M3 Circular Motion - Vertical Circles

8/9/2019 M3 Circular Motion - Vertical Circles

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1.

A particle P of massm is attached to one end of a light inextensible string of lengtha . The otherend of the string is fixed at the pointO. The particle is initially held withOP horizontal and thestring taut. It is then projected vertically upwards with speedu, whereu2 !ag . "hen OP hasturned through an angleθ the speed of P is v and the tension in the string is T , as shownin the diagram above.

(a) Find, in terms of a , g and θ , an expression for v 2.(3)

(b) Find, in terms of m, g and θ , an expression for T .(4)

(c) Prove that P moves in a complete circle.(3)

(d) Find the maximum speed of P .(2)

(Total 12 marks)

Edexcel Internal Review 1


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2.

One end A of a light inextensible string of length 3 a is attached to a fixed point. particle of mass m is attached to the other end B of the string. !he particle is held ine"uilibrium at a distance 2a below the hori#ontal through A, with the string taut. !heparticle is then pro$ected with speed #$2ag ), in the direction perpendicular to AB, in thevertical plane containing A and B, as shown in the diagram above. %n the subse"uentmotion the string remains taut. &hen AB is at an angle θ below the hori#ontal, thespeed of the particle is v and the tension in the string is T .

(a) 'how that v 2 2 ag (3 sin θ % &'.(5)

(b) Find the range of values of T .(6)

(Total 11 marks)

3. One end of a light inextensible string of length l is attached to a fixed point A. !he other end is attached to a particle P of mass m, which is held at a point B with the string taut

and AP ma ing an angle arccos (&

with the downward vertical. !he particle is releasedfrom rest. &hen AP ma es an angle θ with the downward vertical, the string is taut andthe tension in the string is T .

(a) 'how that

.2

cos) mg mg T −= θ

(6)



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t an instant when AP ma es an angle of *+ * to the downward vertical,P is movingupwards, as shown in the diagram. t this instant the string brea s. t the highest pointreached in the subse"uent motion, P is at a distance d below the hori#ontal through A.

(b) Find d in terms of l .(5)

(Total 11 marks)

4.

particle is pro$ected from the highest point A on the outer surface of a fixed smoothsphere of radius a and centre O. !he lowest point B of the sphere is fixed to a

hori#ontal plane. !he particle is pro$ected hori#ontall from A with speed'$

2&

ga. !he

particle leaves the surface of the sphere at the point C , where ,θ =∠ AOC and stri esthe plane at the point P , as shown in the diagram above.



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(a) 'how that cos θ ()

.(7)

(b) Find the angle that the velocit of the particle ma es with the hori#ontal as it

reaches P .(8)

(Total 15 marks)

5. particle P of mass m is attached to one end of a light inextensible string of length a .

!he other end of the string is attached to a fixed point O. !he particle is released fromrest with the string taut and OP hori#ontal.

(a) Find the tension in the string when OP ma es an angle of *+ * with the downwardvertical.

(6)

particle Q of mass 3 m is at rest at a distance a verticall below O. &hen P stri es Q the particles $oin together and the combined particle of mass - m starts to move in avertical circle with initial speed u.

(b) 'how that

=+

gau

.(3)

!he combined particle comes to instantaneous rest at A.

(c) Find

(i) the angle that the string ma es with the downward vertical when thecombined particle is at A,

(ii) the tension in the string when the combined particle is at A.(6)

(Total 15 marks)



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6.

A

T

P

v

O

√$ !2 g a'

a

particle P of mass m is attached to one end of a light inextensible string of length a .!he other end of the string is attached to a fixed point O. t time t +, P is pro$ected

verticall downwards with speed ga

2!

from a point A which is at the same level as Oand a distance a from O. &hen the string has turned through an angle θ and the stringis still taut, the speed of P is v and the tension in the string is T , as shown in thediagram above.

(a) 'how that'sin(!$

22 θ += gav

(3)

(b) Find T in terms of m, g and θ .(3)

!he string becomes slac when θ = α .

(c) Find the value of α.(3)

!he particle is pro$ected again from A with the same velocit as before. &hen P is atthe same level as O for the first time after leaving A, the string meets a small smooth

peg B which has been fixed at a distancea

2&

from O. !he particle now moves on an

arc of a circle centre B. iven that the particle reaches the point C , which isa

2&

verticall above the point B, without the string going slac ,

(d) find the tension in the string when P is at the point C .(6)

(Total 15 marks)



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7. particle P is free to move on the smooth inner surface of a fixed thin hollow sphere ofinternal radius a and centre O. !he particle passes through the lowest point of thespherical surface with speed . !he particle loses contact with the surface when OP isinclined at an angle α to the upward vertical.

(a) 'how that 2 ag (2 / 3cos α).(7)

!he particle has speed ! as it passes through the level of O. iven that cos α =,

)&

(b) show that .)2 ag W =

(5)

(Total 12 marks)

8.

O

P

v

A √$ ) a g '

T

particle P of mass m is attached to one end of a light inextensible string of length a .!he other end of the string is attached to a point O. !he point A is verticall below O,and OA a . !he particle is pro$ected hori#ontall from A with speed #$)ag ). &hen OP ma es an angle θ with the upward vertical through O and the string is still taut, thetension in the string is T and the speed of P is v , as shown in the diagram above.

