CAIE FP2 2017 June — Question 3 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2017
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeCollision followed by wall impact
DifficultyChallenging +1.2 This is a standard multi-collision momentum problem requiring systematic application of conservation of momentum and Newton's restitution law across three collisions. While it involves multiple steps and careful bookkeeping of velocities, the techniques are routine for Further Maths students and the problem structure is predictable. The 'show that' in part (i) provides scaffolding for part (ii), reducing the problem-solving demand.
Spec6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact

Two uniform small smooth spheres \(A\) and \(B\) have equal radii and each has mass \(m\). Sphere \(A\) is moving with speed \(u\) on a smooth horizontal surface when it collides directly with sphere \(B\) which is at rest. The coefficient of restitution between the spheres is \(\frac{2}{3}\). Sphere \(B\) is initially at a distance \(d\) from a fixed smooth vertical wall which is perpendicular to the direction of motion of \(A\). The coefficient of restitution between \(B\) and the wall is \(\frac{1}{3}\).
  1. Show that the speed of \(B\) after its collision with the wall is \(\frac{5}{18}u\). [4]
  2. Find the distance of \(B\) from the wall when it collides with \(A\) for the second time. [6]

Question 3:

AnswerMarks Guidance
3(i)mv + mv = mu (AEF)
A B*M1 Use conservation of momentum (allow v + v = u)
A B
v – v = ⅔ u
AnswerMarks Guidance
B A*M1 Use Newton’s restitution law (consistent LHS signs)
v = 5u/6
AnswerMarks Guidance
BA1 Combine to find v
B
w = ⅓ v = 5u/18 AG
AnswerMarks Guidance
B BB1 Verify speed w of B after collision with wall (ignore
B
sign)
AnswerMarks Guidance
Total:4
QuestionAnswer Marks

AnswerMarks Guidance
3(ii)v = u / 6
ADA1 Find v (dependent on above *M1 *M1)
A
EITHER:
(d – x) / v = d / v + x / w (AEF)
AnswerMarks Guidance
A B B(M1 A1 EITHER: Equate times in terms of reqd. distance x
6(d – x) = 1⋅2 d + 3⋅6 xM1 A1) Substitute for speeds to formulate an eqn. in x
OR:
x = (d/v ) v = (6d/5u) u/6 = 0⋅2 d
AnswerMarks Guidance
A B A(M1 OR: Find dist. x moved by A when B reaches wall
A
t = (0⋅8 d) / (v + w ) = 9d/5u
AnswerMarks Guidance
2 A BM1 A1 Find remaining time t
2
y = v t = 0⋅3 d or y = w t = 0⋅5 d
AnswerMarks Guidance
A A 2 B B 2A1) Find remaining distance moved by A or B
OR2:
x = (d/v ) v = (6d/5u) u/6 = 0⋅2 d
AnswerMarks Guidance
A B A(M1 OR2: Find dist. x moved by A when B reaches wall
A
(0⋅8 d – x) / v = x / w or 0⋅8 d/(v + w ) = x/w
AnswerMarks Guidance
A B A B BM1 A1 Equate remaining times to formulate an eqn. in x
4⋅8 d – 6 x = 3⋅6 x or 1⋅8 d = 3⋅6 xA1)
x = ½ dA1 Find x
Total:6
Question 3:
--- 3(i) ---
3(i) | mv + mv = mu (AEF)
A B | *M1 | Use conservation of momentum (allow v + v = u)
A B
v – v = ⅔ u
B A | *M1 | Use Newton’s restitution law (consistent LHS signs)
v = 5u/6
B | A1 | Combine to find v
B
w = ⅓ v = 5u/18 AG
B B | B1 | Verify speed w of B after collision with wall (ignore
B
sign)
Total: | 4
Question | Answer | Marks | Guidance
--- 3(ii) ---
3(ii) | v = u / 6
A | DA1 | Find v (dependent on above *M1 *M1)
A
EITHER:
(d – x) / v = d / v + x / w (AEF)
A B B | (M1 A1 | EITHER: Equate times in terms of reqd. distance x
6(d – x) = 1⋅2 d + 3⋅6 x | M1 A1) | Substitute for speeds to formulate an eqn. in x
OR:
x = (d/v ) v = (6d/5u) u/6 = 0⋅2 d
A B A | (M1 | OR: Find dist. x moved by A when B reaches wall
A
t = (0⋅8 d) / (v + w ) = 9d/5u
2 A B | M1 A1 | Find remaining time t
2
y = v t = 0⋅3 d or y = w t = 0⋅5 d
A A 2 B B 2 | A1) | Find remaining distance moved by A or B
OR2:
x = (d/v ) v = (6d/5u) u/6 = 0⋅2 d
A B A | (M1 | OR2: Find dist. x moved by A when B reaches wall
A
(0⋅8 d – x) / v = x / w or 0⋅8 d/(v + w ) = x/w
A B A B B | M1 A1 | Equate remaining times to formulate an eqn. in x
4⋅8 d – 6 x = 3⋅6 x or 1⋅8 d = 3⋅6 x | A1)
x = ½ d | A1 | Find x
Total: | 6
Two uniform small smooth spheres $A$ and $B$ have equal radii and each has mass $m$. Sphere $A$ is moving with speed $u$ on a smooth horizontal surface when it collides directly with sphere $B$ which is at rest. The coefficient of restitution between the spheres is $\frac{2}{3}$. Sphere $B$ is initially at a distance $d$ from a fixed smooth vertical wall which is perpendicular to the direction of motion of $A$. The coefficient of restitution between $B$ and the wall is $\frac{1}{3}$.

\begin{enumerate}[label=(\roman*)]
\item Show that the speed of $B$ after its collision with the wall is $\frac{5}{18}u$. [4]
\item Find the distance of $B$ from the wall when it collides with $A$ for the second time. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP2 2017 Q3 [10]}}