| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Coalescence collision |
| Difficulty | Moderate -0.8 This is a straightforward M1 momentum question requiring standard application of conservation of momentum for coalescence, then a variant scenario. Part (i) is direct formula application, part (ii) involves basic algebraic manipulation and kinematic calculations with constant velocity. The sketch requires understanding but no complex reasoning. Below average difficulty for A-level as it's purely procedural with no novel problem-solving. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3\times4 - 0.2\times5 = +/-(0.3+0.2)v\) | M1 | Cons of momentum, no \(g\)*, common \(v\) "after" term |
| A1 | \(0.3\times4 + 0.2\times5 = +/-(0.3+0.2)v\) is M1A0A0 | |
| \(v = 0.4\) m s\(^{-1}\) | A1 | Must be positive |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Q\) (or \(P\) at rest) | B1 | If \(P\) moves, allow \(0.3v\) when considering M1 |
| \(0.3\times4 - 0.2\times5 = 0.2v\) | M1, A1 | \(0.3\times4 + 0.2\times5 = 0.2v\) is M1A0A0 |
| \(v = 1\) m s\(^{-1}\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4t + 5t = 3.6\) | M1 | Or \(9t = 3.6\), Or both \(3.6 - x = 4t\) and \(x = 5t\) |
| \(t = 0.4\) | A1 | |
| \(x_Q = 5\times0.4\ (=2)\) | A1 | Finds initial \(Q\) distance. \(3.6 \times 5/(4+5)\) is M1A1A1 |
| \(T = (2/1 =)\ 2\) s | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph: one horizontal, \(+\)ve \(v\) intercept | B1 | |
| One horizontal, \(-\)ve \(v\) intercept, terminates at same \(t\) | B1 | |
| One along \(t\)-axis, starts at same \(t\) as \(+\)ve line ends, label P | B1 | |
| One horizontal above \(t\)-axis, starts at same \(t\) as \(-\)ve line ends | B1 | (Ignore any values put on graphs) |
| [4] |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3\times4 - 0.2\times5 = +/-(0.3+0.2)v$ | M1 | Cons of momentum, no $g$*, common $v$ "after" term |
| | A1 | $0.3\times4 + 0.2\times5 = +/-(0.3+0.2)v$ is M1A0A0 |
| $v = 0.4$ m s$^{-1}$ | A1 | Must be positive |
| **[3]** | | |
## Question 6(ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Q$ (or $P$ at rest) | B1 | If $P$ moves, allow $0.3v$ when considering M1 |
| $0.3\times4 - 0.2\times5 = 0.2v$ | M1, A1 | $0.3\times4 + 0.2\times5 = 0.2v$ is M1A0A0 |
| $v = 1$ m s$^{-1}$ | A1 | |
| **[4]** | | |
## Question 6(ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4t + 5t = 3.6$ | M1 | Or $9t = 3.6$, Or both $3.6 - x = 4t$ and $x = 5t$ |
| $t = 0.4$ | A1 | |
| $x_Q = 5\times0.4\ (=2)$ | A1 | Finds initial $Q$ distance. $3.6 \times 5/(4+5)$ is M1A1A1 |
| $T = (2/1 =)\ 2$ s | A1 | |
| **[4]** | | |
## Question 6(ii)(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph: one horizontal, $+$ve $v$ intercept | B1 | |
| One horizontal, $-$ve $v$ intercept, terminates at same $t$ | B1 | |
| One along $t$-axis, starts at same $t$ as $+$ve line ends, label P | B1 | |
| One horizontal above $t$-axis, starts at same $t$ as $-$ve line ends | B1 | (Ignore any values put on graphs) |
| **[4]** | | |6 Particle $P$ of mass 0.3 kg and particle $Q$ of mass 0.2 kg are 3.6 m apart on a smooth horizontal surface. $P$ and $Q$ are simultaneously projected directly towards each other along a straight line. Before the particles collide $P$ has speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $Q$ has speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Given that the particles coalesce in the collision, calculate their common speed after they collide.\\
(ii) It is given instead that one particle is at rest immediately after the collision.
\begin{enumerate}[label=(\alph*)]
\item State which particle is in motion after the collision and find the speed of this particle.
\item Find the time taken after the collision for the moving particle to return to its initial position.
\item On a single diagram sketch the $( t , v )$ graphs for the two particles, with $t = 0$ as the instant of their initial projection.\\
$7 \quad A$ and $B$ are two points on a line of greatest slope of a plane inclined at $45 ^ { \circ }$ to the horizontal and $A B = 2 \mathrm {~m}$. A particle $P$ of mass 0.4 kg is projected from $A$ towards $B$ with speed $5 \mathrm {~ms} ^ { - 1 }$. The coefficient of friction between the plane and $P$ is 0.2 .
\begin{enumerate}[label=(\roman*)]
\item Given that the level of $A$ is above the level of $B$, calculate the speed of $P$ when it passes through the point $B$, and the time taken to travel from $A$ to $B$.
\item Given instead that the level of $A$ is below the level of $B$,\\
(a) show that $P$ does not reach $B$,\\
(b) calculate the difference in the momentum of $P$ for the two occasions when it is at $A$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR M1 2013 Q6 [15]}}