| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Coalescence collision |
| Difficulty | Standard +0.3 This is a standard M1 momentum question with straightforward kinematics followed by conservation of momentum. Part (i) uses basic SUVAT equations and momentum conservation for coalescence. Part (ii) requires recognizing that particles may stop before collision and sketching velocity-time graphs, which adds mild complexity but remains routine for M1 level. The multi-part structure and combination of kinematics with momentum is typical but slightly above average difficulty. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| M1 | Use of \(v = u - 0.4t\) | |
| \(p = 4 - 2 \times 0.4 \, (=3.2)\) | A1 | |
| \(q = 1 - 2 \times 0.4 \, (=0.2)\) | A1 | Accept \(q = -0.2\) from \(-1 + 2 \times 0.4\) |
| \(0.7 \times 3.2 - 0.3 \times 0.2 = (1 \times) v\) | M1 | Uses CoLM on reduced velocities |
| \(v = 2.18 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| B1 | Straight line with larger \(y\) intercept slopes towards \(t\) axis, but does not reach it | |
| B1 | Straight line with negative \(y\) intercept slopes towards \(t\) axis | |
| B1 | and gets to \(t\) axis before other line ends. SR if \(t=2\) in ii give B1 if line stops before axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = 1 - 0.4t\) | M1 | Finds when Q comes to rest (any method) |
| \(t = 2.5\) s | A1 | |
| M1 | Uses \(s = ut - 0.4t^2/2\) | |
| \(P = 4 \times 3 - 0.5 \times 0.4 \times 3^2\) | A1 | |
| \(Q = 1 \times 2.5 - 0.5 \times 0.4 \times 2.5^2\) | A1 | nb \(0^{(2)} = 1^{(2)} - 0.4Q^2/2\) B1; convincing evidence (graph to scale, or calculation that Q comes to rest and remains at rest at \(t\) less than 3, M1A1; graph A1 needs \(-ve\) \(v\) intercept) SR if \(t=2\) in iib, allow M1 for \(s = ut - 0.4t^2/2\) |
| \(PQ = 10.2 + 1.25 = 11.45\) m | A1 | And A1 for PQ=8.4 |
# Question 7:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | Use of $v = u - 0.4t$ |
| $p = 4 - 2 \times 0.4 \, (=3.2)$ | A1 | |
| $q = 1 - 2 \times 0.4 \, (=0.2)$ | A1 | Accept $q = -0.2$ from $-1 + 2 \times 0.4$ |
| $0.7 \times 3.2 - 0.3 \times 0.2 = (1 \times) v$ | M1 | Uses CoLM on reduced velocities |
| $v = 2.18 \text{ ms}^{-1}$ | A1 | |
## Part (ii)a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| | B1 | Straight line with larger $y$ intercept slopes towards $t$ axis, but does not reach it |
| | B1 | Straight line with negative $y$ intercept slopes towards $t$ axis |
| | B1 | and gets to $t$ axis before other line ends. **SR** if $t=2$ in ii give B1 if line stops before axis |
## Part (ii)b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = 1 - 0.4t$ | M1 | Finds when Q comes to rest (any method) |
| $t = 2.5$ s | A1 | |
| | M1 | Uses $s = ut - 0.4t^2/2$ |
| $P = 4 \times 3 - 0.5 \times 0.4 \times 3^2$ | A1 | |
| $Q = 1 \times 2.5 - 0.5 \times 0.4 \times 2.5^2$ | A1 | nb $0^{(2)} = 1^{(2)} - 0.4Q^2/2$ B1; convincing evidence (graph to scale, or calculation that Q comes to rest and remains at rest at $t$ less than 3, M1A1; graph A1 needs $-ve$ $v$ intercept) **SR** if $t=2$ in iib, allow M1 for $s = ut - 0.4t^2/2$ |
| $PQ = 10.2 + 1.25 = 11.45$ m | A1 | And A1 for PQ=8.4 |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{470e70de-66ba-4dcc-a205-0c92f29471b1-4_227_901_1352_623}
Two particles $P$ and $Q$ have masses 0.7 kg and 0.3 kg respectively. $P$ and $Q$ are simultaneously projected towards each other in the same straight line on a horizontal surface with initial speeds of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively (see diagram). Before $P$ and $Q$ collide the only horizontal force acting on each particle is friction and each particle decelerates at $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The particles coalesce when they collide.\\
(i) Given that $P$ and $Q$ collide 2 s after projection, calculate the speed of each particle immediately before the collision, and the speed of the combined particle immediately after the collision.\\
(ii) Given instead that $P$ and $Q$ collide 3 s after projection,
\begin{enumerate}[label=(\alph*)]
\item sketch on a single diagram the $( t , v )$ graphs for the two particles in the interval $0 \leqslant t < 3$,
\item calculate the distance between the two particles at the instant when they are projected.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2009 Q7 [15]}}