Standard +0.8 This is a multi-stage collision problem requiring conservation of momentum, Newton's restitution law, and careful tracking of multiple collisions. While the individual principles are standard A-level mechanics, the three-part structure with the second collision introducing a constraint to find e, followed by analyzing when spheres meet again, requires sustained problem-solving across 10 marks. The final part demands setting up and solving equations for when the spheres collide again after B bounces back from the wall, which is more sophisticated than typical single-collision questions.
A small uniform sphere \(A\), of mass \(2m\), is moving with speed \(u\) on a smooth horizontal surface when it collides directly with a small uniform sphere \(B\), of mass \(m\), which is at rest. The spheres have equal radii and the coefficient of restitution between them is \(e\). Find expressions for the speeds of \(A\) and \(B\) immediately after the collision. [4]
Subsequently \(B\) collides with a vertical wall which is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the wall is \(0.4\). After \(B\) has collided with the wall, the speeds of \(A\) and \(B\) are equal. Find \(e\). [2]
Initially \(B\) is at a distance \(d\) from the wall. Find the distance of \(B\) from the wall when it next collides with \(A\). [4]
For A & B use conservation of momentum, e.g.:2mvA + mvB = 2mu
(allow 2vA + vB = 2u) M1
Use Newton’s law of restitution (consistent signs): vB – vA = eu M1
Combine to find v A and vB : vA = (2 – e) u/3, vB = 2(1 + e) u/3 A1, A1
Find e from vA = vB ′ with vB ′ = [–] 0⋅4 vB: (2 – e) = 0⋅8 (1 + e), e = 2/3 M1 A1
EITHER: Equate times in terms of reqd. distance x: (d – x)/ vA = d/vB + x/vB ′ (AEF) M1 A1
[speeds need not be found: vA = vB ′ = 4u/9, vB = 10u/9]
Use vA = vB ′ = 0⋅4 vB to solve for x: d – x = 0⋅4 d + x, x = 0⋅3 d M1 A1
OR: Find dist. moved by A when B reaches wall:
dA = (d/vB) vA = 0⋅4 d (M1 A1)
Answer
Marks
Find reqd. distance x: x = ½ (d – dA) = 0⋅3 d (M1 A1)
4
2
Answer
Marks
Guidance
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10
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Syllabus
Cambridge International A Level – October/November 2015
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Question 2:
2 | For A & B use conservation of momentum, e.g.:2mvA + mvB = 2mu
(allow 2vA + vB = 2u) M1
Use Newton’s law of restitution (consistent signs): vB – vA = eu M1
Combine to find v A and vB : vA = (2 – e) u/3, vB = 2(1 + e) u/3 A1, A1
Find e from vA = vB ′ with vB ′ = [–] 0⋅4 vB: (2 – e) = 0⋅8 (1 + e), e = 2/3 M1 A1
EITHER: Equate times in terms of reqd. distance x: (d – x)/ vA = d/vB + x/vB ′ (AEF) M1 A1
[speeds need not be found: vA = vB ′ = 4u/9, vB = 10u/9]
Use vA = vB ′ = 0⋅4 vB to solve for x: d – x = 0⋅4 d + x, x = 0⋅3 d M1 A1
OR: Find dist. moved by A when B reaches wall:
dA = (d/vB) vA = 0⋅4 d (M1 A1)
Find reqd. distance x: x = ½ (d – dA) = 0⋅3 d (M1 A1) | 4
2
4 | 10
Page 3 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 21
A small uniform sphere $A$, of mass $2m$, is moving with speed $u$ on a smooth horizontal surface when it collides directly with a small uniform sphere $B$, of mass $m$, which is at rest. The spheres have equal radii and the coefficient of restitution between them is $e$. Find expressions for the speeds of $A$ and $B$ immediately after the collision. [4]
Subsequently $B$ collides with a vertical wall which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the wall is $0.4$. After $B$ has collided with the wall, the speeds of $A$ and $B$ are equal. Find $e$. [2]
Initially $B$ is at a distance $d$ from the wall. Find the distance of $B$ from the wall when it next collides with $A$. [4]
\hfill \mbox{\textit{CAIE FP2 2015 Q2 [10]}}