Challenging +1.2 This is a standard circular motion problem requiring application of Newton's second law in circular motion and energy conservation. While it involves multiple steps (finding reaction forces at two points, using the given ratio, then finding the point of losing contact), the techniques are routine for Further Maths students. The algebra is straightforward with the given cos α = 3/5, and the problem follows a predictable structure typical of FP2 circular motion questions.
\includegraphics{figure_3}
A smooth cylinder of radius \(a\) is fixed with its axis horizontal. The point \(O\) is the centre of a circular cross-section of the cylinder. The line \(AOB\) is a diameter of this circular cross-section and the radius \(OA\) makes an angle \(\alpha\) with the upward vertical (see diagram). It is given that \(\cos \alpha = \frac{3}{5}\). A particle \(P\) of mass \(m\) moves on the inner surface of the cylinder in the plane of the cross-section. The particle passes through \(A\) with speed \(u\) along the surface in the downwards direction. The magnitude of the reaction between \(P\) and the inner surface of the sphere is \(R_A\) when \(P\) is at \(A\), and is \(R_B\) when \(P\) is at \(B\). It is given that \(R_B = 10R_A\). Show that \(u^2 = ag\). [6]
The particle loses contact with the surface of the cylinder when \(OP\) makes an angle \(\theta\) with the upward vertical. Find the value of \(\cos \theta\). [4]
Use conservation of energy: mvB 2 = mu 2 + 2mga cosα B1
2 2
12ag
2 2
[vB = u + ]
5
mu2
Use F = ma radially at A and B (B1 for either): R = −mgcosα B1
A
a
mv 2
R = B + mgcosα
B
a
mv 2 mu2
Equate RB to 10 RA: B + mgcosα =10 −mgcosα
a a
M1 A1
Eliminate vB 2 : u 2 + 4ag cos α = 10u 2 – 11ag cosα
17ag
[v 2 = ]
B 5
5ag
u 2 = ( ) cosα = ag A.G. M1 A1
Answer
Marks
Guidance
3
6
Page 5
Mark Scheme
Syllabus
Cambridge International A Level – October/November 2014
9231
22
1 1
Use conservation of energy for loss of contact: mv 2 = mu 2 + mga (cosα – cosθ)
2 2
1
or mvB 2 – mga (cosα + cosθ) B1
2
mv2
Use F = ma radially with R = 0: = mg cosθ B1
a
Eliminate v 2 with u 2 = ag to find cosθ: ga + 2ga (cosα – cosθ) = ga cosθ
1 11
cosθ = (2cosα + 1) = M1 A1
Answer
Marks
Guidance
3 15
4
[1 0]
Question 3:
3 | 1 1
Use conservation of energy: mvB 2 = mu 2 + 2mga cosα B1
2 2
12ag
2 2
[vB = u + ]
5
mu2
Use F = ma radially at A and B (B1 for either): R = −mgcosα B1
A
a
mv 2
R = B + mgcosα
B
a
mv 2 mu2
Equate RB to 10 RA: B + mgcosα =10 −mgcosα
a a
M1 A1
Eliminate vB 2 : u 2 + 4ag cos α = 10u 2 – 11ag cosα
17ag
[v 2 = ]
B 5
5ag
u 2 = ( ) cosα = ag A.G. M1 A1
3 | 6
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22
1 1
Use conservation of energy for loss of contact: mv 2 = mu 2 + mga (cosα – cosθ)
2 2
1
or mvB 2 – mga (cosα + cosθ) B1
2
mv2
Use F = ma radially with R = 0: = mg cosθ B1
a
Eliminate v 2 with u 2 = ag to find cosθ: ga + 2ga (cosα – cosθ) = ga cosθ
1 11
cosθ = (2cosα + 1) = M1 A1
3 15 | 4 | [1 0]
\includegraphics{figure_3}
A smooth cylinder of radius $a$ is fixed with its axis horizontal. The point $O$ is the centre of a circular cross-section of the cylinder. The line $AOB$ is a diameter of this circular cross-section and the radius $OA$ makes an angle $\alpha$ with the upward vertical (see diagram). It is given that $\cos \alpha = \frac{3}{5}$. A particle $P$ of mass $m$ moves on the inner surface of the cylinder in the plane of the cross-section. The particle passes through $A$ with speed $u$ along the surface in the downwards direction. The magnitude of the reaction between $P$ and the inner surface of the sphere is $R_A$ when $P$ is at $A$, and is $R_B$ when $P$ is at $B$. It is given that $R_B = 10R_A$. Show that $u^2 = ag$. [6]
The particle loses contact with the surface of the cylinder when $OP$ makes an angle $\theta$ with the upward vertical. Find the value of $\cos \theta$. [4]
\hfill \mbox{\textit{CAIE FP2 2014 Q3 [10]}}