(a) Find, in terms of a , g and , an expression forv 2.(3)

(b) 'how that T (0 % ) cosθ )mg .(3)

Edexcel Internal Review "


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!he string becomes slac when P is at the point B.

(c) Find, in terms of a , the vertical height of B above A.(2)

fter the string becomes slac , the highest point reached b P is C .

(d) Find, in terms of a , the vertical height of C above B.(5)

(Total 13 marks)

9. One end of a light inextensible string of length l is attached to a particle P of mass m.!he other end is attached to a fixed point A. !he particle is hanging freel at rest with

the string vertical when it is pro$ected hori#ontall with speed.

2! gl

(a) Find the speed of P when the string is hori#ontal.(4)

&hen the string is hori#ontal it comes into contact with a small smooth fixed peg whichis at the point B, where AB is hori#ontal, and AB 1 l . iven that the particle thendescribes a complete semicircle with centre B,

(b) find the least possible value of the length AB.(9)

(Total 13 marks)

10. One end of a light inextensible string of length l is attached to a fixed point A. !he other end is attached to a particle P of mass m which is hanging freel at rest at a point B.!he particle P is pro$ected hori#ontall from B with speed √(3gl ). &hen AP ma es anangle θ with the downward vertical and the string remains taut, the tension in the stringis T#

(a) 'how that T mg(0 / 3 cos θ ).(6)

(b) Find the speed of P at the instant when the string becomes slac .(3)

Edexcel Internal Review $


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(c) Find the maximum height above the level of B reached b P .(5)

(Total 14 marks)

11. smooth solid sphere, with centre O and radius a , is fixed to the upper surface of ahori#ontal table. particle P is placed on the surface of the sphere at a point A, where

OA ma es an angle a with the upward vertical, and + 1 a 1 2π

. !he particle is releasedfrom rest. &hen OP ma es an angle % with the upward vertical, and P is still on thesurface of the sphere, the speed of P is v .

(a) 'how that v 2 2 ga (cos a % cos% ).(4)

iven that cos a ()

, find

(b) the value of % when P loses contact with the sphere,(5)

(c) the speed of P as it hits the table.(4)

(Total 13 marks)

12.

O

Q P A

! m & 2 - .

Edexcel Internal Review &


9/41

trape#e artiste of mass *+ g is attached to the end A of a light inextensible rope OA of length m. !he artiste must swing in an arc of a vertical circle, centre O, from aplatform P to another platform Q, where PQ is hori#ontal. !he other end of the rope isattached to the fixed point O which lies in the vertical plane containing PQ , with ∠POQ

02+ * andOP OQ m, as shown in the diagram above.

As part of her act, the artiste projects herself fromP with speed √0 m s %& in a directionperpendicular to the rope OA and in the plane POQ . 'he moves in a circular arctowards Q. t the lowest point of her path she catches a ball of mass m g which istravelling towards her with speed 3 m s %& and parallel to QP . fter catching the ball, shecomes to rest at the point Q.

modelling the artiste and the ball as particles and ignoring her air resistance, find

(a) the speed of the artiste immediatel before she catches the ball,(4)

(b) the value of m,(7)

(c) the tension in the rope immediatel after she catches the ball.(3)

(Total 14 marks)

13.

! - m

B

A

C

D

/ -

) -

!he diagram above represents the path of a s ier of mass 4+ g moving on a s i5slope ABC' . !he path lies in a vertical plane. From A to B, the path is modelled as a straightline inclined at *+ * to the horizontal. 0romB to ' , the path is modelled as an arc of avertical circle of radius + m. !he lowest point of the arc B' is C .

Edexcel Internal Review (


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t B, the s ier is moving downwards with speed 2+ m s %&. t ' , the path is inclined at3+* to the horizontal and the s1ier is moving upwards. y modelling the slope as smooth andthe s1ier as a particle, find

(a) the speed of the s ier at C ,(3)

(b) the normal reaction of the slope on the s ier at C ,(3)

(c) the speed of the s ier at ' ,(3)

(d) the change in the normal reaction of the slope on the s ier as she passes B.(4)

!he model is refined to allow for the influence of friction on the motion of the s ier.

(e) 'tate briefl , with a reason, how the answer to part (b) would be affected b usingsuch a model. (6o further calculations are expected.)

(2)(Total 15 marks)

Edexcel Internal Review 1)


11/41

14.

O

T

P

v

particle P of mass m is attached to one end of a light inextensible string of length a .!he other end of the string is fixed at a point O. !he particle is held with the string taut

and OP hori#ontal. %t is then projected vertically downwards with speedu, where u2 * 2)

ga .&hen OP has turned through an angle % and the string is still taut, the speed of P is v

and the tension in the string is T+as shown in the diagram above.

(a) Find an expression for v 2 in terms of a , g and %#(2)

(b) Find an expression for T in terms of m, g and %#(3)

(c) Prove that the string becomes slac when % 20+ *. (2)

(d) 'tate, with a reason, whether P would complete a vertical circle if the string werereplaced b a light rod.

(2)



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fter the string becomes slac , P moves freel under gravit and is at the same levelas O when it is at the point A.

(e) 7xplain briefl wh the speed of P at A is √( ) ga2)

.

(1)

!he direction of motion of P at A ma es an angle , with the hori#ontal.

(f) Find , .(4)

(Total 14 marks)

15.

