| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Coalescence collision |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring standard application of conservation of momentum for coalescence, basic kinematics with given velocity function (integration for distance, differentiation for acceleration), and a simple collision equation. All parts follow routine procedures with no problem-solving insight needed, making it easier than average A-level maths. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Before mom \(= 0.2\times4 + 0.3\times2.5\) | B1 | Accept with \(g\) |
| \(0.2\times4 + 0.3\times2.5 = (0.2+0.3)v\) | M1 | Accept with \(g\) |
| \(v = 3.1 \text{ ms}^{-1}\) | A1 | Exact. Award if \(g\) used and cancelled |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V_0 = 3.1\) | B1 FT | FT \(\text{cv}(v(\text{i}))\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int 3.1 - 3t^2 \, dt\) | M1* | Uses integration of velocity\((t)\) |
| \(s = 3.1t - 3t^3/3\) \((+c)\) | A1 FT | FT \(\text{cv}(v(\text{i}))\) or \(\text{cv}(V_0(\text{iia}))\) |
| \(CR = [3.1t - t^3]_0^{0.3}\) | D*M1 | Uses their \(s(0.3)\). Award if \(+c\) never shown or assumed \(= 0\) |
| \(CR = 0.903 \text{ m}\) | A1 | Ans not given, so explicit substitution not needed. Allow 0.90, not 0.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = d(V_0 - 3t^2)/dt\) | M1* | Uses differentiation of \(v\) |
| \(a = -6\times0.3\) | D*M1 | Substitutes \(t = 0.3\) (no other value acceptable) |
| \(a = -1.8 \text{ ms}^{-2}\) | A1 | Exact. Must be negative (accept deceleration is -1.8). Award if \(V_0\) wrong but not if \(V_0\) omitted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mom \(C = (0.2+0.3)(3.1 - 3\times0.3^2)\) | B1 | 1.415 |
| Conservation of momentum used, no \(g\) | M1 | Before momentum must be numerical, after momentum needs two terms in \(v\) (accept \(2v\) or \(v\)) |
| \((0.2+0.3)(3.1 - 3\times0.3^2) = 1.5v - 0.5v\) | A1 FT | FT cv(before momentum) |
| \(v = 1.415 \text{ ms}^{-1}\) | A1 | Exact. Accept 1.41 or 1.42 |
# Question 7:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Before mom $= 0.2\times4 + 0.3\times2.5$ | B1 | Accept with $g$ |
| $0.2\times4 + 0.3\times2.5 = (0.2+0.3)v$ | M1 | Accept with $g$ |
| $v = 3.1 \text{ ms}^{-1}$ | A1 | Exact. Award if $g$ used and cancelled |
## Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V_0 = 3.1$ | B1 FT | FT $\text{cv}(v(\text{i}))$ |
## Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int 3.1 - 3t^2 \, dt$ | M1* | Uses integration of velocity$(t)$ |
| $s = 3.1t - 3t^3/3$ $(+c)$ | A1 FT | FT $\text{cv}(v(\text{i}))$ or $\text{cv}(V_0(\text{iia}))$ |
| $CR = [3.1t - t^3]_0^{0.3}$ | D*M1 | Uses their $s(0.3)$. Award if $+c$ never shown or assumed $= 0$ |
| $CR = 0.903 \text{ m}$ | A1 | Ans not given, so explicit substitution not needed. Allow 0.90, not 0.9 |
## Part (ii)(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = d(V_0 - 3t^2)/dt$ | M1* | Uses differentiation of $v$ |
| $a = -6\times0.3$ | D*M1 | Substitutes $t = 0.3$ (no other value acceptable) |
| $a = -1.8 \text{ ms}^{-2}$ | A1 | Exact. Must be negative (accept deceleration is -1.8). Award if $V_0$ wrong but not if $V_0$ omitted |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mom $C = (0.2+0.3)(3.1 - 3\times0.3^2)$ | B1 | 1.415 |
| Conservation of momentum used, no $g$ | M1 | Before momentum must be numerical, after momentum needs two terms in $v$ (accept $2v$ or $v$) |
| $(0.2+0.3)(3.1 - 3\times0.3^2) = 1.5v - 0.5v$ | A1 FT | FT cv(before momentum) |
| $v = 1.415 \text{ ms}^{-1}$ | A1 | Exact. Accept 1.41 or 1.42 |
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\includegraphics[max width=\textwidth, alt={}, center]{f0813713-d677-4ed7-87e1-971a64bdb6ff-4_122_255_1503_561}
The diagram shows two particles $P$ and $Q$, of masses 0.2 kg and 0.3 kg respectively, which move on a horizontal surface in the same direction along a straight line. A stationary particle $R$ of mass 1.5 kg also lies on this line. $P$ and $Q$ collide and coalesce to form a combined particle $C$. Immediately before this collision $P$ has velocity $4 \mathrm {~ms} ^ { - 1 }$ and $Q$ has velocity $2.5 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the velocity of $C$ immediately after this collision.
At time $t \mathrm {~s}$ after this collision the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of $C$ is given by $v = V _ { 0 } - 3 t ^ { 2 }$ for $0 < t \leqslant 0.3$. $C$ strikes $R$ when $t = 0.3$.
\item (a) State the value of $V _ { 0 }$.\\
(b) Calculate the distance $C$ moves before it strikes $R$.\\
(c) Find the acceleration of $C$ immediately before it strikes $R$.
Immediately after $C$ strikes $R$, the particles have equal speeds but move in opposite directions.
\item Find the speed of $C$ immediately after it strikes $R$.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2012 Q7 [15]}}