9702 November 2009 Paper 11 Worked Solutions | A-Level Physics
Paper 11
1 C 11
C 21 C 31 C
2 A 12
B 22 B 32 D
3 B 13
D 23 B 33 A
4 A 14
A 24 B 34 A
5 A 15
D 25 D 35 B
6 B 16
C 26 D 36 C
7 C 17
C 27 D 37 C
8 D 18
D 28 B 38 D
9 A 19
B 29 A 39 C
10 C 20
D 30 C 40 B
Notes for some specific questions:
5 – Ans: A. {Detailed explanations for this question is
available as Solution 221 at Physics 9702 Doubts | Help Page 36 - http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-36.html}
7 - Ans: C. {Detailed explanations for this question is available as Solution 969 at Physics 9702 Doubts | Help Page 200 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-200.html}
8 - Ans: D. {Detailed explanations for this question is available as Solution 945 at Physics 9702 Doubts | Help Page 195 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-195.html}
9 – Ans: A. {Detailed explanations for this question is
available as Solution 138 at Physics 9702 Doubts | Help Page 25 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-25.html}
10 - Ans: C. {Detailed explanations for this question is available as Solution 98 at Physics 9702 Doubts | Help Page 18 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-18.html}
12 - Ans: B. {Detailed explanations for this question is available as Solution 152 at Physics 9702 Doubts | Help Page 26 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-26.html}
13 – Ans: D. {Detailed explanations for this question is
available as Solution 127 at Physics 9702 Doubts | Help Page 23 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-23.html}
14 – Ans: A. Let initial velocity =
v. At highest point, projectile only has a horizontal component of velocity
which is vcos(45). Vertical component of
velocity is zero. Kinetic energy is proportional to v2. So, at
highest point, kinetic energy proportional to [vcos(45)]2 = 0.5v2.
So, KE is halved.
15 – Ans: D. Sum of
momentum before collision = sum of momentum after collision. Before collision,
momentum is zero since trolleys are stationary. So, after collision, 2kg(2ms-1)
+ 1kg(v) = 0. So, v = -4ms-1 = 4ms-1 (considering
magnitude only). So, energy stored = sum of kinetic energies of trolleys = ½
(2)(2)2 + ½ (1)(4)2 = 12J
16 - Ans: C. {Detailed explanations for this question is available as Solution 949 at Physics 9702 Doubts | Help Page 196 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-196.html}
17 – Ans: C. {Detailed explanations for this question is available as Solution 979 at Physics 9702 Doubts | Help Page 202 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-202.html}
19 - Ans: B. {Detailed explanations for this question is available as Solution 1071 at Physics 9702 Doubts | Help Page 225 - http://physics-ref.blogspot.com/2015/11/physics-9702-doubts-help-page-225.html}
20 - Ans: D.
Which row best defines elastic and plastic behaviour of a material?
21 – Ans: C. For linear
force-extension curve, strain energy = ½ Fx = ½ (100)(2x10-3) =
0.1J. But shape of graph implies an answer 10% greater than this straight line
graph value. So, ans = 0.11J. 0.2J is too big.
22 - Ans: B. {Detailed explanations for this question is available as Solution 796 at Physics 9702 Doubts | Help Page 160 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-160.html}
23 – Ans: B. {Detailed explanations for this question is available as Solution 638 at Physics 9702 Doubts | Help Page 126 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-126.html}
25 - Ans: D. {Detailed explanations for this question is available as Solution 393 at Physics 9702 Doubts | Help Page 73 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-73.html}
26 – Ans: D. dsinθn = nλ. Diffraction grating has
300lines/mm. So, d = 1x10-3 / 300 = 3x10-6 m. Consider
the largest angle possible: θn = 90o. sinθn =
1. Total number of transmitted maxima on 1 side, n = d / λ = (3x10-6)
/ (450x10-9) = 7.4074 = 7 since n is an integer. So, considering the
other side in additional to the central maxima, total number of transmitted
maxima = 7 + 7 + 1 =15.
27 - Ans: D. {Detailed explanations for this question is available as Solution 596 at Physics 9702 Doubts | Help Page 117 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-117.html}
28 - Ans: B. {Detailed explanations for this question is available as Solution 399 at Physics 9702 Doubts | Help Page 74 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-74.html}
29 - Ans: A. {Detailed explanations for this question is available as Solution 403 at Physics 9702 Doubts | Help Page 75 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-75.html}
30 - Ans: C. {Detailed explanations for this question is available as Solution 407 at Physics 9702 Doubts | Help Page 76 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-76.html}
33 - Ans: A. {Detailed explanations for this question is available as Solution 1111 at Physics 9702 Doubts | Help Page 238 - http://physics-ref.blogspot.com/2016/05/physics-9702-doubts-help-page-238.html}
40 – Ans: B. More deflection occurs
for the alpha-particle which is closer to the gold atom.
Can you explain June 2009 paper 1 question 18? Thank you.
ReplyDeleteCheck solution 284 at
Deletehttp://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-48.html
22 pls
ReplyDeleteAdded
DeleteCan you explain q16 11/o/n/09
ReplyDeleteQ16 has been explained
Delete19 pls
ReplyDeleteQ19 is now explained
DeleteQ36 PLEASE
ReplyDeletePlease help on nov 2010 p43 Q10 (b) (i). Thank you
ReplyDeletePlz explain Q18/p11/N10,
ReplyDeleteIt would be so kind on your part,
Thanks in advance
see solution 466 at
Deletehttp://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-90.html
Q20 please why can't it be A
ReplyDeleteexplanation added
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