CAIE FP2 2019 June — Question 3 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeThree-particle sequential collisions
DifficultyStandard +0.3 This is a standard multi-collision mechanics problem requiring conservation of momentum and Newton's restitution law applied twice, followed by energy calculation. While it involves multiple steps and careful bookkeeping across two collisions, the techniques are routine for Further Maths students and the problem structure is clearly signposted. The algebraic manipulation is straightforward, making it slightly easier than average for FP2.
Spec6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact
Three uniform small spheres \(A\), \(B\) and \(C\) have equal radii and masses \(2m\), \(4m\) and \(m\) respectively. The spheres are moving in a straight line on a smooth horizontal surface, with \(B\) between \(A\) and \(C\). The coefficient of restitution between each pair of spheres is \(e\). Spheres \(A\) and \(B\) are moving towards each other with speeds \(2u\) and \(u\) respectively. The first collision is between \(A\) and \(B\).
  1. Find the velocities of \(A\) and \(B\) after this collision. [3] Sphere \(C\) is moving towards \(B\) with speed \(\frac{1}{2}u\) and now collides with it. As a result of this collision, \(B\) is brought to rest.
  2. Find the value of \(e\). [4]
  3. Find the total kinetic energy lost by the three spheres as a result of the two collisions. [3]
Question 3:
AnswerMarks Guidance
3(i)2mv + 4mv = 4mu – 4mu [v + 2v = 0] (AEF)
A B A BM1 Use conservation of momentum for A and B (m may be omitted)
v – v = e (2u + u) [v – v = 3eu]
AnswerMarks Guidance
B A B AM1 Use Newton’s restitution law with consistent LHS signs
v = – 2eu and v = eu
AnswerMarks Guidance
A BA1 Combine to find v and v (A0 if directions unclear)
A B
3
AnswerMarks Guidance
3(ii)[4mv ′] + mv = 4mv – (4/3) mu (AEF)
B C BM1 Use conservation of momentum for B & C (m may be omitted)
v [– v ′] = e (v + 4u/3)
AnswerMarks Guidance
C B BM1 Use Newton’s restitution law
4v – (4/3) u = ev + 4eu/3, 4e – 4/3 = e2 + 4e/3
AnswerMarks Guidance
B BM1 Combine to find quadratic equation for e using v ′ = 0
B
3e2 – 8e + 4 = 0, e = 2/3 [v = 4u/3]
AnswerMarks Guidance
CA1 Find value of e, (implicitly) rejecting e = 2
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
3(iii)For A: Loss = ½ 2m(2u)2 - ½ 2m(4/3u)2 = (20/9) mu2
For B: Loss = ½ 4m u2 = ½ mu2
AnswerMarks Guidance
For C: Loss = 0M1 v = – (4/3) u , [v = (2/3) u], v = (4/3) u
A B C
One correct
AnswerMarks
M1Other two correct
E – E or L + L = (38/9) mu2
AnswerMarks Guidance
initial final 1 2A1 Hence find loss in KE
Alternative method for question 3(iii)
E = ½ 2m(2u)2 + ½ 4mu2 + ½ m(4u/3)2
initial
AnswerMarks Guidance
= 4mu2 + 2mu2 + (8/9) mu2 = (62/9) mu2M1 Find initial KE of 3 particles in terms of m and u
E = ½ 2mv 2 [+ ½ 4mv ′ 2 ] + ½ mv 2
final A B C
AnswerMarks Guidance
= (16/9) mu2 + (8/9) mu2 = (24/9) mu2M1 Find final KE of 3 particles in terms of m and u
E – E or L + L = (38/9) mu2
AnswerMarks Guidance
initial final 1 2A1 Hence find loss in KE
Alternative method for question 3(iii)
L = ½ 2m(2u)2 + ½ 4mu2 – ½ 2mv 2 – ½ 4mv 2
