CAIE FP2 2013 June — Question 2 11 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeSuccessive collisions, three particles in line
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem involving successive collisions with coefficient of restitution. While it requires multiple steps (conservation of momentum, Newton's law of restitution, kinetic energy condition, and checking for further collisions), the techniques are routine for FM students and the problem structure is typical of exam questions. The 'show that' part provides scaffolding, and the logic for deducing no further collisions is straightforward velocity comparison. More challenging than average A-level but standard for Further Maths collision problems.
Spec6.03b Conservation of momentum: 1D two particles

2 Three uniform small smooth spheres, \(A , B\) and \(C\), have equal radii. Their masses are \(4 m , 2 m\) and \(m\) respectively. They lie in a straight line on a smooth horizontal surface with \(B\) between \(A\) and \(C\). Initially \(A\) is moving towards \(B\) with speed \(u , B\) is at rest and \(C\) is moving in the same direction as \(A\) with speed \(\frac { 1 } { 2 } u\). The coefficient of restitution between any two of the spheres is \(e\). The first collision is between \(A\) and \(B\). In this collision sphere \(A\) loses three-quarters of its kinetic energy. Show that \(e = \frac { 1 } { 2 }\). Find the speed of \(B\) after its collision with \(C\) and deduce that there are no further collisions between the spheres.

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(4mv_A + 2mv_B = 4mu\)M1 Use conservation of momentum
\(v_A - v_B = -eu\)M1 Newton's law of restitution (consistent signs)
\(\frac{1}{2}4mv_A^2 = \frac{1}{4}\cdot\frac{1}{2}4mu^2\) \([v_A^2 = \frac{1}{4}u^2]\)M1 Relate \(v_A\) to \(u\) using K.E.
\(v_A = \frac{1}{2}u\), \(v_B = u\), \(e = \frac{1}{2}\)B1 Consider one possible value of \(v_A\)
\(v_A = -\frac{1}{2}u\), \(v_B = 3u\), \(e = 3\frac{1}{2}\)(B1) Consider other value of \(v_A\)
\(v_A = \frac{1}{5}(2-e)u\), \([v_B = \frac{2}{3}(1+e)u]\) Combine first 2 equations to find \(v_A\)
\((2-e)^2 = 9/4\), \(e^2 - 4e + 7/4 = 0\) Find 2 possible values of \(e\) from K.E.
\(e = \frac{1}{2}\) or \(3\frac{1}{2}\)(B1)
\(e \leq 1\) (or \(< 1\)) so \(e = \frac{1}{2}\) A.G.B1 Select one value with reason
\(2mv_B' + mv_C = 2mv_B + \frac{1}{2}mu\)M1 Use conservation of momentum
\(v_B' - v_C = -e(v_B - \frac{1}{2}u)\)M1 Newton's law of restitution (consistent signs)
\(2v_B' + v_C = 5u/2\) and \(v_B' - v_C = -\frac{1}{4}u\) so \(v_B' = \frac{3}{4}u\)M1 A1 Substitute \(v_B = u\) and solve for \(v_B'\)
\(\frac{3}{4}u > \frac{1}{2}u\) \([v_C = u\) not required]B1\(\sqrt{}\) State why no further collisions (\(\sqrt{}\) on \(v_B'\), \(v_A\) provided \(v_B' > v_A\))
Total: 11 marks
## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4mv_A + 2mv_B = 4mu$ | M1 | Use conservation of momentum |
| $v_A - v_B = -eu$ | M1 | Newton's law of restitution (consistent signs) |
| $\frac{1}{2}4mv_A^2 = \frac{1}{4}\cdot\frac{1}{2}4mu^2$ $[v_A^2 = \frac{1}{4}u^2]$ | M1 | Relate $v_A$ to $u$ using K.E. |
| $v_A = \frac{1}{2}u$, $v_B = u$, $e = \frac{1}{2}$ | B1 | Consider one possible value of $v_A$ |
| $v_A = -\frac{1}{2}u$, $v_B = 3u$, $e = 3\frac{1}{2}$ | (B1) | Consider other value of $v_A$ |
| $v_A = \frac{1}{5}(2-e)u$, $[v_B = \frac{2}{3}(1+e)u]$ | | Combine first 2 equations to find $v_A$ |
| $(2-e)^2 = 9/4$, $e^2 - 4e + 7/4 = 0$ | | Find 2 possible values of $e$ from K.E. |
| $e = \frac{1}{2}$ or $3\frac{1}{2}$ | (B1) | |
| $e \leq 1$ (or $< 1$) so $e = \frac{1}{2}$ **A.G.** | B1 | Select one value with reason |
| $2mv_B' + mv_C = 2mv_B + \frac{1}{2}mu$ | M1 | Use conservation of momentum |
| $v_B' - v_C = -e(v_B - \frac{1}{2}u)$ | M1 | Newton's law of restitution (consistent signs) |
| $2v_B' + v_C = 5u/2$ and $v_B' - v_C = -\frac{1}{4}u$ so $v_B' = \frac{3}{4}u$ | M1 A1 | Substitute $v_B = u$ and solve for $v_B'$ |
| $\frac{3}{4}u > \frac{1}{2}u$ $[v_C = u$ not required] | B1$\sqrt{}$ | State why no further collisions ($\sqrt{}$ on $v_B'$, $v_A$ provided $v_B' > v_A$) |

**Total: 11 marks**

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2 Three uniform small smooth spheres, $A , B$ and $C$, have equal radii. Their masses are $4 m , 2 m$ and $m$ respectively. They lie in a straight line on a smooth horizontal surface with $B$ between $A$ and $C$. Initially $A$ is moving towards $B$ with speed $u , B$ is at rest and $C$ is moving in the same direction as $A$ with speed $\frac { 1 } { 2 } u$. The coefficient of restitution between any two of the spheres is $e$. The first collision is between $A$ and $B$. In this collision sphere $A$ loses three-quarters of its kinetic energy. Show that $e = \frac { 1 } { 2 }$.

Find the speed of $B$ after its collision with $C$ and deduce that there are no further collisions between the spheres.

\hfill \mbox{\textit{CAIE FP2 2013 Q2 [11]}}