A

B

C

O

a

particle is at the highest point A on the outer surface of a fixed smooth sphere ofradius a and centre O. !he lowest point B of the sphere is fixed to a hori#ontal plane.!he particle is pro$ected hori#ontall from A with speed u, where u 1 √(ag ). !he particleleaves the sphere at the point C , where OC ma es an angle % with the upward vertical,as shown in the diagram above.



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(a) Find an expression for cos % in terms of u, g and a .(7)

!he particle stri es the plane with speed

2

3ag

.

(b) Find, to the nearest degree, the value of %.(7)

(Total 14 marks)

16.

C

A

B

O

a& 2 - *

Part of a hollow spherical shell, centre O and radius a , is removed to form a bowl with aplane circular rim. !he bowl is fixed with the circular rim uppermost and hori#ontal. !he

point A is the lowest point of the bowl. !he point B is on the rim of the bowl and ∠ AOB 02+ °, as shown in the diagram above. smooth small marble of mass m is placed

inside the bowl at A and given an initial hori#ontal speed u. !he direction of motion ofthe marble lies in the vertical plane AOB . !he marble sta s in contact with the bowluntil it reaches B. &hen the marble reaches B, its speed is v .

(a) Find an expression for v 2.(3)

(b) For the case when u2 * ga , find the normal reaction of the bowl on the marbleas the marble reaches B.

(3)

(c) Find the least possible value of u for the marble to reach B.(3)

!he point C is the other point on the rim of the bowl l ing in the vertical plane OAB .

(d) Find the value of u which will enable the marble to leave the bowl at B and meet itagain at the point C .

(7)(Total 16 marks)



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17. smooth solid hemisphere is fixed with its plane face on a hori#ontal table and itscurved surface uppermost. !he plane face of the hemisphere has centre O and radiusa . !he point A is the highest point on the hemisphere. A particleP is placed on the

hemisphere at A. %t is then given an initial horizontal speedu, where u2 2&

(ag ). &hen

OP ma es an angle % with OA, and while P remains on the hemisphere, the speed of P is v .

(a) Find an expression for v 2.(3)

(b) 'how that, when % arccos +.8, P is still on the hemisphere.(5)

(c) Find the value of cos % when P leaves the hemisphere.(2)

(d) Find the value of v when P leaves the hemisphere.(2)

fter leaving the hemisphere P stri es the table at B.

(e) Find the speed of P at B.(2)

(f) Find the angle at which P stri es the table. (3)(Total 17 marks)

18. (a)

2& &4nergy5 sin !2 2

mga m ag mvθ = × −90 0

2 ! 2 sinv ag ag θ = − 0 3



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16/41

(b) 46n of motion along radius5

2

sin mv

T mg a

θ + =90 0

( )! 2 sin sinmT ag ag mg a

θ θ = − −90

( )! )sinT mg θ = − 0 -

(c) At , 3-C θ = °

( )! ) 2T mg mg = − =90 0

- reachesT P C ∴> 0 3

(d) 7ax speed at lowest point

( 28-θ = °:2 ! 2 sin 28-v ag ag = − ) 902 ! 2v ag ag = +

v #$8ag ) 0 2[12]

19. (a)

( )θ sin) % 22&

% 22& 2 aamg mvag m =×

90 0 0

leading to v 2 2 ga (3sin θ %&' 9 cso 7& A& !

Edexcel Internal Review 1"


17/41

(b) minimum value of T is when v + ⇒ )&

sin =θ 0

)sin

mg mg T == θ

90 0

maximum value of T is when 2π

θ =(v 2 - ag )

:mg

amv

T +=) 90 0

)8mg =

0 *

≤≤

)8

)mg

T mg

[11]

20. (a)

7nerg ;

22

2&

(&

% cos2&

mvmgl mu =

+ θ

90 0


18/41

(b)

'(&

% 2&

$2/- 2

mgl mv =⇒°=θ 90

22 gl v =⇒

vertical motion under gravit ;

; - $v cos3+ *'2 % 2g- 90

&/)

2 % 2)

2-

l s gs

gl =⇒×= 0

>istance below &/!

&/) %

2l l l = 90 0

lternative

( ) mgd vmmgl mv % /-cos2&

/-cos % 2& 22 =

90 0

gd gl gl gl

% (&

(2 %

(×=

90

&/!

&/+( % & l

l d =+= 0

[11]

Edexcel Internal Review 1&


19/41

21. (a) ?et speed at C be u

@7'cos % &$

(2&

% 2& 2 θ mga

ag mmu =

90 0

θ cos2 % (32 ga gau =

amu

Rmg 2

'$cos =+θ 90 0

θ θ cos2 % (

3cos mg

mg mg =

eliminating u 90

?eading to∗=

()

cos θ 90 0 4

(b) t C ga ga

gau

()

()

2 % (

32 =×=0

( )→ a

ga gauu x -)).2/(

28()

()

cos =

=×

== θ

90 0ft

( )↓ a ga ga

uu y 832.&/(2&

(8

()

cos =

=×

== θ

90

gaa g gav ghuv y y y /(2(!

(8

2/(2&

2 222 =×+=⇒+= 90 0

...-&2.)28

2(!tan ≈

==

x

y

u

vψ

90

82≈ψ awrt 42 * A& +


20/41

Alternative ./r t0e la-t .ive mar -

?et speed at P be v .

