Physics 9702 Doubts | Help Page 99
Question 508: [Dynamics]
Stone is projected horizontally in a
vacuum and moves along path shown.
X is a point on this path. XV and XH
are vertical and horizontal lines respectively through X. XT is the tangent to
path at X.
Along which directions do forces act
on the stone at X?
A XV only B XH only C
XV and XH D XT only
Reference: Past Exam Paper – June 2011 Paper 12 Q14
Solution 508:
Answer: A.
Since the stone is projected in a
vacuum, there is no air. Therefore, there is no air resistance.
The force of gravity acts downwards.
So, at any point along the path, there will be a force downwards – this will
along XV in the diagram given.
As this is in a vacuum, there is no
air resistance, nor any other forms of friction (forces that oppose motion).
XT represents the direction of
motion of the stone at a point on the path.
XH represents the horizontal
component of the motion (speed).
Question 509: [Current
of Electricity > Potentiometer]
Diagram shows a potentiometer
circuit.
Contact T is placed on the wire and
moved along wire until the galvanometer reading is zero. The length XT is then
noted.
In order to calculate potential
difference per unit length on the wire XY, which value must also be known?
A the e.m.f. of the cell E1
B the e.m.f. of the cell E2
C the resistance of resistor R
D the resistance of the wire XY
Reference: Past Exam Paper – November 2008 Paper 1 Q37 & November 2012 Paper 11 Q37
Solution 509:
Answer: B.
The potentiometer circuit works as
follows.
When the galvanometer reading is
zero, this means that p.d. across XT of the wire is the same as the e.m.f. E2
connected to the galvanometer.
To calculate the potential
difference per unit length on the wire XY, we need to know the length of the
wire, and the p.d. across it.
The p.d. across XT is known to be equal to the e.m.f. E2 when the galvanometer reading is zero. So, we need to know the e.m.f. of the cell E2 to know the p.d. across the wire.
Question 510: [Simple harmonic motion]
Ball of mass 37 g is held between
two fixed points A and B by two stretched helical springs, as shown in Fig.
Ball oscillates along line AB with
simple harmonic motion of frequency 3.5 Hz and amplitude 2.8 cm.
(a) Show that total energy of the oscillations is 7.0 mJ.
(b) At two points in oscillation of the ball, its kinetic energy is
equal to the potential energy stored in the springs. Calculate magnitude of the
displacement at which this occurs
(c) On axes of Fig. and using answers in (a) and (b), sketch graph to
show the variation with displacement x of
(i) total energy of the system (line
T):
(ii) kinetic energy of ball (line
K):
(iii) potential energy stored in the
springs (line P):
(d) Arrangement in Fig is now rotated through 90° so that line AB is
vertical and the ball oscillates in a vertical plane.
Suggest one form of energy, other
than those in (c), that must be taken into consideration when plotting new
graphs to show energy changes with displacement.
Reference: Past Exam Paper – June 2012 Paper 42 Q2
Solution 510:
(a)
Energy of oscillations = ½ mω2a2 where ω = 2Ï€f
Energy of oscillations = ½ (37x10-3)(2Ï€ x
3.5)2(2.8x10-2)2
Energy of oscillations = 7.0x10-3J
(b)
Kinetic energy = Potential energy
Either
{The total energy is
7.0mJ. So, for KE = PE, they should both be halved the value of the total
energy.}
KE or PE = 3.5mJ
{In this method, we are
using the formula for either KE or PE which are both 3.5mJ to obtained x.
Total energy = ½ mω2a2 ; Kinetic
energy = at displacement x = ½ mω2(a2 – x2)
And Potential energy at
displacement x = ½ mω2x2.}
KE = ½ mω2(a2 – x2) or PE = ½ mω2x2
Displacement x = 2.0cm
(KE or PE = 7.0mJ scores 0/3)
OR
{Here, since KE = PE, by
equating them we can obtain x}
½ mω2(a2
– x2) = ½ mω2x2
x = a / (20.5) = 2.8 / (20.5)
= 2.0cm
OR
Allow:
ω =
(k/m)0.5 So, k = ω2m = (2π x 3.5)2(0.037) = 17.9
E = ½ kx2 = 3.5x10-3
J
x = 2.0cm
(c)
(i)
{amplitude is 2.8cm}
Graph: horizontal line with
y-intercept 7.0mJ with end-points of line at +2.8cm and -2.8cm
(ii)
Graph: reasonable curve
with maximum at (0,7.0) and
end-points of line at (-2.8,0) and (+2.8,0)
(iii) potential energy stored in the
springs (line P):
{As calculated in (b),
when KE = PE = 3.5mJ, the displacement is -2.0cm or +2.0cm.}
Graph: inverted version of part (ii)
with intersections at (-2.0,3.5) and
(+2.0,3.5)
(d) Gravitational potential
energy
Question 511: [Matter
> Hooke’s law]
Two springs P and Q both obey
Hooke’s law. They have spring constants 2k and k respectively.
