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7/28/2019 2012 T2 Exam Paper 1 and 2 Answers
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May 2012. Term Two Physics AS examination. MC answers.1 A 21 C2 C 22 C3 C 23 C4 B 24 C
5 D 25 C6 C 26 B7 B 27 A8 A 28 C9 B 29 B10 C 30 B11 C 31 A12 A 32 C13 C 33 D14 A
15 D16 B17 A18 C19 D
20 A
1. (a) (i)
F1 weight /mg(1)
F2 reaction or normal contact force(1)
F3 driving force(1)
F4 friction or air resistance(1)
(ii) zero acceleration (1)
zero resultant force(1) max 5
(b) (P =Fvgives) 18 103 =F 10(1) (andF =1.8 103 N) 1
(c) (i) 1800 250 =1.6 103 N (1) (1.55 103 N)
(ii) force =4 1.55 103 =6.2 103 N (1)(allow e.c.f. from(i))
(iii) total force =6200 +250(N) (1) (=6.45 103(N))
(P =Fvgives) P =6.45 103 20 =1.3 105 W (1) (1.29 105 W)(allow e.c.f. for value of total force) 4 [10]
2. (a) scalars havemagnitude(or size) (1)vectors havemagnitudeanddirection(1) 2
7/28/2019 2012 T2 Exam Paper 1 and 2 Answers
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(b) (i) s=t(1)
s=100 =5 km(1)
(ii) 1.59 (1) km (or other correct unit) (1) 4[6]
3. (a) 2
(b) (i) taking moments (about A) (1)(400 4.0) +(300 2) =FB 1.6(1)
=FB soFB =1375(N) (1)
(ii) FA =1375 700 =675 (1) N (1) allow e.c.f. fromFB
(no marks if weight of board not used) max 4[6]
4. accept mirror image for (a) and (b)
(a) (b)
straight line sloping up (1) constant value shown(1)sudden change to negative velocity (1)smaller negative velocity (1)same gradient as positive line(1) 4
(c) (i) vertically down at P (1) (ii) vertically down at Q (1)
(iii) along tangent at P (1) (iv) along tangent at Q (1) 4
(d) horizontal component of velocity at Q = 15 cos 50(1) =9.64 m s1(1)
momentum at Q =0.15 9.64 =1.45 (1) Ns (or kg m s1) (horizontally) (1) 4[13]
60
3
F
F
A
B(1)
(1)
6.1
2200
velocity
time
acceleration
time
7/28/2019 2012 T2 Exam Paper 1 and 2 Answers
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5. (a) (i)
B must make an appreciable angle with wall and bar
(ii) A weight of sign and bar (accept gravity) (1)B reaction of wall (1)C tension in wire(1) max 5
(b)
use ofmg(1) clockwise moments 118 0.375 (1)
=anticlockwise moments (Tcos40 (1)) 0.750(1) T=77 N (1) max 4[9]
6. (a) (i) equation showing momentum before =momentum after (1)
correct use of sign(1)
(ii) no external forces (on any system of colliding bodies) (1) 3
(b) (i) (by conservation of momentumm11 +m22 =0)
0.25 2.2 =()0.502(1)
2 =()1.1(0)msl(1)
(ii) =total k.e. = 0.25 2.22 + 0.5 1.12(1)
=0.91J (1) 4 [7]
7. (a) (i) gradient = =3.0 ms2(1)
(ii) distance is area under graph (to t=0.1 s)
or 0.7 2.1 0.3 (1) =1.4(2) m(1) 3
(b) (i) T mg=ma [orT=1500(9.8+3.0)] (1)
=1.9 104 N (1)
T=mg=l.5 104N (1)
(ii) EF (1) 4
(c) power =F or l.5 1042.5(1) =3.7[3.8] 104 W (1) 2[9]
B
C
A
(1)
(1)
(1)
0.375m
118N
40
2
1
2
1
7.0
1.2
2
1
+
2
5.21.2
7/28/2019 2012 T2 Exam Paper 1 and 2 Answers
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8. (i) a= =11 ms2(1) F =ma=1.1 105 N (1) (ii) =236 m s1
a= =29.5 ms2(1)(iii) sone= t= 4.0 =88m(1)
stwo = t= 8.0(1) =1296(m) (1) total distance =1384 m(1)
[6]
9. (a) sum of clockwise moments equals sum of anticlockwise moments(1)for a body in equilibrium(1) 2
(b) point in the body through which the weight/mass (appears to) acts[or point where resultant torque/moment is zero]
[or point where body would balance] (1) 1
(c) (i) towards A (1)so that weight of ruler (1)provides balancing moment (1)
(ii) (moments about pivot give) 1.0 (0.30 d) =0.50 d(1)1.5d=0.30 andd=0.20 m(1) 5
[8]
10. (a) (i) (use ofv2 =u2 +2asgives) 0 =252 - 2 9.81 s(1)19.6s=625 ands=32 m(1)
(ii) t= =2.5 s(1)
(iii) (use ofv2 =u2 +2asgives) v2 =252 2 9.81 16(1)
(allow C.E. from (a)(i)) andv=18 m s1(1) max 4
(b) time to stop the ball is greater (1) rate of change of momentum is less(1)
[or work done on ball is the same but greater distance(1) less force(1) ] 2[6]
11. (a) kinetic energy not conserved (1)[or velocity of approach is not equal to velocity of separation] 1
(b) (i) (use ofp=mvgives) p=4.5 102
60 (1)=2.7 kg m s1 (1)
(ii) (use ofF = gives) F = (1)
=180 N (1)
[or a= = =4000 (ms1)
F =(ma) =4.5 102 4000 =180 N] 4
[5]
0.4
44
0.8
236
av
+
2
044
av
+
2
44280
81.9
25
t
mv
)(31015
7.2
t
uv31015
60