@7amg

ag mmv 2

(2&

% 2& 2 ×=

or e"uivalent 90

(&82 mgav =

90 0

-.)&!28228

&8(

/(28

cos ≈

=

×==

vu xψ

90

ψ = 82* awrt 82* A&Note

T0e time /. .lig0t .r/m C t/ P i-

≈

g a

g a

&.)+)8)+2& % 2)!

[15]

22. (a)

Tu -

va

m a

2

7nerg 2&

mv 2 mga cos % 90 0v 2 2 ga cos θ

ma T % mg cos % amv 2

90 0

'ub for av 2

; T mg cos θ / 2 mg cos θ ; θ *+ AT )&

mg 90 0 *

(b) 'peed of P before impact ga2 0

→ ga2 → + → u

P@?9; • • • Au 9

+(

2 ga ga =90 0cso 3

m 3m -m



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(c) (i)

A

A t A -v

t A v + so conservation of energ gives;

'cos&$((2& 2 θ −= gamum

90 0

'cos&$&/

θ −= ga ga90

°== 2-,&/&!

cos θ θ 0

(ii) t A T - mg cos % (&!mg

(accept 3.4 mg) 90 0 *[15]

23. (a) 7nerg e"uation with two terms on


22/41

(d) t top of small circle, 22!

.2&

2& 2 mga gammv −=

90 0 0(90 for energ e"uation with 3 terms)

⇒ v 2

ga

2

)

0-.4 a


23/41

Alternative u-ing c/n-ervati/n /. energ .r/m t0e /int w0ere P l/-e- c/ntact wit0 -ur.ace#

==

)cos2 ga

ag V α

7nergα cos'$

2& 22 mgaV W m =−

90 0

)

&

)

&2& 2 ×=

− mgaag W m

0?eading to ! 2 ag √3 = cso 90 0

Alternative u-ing r/,ectile m/ti/n .r/m t0e /int w0ere P l/-e-c/ntact wit0 -ur.ace#

2 ag cos a ) ga

↓ α α cos2sin222 gaV W y +=

ag gaag 3

)+

)

&2

)&

&)

& =×+

−

90 0¬ x cos a 0

9))&

))&

3)+222 ag ag ag V W W x =×+=+= γ

cso 90 0[12]

25. (a) 7nerg ;'cos&$

2&

).2& 2 θ +=− mgamvag m

90 0v2 ag (0 – 2 cos %) (o.e.) 0 3

(b) T / mg cos % av

m2

90 0Bence T (0 %) cos%)mg (=) 0 cso 3

(c) Csing T + to find cos θ 90

Bence height above Aa

)(

ccept 0.33 a (but must have 3/ s.f.) 0 2



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(d) v 2 )&

ag (o.e. ) f.t. using cos % )&

in v 2 0ftconsider vert motion; ( v sin %)2 2g0 (with v resolved) 90 0

sin 2 % 3

+

(or % 4+. 3, sin % +.8-3) and solve for 0 (as a) 90

0a

28(

or +.0-D a (awrt) 0

OR consider energ ; 2&

m(v cos %)2 6 mg0 2&

mv 2 (3 non5#ero terms)90 0'ub for v+θ and solve for 0 90

0a

28(

or +.0-D a (awrt) 0[13]

26. (a) A

P

u

! g l 2√

@onservation of 7nerg

mgl u gl

m =

− 22

!2&

90 0 0

?eading to u √

2

gl

0 -



25/41

(b)v

ur B A

T m g

@onservation of 7nerg

mgr vum =− '$2& 22

90 0

gr uv 222 −=

( )r

mvmg T R

2=+↓

90 0

mg gr ur m

T −−= '2$ 290

mg r

mu)

2

−= 0

mg r

mgl )

2−=

90

mg r

mgl T )

2- ≥⇒≥

90r ≥⇒

/&

/! l

AB M ! = 0 8

[13]



26/41

27. (a) A

B

C l v

u √$ ) ' g l

7nerg 2&

m(u2 % v 2) mgl (0 % cosθ ) 90 0

Ev 2 gl / 2 gl cos θ

62? T – mg cos θ * l mv2

90 0

m g l $ & > 2 c o s 'l 90

T mg (0 / 3 cos θ ) = cso 0 *

(b) T + ⇒ cos θ % )&

0

v 2 gl %)2

gl ⇒ v

2&

)

gl

90 0 3

(c) i v

2&

)

gl

sin θ

√

)22

)2

& gl

90

v

yv

v 2 u2 % 2g0 ⇒ 2 g0 3+.

)

gl

⇒ 0 28

( l

90 0

7 l (0 % cosθ ) / 28( l

28(- l

90 0[14]



27/41

28.