Springs are stretched, separately,
by force that is gradually increased from zero up to a certain maximum value,
the same for each spring. Work done in stretching spring P is WP, and
work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP = ¼ WQ B WP = ½ WQ
C WP = 2WQ
D WP = 4WQ
Reference: Past Exam Paper – November 2002 Paper 1 Q24
Solution 511:
Answer: B.
Hooke’s law: F = ke where F: force, k: spring constant and e:
extension
Extension e = F / k
The work done in stretching the
springs is given by the area under the force-extension graph of the spring.
Since both springs obey Hooke’s law, the graphs would be straight lines with
positive gradient and passing through the origin.
The area under the graph would be
that of a triangle with the base being the extension and the height being the
force applied.
Consider spring P. Spring constant =
2k
Let the force applied be F.
Extension, e = F / 2k = 0.5 (F/k)
Consider spring Q. Spring constant =
k
Let the force applied be F.
Extension, e = F / k = 1 (F/k)
It can be observed that for any
force F, the extension for spring Q is twice that of spring P. This means that
the area under the graph (which represents the work done) for spring Q is twice
that for spring P.
So, WQ = 2WP
giving WP = ½ WQ
Question 512: [Kinematics]
Car is stationary at traffic lights. When traffic lights go green, the driver presses down sharply on the accelerator. The resultant horizontal force acting on car varies with time as shown.
Which graph shows variation with time of the speed of the car?
Reference: Past Exam Paper – June 2013 Paper 11 Q9
Solution 512:
Answer: A.
From the force-time graph given, the
force on the accelerator increases sharply to a constant value (indicated by
the horizontal line in the graph) at a point. This causes a constant acceleration
since F = ma.
Acceleration is the rate of change
of velocity.
Since the acceleration is constant,
the increase in velocity in uniform with time. This is represented by a straight
line graph, showing that the gradient of the speed-time graph (which gives the
acceleration) is constant.
in vacuum how is gravity there?
ReplyDeleteGravitational force is due to masses. examples: Earth, Sun, planets, any mass (the bigger the mass, the more significant is the force)
DeleteVacuum might be thought as emptiness. e.g. absence of air - in this case, there is no air resistance
So Why does a Rock and a Feather falls downwards at same velocity in Vaccum?
Deletein vacuum, there is no air resistance.
Deletethey both fall with the same acceleration due to gravity (= 9.81 ms^-2)
in q 508 the horizontal component accounts for the air resistance and is the always the case regardless of air or vacuum?
ReplyDeleteNo, in vacuum, there is no air resistance.
DeleteIf there was air, then there would be a force along XH (the force being the air resistance). But this is not the case here.
thank u
Deleteq8 paper 11 2009 nov pls
ReplyDeletedoesn't the velocity change when the ball rebounds?
Go to
Deletehttp://physics-ref.blogspot.com/2014/10/9702-november-2009-paper-11-worked.html
Sir, how to do 9702/12/ON/12 no 36? Tks
ReplyDeleteHow to do 9702/12/ON/12 no 36? Thanks.
ReplyDeleteGo to
Deletehttp://physics-ref.blogspot.com/2014/08/9702-november-2012-paper-12-worked.html
Heloo
ReplyDeleteCan you please help me in solving the 18th question in 2004 oct/nov paper 1
See solution 1094 at
Deletehttp://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-233.html
Sir? Oct/Nov 2012 qp11 question 23? Please?
ReplyDeleteSee solution 1113 at
Deletehttp://physics-ref.blogspot.com/2016/06/physics-9702-doubts-help-page-239.html