Standard +0.3 This is a standard A-level mechanics collision problem requiring application of conservation of momentum and Newton's law of restitution. The algebra is straightforward (finding when velocity changes sign), and the setup is typical textbook material. While it requires careful sign conventions and solving a linear inequality, it involves no novel insight or complex multi-step reasoning beyond standard collision mechanics.
Two smooth spheres \(A\) and \(B\), of equal radii and masses \(2m\) and \(m\) respectively, lie at rest on a smooth horizontal table. The spheres \(A\) and \(B\) are projected directly towards each other with speeds \(4u\) and \(3u\) respectively. The coefficient of restitution between the spheres is \(e\). Find the set of values of \(e\) for which the direction of motion of \(A\) is reversed in the collision. [5]
Use conservation of momentum, e.g.: 2mvA + mvB = 8mu – 3mu B1
Use restitution
(must be consistent with prev. eqn.): v A – vB = –e (4u + 3u) B1
1
Solve for vA (or 3vA): vA = (5−7e)u M1
3
1
[vB = (5+14e)u ]
3
5
Find lower limit on e for which vA < 0: 5 – 7e < 0, e > or 0⋅714 M1 A1
Answer
Marks
Guidance
7
5
[5]
Question 1:
1 | Use conservation of momentum, e.g.: 2mvA + mvB = 8mu – 3mu B1
Use restitution
(must be consistent with prev. eqn.): v A – vB = –e (4u + 3u) B1
1
Solve for vA (or 3vA): vA = (5−7e)u M1
3
1
[vB = (5+14e)u ]
3
5
Find lower limit on e for which vA < 0: 5 – 7e < 0, e > or 0⋅714 M1 A1
7 | 5 | [5]
Two smooth spheres $A$ and $B$, of equal radii and masses $2m$ and $m$ respectively, lie at rest on a smooth horizontal table. The spheres $A$ and $B$ are projected directly towards each other with speeds $4u$ and $3u$ respectively. The coefficient of restitution between the spheres is $e$. Find the set of values of $e$ for which the direction of motion of $A$ is reversed in the collision. [5]
\hfill \mbox{\textit{CAIE FP2 2014 Q1 [5]}}