Edexcel M3 2012 June — Question 3 10 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeTwo strings, two fixed points
DifficultyStandard +0.3 This is a standard M3 circular motion problem with two strings. Students must resolve forces horizontally (centripetal) and vertically (equilibrium), then use geometry to find angles. While it requires careful setup and simultaneous equations, it follows a well-practiced template with no novel insight needed—slightly easier than average for M3 level.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
\includegraphics{figure_1} A particle \(Q\) of mass 5 kg is attached by two light inextensible strings to two fixed points \(A\) and \(B\) on a vertical pole. Each string has length 0.6 m and \(A\) is 0.4 m vertically above \(B\), as shown in Figure 1. Both strings are taut and \(Q\) is moving in a horizontal circle with constant angular speed 10 rad s\(^{-1}\). Find the tension in
  1. \(AQ\),
  2. \(BQ\). [10]
Question 3:
AnswerMarks
3A
q 0.6 m
T
A
0.4 m
Q
T
B
0.6 m
B mg
0.2 1
cosq = = 
0.6 3
Resolve vertically:
( )
T coqs =T cosq +mg T =T +3mg
A B A B
Acceleration towards the centre:
 3 
T sinq+ T sin=q m· 0.6siqn · w 2 T+ T= 5· · 100 =300
A B  A B 5 
Substitute values for w and trig functions and solve to find T or T
A B
T +147+T =300, 2T= 300- 147 =153
B B B
T = 223.5(N) , T =76.5(N)
A B
T =224 or 220 T =76
A B
T =76.5 or 77 T =223
AnswerMarks
B AB1
M1
A2,1,0
M1
A2,1,0
M1
A1,A1
(10)
10
4
(a)
·
AnswerMarks
(b)volume Mass ratio C of M from V
Large 1 2 2 3 3
pa2.2a = pa3 · 2a = a
cone
3 3 4 2
Small 1 1 1 3 7
p a2.a = pa3 a+ a = a
cone
3 3 4 4
S 1 1 1 D
p a2.a = pa3
3 3
3 7
1=D · 2 a- 1· a
2 4
12- 7 5
= a = a **
4 4
V
45° 81.87°
26.6°
C
5a
A
Mg
B
k Mg
( ) ( )
45(cid:176) +26.6(cid:176) =71.6(cid:176) , 81.8698......= 81.9(cid:176)
Take moments about V:
5
Mg· a· cos71=.6 kM· g · 5a cos81.9
4
5cos71.6
k = =1.25
AnswerMarks
4 5cos81.9B1, B1
M1A1
A1
(5)
M1
A2
M1A1
(5)
10
AnswerMarks Guidance
volumeMass ratio C of M from V
Large
AnswerMarks
cone1 2
pa2.2a = pa3
AnswerMarks Guidance
3 32 3 3
· 2a = a
4 2
Small
AnswerMarks
cone1 1
p a2.a = pa3
AnswerMarks Guidance
3 31 3 7
a+ a = a
4 4
AnswerMarks
S1 1
p a2.a = pa3
AnswerMarks Guidance
3 31 D
5(a)
AnswerMarks
(b)A
R
P
α
a
θ
mg
O
Conservation of energy : Loss in GPE = gain in KE
( ) 1
mga coas - cosq = mv2
2
Substitute for cosa and rearrange to given answer:
2mga3  2ga( ) *
v2 =  - cosq = 3- 5cosq
m 5  5
Considering the acceleration towards the centre of the hemisphere:
mv2
mgcosq - R =
a
Substitute for v2 to form expression for R:
mv2 ( )  6
R = mgcosq - = mg 3coqs - 2cosa  = mg3cosq -   
a   5
Loses contact with the surface when R = 0
2
cosq =
5
2ga 2ga
v2 = , v =
AnswerMarks
5 5M1
A2,1,0
A1
(4)
M1
A2,1,0
DM1
A1
M1
A1
A1
(8)
12
AnswerMarks
Alt:mn 2
R = 0 ⇒mgcosq =
a
n 2
cosq =
ga
2ga n 2 
Substitute in given (a) n 2 = 3- 5 
5  ga
6ga 6ga
n 2 = - n2 2, 3n 2 =
5 5
2ga
v =
AnswerMarks
5DM1
A1
M1
A1
A1
2a3
D = ( 3 ) = 2a **
3- p a2 3 3- p
AnswerMarks
3A1
(6)
12
7(a)
(b)
(c)
AnswerMarks
(d)lx
Use of T = = mg
a
24.5x
T = = 0.5g
0.75
0.75· 0.5g
x = = 0.15, AE = 0.75+0.15=0.9(m) (**)
24.5
Using gain in EPE = loss in GPE
lx2 24.5x2
= =.....
2a 1.5
….. = 0.5g(0.75 + x)
Form quadratic in x and attempt to solve for x :
24.5x2 =5.5125+7.35x, 24.5x2- 7.35x- 5.5125= 0,
7.3– 5 7.35+2 4· 24.5· 5.5125
x =
49
12– 144+3600
