| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | String becomes taut problem |
| Difficulty | Moderate -0.3 This is a standard M1 impulse-momentum question requiring conservation of momentum for inelastic collision and impulse calculation. The setup is straightforward with given masses and initial velocity, requiring only direct application of formulas with minimal problem-solving insight. Slightly easier than average due to its routine nature, though the two-particle system and impulse calculation prevent it from being trivial. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) cons. of mom: \(2(7) + 0 = (2 + 1.5)v\), \(v = 4 \text{ ms}^{-1}\) | M2 A1 | |
| (b) impulse \(= \Delta \text{ mom}\) i.e. for \(Q: 1.5(4 - 0) = 6 \text{ Ns}\) | M1 A1 | (5) |
**(a)** cons. of mom: $2(7) + 0 = (2 + 1.5)v$, $v = 4 \text{ ms}^{-1}$ | M2 A1 |
**(b)** impulse $= \Delta \text{ mom}$ i.e. for $Q: 1.5(4 - 0) = 6 \text{ Ns}$ | M1 A1 | (5)
---Two particles, $P$ and $Q$, of mass 2 kg and 1.5 kg respectively are at rest on a smooth, horizontal surface. They are connected by a light, inelastic string which is initially slack. Particle $P$ is projected away from $Q$ with a speed of 7 ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the common speed of the particles after the string becomes taut. [3 marks]
\item Calculate the impulse in the string when it jerks tight. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q1 [5]}}