# Physics 9702 Doubts | Help Page 25

__Question 138: [Dynamics > Momentum]__Diagram shows two spherical masses approaching each other head-on at an equal speed u. One has mass 2m and other has mass m.

Which diagram, showing situation after the collision, shows result of an elastic collision?

**Reference:**

*Past Exam Paper – November 2009 Paper 12 Q8 & November 2009 Paper 11 Q9*

__Solution 138:__**Answer: A.**

__For an elastic collision__,

**Velocity of approach (before collision) = Velocity of separation (after collision)**

{The above result can be obtained by
considering that for elastic collision, both momentum and kinetic energy is
conserved. Momentum, p = mv. Kinetic energy = ½mv

^{2}. By equating the sum of momentum before collision to that after collision and by equating the sum of KE before collision to that after collision, 2 equations are obtained which can be simplified into the above stated result: Velocity of approach (before collision) = Velocity of separation (after collision). The proof will not be shown here}
{Approach means that the 2
sphere are coming towards each other and separation means that they are moving
away from each other}

**Before collision, velocity of approach = u + u = 2u**

Consider A:

Velocity of separation = (u/3) + (5u/3)
= 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3)
= 5u/6

Consider C:

Both spheres are moving in the same
direction. So, speed of separation is the difference in the 2 speed here/

Velocity of separation = (2u/3) – (u/6)
= 3u/6 = 0.5u

Consider D:

Since the spheres stick together,
they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of
separation = 2u.

__Question 139: [Dynamics > Inelastic collision]__Which quantities are conserved in inelastic collision?

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q17*

__Solution 139:__**Answer: C.**

The laws of conservation
of energy and conservation of momentum state that, in a system, the

__total__energy and__total__momentum is always conserved.
Momentum in a system is always
conserved in any collision. Linear
momentum of the momentum of a body in 1 dimension (linear motion).

In an inelastic collision, kinetic energy is NOT conserved. Some of the initial kinetic energy is converted into other forms of energy. So, the total energy is still conserved.

__Question 140: [Waves > Polarization > Intensity]__When plane-polarised light of amplitude A is passed through polarising filter as shown, amplitude of the light emerging is A cosθ.

Intensity of initial beam is I.

What is the intensity of emerging light when θ is 60.0°?

A 0.250 I B 0.500 I C 0.750 I D 0.866 I

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q25 & Paper 13 Q24 & November 2013 Paper 11 & 12 Q30*

__Solution 140:__**Answer: A.**

The intensity, I of the light is proportional to the square of the amplitude, A.

Amplitude of emerging light = A cos 60°

Intensity of emerging light is proportional to A

^{2 }cos

^{2}(60°) = 0.250A

^{2}

[Since A

^{2}∝ I,] Thus, the intensity of emerging light is equal to 0.250I.

__Question 141: [Current of Electricity > Charge]__Current in a component is reduced uniformly from 100 mA to 20 mA over period of 8.0s.

What is the charge that flows during this time?

A 160 mC B 320 mC C 480 mC D 640 mC

**Reference:**

*Past Exam Paper – June 2003 Paper 1 Q30 & November 2013 Paper 13 Q32*

__Solution 141:__**Answer: C.**

Charge, Q = It where I is the
current and t is the time

Since the current I is changing, the
average current should be calculated. {The fact that
the average current needs to be calculated would have been deduced if a quick
sketch graph was drawn.}

Average current = (100 + 20) /2 =
60mA

Charge, Q = 60 x 8 = 480mC

__Question 142: [Current of Electricity > Resistance > Non-uniform wire]__Circular cross-sectional area of metal wire varies along its length. There is current in the wire. Narrow end of wire is at a reference potential of zero.

Which graph best represents variation with distance x along wire of the potential difference V relative to the reference zero?

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q32*

__Solution 142:__**Answer: C.**

We need the

*(the cross-sectional area of which varies with length)*__variation with distance x along the wire__*(so, at x = 0, the potential is maximum {at x = 0, we are not at zero potential, but at the start of the wide cross-sectional area}). All the 4 graphs correctly indicate the potential difference to be maximum at x = 0.*__of the potential difference V relative to the reference zero__
Ohm’s law:

**V = IR**
The current is constant along the
wire. Only the resistance changes and this causes a change in the potential difference.

Resistance of wire,

**R =****ρ****L/A**
This requires careful thinking.

First, assume that the wire was uniform,
with the diameter not changing. When the length of wire L is furthest away from
the zero potential reference, the potential difference would be maximum since length
L is highest, causing resistance R to maximum and thus, potential difference V
= IR is also the highest. As distance x increases, the length L (from the zero
potential reference) decreases, and so, R and finally V also decreases. [A is incorrect]

So, as explained above, as x
increases, V should be decreasing.

