| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Standard +0.3 This is a straightforward FM1 mechanics question requiring application of the power equation P=Fv and Newton's second law. Part (a) uses P=Fv directly at constant speed (equilibrium). Part (b) requires finding acceleration of the system then using F=ma to find tension - standard two-body problem with clear setup and routine calculations, slightly above average due to being Further Maths content but mechanically straightforward. |
| Spec | 3.03d Newton's second law: 2D vectors3.03o Advanced connected particles: and pulleys6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{16000}{v}\) | M1 | Correct use of \(P = Fv\). Expression \(\frac{16000}{v}\) may be on diagram or embedded in \(F = ma\). Condone use of 16000 or 16 for method mark |
| Equation of motion: \(F - 400 = 0\) | M1 | Correct unsimplified equation of motion with \(a = 0\) or equilibrium equation. \(F\) does not need to be substituted |
| \(U = 40\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{16000}{\left(\frac{20}{3}\right)}\) | M1 | Correct use of \(P = Fv\) with \(v = \frac{20}{3}\). Condone use of 16000 or 16 for method mark |
| Equation of motion for system or car or trailer | M1 | Must have all terms, dimensionally correct. Condone sign errors. M0 if \(a=0\). Full marks available if consistent extra \(g\)'s present in both \(ma\) terms |
| \(F - 700 = 1600a\) or \(F - 400 - T = 1000a\) or \(T - 300 = 600a\) | A1 | One correct unsimplified equation |
| Second equation of motion | A1 | Two correct unsimplified equations. Note: \(a = \frac{17}{16}\) but does not need to be seen |
| \(T = 940\) or \(938\) or \(937.5\) or \(\frac{1875}{2}\) oe (N) | A1 | Correct answer |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{16000}{v}$ | M1 | Correct use of $P = Fv$. Expression $\frac{16000}{v}$ may be on diagram or embedded in $F = ma$. Condone use of 16000 or 16 for method mark |
| Equation of motion: $F - 400 = 0$ | M1 | Correct unsimplified equation of motion with $a = 0$ or equilibrium equation. $F$ does not need to be substituted |
| $U = 40$ | A1 | cao |
---
# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{16000}{\left(\frac{20}{3}\right)}$ | M1 | Correct use of $P = Fv$ with $v = \frac{20}{3}$. Condone use of 16000 or 16 for method mark |
| Equation of motion for system or car or trailer | M1 | Must have all terms, dimensionally correct. Condone sign errors. M0 if $a=0$. Full marks available if consistent extra $g$'s present in both $ma$ terms |
| $F - 700 = 1600a$ or $F - 400 - T = 1000a$ or $T - 300 = 600a$ | A1 | One correct unsimplified equation |
| Second equation of motion | A1 | Two correct unsimplified equations. Note: $a = \frac{17}{16}$ but does not need to be seen |
| $T = 940$ or $938$ or $937.5$ or $\frac{1875}{2}$ oe (N) | A1 | Correct answer |
---\begin{enumerate}
\item A car of mass 1000 kg moves in a straight line along a horizontal road at a constant speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The resistance to the motion of the car is a constant force of magnitude 400 N.
\end{enumerate}
The engine of the car is working at a constant rate of 16 kW .\\
(a) Find the value of $U$.
The car now pulls a trailer of mass 600 kg in a straight line along the road using a tow rope which is parallel to the direction of motion. The resistance to the motion of the car is again a constant force of magnitude 400 N . The resistance to the motion of the trailer is a constant force of magnitude 300 N .
The engine of the car is working at a constant rate of 16 kW .\\
The tow rope is modelled as being light and inextensible.\\
Using the model,\\
(b) find the tension in the tow rope at the instant when the speed of the car is $\frac { 20 } { 3 } \mathrm {~ms} ^ { - 1 }$
\hfill \mbox{\textit{Edexcel FM1 2023 Q2 [8]}}