CAIE FP2 2012 June — Question 1 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeDirect collision with energy loss
DifficultyStandard +0.3 This is a standard direct collision problem requiring application of conservation of momentum and Newton's restitution law, followed by algebraic manipulation to find energy loss. While it involves multiple steps and algebraic work, it follows a well-established procedure taught in mechanics courses with no novel insight required—slightly easier than average for A-level Further Maths.
Spec6.03b Conservation of momentum: 1D two particles

1 Two smooth spheres \(A\) and \(B\), of equal radii and of masses \(3 m\) and \(6 m\) respectively, are at rest on a smooth horizontal surface. Sphere \(A\) is projected directly towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(e\). Show that the kinetic energy lost in the collision between \(A\) and \(B\) is \(m u ^ { 2 } \left( 1 - e ^ { 2 } \right)\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3mv_A + 6mv_B = 3mu\)M1 Conservation of momentum
\(v_A - v_B = -eu\)M1 A1 Newton's law of restitution
\(v_A = \frac{1}{3}(1-2e)u\), \(v_B = \frac{1}{3}(1+e)u\)M1 A1 Solve for \(v_A\) and \(v_B\) (AEF; A0 if any dirn. error)
\(\frac{1}{2}m(3u^2 - 3v_A^2 - 6v_B^2)\) Substitute for \(v_A\), \(v_B\)
\(= mu^2\{3/2 - (1-4e+4e^2)/6 - (1+2e+e^2)/3\}\)
\(= mu^2\{3/2 - \frac{1}{2}(1+2e^2)\}\) Some working needed to earn A1
\(= mu^2(1-e^2)\) A.G.M1 A1 Ignore any earlier dirn. error
Total: 7 marks
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3mv_A + 6mv_B = 3mu$ | M1 | Conservation of momentum |
| $v_A - v_B = -eu$ | M1 A1 | Newton's law of restitution |
| $v_A = \frac{1}{3}(1-2e)u$, $v_B = \frac{1}{3}(1+e)u$ | M1 A1 | Solve for $v_A$ and $v_B$ (AEF; A0 if any dirn. error) |
| $\frac{1}{2}m(3u^2 - 3v_A^2 - 6v_B^2)$ | | Substitute for $v_A$, $v_B$ |
| $= mu^2\{3/2 - (1-4e+4e^2)/6 - (1+2e+e^2)/3\}$ | | |
| $= mu^2\{3/2 - \frac{1}{2}(1+2e^2)\}$ | | Some working needed to earn A1 |
| $= mu^2(1-e^2)$ **A.G.** | M1 A1 | Ignore any earlier dirn. error |

**Total: 7 marks**

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1 Two smooth spheres $A$ and $B$, of equal radii and of masses $3 m$ and $6 m$ respectively, are at rest on a smooth horizontal surface. Sphere $A$ is projected directly towards $B$ with speed $u$. The coefficient of restitution between $A$ and $B$ is $e$. Show that the kinetic energy lost in the collision between $A$ and $B$ is $m u ^ { 2 } \left( 1 - e ^ { 2 } \right)$.

\hfill \mbox{\textit{CAIE FP2 2012 Q1 [7]}}