| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Moderate -0.3 This is a standard M1 collision problem requiring conservation of momentum and given constraints to form simultaneous equations. While it involves multiple steps (setting up momentum equation, using the speed difference condition, solving), these are routine techniques practiced extensively in M1. The problem is slightly easier than average A-level because it's a textbook application with no conceptual surprises, though the algebra requires care. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks |
|---|---|
| CLM: \(3x3 - 2x2 = 3v + 2(v+1)\) | M1 A1 |
| \(v_F = 0.6 \text{ m s}^{-1}\); \(v_Q = 1.6 \text{ m s}^{-1}\) | M1 A1 (A1 ft) |
| (5) |
| Answer | Marks |
|---|---|
| \(3(v - 3) = 7.2 \text{ N}\) OR \(2(v + 1 - (-2)) = 7.2 \text{ N}\) | M1 A1 ft |
| \(= 7.2 \text{ N}\) | A1 |
| (3) | |
| 8 |
## (a)
CLM: $3x3 - 2x2 = 3v + 2(v+1)$ | M1 A1 |
$v_F = 0.6 \text{ m s}^{-1}$; $v_Q = 1.6 \text{ m s}^{-1}$ | M1 A1 (A1 ft) |
| (5) |
## (b)
$3(v - 3) = 7.2 \text{ N}$ OR $2(v + 1 - (-2)) = 7.2 \text{ N}$ | M1 A1 ft |
$= 7.2 \text{ N}$ | A1 |
| (3) |
| **8** |
---Particle $P$ has mass 3 kg and particle $Q$ has mass 2 kg. The particles are moving in opposite directions on a smooth horizontal plane when they collide directly. Immediately before the collision, $P$ has speed 3 m s$^{-1}$ and $Q$ has speed 2 m s$^{-1}$. Immediately after the collision, both particles move in the same direction and the difference in their speeds is 1 m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of each particle after the collision. [5]
\item Find the magnitude of the impulse exerted on $P$ by $Q$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2011 Q2 [8]}}