CAIE FP2 2013 June — Question 3 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeDirect collision with energy loss
DifficultyChallenging +1.2 This is a two-collision momentum problem requiring systematic application of conservation of momentum, Newton's restitution law, and energy considerations across multiple stages. While it involves several steps and careful bookkeeping of velocities through two collisions, the techniques are standard for Further Maths mechanics and the problem structure is clearly signposted. The algebraic manipulation is moderate but straightforward.
Spec6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact

3 Two uniform small smooth spheres \(A\) and \(B\), of masses \(m\) and \(2 m\) respectively, and with equal radii, are at rest on a smooth horizontal surface. Sphere \(A\) is projected directly towards \(B\) with speed \(u\), and collides with \(B\). After this collision, sphere \(B\) collides directly with a fixed smooth vertical barrier. The total kinetic energy of the spheres after this second collision is equal to one-ninth of its value before the first collision. Given that the coefficient of restitution between \(B\) and the barrier is 0.5 , find the coefficient of restitution between \(A\) and \(B\).

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(mv_A + 2mv_B = mu\)B1 Use conservation of momentum
\(v_A - v_B = -eu\)B1 Use restitution (must be consistent with prev. eqn.)
\(v_B' = \frac{1}{2}v_B\)M1 Find speed of \(B\) after striking barrier (ignore sign)
\((\frac{1}{2}mu^2)/9 = \frac{1}{2}mv_A^2 + \frac{1}{2}(2m)v_B'^2\)M1 Relate K.E. before and after collision
\(v_A = \frac{1}{3}(1-2e)u,\ v_B = \frac{1}{3}(1+e)u\)M1 A1 *EITHER*: Solve first two eqns for \(v_A\) and \(v_B\) (A.E.F.)
\(u^2/9 = (1-2e)^2u^2/9 + \frac{1}{2}(1+e)^2u^2/9\)A1 Substitute for \(v_A\), \(v_B'\) in KE eqn
\(9e^2 - 6e + 1 = 0,\ e = \frac{1}{3}\)M1 A1 Simplify and solve for \(e\)
\(81v_A^2 - 18uv_A + u^2 = 0\) or \(81v_B^2 - 72uv_B + 16u^2 = 0\)(M1 A1) *OR*: Use \(v_A + 2v_B = u\) in KE eqn
\(v_A = u/9\) and \(v_B = 4u/9\)(A1) Solve for \(v_A\) and \(v_B\)
\(e = (4u/9 - u/9)/u = \frac{1}{3}\)(M1 A1) Find \(e\) from restitution eqn
Total: 9 marks
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $mv_A + 2mv_B = mu$ | B1 | Use conservation of momentum |
| $v_A - v_B = -eu$ | B1 | Use restitution (must be consistent with prev. eqn.) |
| $v_B' = \frac{1}{2}v_B$ | M1 | Find speed of $B$ after striking barrier (ignore sign) |
| $(\frac{1}{2}mu^2)/9 = \frac{1}{2}mv_A^2 + \frac{1}{2}(2m)v_B'^2$ | M1 | Relate K.E. before and after collision |
| $v_A = \frac{1}{3}(1-2e)u,\ v_B = \frac{1}{3}(1+e)u$ | M1 A1 | *EITHER*: Solve first two eqns for $v_A$ and $v_B$ (A.E.F.) |
| $u^2/9 = (1-2e)^2u^2/9 + \frac{1}{2}(1+e)^2u^2/9$ | A1 | Substitute for $v_A$, $v_B'$ in KE eqn |
| $9e^2 - 6e + 1 = 0,\ e = \frac{1}{3}$ | M1 A1 | Simplify and solve for $e$ |
| $81v_A^2 - 18uv_A + u^2 = 0$ or $81v_B^2 - 72uv_B + 16u^2 = 0$ | (M1 A1) | *OR*: Use $v_A + 2v_B = u$ in KE eqn |
| $v_A = u/9$ and $v_B = 4u/9$ | (A1) | Solve for $v_A$ and $v_B$ |
| $e = (4u/9 - u/9)/u = \frac{1}{3}$ | (M1 A1) | Find $e$ from restitution eqn |

**Total: 9 marks**

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3 Two uniform small smooth spheres $A$ and $B$, of masses $m$ and $2 m$ respectively, and with equal radii, are at rest on a smooth horizontal surface. Sphere $A$ is projected directly towards $B$ with speed $u$, and collides with $B$. After this collision, sphere $B$ collides directly with a fixed smooth vertical barrier. The total kinetic energy of the spheres after this second collision is equal to one-ninth of its value before the first collision. Given that the coefficient of restitution between $B$ and the barrier is 0.5 , find the coefficient of restitution between $A$ and $B$.

\hfill \mbox{\textit{CAIE FP2 2013 Q3 [9]}}