Physics 9702 Doubts | Help Page 192
Question 932: [Dynamics
> Collisions]
Two spheres approach each other
along the same straight line. Their speeds are u1 and u2
before collision. After collision, the spheres separate with speeds v1
and v2 in the directions shown below.
Which equation must be correct if
the collision is perfectly elastic?
A u1 – u2 = v2
+ v1
B u1 – u2 = v2
– v1
C u1 + u2 = v2
+ v1
D u1 + u2 = v2
– v1
Reference: Past Exam Paper – November 2008 Paper 1 Q10
& June 2012 Paper 12 Q12 & June 2016 Paper 11 Q10
Solution 932:
Answer: D.
For a perfectly elastic collision, both
KE and momentum are conserved.
The relative speed of approach
before collision should be equal to the relative speed of separation
after collision in an elastic collision.
Since before collision, the 2 spheres
are moving towards each other (they are APPROACHing each other), the relative
speed of approach is the sum of the 2 speeds.
After the collision, the 1st
sphere is moving towards the 2nd one while the 2nd sphere
is moving away from the 1st one. If the speed of the 1st
sphere (v1) is greater than v2, it would actually be
approaching the 2nd sphere, and if the speed of the 2nd
sphere (v2) is greater than v1, it would actually be
moving away (separating) from the 1st one.
We want to relative speed of
SEPARATION (that is, v2 is assumed to be greater than v1).
The relative speed of separation is v2 – v1 since both are moving in the same
direction.
Question 933: [Electric
field]
Two charged parallel metal plates
produce an electric field.
A charged particle moves from X to
Y.
Which graph shows variation of the
force on the particle with distance from X along the line XY?
Reference: Past Exam Paper – November 2013 Paper 13
Q30
Solution 933:
Answer: C.The electric field strength E (= V / d) between the parallel metal plates is constant.
Force on the particle = Eq where q is the charge of the particle
The force does not depend on the distance of the particle from the plates. So, the electric force is constant.
Question 934: [Simple
harmonic motion]
A student sets up the apparatus
illustrated in Fig.1 in order to investigate the oscillations of a metal cube
suspended on a spring.
Amplitude of the vibrations produced
by the oscillator is constant.
The variation with frequency of the
amplitude of the oscillations of the metal cube is shown in Fig.2.
(a)
(i) State the phenomenon illustrated
in Fig.2.
(ii) For the maximum amplitude of
vibration, state the magnitudes of the amplitude and the frequency.
(b) The oscillations of the metal cube of mass 150 g may be assumed to
be simple harmonic.
Use answers in (a)(ii) to determine,
for the metal cube,
(i) its maximum acceleration,
(ii) maximum resultant force on the
cube
(c) Some very light feathers are attached to the top surface of the
cube so that the feathers extend outwards, beyond the vertical sides of the
cube.
The investigation is now repeated.
On Fig.2, draw a line to show the
new variation with frequency of the amplitude of vibration for frequencies
between 2 Hz and 10 Hz.
Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q3
Solution 934:
(a)
(i) Resonance
(ii) Amplitude = 16mm and Frequency = 4.6Hz
(b)
(i)
a = (-)ω2x and ω
= 2Ï€f
So, a = (2Ï€ × 4.6)2 × (16×10-3) =
13.4ms-2
(ii) F = ma = (150×10-3)
×
13.4 = 2.0N
(c) The line is always ‘below’ the given line in the graph and never
zero. Its peak is at 4.6Hz (or slightly lower frequency) and it is flatter
(peak is not as sharp as the one given).
Question 935: [Current
of Electricity > Potential divider]
Circuit is designed to trigger an
alarm system when the input voltage exceeds some preset value. It does this by
comparing Vout with a fixed reference voltage, which is set at 4.8
V.
Vout is equal to 4.8 V.
What is the input voltage Vin?
A 4.8 V B 7.2 V C
9.6 V D 12 V
Reference: Past Exam Paper – November 2007 Paper 1 Q33
Solution 935:
Answer: D.
This is a potential divider circuit.
The potential divider equation can be written as follows:
Vout = [10 / (10 + 15)] × Vin = 10Vin / 25
Vin = 25Vout /
10
When Vout = 4.8V,
Vin = (25 × 4.8)
/ 10 = 12V
May I know the explaination(reasons) for Question 934 (c)?
ReplyDeletethe feathers cause light damping and the new graph drawn corresponds to this.
Deletehow
Deletethe feathers increases the effect of air resistance in the motion, which causes damping.
DeleteFor Q 932 I didn't understand that last paragraph. Can someone please explain
ReplyDeleteFor Q 932 can I know the explanation for the last line
ReplyDeletefor the spheres to be separating, v2 should be greater than v1. If v1 is greater than v2, the 2 spheres would be approaching
DeleteAbsolutely stunned to find such a helpful website,well done mate.I wanted to ask a question. (Reference M-J/2012/12),i am confused in choosing A and D Q. A mass is dropped from rest, and falls through a distance of 2.0 m in a vacuum. An observer
ReplyDeleterecords the time taken for the mass to fall through this distance using a manually operated
stopwatch and repeats the measurements a further two times. The average result of these
measured times, displayed in the table below, was used to determine a value for the acceleration
of free fall. This was calculated to be 9.8ms–2.
first measurement second measurement third measurement average
time/ s 0.6 0.73 0.59 0.64
Which statement best relates to the experiment?
A The measurements are precise and accurate with no evidence of random errors.
B The measurements are not accurate and not always recorded to the degree of precision of
the measuring device but the calculated experimental result is accurate.
C The measurements are not always recorded to the degree of precision of the measuring
device but are accurate. Systematic errors may be present.
D The range of results shows that there were random errors made but the calculated value is
correct so the experiment was successful.
Thanks. Glad to know that it is helpful to you
DeleteCan you explain as to which option, B or D would we choose?
ReplyDeletego to
Deletehttps://physics-ref.blogspot.com/2014/06/9702-june-2012-paper-12-worked.html