A

P R

m g

B

(a) 2&

mv 2 mg (a cos a %a cos %) 90 0 0v 2 2 ga (cos a % cos%) (=) cso 0 -

(b) Emg cos % ( % R ) amv2

(R +) 90 0 0

g cos % 2 g

− θ cos()

90

cos % 2&

⇒ % )π

0acce t ") "

(c) From A to B 2&

mw 2 mg (a / a cos a ) 90 0 0

w 2 2 ga

+

()

& ⇒ w

2&

28

ga

0 -

Alternative

r/m P t/ C

22

&

(

)22

ga gav P =

−=

− 22

&2& 2 gamm#

mg (a / a cos %) 90 0 0

w 2 %

+=

2&

&22

mga ga

⇒ w

2&

28

ga

0 -

Edexcel Internal Review 2$


28/41

Alternative- u-ing r/,ectile m/ti/n .r/m P

v P

2&

2

ga

, as above

↓ u 2&

2

ga

sin*+ *

2&

+)

ga

↓ +28

,2

)222

gaa g uv y y =×+=

90, 0

→ u x 2&

2

ga

cos*+ *

2&

+

ga

0

w 2

2

8+

28+

22 ga ga gavu y x =+=+ ⇒ w

2&

28

ga

0 -

T0ere are al-/ l/nger r/,ectile met0/d- u-ing time /. .lig0t

%n outline, solving

22&

2&

+)

2)

g$ $ gaa +

=

gives t

2&

2)

g a

,

then, using v u / at gives v

2&

2&

2&

+28

2)

+)

=

+

ga

g a

g ga

,then as before. 90 0

[13]

29. (a) 2&

m(v 2 % &!' mg (0 % cos/-*' 7& A& A&v D ms %& 0 -

(b) 2&

8w 2 8g (0 % cos/-*' 7& A&@?9; *+ ? + % )m (*+ / m)w 90 0ft 0-D+ % )m -2+ / 4 m

*+ 0+ m 90 (solving form)* m 0 4



29/41

(c) T % //g !8// 2×

90 0ftT 032 g

@&)--

@&23-

0 3[14]

30. All 8 mar - re%uire c/rrect num9er /. term- wit0 a r/ riate term- re-/lved

(a) B to C ; 2&

mv 2 5 2&

m2+ 2 mg ? !-$&5 sin 3+ °) 90 0v 3+ ms 50 (28.D) 0 3

(b) () at C , R 5 mg m !-+3-

90 0 ftR 08++ 6 (083+ 6) 0 3

(c) C to ' ; 2&

m D8+ 5 2&

mw 2 mg ? !- $&5 cos 3+ °) 90 0 ftw 2D ms 50 (24. ) 0 3

(d) efore; R mg cos % 0

fter; R mg cos % / m !-2- 2

90 0

@hange 4+ ? !-2- 2

*+ 6 0 c.s.o -

(e) ?ower speed at C ⇒ < reduced 90 0 2[15]

31. (a) 7nerg ; mv 2 % mu 2 mga sin % 90

v 2 2)

ga + 2 ga sin" 0 2

(b)


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(c) 'etting T + and solving trig. e"uation: (sin % % '⇒ % 20+ ° =90: 0 2(d) 'etting v = 0 in (i) and solving for % 90

sin % %B so not complete circle A& 2or $ii' loo1ing at energy at top $mga )and at start ( Bmga ) so not possible

or substituting % 24+ ° in (a): v 2 1 + so not possible to complete

(e) 6o change in P7 ⇒ no change in G7 (@of 7) so v u 0 0

(f) &hen string becomes slac , 2 ga Esin% % in $a'C &ft

"or1ing horizontally5 2 ga

cos*+ * = 2) ga

cos . 90 0

or verticall ; 2)

ag sin 2 . 2&

ag .sin 2 *+ * >ag ( +&&ag

)or finding H a ! B and using to find tan . . 73 ° or 73.2 ° 0 -

[14]

32. (a) 7nerg ; 2&

mv 2 5 2&

mu 2 mga (0 % cos%) 90 0 0


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(b) 7nerg ( C and ground); 2&

m

2

3ag

5 2&

mv 2 ( m)ga (0 / cos %) 90 0'ubstituting v 2 g a cos % 90 0

Finding cos % cos % /!

90 0 ft:.t /n c/- % /nl i. ) ; c/- % ; 1<

% 3- ° 0ca/

EOr energ ( A and ground); 2&

m

2

3ag

5 2&

mu 2 2 mga 90 0

'ubstituting u2 from (a)

Csing with (a) to find cos % /!

90 0 ft: % 3- ° 0 4

Pr/,ectile a r/ac0 ; x v cos % : 2 (v sin %)2 / 2 ga (0 / cos %)

2

3ag

x 2 /

2 ⇒

2

3ag

5 v 2 2 ga (0 / cos %) % 7& A&, then as schemeC[14]

33. (a) 2&

mu 2 5 2&

mv 2 mga (0 / cos *+ °) 90 0v 2 * u2 = 3 ga 0 3

(b) R 6 mg cos *+ ° amv2

90 0

R am

(*ga 5 3ga ) 5 2mg

2! mg

0 3

(c) R + at B ⇒ 2mg

amv2

⇒ v 2 2&

ag 90

⇒ u2 28 ga

⇒ u 28 ga

90 0 3



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(d) a a

/ - * BC

v

) )2 2

(→) B to C ; v cos *+ ° ? t a ) 90 0

t va )2

() B to C ; + v sin *+t 5 2&

gt 2 90 0

⇒ t g v °/-sin2

g v )

A va )2

g v )

⇒

v 2 2 ga 90 0

⇒ u2 ga

⇒ u ga! 0 4[16]

34. (a)

m g

R

v

7nerg 2&

mv 2 2&

mu 2 / mga (0 % cos%) 90 0

v 2 2&

ga / 2 ga (0 % cos%)

2&

ga ( % ( cos%) 0 3



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(b) R ( ); mg cos % 5 R amv 2

90 0

so R mg (3 cos % 5 2!

