| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: inclined road |
| Difficulty | Standard +0.3 This is a standard M2 power-force-motion question with two straightforward parts: (a) uses P=Fv to find driving force then applies F=ma to a two-body system on an incline, and (b) applies work-energy principle with given resistances. Both parts follow routine procedures taught in M2 with no novel insight required, making it slightly easier than average. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | Equation of motion for whole system. | M1 |
| F 6 4 0 1 1 0 0 g s i n 1 1 0 0 a − − = | A1 A1 |
| Answer | Marks |
|---|---|
| 1 2 | M1 |
| a=0.16 ( ms−2 ) or a=0.163 ( ms−2 ) | A1 |
| Answer | Marks |
|---|---|
| M1 | Equation of motion for whole system. All terms required. |
| Answer | Marks |
|---|---|
| A1 | Unsimplified equation in F or P with at most one error. Missing g |
| Answer | Marks |
|---|---|
| A1 | Correct unsimplified equation in F or P |
| M1 | Use of P = F v . Condone slip with zeroes for M mark. |
| A1 | Correct answer, 2 sf or 3 sf. |
| Answer | Marks | Guidance |
|---|---|---|
| 3(b) | Work-energy equation for the trailer. | M1 |
| Answer | Marks |
|---|---|
| 2 2 5 | A1 A1 |
| ( A B = ) 6 2 ( m ) o r ( A B = ) 6 1 . 6 ( m ) | A1 |
| Answer | Marks |
|---|---|
| M1 | Work-energy equation. Condone slip with zeroes for M mark. |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | Unsimplified equation with at most one error. | |
| A1 | Correct unsimplified equation with mass replaced. | |
| A1 | Correct answer, 2 sf or 3 sf No need for AB = | |
| Question | Scheme | Marks |
Question 3:
--- 3(a) ---
3(a) | Equation of motion for whole system. | M1
F 6 4 0 1 1 0 0 g s i n 1 1 0 0 a − − = | A1 A1
1 5 0 0 0
Use of P=Fv F = ( = 1 2 5 0 )
1 2 | M1
a=0.16 ( ms−2 ) or a=0.163 ( ms−2 ) | A1
(5)
3(a)
Over-accuracy or under-accuracy is penalised max once per complete
question. Penalise final A mark in the appropriate part.
Use of g = 9.81 or 10 is penalised max once per complete question.
Penalise final A mark in the appropriate part.
If both errors occur, could lose the final two A marks.
M1 | Equation of motion for whole system. All terms required.
Dimensionally correct. Condone sign errors and sin/cos confusion.
May form two equations of motion (van and trailer) and combine to
eliminate T. Condone slip with zeroes for M mark. The forces on the
LHS must be consistent with the mass in the ‘ma’.
1
Note that s i n is an accuracy error, not a method error.
2 5
A1 | Unsimplified equation in F or P with at most one error. Missing g
from both weight terms counts as 1 error.
A1 | Correct unsimplified equation in F or P
M1 | Use of P = F v . Condone slip with zeroes for M mark.
A1 | Correct answer, 2 sf or 3 sf.
--- 3(b) ---
3(b) | Work-energy equation for the trailer. | M1
1 1
2 0 0 1 4 2 = 2 4 0 d + 2 0 0 g d
2 2 5 | A1 A1
( A B = ) 6 2 ( m ) o r ( A B = ) 6 1 . 6 ( m ) | A1
(4)
(9)
Notes
3(b)
M1 | Work-energy equation. Condone slip with zeroes for M mark.
Dimensionally correct, all terms required and with correct structure:
KE, GPE, WD. No repeats. Must use mass of trailer only. Condone
sign errors and sin / cos confusion. M0 for suvat.
1
Note that s i n is an accuracy error, not a method error.
2 5
A1 | Unsimplified equation with at most one error.
A1 | Correct unsimplified equation with mass replaced.
A1 | Correct answer, 2 sf or 3 sf No need for AB =
Question | Scheme | Marks
3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-06_275_1143_303_461}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A van of mass 900 kg is moving up a straight road inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 25 }$. The van is towing a trailer of mass 200 kg . The trailer is attached to the van by a rigid towbar which is parallel to the direction of motion of the van and the trailer, as shown in Figure 1.
The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude 400 N .\\
The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 240 N .
The towbar is modelled as a light rod.\\
The engine of the van is working at a constant rate of 15 kW .
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the van at the instant when the speed of the van is $12 \mathrm {~ms} ^ { - 1 }$
At the instant when the speed of the van is $14 \mathrm {~ms} ^ { - 1 }$, the trailer is passing the point $A$ on the slope and the towbar breaks.
The trailer continues to move up the slope until it comes to rest at the point $B$.\\
The resistance to the motion of the trailer from non-gravitational forces is still modelled as a constant force of magnitude 240 N .
\item Use the work-energy principle to find the distance $A B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2024 Q3}}