| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| 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 work-energy question with friction on an inclined plane. It requires routine application of work-energy principles, resolving forces to find friction, and calculating work done. The multi-part structure and numerical values (sin θ = 24/25) are typical M2 fare, requiring careful but straightforward calculation rather than novel insight. Slightly above average due to the three-part structure and need to handle both upward and downward motion. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = \frac{5}{4}\) | M1 | 1.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r} = (2t^2 - 5t)\mathbf{i} + 3t\mathbf{j} (+\mathbf{c})\) | M1 | Integrate the velocity vector |
| A1 | NB Also correct to use suvat with \(\mathbf{a} = 4\mathbf{i}\) and \(\mathbf{u} = -5\mathbf{i} + 3\mathbf{j}\) | |
| \(t = 0\): \(2\mathbf{i} + 5\mathbf{j} = \mathbf{c}\) | DM1 | Correct |
| \(\mathbf{r} = (2t^2 - 5t)\mathbf{i} + 3t\mathbf{j} + (2\mathbf{i} + 5\mathbf{j})\) | A1 | Use \(\mathbf{r}_0\) to find \(C\) |
| \((2t^2 - 5t + 2)\mathbf{i} + (3t + 5)\mathbf{j}\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r}_Q = 11\mathbf{i} + 2\mathbf{j} - 2t\mathbf{i} + ct\mathbf{j}\) | ||
| \((11 - 2t)\mathbf{i} + (2 + ct)\mathbf{j}\) | ||
| \(\mathbf{r}_P = (2t^2 - 5t + 2)\mathbf{i} + (3t + 5)\mathbf{j}\) | Correct \(\mathbf{j}\) component of \(\mathbf{r}_Q\). Do not actually require the whole thing | |
| \(\mathbf{r}_Q = \mathbf{r}_P = d\mathbf{i} + 14\mathbf{j}\) | ||
| \(3t + 5 = 14 \Rightarrow\) and \(2t^2 - 3t - 9 = 0\), \((2t+3)(t-3)=0\), \(t = 3\) | M1 | Form an equation in \(t\) only |
| \(t = 3\) | A1 ft | Their \(t\) |
| \(2 + ct = 14 \Rightarrow c = 4\) | A1 ft | Their \(t\) |
| \(d = 11 - 2 \times 3 = 5\) or \(d = 2\times3^2 - 5\times3 + 2 \Rightarrow d = 5\) | A1 ft | Their \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N = 2g\cos\theta = \frac{14}{25}g\) | M1 | Resolve perpendicular to plane. Condone trig confusion |
| \(F = \mu N = \frac{5}{12} \times \frac{14}{25}g = \frac{7g}{30}\) | B1 | Correct value of \(F\) seen or implied |
| Work done \(= \frac{7}{30}g \times 1.5 = 3.43... = 3.4 \text{ J}\) | DM1 | Their \(F \times 1.5\) |
| A1 | \(\frac{7g}{20}\), 3.4 or 3.43 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3.43 + 2g\sin\theta \times 1.5 = \frac{1}{2} \times 2U^2\) | M1 | Energy equation - needs all 3 terms, but condone sign errors & trig confusion. Must have expression for vertical height |
| A1 | Correct with one slip for their WD | |
| A1 | All correct for their WD | |
| \(U = 5.626... = 5.6\) | A1 | 5.6 & 5.63 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2g\sin\theta \times 1.5 = 3.43 + \frac{1}{2} \times 2v^2\) | M1 | Energy equation - needs all three terms. Condone sign errors & trig confusion. Extra terms give M0 |
| \(v^2 = 3g\sin\theta - 3.43\) | A1 | All correct (their WD & \(U\)) |
| Speed \(= 5.0 \text{ ms}^{-1}\) | A1 | Accept 4.98 |
| Alt (c): \(mg\sin\theta - F = ma\) and \(v^2 = (u^2) + 2as\) | M1 | Equation of motion - needs all three terms. Together with suvat |
| \(2g\sin\theta - \frac{7g}{30} = \frac{48g}{25} - \frac{7g}{30} = 2a\), \(a = \frac{253g}{300} = 8.26...\) | A1 | |
| \(v^2 = 24.794\), \(v = 5.0\) | A1 | Accept 4.98 |
