Post on 15-Oct-2015
description
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]