) 0

cos % +.8 ⇒ R mg (2.4 % 2.!' 7& -.2mg I + ⇒ P still on hemisphere

P leaves hemisphere when R + ⇒ 3 cos % 5 2!

+ ⇒ cos % /!

0

cos % /!

⇒ v 2 2&

ga ( % ( ? /!

) 90 0 2

/! ga

, v /! ga

0 2

t B, speed v is given b v 2 u2 / 2 ga ga2!

, v 2! ga

0 2

fter leaving hemisphere, hori#ontal component of velocit remains

constant /!

/! ga

0

cos . 2!

/!

/!

ga

ga

)/! v

90

⇒ . *0.2 ° or *0 ° to hori#ontal 0 3[17]



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35. 9ost candidates were able to ma e a good start on this "uestion and a considerablenumber scored highl . Part (a) was done well b the vast ma$orit of students. fewused the bottom of the circle as their reference point for P7, which did not necessarilproduce an incorrect solution, but did re"uire more wor . !he rare errors were signerrors, using cosine or failing to eliminate the half from the G7 correctl . Part (b) was

also straightforward for the ma$orit , although some failed to include the component ofthe weight in their e"uation of motion and others made sign errors once again. !hema$orit of students reali#ed the use of the result from (b) was re"uired to prove thatthe particle moved in complete circles and produced a convincing argument, though afew started from scratch. lot of students thought that having v 2 greater than #ero atthe top was the re"uired condition and man added this to an alread correct solution:perhaps some thought that both T I + and v 2 I + were needed. !he ma$orit ofstudents reali#ed that θ 8+ * was re6uired, but some students produced convincingarguments based on the range of values for sinθ . Part (d) was done well b most students

gain most used their result from (a), with a few solutions wor ed out from scratchusing energ . !he common error was to maximi#e v b simpl stating θ 8+ * or sinθ

+. !his result not onl ignores the mechanics but also shows a lac of imagination.

36. Part (a) was a straightforward application of the principle of conservation of energ andwas done well b man candidates, although some needed several attempts beforethe arrived at the given answer. !hose who tried to "uote a single potential energterm involving onl the general position, without an reference to the starting point,were misled into thin ing the had obtained a correct result. Finding the heightdifference between the general position and the starting level proved to be much morechallenging than it should have been and some candidates were thrown b having todeal with an angle with the hori#ontal rather than the more usual angle with the vertical.Obtaining a complete solution to part (b) proved impossible for man candidates.Csuall onl the maximum value for T was found correctl , b using sin θ 0: manthought that the minimum value was +. 'ome candidates found the tension at the top of the DcircleE, even though their calculations showed thatv 2 was negative at this pointJ %t wasrare to find a candidate who appreciated that the minimum tension would occur when v

+ .

37. Part (a) of this "uestion was a fairl standard vertical circular motion "uestion andman candidates could produce two e"uations b considering change in inetic andpotential energies and resolving along the radius. 'ome tried to resolve verticall withlittle success. !he need to find a difference in potential energ between two pointscreated problems for some candidates who seemed to thin the particle either startedat the top of the circle or from the hori#ontal level of A. !he presence of a printedanswer did enable some candidates to retrace their steps and correct the signs in theirwor ing. !his was not a problem as long as the result was consistentl correct. Part (b)produced a different set of problems. 'ome candidates thought that the string bro e atthe point where T + and so found an incorrect value for the initial velocit of thepro$ectile. !he pro$ectile problem could be solved b using vertical motion under gravitor b energ considerations. 9an who opted for energ forgot that the particle wasstill moving (hori#ontall ) at the highest point of its path: some of those who opted forvertical motion under gravit forgot that the initial velocit the had found needed to beresolved verticall before use in the appropriate e"uation. 9ost who had wor ed to thispoint in the "uestion remembered to finish off b finding the distance below thehori#ontal through A.



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38. Hertical circle "uestions usuall present problems for man candidates and this onewas no exception. 6ot all seem to be aware that an energ e"uation and an e"uationof motion along a radius (in this case at C ) should be sufficient to ma e a sound starton the "uestion. !here were incorrect signs in the energ e"uation which were thenad$usted later to arrive at the given result. 'imilarl , incorrect trigonometric functions

became correct ones. Part (b) was attempted b most of those who had achievedsuccess with (a) but the pro$ectile motion defeated man . 'ome could not relate θ correctl to the hori#ontal and vertical components of the velocit at C . variet ofmethods were seen in (b). 'ome used the separate components, finding thecomponents at P and finishing off with the tangent of the re"uired angle. Others usedan energ approach, wor ing from either A or C to obtain the final velocit at P andfinished off with the hori#ontal component and the cosine. 'ome made extra wor forthemselves b finding the final velocit and the final vertical component, finishing offwith the sine of the re"uired angle. !he usual mista es such as using v 2 u2 % 2a- when energ was re"uired occurred.

