| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | String becomes taut problem |
| Difficulty | Standard +0.8 This is a multi-stage mechanics problem requiring conservation of momentum for the taut rope (standard M1), impulse calculation (routine), then a more demanding kinematics analysis involving reaction time and inequality constraints. Part (c) requires careful consideration of distances, timing, and setting up an inequality—this elevates it above typical M1 questions but remains within syllabus scope with methodical application. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks |
|---|---|
| Cons. of mom.: \(1500(2) + 0 = (1500 + 750)V\) | M1 |
| \(3000 = 2250V \therefore V = \frac{4}{3}\) | M1 A1 |
| Answer | Marks |
|---|---|
| Impulse = \(\Delta\) mom. = \(750(\frac{4}{3} - 0) = 1000\) Ns | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Car has \((27 + 9)\) m in which to stop and travels 18 m in first second must stop from 18 m\(s^{-1}\) in 18 m | M1 A1 | |
| \(u = 18, s = 18, v = 0, a = f\) | M1 | |
| \(v^2 = u^2 + 2as\), so \(0 = 324 - 36f\) | M1 | |
| \(f = 9\) so to stop before hitting other car, \(f > 9\) | A1 | (10) |
**Part (a)**
Cons. of mom.: $1500(2) + 0 = (1500 + 750)V$ | M1 |
$3000 = 2250V \therefore V = \frac{4}{3}$ | M1 A1 |
**Part (b)**
Impulse = $\Delta$ mom. = $750(\frac{4}{3} - 0) = 1000$ Ns | M1 A1 |
**Part (c)**
Car has $(27 + 9)$ m in which to stop and travels 18 m in first second must stop from 18 m$s^{-1}$ in 18 m | M1 A1 |
$u = 18, s = 18, v = 0, a = f$ | M1 |
$v^2 = u^2 + 2as$, so $0 = 324 - 36f$ | M1 |
$f = 9$ so to stop before hitting other car, $f > 9$ | A1 | (10)3. A small van of mass 1500 kg is used to tow a car of mass 750 kg by means of a rope of length 9 m joined to both vehicles. The van sets off with the rope slack and reaches a speed of $2 \mathrm {~ms} ^ { - 1 }$ just before the rope becomes taut and jerks the car into motion. Immediately after the rope becomes taut, the van and car travel with common speed $V \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \frac { 4 } { 3 }$.
\item Calculate the magnitude of the impulse on the car when the rope tightens.
The van and car eventually reach a steady speed of $18 \mathrm {~ms} ^ { - 1 }$ with the rope taut when a child runs out into the road, 30 m in front of the van. The van driver brakes sharply and decelerates uniformly to rest in a distance of 27 m .
It takes the driver of the car 1 second to react to the van starting to brake. He then brakes and the car decelerates uniformly at $f \mathrm {~m} \mathrm {~s} ^ { - 2 }$, coming to rest before colliding with the van.
\item Find the set of possible values of $f$.\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q3 [10]}}