| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Two connected particles, horizontal surface |
| Difficulty | Moderate -0.3 This is a standard M1 connected particles question requiring Newton's second law applied to a system and individual particles. While it has multiple parts (5 sub-questions), each part follows routine procedures: finding acceleration from F=ma for the system, using kinematics for speed, finding tension/thrust by considering individual particles. The calculations are straightforward with no conceptual surprises, making it slightly easier than average but still requiring systematic application of mechanics principles. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For system N2L: \(4 - 3 = 0.8a\) | M1 A1 | M1 for resolving horizontally to produce equation in \(a\) ONLY; A1 for correct equation |
| \(a = 1.25 \text{ ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = u + at \Rightarrow v = 0 + 1.25 \times 6 = 7.5 \text{ ms}^{-1}\) | M1 A1 | M1 for complete method to find speed; A1 cao \(7.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(P\), N2L: \(T - 1 = 0.3 \times 1.25\) | M1 A1ft | M1 for resolving horizontally for either \(P\) or \(Q\) to give equation in \(T\) only; ft on their \(a\) |
| \(T = 1.375\) N, \(1.38\), \(1.4\) | A1 | cao |
| OR for \(Q\), N2L: \(4 - 2 - T = 0.5 \times 1.25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For system N2L: \(-3 = 0.8a \Rightarrow a = -3.75\) | M1 A1 | M1 for resolving horizontally, equation in \(a\) ONLY; A1 cao \(-3.75\) (or \(3.75\)) |
| \(v^2 = u^2 + 2as \Rightarrow 0^2 = 7.5^2 - 2 \times 3.75s\) | M1 | Use of \(v^2 = u^2 + 2as\) with \(v=0\), \(u =\) their (b), their \(a\); M0 if \(a\) not calculated |
| \(s = 7.5\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(P\), N2L: \(T' + 1 = 0.3 \times 3.75\) | M1 A1 | M1 for resolving horizontally for either \(P\) or \(Q\) to give equation in \(T'\) only; M0 if \(a\) not calculated; A1 for correct equation |
| \(T' = 0.125\) N, \(0.13\) | A1 | cao; must be positive |
| Alternative for \(Q\): \(2 - T' = 0.5 \times 3.75 \Rightarrow T' = 0.125\) N | M1 A1 A1 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For system N2L: $4 - 3 = 0.8a$ | M1 A1 | M1 for resolving horizontally to produce equation in $a$ ONLY; A1 for correct equation |
| $a = 1.25 \text{ ms}^{-2}$ | A1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = u + at \Rightarrow v = 0 + 1.25 \times 6 = 7.5 \text{ ms}^{-1}$ | M1 A1 | M1 for complete method to find speed; A1 cao $7.5$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $P$, N2L: $T - 1 = 0.3 \times 1.25$ | M1 A1ft | M1 for resolving horizontally for either $P$ or $Q$ to give equation in $T$ only; ft on their $a$ |
| $T = 1.375$ N, $1.38$, $1.4$ | A1 | cao |
| OR for $Q$, N2L: $4 - 2 - T = 0.5 \times 1.25$ | | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For system N2L: $-3 = 0.8a \Rightarrow a = -3.75$ | M1 A1 | M1 for resolving horizontally, equation in $a$ ONLY; A1 cao $-3.75$ (or $3.75$) |
| $v^2 = u^2 + 2as \Rightarrow 0^2 = 7.5^2 - 2 \times 3.75s$ | M1 | Use of $v^2 = u^2 + 2as$ with $v=0$, $u =$ their (b), their $a$; M0 if $a$ not calculated |
| $s = 7.5$ m | A1 | |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $P$, N2L: $T' + 1 = 0.3 \times 3.75$ | M1 A1 | M1 for resolving horizontally for either $P$ or $Q$ to give equation in $T'$ only; M0 if $a$ not calculated; A1 for correct equation |
| $T' = 0.125$ N, $0.13$ | A1 | cao; must be positive |
| Alternative for $Q$: $2 - T' = 0.5 \times 3.75 \Rightarrow T' = 0.125$ N | M1 A1 A1 | |
The images you've shared appear to be blank/empty pages — one is completely white and the other shows only publication/contact information for Edexcel (back cover details).
There is no mark scheme content visible in these images to extract. Could you please share the actual mark scheme pages that contain the questions and marking criteria? I'd be happy to format those for you once I can see the content.7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c908e75-73df-46be-93bb-09dba2cb3b7e-12_150_1104_255_422}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two particles $P$ and $Q$, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod. The system of the particles and the rod is at rest on a horizontal plane. At time $t = 0$, a constant force $\mathbf { F }$ of magnitude 4 N is applied to $Q$ in the direction $P Q$, as shown in Figure 3. The system moves under the action of this force until $t = 6 \mathrm {~s}$. During the motion, the resistance to the motion of $P$ has constant magnitude 1 N and the resistance to the motion of $Q$ has constant magnitude 2 N .
Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the particles as the system moves under the action of $\mathbf { F }$,
\item the speed of the particles at $t = 6 \mathrm {~s}$,
\item the tension in the rod as the system moves under the action of $\mathbf { F }$.
At $t = 6 \mathrm {~s} , \mathbf { F }$ is removed and the system decelerates to rest. The resistances to motion are unchanged. Find
\item the distance moved by $P$ as the system decelerates,
\item the thrust in the rod as the system decelerates.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2012 Q7 [15]}}