| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces and application of circular motion equations. Part (i) involves resolving vertically (T cos θ = mg) and using the given T = 3mg to find cos θ, then using horizontal equation to find length. Part (ii) uses v = rω with the radius found from part (i). The problem is slightly easier than average as it's a textbook application of standard techniques with clear given information and straightforward algebra. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| (ii) | mω2 |
| Answer | Marks |
|---|---|
| v = 5.66 ms | M1 |
| Answer | Marks |
|---|---|
| A1 | [3] |
| [4] | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International A Level – May/June 2015 | 9709 | 51 |
| Answer | Marks |
|---|---|
| 6 (i) | x' = 15cos30 ( = 12.990..) |
| Answer | Marks |
|---|---|
| k = –2 | B1 |
| Answer | Marks |
|---|---|
| A1* | [5] |
| Answer | Marks |
|---|---|
| [2] | Or with signs reversed |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme | Syllabus |
| Cambridge International A Level – May/June 2015 | 9709 | 51 |
| Answer | Marks |
|---|---|
| (iii) | ∫vdv ∫−2e−x |
| Answer | Marks |
|---|---|
| r r | M1 |
| Answer | Marks |
|---|---|
| A1 | [4] |
| Answer | Marks |
|---|---|
| [5] | Integrates |
Question 3:
--- 3 (i)
(ii) ---
3 (i)
(ii) | mω2
Tsinθ = r
ω
3ωsinθ = 52 (Lsinθ)
g
L = 1.2 m AG
3ω cosθ = ω
θ = 70.53°
v = 5×1.2sinθ
–1
v = 5.66 ms | M1
A1
A1
M1
A1
M1
A1 | [3]
[4] | 2
Newton’s 2nd law, acceleration = 5 r
and component of T
2
3mgsinθ = m 5 (Lsinθ)
Resolves vertically for P
1
OR θ = cos –1 ,
3
8
–1
θ = sin
9
etc.
v = ωr
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – May/June 2015 | 9709 | 51
4 (i)
(ii)
5 (i)
(ii) (a)
(ii) (b)
6 (i) | x' = 15cos30 ( = 12.990..)
y' = 15sin30 – 3g ( = – 22.5 )
2 2 2
v = (15cos30) + (15sin30 – 3g) or
3g – 15sin30
tanθ =
15cos30
–1
v = 26(.0) ms
θ = 60° to the horizontal
32
g
y = 3(15sin30) –
2
Height = 22.5 m
OR
2 2
( – 22.5) = (15sin30) – 2 × 10y
Height = 22.5 m
18e
0.3g =
0.9
e = 0.15 m
18ext
12 = and ht = 3 – 0.9 – ext
0.9
ht = 1.5 m
0.3×62 0.3u 2
– + 0.3g(0.6 – 0.15)
2 2
18×0.62 18×0.152
= –
2×0.9 2×0.9
0.3u2
=3.375
2
2 2
0.3v = 0.3u + 0.3g(3 – 0.6 – 0.9)
2 2
OR v = u + 2g(3 – 0.6 – 0.9)
–1
v = 7.25 ms
dv
–x
0.1v = – 0.2 e
dx
dv
–x
v = – 2e
dx
k = –2 | B1
B1
M1
A1
A1
M1
A1
M1
A1
M1
A1
M1
A1
M1
A1
M1
A1
M1
A1* | [5]
[2]
[2]
[2]
4
[2] | Or with signs reversed
Or 30° to the vertical
Uses s = ut +1at 2
2
N.B. this is also y'
2 2
Uses v = u + 2as
λx
Uses T =
l
Both ideas needed, ext = 0.6
KE/PE/EE balance up to string
breaking
2
u = 22.5
KE/PE balance after string breaks or
v 2 = u 2 + 2g(ht) using ht from (ii)(a)
7.2456
Newton’s 2nd law, 1 force
Must have negative coefficient
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – May/June 2015 | 9709 | 51
(ii)
(iii)
7 (i)
(ii) (a)
(ii) (b)
(iii) | ∫vdv ∫−2e−x
= dx
v2
–x
= 2e ( + c )
2
2.22
c= −2e0 =0.42
2
2
2
2e−x
= + 0.42
2
x = 0.236
v2
2e−∞
= + 0.42
2
–1
v = 0.917 ms AG
d(0.6 × 0.8 + 0.6 2 ) =
0.4(0.6 × 0.8) – (0.6/3) × 0.6 2
d = 0.143 m
21 × 0.143 = 1.2 P
P = 2.5(0)
F = 21sin45 – 2.5cos45 and
r
R = 21cos45 + 2.5sin45
µ = 0.787
P × 0.6 = 21 × (0.143 + 0.6)
P = 26(.0)
Required F = 26sin45 – 21sin45
r
Max F = 26(.155)
r
As actual F < maxF , no sliding
r r | M1
D*A1
M1
A1
M1
A1
M1
A1
A1
M1
A1
B1
M1
A1
M1
A1
M1
A1
A1 | [4]
[2]
[3]
[2]
[3]
[5] | Integrates
Needs first A1 in (i)
Or uses limits of 2 and 2.2 for v and x
and 0 for x
OR finds x when v = 0.9165
No solution when v = 0.917
Moments about BAD
Exact 1/7
Moments about B
For using F = µR
r
Moments about C
3.5355..
0.787 × (26cos45 + 21cos45)\includegraphics{figure_3}
One end of a light inextensible string is attached to a fixed point $A$ and the other end of the string is attached to a particle $P$. The particle $P$ moves with constant angular speed $5$ rad s$^{-1}$ in a horizontal circle which has its centre $O$ vertically below $A$. The string makes an angle $\theta$ with the vertical (see diagram). The tension in the string is three times the weight of $P$.
\begin{enumerate}[label=(\roman*)]
\item Show that the length of the string is $1.2$ m. [3]
\item Find the speed of $P$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2015 Q3 [7]}}