39. !here were a lot of completel , or nearl completel , correct solutions to this "uestion.'ome lost the final mar onl for using m instead of - m. 9ost candidates who madean attempt at this "uestion scored highl on it. !he seemed well versed in the twosignificant principles of using @onservation of 7nerg and 6ewton Fs Gecond Haw towardsthe centre. nfortunately again it may have been a case of results learnt rather than mechanics principles understood. In part $a', eitherv 2 u2 / 2 ag (0 % cos /-' orv 2 2 ag (0 % cos /-' wasused in the J of 4 e6uation rather than identifying a correct trigonometric expression to lin1 theKL4 at the two positions. This happened again in part $c'$i' when they usedmg0 as change in

P7 and then hadcos θ 0K0* as their result for finding θ . 'imilarl in part (c)(ii) where v + was thecondition, too man thought that meant no acceleration and hence resolved verticall .%n both (c)(i) and (ii) - m was not used consistentl .

40. Hertical circle solutions have continued to improve and candidates understood whatwas re"uired. Part (a)was usuall completed successfull with the printed answerproviding a useful chec . Part (b) caused more problems and was a good discriminator5 some forgot the weight component, others resolved as mg cos θ instead of mg sin θ .%n part(c) man who had answered part (b) correctl achieved the first two mar s but asignificant group could not produce the correct angle.

Part (d) was often omitted. 'ome did not realise that the speed going up when thestring was hori#ontal was the same as the original speed and wasted time on anotherenerg e"uation. 'ome wor ed with a general position until the end which made thealgebra more difficult.

9ista es involved not using the top of the circle, having a wrong radius and having anegative mg in their Force mass acceleration e"uation



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41. 9ost candidates tac led this "uestion well and, usuall , if the were successful in part(a), the were also successful in part (b). s in L2, however, part (b) was correctlanswered more often than part (a). %n part (a), among the less successful candidates,most were aware that the had to use an energ e"uation but a fairl common mista ewas assuming the velocit to be #ero at the point of leaving the surface. 'ome

candidates used memorised formulae involving an acute angle measured with thedownward vertical. A few of these managed successfully to justif the changes needed toproduce the re"uired result for angle α but more tried simpl to manipulate their signs.%n part (b), it was pleasing to see that the ma$orit found the most efficient wa ofcompleting the "uestion, considering the change of energ between the lowest pointand the point on the hori#ontal through O. significant minorit too the approach offinding the velocit where the particle left the surface of the sphere and wor ing fromthat point to the point on the hori#ontal through O. Often, though, the did not ma e itver clear that the were doing this and some lost trac of their #ero level for thegravitational potential energ .!here were ver few valid pro$ectile attempts. 9ost candidates attempting this methodforgot that the had to consider two components and, although a few managed tocalculate the correct vertical component of ! , successful attempts to combine this witha hori#ontal component were ver rare.

42. Parts (a) and (b) were generall well done with the relevant principles clearl wellnown, and part (c) caused little problem. %n part (d), onl the better candidates

realised that the particle still had some speed at its highest point, and several simplassumed that the had to find when it came to rest (which it never doesJ).

43. Part (a) was ver well done. !he onl error that was seen at all commonl was having

the difference in inetic energies the wrong wa round. %n part (b) the error of thin ingthat v + at the highest point of the semi5circle was widespread. !his lead to a verbrief solution which could gain at most two of the nine mar s. !hose who did write

down a correct e"uation for the tension in the string often arrived at AB /! l

but hadoften assumed that T + and were unable to give a $ustification for their result being aminimum. %ne"ualities remain unpopular with candidates in mechanics. considerablenumber of candidates appeared to rel on memorised formulae involving an angle for both the energy and the tension without any clear idea of what was in this 6uestion. Theintroduction of such an angle is an unnecessary complication in this 6uestion, where only thehighest point needed to be considered.

Edexcel Internal Review 3$


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44. Part (a) was done either ver well or ver badl . 9an candidates were completelcomfortable with the theor and produced clear and concise derivations of T# 6earlever one else managed to arrive at the same formula using some ind of fudged

argument.. favourite approach wasmg

l mv

T +=2

and gl v )2 = at B, so the tension in

new position is mg(0 / 3cos %). !here were man variations on this. gain, there wereversions that ma have been correctl reasoned but weren Ft expressed clearly enough forthe examiners. 0or instance, one proof was perfect up to the point where it statedma for

circular motion'cos2&$

2θ +== mg

l mv

and then said D7ust addmg cos % to account forthe weight E. Jandidates should be under no illusion that this type of argument will receive nocredit whatsoever.

Lart $b' followed on nicely from $a' for the majority who used a result obtained forv M in $a'with cos %N.Lart $c' did discriminate between the top few candidates and the rest. Oery few recognized theneed to usev , either as the initial velocit in v M uM > 2a- , or + (with v x still present)when using energ . !oo man thought that finding the height when the string went slacor to whenv + was sufficient to answer this part.

45. Part (a) was well done with the printed answer enabling man candidates to correctincorrect signs. %n such "uestions the examiners re"uire the clear use of the principle of

conservation of energ and the constant acceleration formula v 2

u2

/ 2 a- is notaccepted, as it stands, without some evidence that inetic and potential energies havebeen used.

%n part (b), a few thought that the particle left the sphere when the velocit was #ero butthe ma$orit realised that the normal reaction between the sphere and the particle was#ero and, apart from some confusion which was not infre"uentl seen between θ and α,could complete the "uestion.

