| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Moderate -0.8 This is a straightforward two-part momentum conservation problem requiring only standard M1 techniques: apply conservation of momentum to find B's final speed, then use impulse-momentum theorem (I = Δp) to find m. Both parts involve direct substitution into well-practiced formulas with no conceptual challenges or problem-solving insight required. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| CLM: \(5m \times 3 - 2m \times 4 = 5m \times 0.8 + 2mv\) | M1 A1 | M1 for attempt at CLM equation, correct no. of terms, correct masses, dimensionally consistent. Allow consistent extra \(g\)'s, consistent missing \(m\)'s and sign errors. M0 if masses not paired with correct speeds. First A1 for correct equation. |
| \(v = 1.5\) (Speed is \(1.5 \text{ ms}^{-1}\)) | A1 | Second A1 for \(v = 1.5\). (\(-1.5\) scores A0). N.B. Allow M1 for attempt to equate impulses, must have \(5m(0.8-3)\) or \(5m(3-0.8)\) on one side and \(2m(\pm v \pm 4)\) on other. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse for \(A\): \(5m(0.8 - 3) = -3.3\) | M1 A1 | M1 for attempt at impulse = difference in momenta, for either particle (must be considering one particle). M0 if \(g\)'s included or mass omitted or just \(m\) used. Allow Initial Momentum \(-\) Final Momentum. A1 cao (no ft on their \(v\)) for correct equation in \(m\) only. |
| \(m = 0.3\) | A1 | A1 for \(m = 0.3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse for \(B\): \(2m(1.5 - {-4}) = 3.3\) | M1 A1 | |
| \(m = 0.3\) | A1 |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| CLM: $5m \times 3 - 2m \times 4 = 5m \times 0.8 + 2mv$ | M1 A1 | M1 for attempt at CLM equation, correct no. of terms, correct masses, dimensionally consistent. Allow consistent extra $g$'s, consistent missing $m$'s and sign errors. M0 if masses not paired with correct speeds. First A1 for correct equation. |
| $v = 1.5$ (Speed is $1.5 \text{ ms}^{-1}$) | A1 | Second A1 for $v = 1.5$. ($-1.5$ scores A0). N.B. Allow M1 for attempt to equate impulses, must have $5m(0.8-3)$ or $5m(3-0.8)$ on one side and $2m(\pm v \pm 4)$ on other. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse for $A$: $5m(0.8 - 3) = -3.3$ | M1 A1 | M1 for attempt at impulse = difference in momenta, for either particle (must be considering one particle). M0 if $g$'s included or mass omitted or just $m$ used. Allow Initial Momentum $-$ Final Momentum. A1 cao (no ft on their $v$) for correct equation in $m$ only. |
| $m = 0.3$ | A1 | A1 for $m = 0.3$ |
### Alternative for (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse for $B$: $2m(1.5 - {-4}) = 3.3$ | M1 A1 | |
| $m = 0.3$ | A1 | |\begin{enumerate}
\item Two particles $A$ and $B$, of mass $5 m \mathrm {~kg}$ and $2 m \mathrm {~kg}$ respectively, are moving in opposite directions along the same straight horizontal line. The particles collide directly. Immediately before the collision, the speeds of $A$ and $B$ are $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. The direction of motion of $A$ is unchanged by the collision. Immediately after the collision, the speed of $A$ is $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Find the speed of $B$ immediately after the collision.
\end{enumerate}
In the collision, the magnitude of the impulse exerted on $A$ by $B$ is 3.3 N s .\\
(b) Find the value of $m$.\\
\hfill \mbox{\textit{Edexcel M1 2012 Q1 [6]}}