| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Particle-wall perpendicular collision |
| Difficulty | Moderate -0.8 This is a straightforward application of standard mechanics formulas (coefficient of restitution e = speed after/speed before, impulse = change in momentum, KE loss calculation) with no problem-solving required. All parts are direct substitutions into well-rehearsed formulas, making it easier than a typical A-level question which would require some multi-step reasoning or connection between concepts. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03e Impulse: by a force6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.6 = e \times 2.4 \Rightarrow e = \frac{2}{3}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4.5 \times -1.6 - 4.5 \times 2.4\) | M1 | Attempt at \(mv - mu\); allow sign confusion e.g. \(1.6 - 2.4\); ignore missing units |
| \(= -18\) so \(18\) Ns (or kg ms\(^{-1}\)) | A1 | Allow \(\pm 18\) |
| ...in the final direction of motion of \(P\) | B1 | Could be shown on a diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 4.5 \times 2.4^2 - \frac{1}{2} \times 4.5 \times 1.6^2\) | M1 | Attempt at \(\pm\left(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\right)\) |
| \(7.2\) J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not perfectly elastic since KE is lost (due to the collision) | B1 | Or \(e < 1\) but valid reason must be given. Must mention KE or collision, not just e.g. "energy lost" |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.6 = e \times 2.4 \Rightarrow e = \frac{2}{3}$ | B1 | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4.5 \times -1.6 - 4.5 \times 2.4$ | M1 | Attempt at $mv - mu$; allow sign confusion e.g. $1.6 - 2.4$; ignore missing units |
| $= -18$ so $18$ Ns (or kg ms$^{-1}$) | A1 | Allow $\pm 18$ |
| ...in the final direction of motion of $P$ | B1 | Could be shown on a diagram |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 4.5 \times 2.4^2 - \frac{1}{2} \times 4.5 \times 1.6^2$ | M1 | Attempt at $\pm\left(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\right)$ |
| $7.2$ J | A1 | |
### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not perfectly elastic since KE is lost (due to the collision) | B1 | Or $e < 1$ but valid reason must be given. Must mention KE or collision, not just e.g. "energy lost" |
---2 A particle $P$ of mass 4.5 kg is moving in a straight line on a smooth horizontal surface at a speed of $2.4 \mathrm {~ms} ^ { - 1 }$ when it strikes a vertical wall directly. It rebounds at a speed of $1.6 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of restitution between $P$ and the wall.
\item Determine the impulse applied to $P$ by the wall, stating its direction.
\item Find the loss of kinetic energy of $P$ as a result of the collision.
\item State, with a reason, whether the collision is perfectly elastic.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2020 Q2 [7]}}