%n part (c), there was a fairl even split between those use energ and those usingpro$ectile methods. !he easiest solution, using conservation of energ from the initialposition, was not often seen but, when it was, it was almost invariabl completelcorrect. !he pro$ectile solutions were longer, harder and less fre"uentl correct. !hose

using pro$ectile methods often dealt onl with the vertical component of velocit .

46. Part (a) was generall well done but the second part proved to be much moredemanding. 'ome thought that energ was conserved in the impact and scored fewmar s and even those that realised that conservation of momentum was re"uired wereunable to complete the "uestion due to poor algebraic s ills % an early simplification gavea linear e6uation rather than a very complex 6uadratic. The method was generally 1nown in part$c' but there were relatively few correct answers, either due to the wrong mass being used orelse a failure to round off the answer to 2 sf or ) sf because of the use of g 3.+.



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47. !his section of the s llabus seems to be poorl understood b man and there wasmuch evidence that some candidates were rel ing on formulae. !he comments madeabove about the use of mv 2Kr as an extra force (L-) are e"uall applicable here. %n (a)and (c) it was common to see 0K2 mv 2 % &P2mu 2 mgr (0 % cosa) "uoted and then usedwithout apparent understanding of where the angle referred to as could be found or

that this version of the formula was not appropriate when the s1ier was travelling upwards after point J. The best solutions to these two parts, and there were a great many of them, showed aclear calculation of the height difference between the two points before consideration of theenergy e6uation. In $b' and $d' the same reliance on formulae was apparent among wea1ercandidates.R – mg cosa mv 2Kr was often "uoted but not adapted for the verticalposition re"uired in (b). R mv 2Kr was also ver common. 7ven so, there were a lot ofcorrect solutions, the vast ma$orit of them losing a mar for giving the inappropriate -figure answer 08326. !here were ver few completel correct solutions to (d), mostcandidates failing to realise that passage through point caused a change in < fromthat re"uired for e"uilibrium perpendicular to the incline to one which would allowmotion in a circle. %t was ver common for candidates to appl R %mg cosa mv 2Kr attwo different points and then to subtract the answers. nswers to part (e) were oftenexcellent. &hile a minorit of candidates argued that the frictional force would have noeffect as it was at all times perpendicular to the reaction (wrong, but at least showingsome grasp of a basic principle), man gave correct and ver lucid explanations of therelationship between friction, wor , energ , speed and finall reaction.

48. !here was some evidence of shortage of time with some scrapp and rushed answerstowards the end of this "uestion. %t is perhaps not surprising then to report that it wasver rare to see full mar s gained here. !he first two parts were familiar and candidates

new the principles, errors tending to be either with signs or wrong trigonometr . %t is

worth reminding candidates, however, that omission of a term, for example mgsin %with part (b), results in no mar s being awarded. %n part (c) man candidates translatedDproveE as DverifyE and lost mar1s.There were some good answers to parts $d' and $e', but as with all 6uestions where reasons andexplanations are re6uired some candidates lost the odd mar1. Oery few candidates successfullynegotiated the last part.

Edexcel Internal Review 3(


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49. !he condition u 1 √(ag ), necessar for the particle to remain on the sphere initiall , wasclearl misunderstood b some candidates who, at some stage in the "uestion,

e"uated2u to ag . !his happened less fre"uentl in part (a) but in part (b) it was a more

common error.

%n part (a) the ma$orit of candidates made some progress, usuall having the energe"uation, and often the radial e"uation of motion, substantiall correct. !his is a fairlcommon piece of boo wor but the manipulation re"uired did catch out a significantnumber of candidates who either made sign errors, set v + at C , or DlostE either g ora#'ection (b) caused ma$or problems and onl the better candidates scored well in thispart. !hose who considered an energ approach rather than a pro$ectile approachusuall fared better, although there was much confused thin ing in either. @andidateswho considered the motion from C to the ground often confused their DuE with that given

in the 6uestionQ the energy e6uation2&

2

g3am

%2&

mu 2 mga (0 / cos θ ) with u2 replacedb the expression in part (a) was common.

!hose candidates who used a pro$ectile approach often made a serious error and verrarel gained man mar s. !he commonest errors were, when wor ing verticall , to use

either 23ag

, or 23ag

sin % instead of 23ag

sin φ , as the final velocit component.

!hose candidates who started their solution with

2

g3a

%v 2 2g a (0 / cos θ ), without

an reference to energ , did lose mar s. lthough it is onl a factor of 2&

m awa from acorrect energ e"uation, the fact that most candidates preceded their e"uation withDusingv 2 u2 / 2 a- E sheds considerable doubt on the thin1ingR It is also a consideration thatcandidates using the projectile approach needed to have done a few steps of wor1ing beforereaching the same point in their solution.

50. %n part (a) most candidates new that energ methods were re"uired but there wereoften sign errors or incorrect terms. 'imilarl , the radial resolution in the second partoften contained an incorrect trig ratio or further sign errors. 9an candidates assumedin part (c) that v + at B was sufficient for the marble to reach B instead of consideringthe reaction. !he final part was a good discriminator with onl a few full correctsolutions. 9ost realised that the marble was now moving as a pro$ectile but onl asmall number were able to write down correct e"uations for the hori#ontal and verticalmotion, eliminate t and solve for u. significant minorit $ust assumed that the marblewould follow the missing part of the sphere and obtained a correct answer but scoredno mar sJ



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51. 6o

Documents

M3 Circular Motion - Vertical Circles