| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Connected particles pulley energy method |
| Difficulty | Standard +0.3 This is a standard M2 connected particles problem using work-energy principle. Part (a) is straightforward bookwork showing PE change, part (b) applies work-energy directly, and part (c) uses basic kinematics. All steps are routine applications of standard methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| Loss in GPE: \(3mgd - 2mg \times d \sin\theta = 3mgd - \frac{4}{5}mgd\) | M1A1 |
| \(d = \frac{11}{5}mgd\) *Given answer* | A1 (3) |
| Notes: M1 for \(\pm(3mgd - 2mgd\sin\theta)\) (allow if cos used instead of sin); First A1 for a correct unsimplified expression in \(\pm mgd\); Second A1 for given positive answer |
| Answer | Marks |
|---|---|
| Must be using work-energy | M1A2 |
| \(\frac{1}{2} \times 2mv^2 + \frac{1}{2} \times 3mv^2 + \frac{3}{5}mgd = \frac{11}{5}mgd\) | |
| \(v^2 = \frac{16}{25}gd\) | A1 (4) |
| Notes: First M1 for a dimensionally correct work-energy equation, with all 4 terms as above, if they use the answer from (a); OR \(\frac{1}{2} \times 2mv^2 + \frac{1}{2} \times 3mv^2 + \frac{3}{5}mgd + 2mgd\sin\theta = 3mgd\) i.e. all 5 terms if they don't, but condone sign errors (M0 if incorrect no. of terms); First A1 and Second A1 for a correct equation, A1A0 if one error; Third A1 for \(\frac{16gd}{25}\) oe or 6.3d or 6.27d |
| Answer | Marks |
|---|---|
| Use of \(s = \frac{u+v}{2}t\) or equivalent: \(1.5 = \frac{0 + \sqrt{\frac{24g}{25}}}{2}T\) | M1 |
| \(T = 0.98\) or 0.978 or \(5\sqrt{30}/28\) | A1 (2) [9] |
| Notes: M1 for a complete method to give an equation in \(T\) only (M0 if they just assume a value for \(a\) e.g. \(a = 0\) or \(g\)); A1 for 0.98 or 0.978 or \(15/\sqrt{(24g)}\) oe |
## 3a
| Loss in GPE: $3mgd - 2mg \times d \sin\theta = 3mgd - \frac{4}{5}mgd$ | M1A1 |
| $d = \frac{11}{5}mgd$ *Given answer* | A1 (3) |
| **Notes:** M1 for $\pm(3mgd - 2mgd\sin\theta)$ (allow if cos used instead of sin); First A1 for a correct unsimplified expression in $\pm mgd$; Second A1 for given positive answer |
## 3b
| Must be using work-energy | M1A2 |
| $\frac{1}{2} \times 2mv^2 + \frac{1}{2} \times 3mv^2 + \frac{3}{5}mgd = \frac{11}{5}mgd$ | |
| $v^2 = \frac{16}{25}gd$ | A1 (4) |
| **Notes:** First M1 for a dimensionally correct work-energy equation, with all 4 terms as above, if they use the answer from (a); OR $\frac{1}{2} \times 2mv^2 + \frac{1}{2} \times 3mv^2 + \frac{3}{5}mgd + 2mgd\sin\theta = 3mgd$ i.e. all 5 terms if they don't, but condone sign errors (M0 if incorrect no. of terms); First A1 and Second A1 for a correct equation, A1A0 if one error; Third A1 for $\frac{16gd}{25}$ oe or 6.3d or 6.27d |
## 3c
| Use of $s = \frac{u+v}{2}t$ or equivalent: $1.5 = \frac{0 + \sqrt{\frac{24g}{25}}}{2}T$ | M1 |
| $T = 0.98$ or 0.978 or $5\sqrt{30}/28$ | A1 (2) [9] |
| **Notes:** M1 for a complete method to give an equation in $T$ only (M0 if they just assume a value for $a$ e.g. $a = 0$ or $g$); A1 for 0.98 or 0.978 or $15/\sqrt{(24g)}$ oe |
---3.
Two particles $P$ and $Q$, of mass $2 m$ and $3 m$ respectively, are connected by a light inextensible string. Initially $P$ is held at rest on a fixed rough plane inclined at $\theta$ to the horizontal ground, where $\sin \theta = \frac { 2 } { 5 }$. The string passes over a small smooth pulley fixed at the top of the plane. The particle $Q$ hangs freely below the pulley, as shown in Figure 1. The part of the string from $P$ to the pulley lies along a line of greatest slope of the plane. At time $t = 0$ the system is released from rest with the string taut. When $P$ moves the friction between $P$ and the plane is modelled as a constant force of magnitude $\frac { 3 } { 5 } m g$. At the instant when each particle has moved a distance $d$, they are both moving with speed $v$, particle $P$ has not reached the pulley and $Q$ has not reached the ground.
\begin{enumerate}[label=(\alph*)]
\item Show that the total potential energy lost by the system when each particle has moved a distance $d$ is $\frac { 11 } { 5 } m g d$.
\item Use the work-energy principle to find $v ^ { 2 }$ in terms of $g$ and $d$.
When $t = T$ seconds, $d = 1.5 \mathrm {~m}$.
\item Find the value of $T$.\\
DO NOT WIRITE IN THIS AREA
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q3 [9]}}