 + (a) + 0.3 \times g \times (15\sin\alpha)\) | A1ft, A1ft | Follow their answer to (a). Correct unsimplified equation with at most one error. Correct unsimplified equation |
| \(U = 27.6 \quad (28)\) | A1 | 3 sf or 2 sf |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Time to ground | M1 | Complete method using suvat to form an equation in \(t\) |
| \(-15\sin\alpha = 7t - \frac{1}{2}gt^2\) | A1ft | Correct unsimplified equation in \(t\) ft their 4.2 |
| \(t = 1.88 \quad (1.9)\) (s) | A1 | 3 sf or 2 sf; \(\frac{5+\sqrt{67}}{7}\) is A0 |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vertical component of speed | M1 | Or use energy to find the speed |
| \(= \pm(7 - (\text{their } t) \times 9.8) \quad (\pm 11.459...)\) | A1ft | Or: \(0.15 \times 625 + .3 \times 9.8 \times \text{their } 4.2 = 0.15v^2\), \((v = 26.59...)\), condone \(v = \frac{7\sqrt{67}}{5}\) |
| Correct use of trig: \(\tan\theta° = \frac{\text{their vertical}}{24}\) | M1 | Or \(\cos\theta° = \frac{24}{\text{their speed}}\) |
| \(\theta = 25.5\ (26)\) | A1 | 3 sf or 2 sf |
| Total | 4 |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal reaction between $P$ and ramp: $(R) = 0.3g\cos\alpha \left(0.3g \times \frac{24}{25} = 2.82...\right)$ | M1 | Seen or implied. Condone sin/cos confusion (implied by use of $\frac{7}{25}$) |
| Work done against friction $= \frac{1}{5}R \times 15$ | M1 | Use of $WD = \mu R \times \text{distance}$ with their $R$ |
| $= 8.47\ (8.5)$ (J) | A1 | 3 sf or 2 sf |
| **Total** | **3** | |
---
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation | M1 | All terms required. Dimensionally correct. Condone sign errors |
| $\frac{1}{2} \times 0.3U^2 = \frac{1}{2} \times 0.3 \times 25^2 + (a) + 0.3 \times g \times (15\sin\alpha)$ | A1ft, A1ft | Follow their answer to (a). Correct unsimplified equation with at most one error. Correct unsimplified equation |
| $U = 27.6 \quad (28)$ | A1 | 3 sf or 2 sf |
| **Total** | **4** | |
---
# Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Time to ground | M1 | Complete method using suvat to form an equation in $t$ |
| $-15\sin\alpha = 7t - \frac{1}{2}gt^2$ | A1ft | Correct unsimplified equation in $t$ ft their 4.2 |
| $t = 1.88 \quad (1.9)$ (s) | A1 | 3 sf or 2 sf; $\frac{5+\sqrt{67}}{7}$ is A0 |
| **Total** | **3** | |
---
# Question 8(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical component of speed | M1 | Or use energy to find the speed |
| $= \pm(7 - (\text{their } t) \times 9.8) \quad (\pm 11.459...)$ | A1ft | Or: $0.15 \times 625 + .3 \times 9.8 \times \text{their } 4.2 = 0.15v^2$, $(v = 26.59...)$, condone $v = \frac{7\sqrt{67}}{5}$ |
| Correct use of trig: $\tan\theta° = \frac{\text{their vertical}}{24}$ | M1 | Or $\cos\theta° = \frac{24}{\text{their speed}}$ |
| $\theta = 25.5\ (26)$ | A1 | 3 sf or 2 sf |
| **Total** | **4** | |8. [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in a vertical plane, with $\mathbf { i }$ being horizontal and $\mathbf { j }$ being vertically upwards.]
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1732eb73-8c16-4a45-8d3b-a88e659e47ea-24_378_1219_347_349}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A rough ramp is fixed to horizontal ground.\\
The ramp is inclined to the ground at an angle $\alpha$, where $\tan \alpha = \frac { 7 } { 24 }$\\
The point $A$ is at the bottom of the ramp and the point $B$ is at the top of the ramp. The line $A B$ is a line of greatest slope of the ramp and $A B = 15 \mathrm {~m}$, as shown in Figure 3.
A particle $P$ of mass 0.3 kg is projected with speed $U \mathrm {~ms} ^ { - 1 }$ from $A$ directly towards $B$. At the instant $P$ reaches the point $B$, the velocity of $P$ is $( 24 \mathbf { i } + 7 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ The particle leaves the ramp at $B$, and moves freely under gravity until it hits the horizontal ground at the point $C$.\\
The coefficient of friction between $P$ and the ramp is $\frac { 1 } { 5 }$
\begin{enumerate}[label=(\alph*)]
\item Find the work done against friction as $P$ moves from $A$ to $B$.
\item Use the work-energy principle to find the value of $U$.
\item Find the time taken by $P$ to move from $B$ to $C$.
At the instant immediately before $P$ hits the ground at $C$, the particle is moving downwards at $\theta ^ { \circ }$ to the horizontal.
\item Find the value of $\theta$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2022 Q8 [14]}}