(or 40x2- 12x- 9 =0, x = )
80
x =0.647...(m) AC » 1.4(m)
Using F = ma and displacement x from E:
24.5(x+0.15)
0.5g - = 0.5 & x &
0.75
196
& x=& - x, so SHM
3
Max speed = their a x their ω
196
= (0.647- 0.15)·
3
AnswerMarks
» 4.0 ms-1 (4.02)M1
A1
A1
(3)
M1
A1
A1
DM1
A1
(5)
M1
A2,1,0
A1
M1
(4)
A1
(2)
14
Question 3:
3 | A
q 0.6 m
T
A
0.4 m
Q
T
B
0.6 m
B mg
0.2 1
cosq = = 
0.6 3
Resolve vertically:
( )
T coqs =T cosq +mg T =T +3mg
A B A B
Acceleration towards the centre:
 3 
T sinq+ T sin=q m· 0.6siqn · w 2 T+ T= 5· · 100 =300
A B  A B 5 
Substitute values for w and trig functions and solve to find T or T
A B
T +147+T =300, 2T= 300- 147 =153
B B B
T = 223.5(N) , T =76.5(N)
A B
T =224 or 220 T =76
A B
T =76.5 or 77 T =223
B A | B1
M1
A2,1,0
M1
A2,1,0
M1
A1,A1
(10)
10
4
(a)
·
(b) | volume Mass ratio C of M from V
Large 1 2 2 3 3
pa2.2a = pa3 · 2a = a
cone
3 3 4 2
Small 1 1 1 3 7
p a2.a = pa3 a+ a = a
cone
3 3 4 4
S 1 1 1 D
p a2.a = pa3
3 3
3 7
1=D · 2 a- 1· a
2 4
12- 7 5
= a = a **
4 4
V
45° 81.87°
26.6°
C
5a
A
Mg
B
k Mg
( ) ( )
45(cid:176) +26.6(cid:176) =71.6(cid:176) , 81.8698......= 81.9(cid:176)
Take moments about V:
5
Mg· a· cos71=.6 kM· g · 5a cos81.9
4
5cos71.6
k = =1.25
4 5cos81.9 | B1, B1
M1A1
A1
(5)
M1
A2
M1A1
(5)
10
volume | Mass ratio | C of M from V
Large
cone | 1 2
pa2.2a = pa3
3 3 | 2 | 3 3
· 2a = a
4 2
Small
cone | 1 1
p a2.a = pa3
3 3 | 1 | 3 7
a+ a = a
4 4
S | 1 1
p a2.a = pa3
3 3 | 1 | D
5(a)
(b) | A
R
P
α
a
θ
mg
O
Conservation of energy : Loss in GPE = gain in KE
( ) 1
mga coas - cosq = mv2
2
Substitute for cosa and rearrange to given answer:
2mga3  2ga( ) *
v2 =  - cosq = 3- 5cosq
m 5  5
Considering the acceleration towards the centre of the hemisphere:
mv2
mgcosq - R =
a
Substitute for v2 to form expression for R:
mv2 ( )  6
R = mgcosq - = mg 3coqs - 2cosa  = mg3cosq -   
a   5
Loses contact with the surface when R = 0
2
cosq =
5
2ga 2ga
v2 = , v =
5 5 | M1
A2,1,0
A1
(4)
M1
A2,1,0
DM1
A1
M1
A1
A1
(8)
12
Alt: | mn 2
R = 0 ⇒mgcosq =
a
n 2
cosq =
ga
2ga n 2 
Substitute in given (a) n 2 = 3- 5 
5  ga
6ga 6ga
n 2 = - n2 2, 3n 2 =
5 5
2ga
v =
5 | DM1
A1
M1
A1
A1
2a3
D = ( 3 ) = 2a **
3- p a2 3 3- p
3 | A1
(6)
12
7(a)
(b)
(c)
(d) | lx
Use of T = = mg
a
24.5x
T = = 0.5g
0.75
0.75· 0.5g
x = = 0.15, AE = 0.75+0.15=0.9(m) (**)
24.5
Using gain in EPE = loss in GPE
lx2 24.5x2
= =.....
2a 1.5
….. = 0.5g(0.75 + x)
Form quadratic in x and attempt to solve for x :
24.5x2 =5.5125+7.35x, 24.5x2- 7.35x- 5.5125= 0,
7.3– 5 7.35+2 4· 24.5· 5.5125
x =
49
12– 144+3600
(or 40x2- 12x- 9 =0, x = )
80
x =0.647...(m) AC » 1.4(m)
Using F = ma and displacement x from E:
24.5(x+0.15)
0.5g - = 0.5 & x &
0.75
196
& x=& - x, so SHM
3
Max speed = their a x their ω
196
= (0.647- 0.15)·
3
» 4.0 ms-1 (4.02) | M1
A1
A1
(3)
M1
A1
A1
DM1
A1
(5)
M1
A2,1,0
A1
M1
(4)
A1
(2)
14
\includegraphics{figure_1}

A particle $Q$ of mass 5 kg is attached by two light inextensible strings to two fixed points $A$ and $B$ on a vertical pole. Each string has length 0.6 m and $A$ is 0.4 m vertically above $B$, as shown in Figure 1.

Both strings are taut and $Q$ is moving in a horizontal circle with constant angular speed 10 rad s$^{-1}$.

Find the tension in
\begin{enumerate}[label=(\roman*)]
\item $AQ$,
\item $BQ$. [10]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2012 Q3 [10]}}
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