## Question 5 (vectors):
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{5}{4}$ | M1 | 1.25 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = (2t^2 - 5t)\mathbf{i} + 3t\mathbf{j} (+\mathbf{c})$ | M1 | Integrate the velocity vector |
| | A1 | NB Also correct to use suvat with $\mathbf{a} = 4\mathbf{i}$ and $\mathbf{u} = -5\mathbf{i} + 3\mathbf{j}$ |
| $t = 0$: $2\mathbf{i} + 5\mathbf{j} = \mathbf{c}$ | DM1 | Correct |
| $\mathbf{r} = (2t^2 - 5t)\mathbf{i} + 3t\mathbf{j} + (2\mathbf{i} + 5\mathbf{j})$ | A1 | Use $\mathbf{r}_0$ to find $C$ |
| $(2t^2 - 5t + 2)\mathbf{i} + (3t + 5)\mathbf{j}$ | B1 | oe |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r}_Q = 11\mathbf{i} + 2\mathbf{j} - 2t\mathbf{i} + ct\mathbf{j}$ | | |
| $(11 - 2t)\mathbf{i} + (2 + ct)\mathbf{j}$ | | |
| $\mathbf{r}_P = (2t^2 - 5t + 2)\mathbf{i} + (3t + 5)\mathbf{j}$ | | Correct $\mathbf{j}$ component of $\mathbf{r}_Q$. Do not actually require the whole thing |
| $\mathbf{r}_Q = \mathbf{r}_P = d\mathbf{i} + 14\mathbf{j}$ | | |
| $3t + 5 = 14 \Rightarrow$ and $2t^2 - 3t - 9 = 0$, $(2t+3)(t-3)=0$, $t = 3$ | M1 | Form an equation in $t$ only |
| $t = 3$ | A1 ft | Their $t$ |
| $2 + ct = 14 \Rightarrow c = 4$ | A1 ft | Their $t$ |
| $d = 11 - 2 \times 3 = 5$ or $d = 2\times3^2 - 5\times3 + 2 \Rightarrow d = 5$ | A1 ft | Their $t$ |
---
## Question 5 (inclined plane):
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 2g\cos\theta = \frac{14}{25}g$ | M1 | Resolve perpendicular to plane. Condone trig confusion |
| $F = \mu N = \frac{5}{12} \times \frac{14}{25}g = \frac{7g}{30}$ | B1 | Correct value of $F$ seen or implied |
| Work done $= \frac{7}{30}g \times 1.5 = 3.43... = 3.4 \text{ J}$ | DM1 | Their $F \times 1.5$ |
| | A1 | $\frac{7g}{20}$, 3.4 or 3.43 only |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3.43 + 2g\sin\theta \times 1.5 = \frac{1}{2} \times 2U^2$ | M1 | Energy equation - needs all 3 terms, but condone sign errors & trig confusion. Must have expression for vertical height |
| | A1 | Correct with one slip for their WD |
| | A1 | All correct for their WD |
| $U = 5.626... = 5.6$ | A1 | 5.6 & 5.63 only |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2g\sin\theta \times 1.5 = 3.43 + \frac{1}{2} \times 2v^2$ | M1 | Energy equation - needs all three terms. Condone sign errors & trig confusion. Extra terms give M0 |
| $v^2 = 3g\sin\theta - 3.43$ | A1 | All correct (their WD & $U$) |
| Speed $= 5.0 \text{ ms}^{-1}$ | A1 | Accept 4.98 |
| **Alt (c):** $mg\sin\theta - F = ma$ and $v^2 = (u^2) + 2as$ | M1 | Equation of motion - needs all three terms. Together with suvat |
| $2g\sin\theta - \frac{7g}{30} = \frac{48g}{25} - \frac{7g}{30} = 2a$, $a = \frac{253g}{300} = 8.26...$ | A1 | |
| $v^2 = 24.794$, $v = 5.0$ | A1 | Accept 4.98 |
---5. The point $A$ lies on a rough plane inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 24 } { 25 }$. A particle $P$ is projected from $A$, up a line of greatest slope of the plane, with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The mass of $P$ is 2 kg and the coefficient of friction between $P$ and the plane is $\frac { 5 } { 12 }$. The particle comes to instantaneous rest at the point $B$ on the plane, where $A B = 1.5 \mathrm {~m}$. It then moves back down the plane to $A$.
\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 speed of $P$ when it returns to $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q5 [11]}}