1 A B
AnswerMarks Guidance
= 4mu2 + 2mu2 – (16/9) mu2 – (8/9) mu2 = (30/9) mu2M1 Find losses in KE in both collisions in terms of m and u
L = ½ 4mv 2 + ½ m(4u/3)2 – [½ 4mv ′ 2 ] – ½ mv 2
2 B B C
AnswerMarks
= (8/9) mu2 + (8/9) mu2 – (8/9) mu2 = (8/9) mu2M1
E – E or L + L = (38/9) mu2
AnswerMarks Guidance
initial final 1 2A1 Hence find loss in KE
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(i) ---
3(i) | 2mv + 4mv = 4mu – 4mu [v + 2v = 0] (AEF)
A B A B | M1 | Use conservation of momentum for A and B (m may be omitted)
v – v = e (2u + u) [v – v = 3eu]
B A B A | M1 | Use Newton’s restitution law with consistent LHS signs
v = – 2eu and v = eu
A B | A1 | Combine to find v and v (A0 if directions unclear)
A B
3
--- 3(ii) ---
3(ii) | [4mv ′] + mv = 4mv – (4/3) mu (AEF)
B C B | M1 | Use conservation of momentum for B & C (m may be omitted)
v [– v ′] = e (v + 4u/3)
C B B | M1 | Use Newton’s restitution law
4v – (4/3) u = ev + 4eu/3, 4e – 4/3 = e2 + 4e/3
B B | M1 | Combine to find quadratic equation for e using v ′ = 0
B
3e2 – 8e + 4 = 0, e = 2/3 [v = 4u/3]
C | A1 | Find value of e, (implicitly) rejecting e = 2
4
Question | Answer | Marks | Guidance
--- 3(iii) ---
3(iii) | For A: Loss = ½ 2m(2u)2 - ½ 2m(4/3u)2 = (20/9) mu2
For B: Loss = ½ 4m u2 = ½ mu2
For C: Loss = 0 | M1 | v = – (4/3) u , [v = (2/3) u], v = (4/3) u
A B C
One correct
M1 | Other two correct
E – E or L + L = (38/9) mu2
initial final 1 2 | A1 | Hence find loss in KE
Alternative method for question 3(iii)
E = ½ 2m(2u)2 + ½ 4mu2 + ½ m(4u/3)2
initial
= 4mu2 + 2mu2 + (8/9) mu2 = (62/9) mu2 | M1 | Find initial KE of 3 particles in terms of m and u
E = ½ 2mv 2 [+ ½ 4mv ′ 2 ] + ½ mv 2
final A B C
= (16/9) mu2 + (8/9) mu2 = (24/9) mu2 | M1 | Find final KE of 3 particles in terms of m and u
E – E or L + L = (38/9) mu2
initial final 1 2 | A1 | Hence find loss in KE
Alternative method for question 3(iii)
L = ½ 2m(2u)2 + ½ 4mu2 – ½ 2mv 2 – ½ 4mv 2
1 A B
= 4mu2 + 2mu2 – (16/9) mu2 – (8/9) mu2 = (30/9) mu2 | M1 | Find losses in KE in both collisions in terms of m and u
L = ½ 4mv 2 + ½ m(4u/3)2 – [½ 4mv ′ 2 ] – ½ mv 2
2 B B C
= (8/9) mu2 + (8/9) mu2 – (8/9) mu2 = (8/9) mu2 | M1
E – E or L + L = (38/9) mu2
initial final 1 2 | A1 | Hence find loss in KE
3
Question | Answer | Marks | Guidance
Three uniform small spheres $A$, $B$ and $C$ have equal radii and masses $2m$, $4m$ and $m$ respectively. The spheres are moving in a straight line on a smooth horizontal surface, with $B$ between $A$ and $C$. The coefficient of restitution between each pair of spheres is $e$. Spheres $A$ and $B$ are moving towards each other with speeds $2u$ and $u$ respectively. The first collision is between $A$ and $B$.

(i) Find the velocities of $A$ and $B$ after this collision. [3]

Sphere $C$ is moving towards $B$ with speed $\frac{1}{2}u$ and now collides with it. As a result of this collision, $B$ is brought to rest.

(ii) Find the value of $e$. [4]

(iii) Find the total kinetic energy lost by the three spheres as a result of the two collisions. [3]

\hfill \mbox{\textit{CAIE FP2 2019 Q3 [10]}}
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