# 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 E

_{1}
B the e.m.f. of the cell E

_{2}
C the resistance of resistor R

D the resistance of the wire XY

**Reference:**

*Past Exam Paper – November 2008 Paper 1 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. E

_{2}connected to the galvanometer.
Note that at any time, the points T
and Y may be considered to be the same since the part TY of the wire can be
neglected (that is, they do not contribute any resistance to the circuit).

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 XY (since Y may be
considered to be at the same potential as T, this is also the p.d. across XT)
is known to be equal to the e.m.f. E

_{2}when the galvanometer reading is zero. So, we need to know the e.m.f. of the cell E_{2}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 mÏ‰

^{2}a^{2}where Ï‰ = 2Ï€f
Energy of oscillations = ½ (37x10

^{-3})(2Ï€ x 3.5)^{2}(2.8x10^{-2})^{2}
Energy of oscillations = 7.0x10

^{-3}J**(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 mÏ‰

^{2}a^{2}; Kinetic energy = at displacement x = ½ mÏ‰^{2}(a^{2}– x^{2})
And Potential energy at
displacement x = ½ mÏ‰

^{2}x^{2}.}
KE = ½ mÏ‰

^{2}(a^{2}– x^{2}) or PE = ½ mÏ‰^{2}x^{2}
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 mÏ‰

^{2}(a^{2}– x^{2}) = ½ mÏ‰^{2}x^{2}
x = a / (2

^{0.5}) = 2.8 / (2^{0.5}) = 2.0cm
OR

Allow:

Ï‰ =
(k/m)

^{0.5}So, k = Ï‰^{2}m = (2Ï€ x 3.5)^{2}(0.037) = 17.9
E = ½ kx

^{2}= 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 W

_{P}, and work done in stretching spring Q is W_{Q}.
How is W

_{P}related to W_{Q}?
A W

_{P}= ¼ W_{Q}B W_{P}= ½ W_{Q}C W_{P}= 2W_{Q}D W_{P}= 4W_{Q}**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, W

_{Q}= 2W_{P}giving W_{P}= ½ W_{Q}

__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

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