Challenging +1.2 This is a standard two-stage collision problem requiring systematic application of conservation of momentum and Newton's restitution law. While it involves multiple collisions and algebraic manipulation to prove the given results, the techniques are routine for Further Maths students and the question provides clear guidance on what to show. The algebra is straightforward once the correct equations are set up, making this moderately above average difficulty but not requiring novel insight.
Two small smooth spheres \(A\) and \(B\) of equal radius have masses \(m\) and \(3m\) respectively. They lie at rest on a smooth horizontal plane with their line of centres perpendicular to a smooth fixed vertical barrier wall \(9\) feet away from the barrier. The coefficient of restitution between \(A\) and \(B\), and between \(B\) and the barrier, is \(e\), where \(e > \frac{1}{4}\). Sphere \(A\) is projected directly towards \(B\) with speed \(u\). Show that after colliding with \(B\) the direction of motion of \(A\) is reversed. [5]
After the impact, \(B\) hits the barrier and rebounds. Show that \(B\) will subsequently collide with \(A\) again unless \(e = 1\). [3]
Use Newton’s law of restitution: vA – vB = – eu M1
Solve for vA : vA = ¼ (1 – 3e) u M1 A1
Use e > ⅓ to find direction of A: 1 < 3e so vA < 0 A.G. B1
Find speed of B before striking barrier: vB = ¼ (1 + e) u A1
2
Find rel. speed or ratio of speeds after collision: e vB + vA = ¼ (1 – e) u
2
or – vA /e vB = 1 – (1 – e) /e(1 + e) M1
Answer
Marks
Guidance
Derive condition for subsequent collision: e vB > –vA unless e = 1 A.G. A1
(5)
(3)
[8]
Page 5
Mark Scheme: Teachers’ version
Syllabus
GCE A LEVEL – October/November 2009
9231
02
Question 3:
3 | Use conservation of momentum: mvA + 3mvB = mu M1
Use Newton’s law of restitution: vA – vB = – eu M1
Solve for vA : vA = ¼ (1 – 3e) u M1 A1
Use e > ⅓ to find direction of A: 1 < 3e so vA < 0 A.G. B1
Find speed of B before striking barrier: vB = ¼ (1 + e) u A1
2
Find rel. speed or ratio of speeds after collision: e vB + vA = ¼ (1 – e) u
2
or – vA /e vB = 1 – (1 – e) /e(1 + e) M1
Derive condition for subsequent collision: e vB > –vA unless e = 1 A.G. A1 | (5)
(3) | [8]
Page 5 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2009 | 9231 | 02
Two small smooth spheres $A$ and $B$ of equal radius have masses $m$ and $3m$ respectively. They lie at rest on a smooth horizontal plane with their line of centres perpendicular to a smooth fixed vertical barrier wall $9$ feet away from the barrier. The coefficient of restitution between $A$ and $B$, and between $B$ and the barrier, is $e$, where $e > \frac{1}{4}$. Sphere $A$ is projected directly towards $B$ with speed $u$. Show that after colliding with $B$ the direction of motion of $A$ is reversed. [5]
After the impact, $B$ hits the barrier and rebounds. Show that $B$ will subsequently collide with $A$ again unless $e = 1$. [3]
\hfill \mbox{\textit{CAIE FP2 2009 Q3 [8]}}