| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 Part (a) is a standard direct collision problem requiring conservation of momentum and Newton's experimental law (coefficient of restitution formula) - routine A-level mechanics with straightforward algebra. Part (b) is a direct application of the impulse-momentum theorem in vector form, requiring only one step of calculation. Both parts are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathrm{C \, of \, M}: \lambda m \times 2u - m \times u = \lambda m \times \frac{11u}{10} + m \times \frac{7u}{2}\) | M1 | In direction of \(B\). 4 terms; condone sign errors. |
| \(\therefore \frac{9.2}{10} = \frac{9}{2}\); \(\therefore \lambda = 5\) | A1 | c.a.o. |
| \(\mathrm{NEL}: e(2u + u) = \frac{7u}{2} - \frac{11u}{10}\) | M1 | Condone sign errors. |
| \(\therefore 3e = \frac{24}{10}\); \(e = \frac{8}{10}\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(3v - 3(2i + 3j - 2k) = 6i - 6j - 9k\) | M1 | Use of 'Impulse = change in mom.' All signs/terms correct. |
| \(\therefore v = (2i + 3j - 2k) + (6i - 6j - 9k)/3 = 4i + j - 5k\) | A1, A1 | All signs/terms correct. \(v\) fully correct. c.a.o. |
### (a)
$\mathrm{C \, of \, M}: \lambda m \times 2u - m \times u = \lambda m \times \frac{11u}{10} + m \times \frac{7u}{2}$ | M1 | In direction of $B$. 4 terms; condone sign errors.
$\therefore \frac{9.2}{10} = \frac{9}{2}$; $\therefore \lambda = 5$ | A1 | c.a.o.
$\mathrm{NEL}: e(2u + u) = \frac{7u}{2} - \frac{11u}{10}$ | M1 | Condone sign errors.
$\therefore 3e = \frac{24}{10}$; $e = \frac{8}{10}$ | A1 | c.a.o. | [4]
### (b)
$3v - 3(2i + 3j - 2k) = 6i - 6j - 9k$ | M1 | Use of 'Impulse = change in mom.' All signs/terms correct.
$\therefore v = (2i + 3j - 2k) + (6i - 6j - 9k)/3 = 4i + j - 5k$ | A1, A1 | All signs/terms correct. $v$ fully correct. c.a.o. | [3]
---\begin{enumerate}[label=(\alph*)]
\item A particle $A$ of mass $m$ travelling with speed $u$ on a smooth horizontal surface collides directly with a particle $B$ of mass $3m$ travelling with speed $\frac{2u}{5}$ in the opposite direction. After the collision, $A$ travels at speed $\frac{2u}{5}$ and $B$ travels at speed $\frac{4u}{15}$, both in the same direction as $B$ before the collision. Find $A$ and the coefficient of restitution between the two particles. [4]
\item A particle of mass 3 kg moving with velocity $(2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) \text{ m s}^{-1}$ receives an impulse of $(6\mathbf{i} - 6\mathbf{j} - 9\mathbf{k})$ N s. Find the velocity of the particle after the impulse. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q10 [7]}}