| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and the coefficient of restitution formula. While it involves simultaneous equations and careful sign conventions for velocities, it follows a well-practiced procedure with no novel insight required. The three-part structure guides students through the solution systematically, making it slightly easier than average for M2 content. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Momentum: \(1 \cdot 2 + 0 \cdot 2u = 0 \cdot 3v + 0 \cdot 4\) | M1 A1 | \(3v - 2u = 8\) |
| Elasticity: \((2 - v)(u - 4) = -\frac{1}{3}\) | M1 A1 | \(3v - u = 2\) |
| Solve: \(u = -6\), \(v = -\frac{4}{3}\) | A1 A1 | |
| (a) \(Q\) before collision: \(6 \text{ ms}^{-1}\) | A1 A1 | |
| (b) \(P\) after collision: \(\frac{4}{3} \text{ ms}^{-1}\) | A1 A1 | |
| (c) K.E. before \(= 0.15(16) + 0.1(36) = 6 \text{ J}\) | M1 A1 | |
| K.E. after \(= 0.15\left(\frac{16}{9}\right) + 0.1(4) = \frac{4}{3} \text{ J}\) | M1 A1 | |
| Loss \(= 5\frac{1}{3} \text{ J}\) | M1 A1 |
Momentum: $1 \cdot 2 + 0 \cdot 2u = 0 \cdot 3v + 0 \cdot 4$ | M1 A1 | $3v - 2u = 8$
Elasticity: $(2 - v)(u - 4) = -\frac{1}{3}$ | M1 A1 | $3v - u = 2$
Solve: $u = -6$, $v = -\frac{4}{3}$ | A1 A1 |
**(a)** $Q$ before collision: $6 \text{ ms}^{-1}$ | A1 A1 |
**(b)** $P$ after collision: $\frac{4}{3} \text{ ms}^{-1}$ | A1 A1 |
**(c)** K.E. before $= 0.15(16) + 0.1(36) = 6 \text{ J}$ | M1 A1 |
K.E. after $= 0.15\left(\frac{16}{9}\right) + 0.1(4) = \frac{4}{3} \text{ J}$ | M1 A1 |
Loss $= 5\frac{1}{3} \text{ J}$ | M1 A1 |
**Total: 12 marks**
---Two particles $P$ and $Q$, of masses 0.3 kg and 0.2 kg respectively, are moving towards each other along a straight line. $P$ has speed 4 ms$^{-1}$. They collide directly. After the collision the direction of motion of both particles has been reversed, and $Q$ has speed 2 ms$^{-1}$. The coefficient of restitution between $P$ and $Q$ is $\frac{1}{3}$. Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $Q$ before the collision, [4 marks]
\item the speed of $P$ after the collision, [4 marks]
\item the kinetic energy, in J, lost in the impact. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [12]}}