5/26/2018 MARK SCHEME VIMA 4 - M2 Centres of Mass Projectiles and Kinematics

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The Sydney Russell School 1

1. (a) Triangle Circle S

Mass ratio 126 9 1269 B1 B1ft(28.3) (97.7)

x 7 5 x

y 4 5 y 4, 7 seen B1

126 7 = 95 + (1269) x ft their table values M1 A1ft

x 7.58 (

9126

45882

) awrt 7.6 A1

126 4 = 95 + (1269) y ft their table values M1 A1ft

y 3.71 (

9126

45504

) awrt 3.7 A1 9

(b) x

y

21tan ft their yx, M1 A1ft

15 A1 3[12]

2. (a) m m I v u

0.5 20 0.5 10 24 i i j M15 12 i j A1

5 12 13 Ns

i j M1 A1 4

(b)

12tan

5 M1

67.38 67.4 A1 2

5/26/2018 MARK SCHEME VIMA 4 - M2 Centres of Mass Projectiles and Kinematics

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The Sydney Russell School 2

(c) 2 2 21 12 2K.E.lost 0.5 10 24 0.5 20

M1 A1

69 J A1 3

[9]

3. (a)

Conservation of momentum: 4mu3mv= 3mkv M1A1

Impact law: vuk v 4

3 M1A1

Eliminate k: 4mu3mv= 3m vu 4

3 DM1

u= 3v(Answer given) A1 6

(b) 3,34

3 kvvk v M1,A1 2

(c) Impact law: (kv+ 2v) e = vC vB(5ve= vC vB) B1

Conservation of momentum : 3 kv1 2v=3vB+ vC (7v= 3vB+ vc) B1

Eliminate vC: 0574

ev

vB hence no

further collision withA. M1 A1 4[12]

4. (a) Horizontal distance: 57.6 =p 3 M1p= 19.2 A1 2

5/26/2018 MARK SCHEME VIMA 4 - M2 Centres of Mass Projectiles and Kinematics

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The Sydney Russell School 3

(b) Use 2

2

1atuts for vertical displacement. M1

232

139.0 gq A1

1.4432

939.0 q

gq

4.143

2.43q * AG * A1 cso 3

(c) initial speed 22 4.14p (with theirp) M1

)sm(24576 1

A1 cao 2

(d) )4

3(

4.14tan

p (with theirp) B1 1

(e) When the ball is 4 m above ground:

used2

11.3

2atut M1

3.1 = 14.4t

2

1 gt2o.e (4.9t214.4t + 3.1 = 0) A1

)9.4(2

)1.3)(9.4(4)4.14(4.14 2t seen or implied M1

2.7and0.23awrt2.70488....or0.023389..8.9

6.1464.14

t A1

duration = 2.70488...0.23389... M1= 2.47 or 2.5 (seconds) A1 6

or

M1A1M1 as above

8.9

6.1464.14 t A1

Duration 2 8.9

6.146 o.e. M1

= 2.47 or 2.5 (seconds) A1 6

5/26/2018 MARK SCHEME VIMA 4 - M2 Centres of Mass Projectiles and Kinematics

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The Sydney Russell School 4

(f) Eg. : Variable g, Air resistance, Speed of wind, Swing of ball,The ball is not a particle. B1 1

[15]

5. (a)12.6

0.1

x

12.6t=x B1

0.1 = 4.9t2 B1

0.1 = 4.9 6.12

2x M1

x= 1.8 m A1 4

(b) )

ucos .t= 2.5 M1 A1

usin.t=2

1gt2 M1 A1

u.25

24t= 2.5

u.25

7= 4.9.

u24

25.5.2

u2=

247

255.29.4 2

u6.75 or 6.8 m s1 M1 A1 6[10]