| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2007 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Difficulty | Challenging +1.2 This is a standard M4 oblique collision problem requiring conservation of momentum (in both directions), Newton's experimental law, and vector calculations. While it involves multiple steps and careful bookkeeping of components, the techniques are routine for this module with no novel insight required. The 13 marks reflect computational length rather than conceptual difficulty, placing it moderately above average. |
| Spec | 6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| CLM: \(2v_2 - v_1 = 1 - 2 = -1\) | M1A1 |
| NIL: \(1 + 1 = \frac{1}{2}(v_1 + v_2)\) | M1A1 |
| \(\therefore v_2 = 1, v_I = 3\) above | DM1 |
| Horizontal components unchanged (i.e. 2 & 3) | A1 |
| Total: 7 marks |
| Answer | Marks |
|---|---|
| Independent of all other marks | |
| \(\mathbf{v}_A = 3\mathbf{i} + \mathbf{j}; \mathbf{v}_B = 2\mathbf{i} - 3\mathbf{j}\) | M1A1 |
| Total: 2 marks |
| Answer | Marks |
|---|---|
| For B: \(I = m(1-(-3)) = 4m\) | M1A1 |
| (Or For A: \(-I = 2m(-1 - 1) \therefore I = 4m\)) | |
| \(\sqrt{\frac{(3)}{(1)}} \sqrt{(3)^2 + (-1)^2} = \sqrt{3^2 + 1^2} \sqrt{3^2 + (-1)^2} \cos\theta\) | M1A1 |
| \(\Rightarrow 8 = 10\cos\theta\) | |
| \(\theta = 37°\) | |
| Total: 4 marks | |
| Alternative: | |
| M1: where \(\tan\theta = \frac{1}{3}\) | |
| A1: required angle is 20 | |
| M1A1: |
## Part (a)
| CLM: $2v_2 - v_1 = 1 - 2 = -1$ | M1A1 |
| NIL: $1 + 1 = \frac{1}{2}(v_1 + v_2)$ | M1A1 |
| $\therefore v_2 = 1, v_I = 3$ above | DM1 |
| Horizontal components unchanged (i.e. 2 & 3) | A1 |
| **Total: 7 marks** | |
## Part (b)
| **Independent of all other marks** | |
| $\mathbf{v}_A = 3\mathbf{i} + \mathbf{j}; \mathbf{v}_B = 2\mathbf{i} - 3\mathbf{j}$ | M1A1 |
| **Total: 2 marks** | |
## Part (c)
| For B: $I = m(1-(-3)) = 4m$ | M1A1 |
| (Or For A: $-I = 2m(-1 - 1) \therefore I = 4m$) | |
| $\sqrt{\frac{(3)}{(1)}} \sqrt{(3)^2 + (-1)^2} = \sqrt{3^2 + 1^2} \sqrt{3^2 + (-1)^2} \cos\theta$ | M1A1 |
| $\Rightarrow 8 = 10\cos\theta$ | |
| $\theta = 37°$ | |
| **Total: 4 marks** | |
| **Alternative:** | |
| M1: where $\tan\theta = \frac{1}{3}$ | |
| A1: required angle is 20 | |
| M1A1: | |
### Guidance:
**a)** M1: Conservation of momentum along the line of centres. Condone sign errors. A1: equation correct. M1: Impact law along the line of centres. e must be used correctly, but condone sign errors. A1: equation correct. The signs need to be consistent between the two equations. M1: Solve the simultaneous equations for their $v_1$ and $v_2$. A1: i components correct – independent mark. A1: $\mathbf{v}_A$ & $\mathbf{v}_B$ correct.
**b)** M1: Impulse = change in momentum for one sphere. Condone order of subtraction. A1: Magnitude correct.
**c)** M1: Any complete method to find the trig ratio of a relevant angle. A1: $\cos\theta = \frac{4}{5}, \tan\frac{\theta}{2} = \frac{1}{3}, ...$ Or M1: find angle of approach to the line of centres and angle after collision. A1: values correct. (both 71.56 .....). M1: solve for $\theta$. A1: 37º (Q specifies nearest degree).
**Special case: candidates who act as if the line of centres is in the direction of i:**
CLM: $u + 2v = 8$
NIL: $v - u = 2$
$u = 4/3, v = 10/3$
$4/3\mathbf{i} + \mathbf{j}; 10/3\mathbf{i} - \mathbf{j}$
Impulse: $2m - 4/3m = 2/3m$
$\frac{10 + 1}{\sqrt{10} \sqrt{\frac{109}{9}}} = \cos\theta$ $\theta = 1.70°$
**Work is equivalent, so treat as MR: M1A0M1A0M1A1A1 M1A1M1A1M1A1M1A1 M1A1M1A1M1A1**
---A smooth uniform sphere $A$ has mass $2m$ kg and another smooth uniform sphere $B$, with the same radius as $A$, has mass $m$ kg. The spheres are moving on a smooth horizontal plane when they collide. At the instant of collision the line joining the centres of the spheres is parallel to $\mathbf{j}$. Immediately after the collision, the velocity of $A$ is $(3\mathbf{i} - \mathbf{j})$ m s$^{-1}$ and the velocity of $B$ is $(2\mathbf{i} + \mathbf{j})$ m s$^{-1}$. The coefficient of restitution between the spheres is $\frac{1}{2}$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocities of the two spheres immediately before the collision. [7]
\item Find the magnitude of the impulse in the collision. [2]
\item Find, to the nearest degree, the angle through which the direction of motion of $A$ is deflected by the collision. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2007 Q5 [13]}}