CAIE FP2 2016 June — Question 2 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeDirect collision with energy loss
DifficultyStandard +0.8 This is a multi-stage collision problem requiring conservation of momentum, Newton's law of restitution applied twice, and an energy condition linking both collisions. Students must track velocities through two separate collision events and solve a non-trivial equation involving e, going beyond standard single-collision exercises.
Spec6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions

2 A small smooth sphere \(A\) of mass \(m\) is moving with speed \(u\) on a smooth horizontal surface when it collides directly with an identical sphere \(B\) which is initially at rest on the surface. The coefficient of restitution between the spheres is \(e\). Sphere \(B\) subsequently collides with a fixed vertical barrier which is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the barrier is \(\frac { 1 } { 2 }\). Given that \(80 \%\) of the initial kinetic energy is lost as a result of the two collisions, find the value of \(e\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(mv_A + mv_B = mu\), \(v_B - v_A = eu\)M1 Use momentum and Newton's law; M0 for inconsistent signs; allow \(v_A + v_B = u\)
\(v_A = \pm(1-e)\,u/2\), \(v_B = \pm(1+e)\,u/2\)A1, A1 Combine to find \(v_A\) and \(v_B\) (may be implied)
\(w_B = \pm \frac{1}{2} v_B [= \pm(1+e)\,u/4]\)B1 M1 A1 Relate speed \(w_B\) of \(B\) after collision with wall to \(v_B\); Relate KEs before and after (AEF); Treat error in 1/5 as M1 A0
\(\frac{1}{2}mu^2 / 5 = \frac{1}{2}mv_A^2 + \frac{1}{2}mv_B^2\)
\(1/5 = (1-e)^2/4 + (1+e)^2/16\)
\(25e^2 - 30e + 9 = (5e-3)^2 = 0\)
\(e = 3/5\) or \(0.6\)M1 A1 Total: [8]
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $mv_A + mv_B = mu$, $v_B - v_A = eu$ | M1 | Use momentum and Newton's law; M0 for inconsistent signs; allow $v_A + v_B = u$ |
| $v_A = \pm(1-e)\,u/2$, $v_B = \pm(1+e)\,u/2$ | A1, A1 | Combine to find $v_A$ and $v_B$ (may be implied) |
| $w_B = \pm \frac{1}{2} v_B [= \pm(1+e)\,u/4]$ | B1 M1 A1 | Relate speed $w_B$ of $B$ after collision with wall to $v_B$; Relate KEs before and after (AEF); Treat error in 1/5 as M1 A0 |
| $\frac{1}{2}mu^2 / 5 = \frac{1}{2}mv_A^2 + \frac{1}{2}mv_B^2$ | | |
| $1/5 = (1-e)^2/4 + (1+e)^2/16$ | | |
| $25e^2 - 30e + 9 = (5e-3)^2 = 0$ | | |
| $e = 3/5$ or $0.6$ | M1 A1 | Total: [8] |

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2 A small smooth sphere $A$ of mass $m$ is moving with speed $u$ on a smooth horizontal surface when it collides directly with an identical sphere $B$ which is initially at rest on the surface. The coefficient of restitution between the spheres is $e$. Sphere $B$ subsequently collides with a fixed vertical barrier which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the barrier is $\frac { 1 } { 2 }$. Given that $80 \%$ of the initial kinetic energy is lost as a result of the two collisions, find the value of $e$.

\hfill \mbox{\textit{CAIE FP2 2016 Q2 [8]}}