Now, the diameter of the wire is
falling linearly (as x increases) but the area
of cross-section, A is not falling linearly [since
cross-sectional area = π(d / 2)

^{2}where d is the diameter].
Since resistance R is inversely
proportional to cross-sectional area A, it is also inversely proportional to d

^{2}(when d is small, R is big). Therefore, there is a greater percentage fall (in resistance, and thus, potential difference) per unit length at narrow end than at wide end (that is, the decrease is not linear, but depends on d^{2}). [B, which represents a linear decrease, is incorrect]
This means that potential difference
per unit length (this is represented by the

__gradient__of the graphs shown which equal to V/x) is less at the wide end than at the narrow end (the gradient at the highest value of x should be greater than the gradient at x = 0 and the gradient keeps on increasing as x increases). [Answer D is wrong since gradient is highest at x = 0]
This makes the correct answer C.

__Question 143: [Kinematics > Projectile motion]__
Projectile is launched at 45° to
horizontal with initial kinetic energy E.

Assuming air resistance to be
negligible, what will be the kinetic energy of projectile when it reaches its
highest point?

A 0.50 E B 0.71 E C
0.87 E D E

**Reference:**

*Past Exam Paper – November 2009 Paper 12 Q13*

__Solution 143:__**Answer: A.**

Let the initial velocity = v

Horizontal component of velocity = v
cos(45)

Vertical component of velocity = v
sin(45)

At the highest point, the vertical
component of velocity of the projectile is zero. The projectile only has a
horizontal component of velocity which is v cos(45) [since
air resistance is negligible, this component is unchanged].

Kinetic energy = ½ mv

^{2}
Kinetic energy is proportional to (speed)

^{2}. Initial kinetic energy = E
Therefore, at the highest point, the
kinetic energy is proportional to [vcos(45)]

^{2}= 0.5v^{2}.
So, kinetic energy is halved at the
highest point.

__Question 144: [Waves > Graph]__
Speed v of waves in deep water is
given by equation v

^{2}= gλ / 2π where λ is the wavelength of waves and g is the acceleration of free fall.
Student measures wavelength λ and
frequency f of a number of these waves.

Which graph should he plot to give a
straight line through origin?

A f

^{2}against λ
B f against λ

^{2}
C f against 1 / λ

D f

^{2}against 1/ λ**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q24*

__Solution 144:__**Answer: D.**

The equation given is:

**v**^{2}= gλ / 2**π**
Speed of wave,

**v = fλ**
Substituting v = fλ in the 1

^{st}equation gives (fλ)^{2}= gλ / 2π which is simplified into**f**= g / 2πλ = (g / 2π)

^{2}**(1 /**

**λ)**

Equation of a straight line:

**y**= m**x**+ c
If the straight line passes through
the origin, the equation becomes:

**y**= m**x**
Therefore, a plot of f

^{2}against 1/λ would give a straight line through the origin.

__Question 145: [Nuclear Physics > Nucleus]__
What is the approximate mass of a
nucleus of uranium?

A 10

^{–15}kg B 10^{–20}kg C 10^{–25}kg D 10^{–30}kg**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q39*

__Solution 145:__**Answer: C.**

The unified atomic mass constant, u
= 1.66x10

^{-27}kg
Uranium contains about 238 nucleons.

Therefore, mass of uranium nucleus =
238 x 1.66x10

^{-27}≈ 10^{-25}kg
The interference patterns from a diffraction grating and a double slit are compared.

ReplyDeleteUsing the diffraction grating, yellow light of the first order is seen at 30° to the normal to the

grating.

The same light produces interference fringes on a screen 1.0 m from the double slit. The slit

separation is 500 times greater than the line spacing of the grating.

What is the fringe separation on the screen?

A 2.5 × 10–7m

B 1.0 × 10–5m

C 1.0 × 10–3m

D 1.0 × 10–1m

Check question 787 at

ReplyDeletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-159.html

please help me with these questions

ReplyDelete2008 o/n P1 Q36

2006 o/n P1 Q27

2011 m/j P1 Q33

2004 m/j P1 Q25

2003 m/j P1 Q30

For 2008 o/n P1 Q36, see solution 801 at

Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-161.html

For 2006 o/n P1 Q27, see solution 787 at

http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-159.html

For 2011 m/j P1 Q33, which variant?

For 2004 m/j P1 Q25, see solution 773 at

http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-156.html

For 2003 m/j P1 Q30, see solution 141 at

http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-25.html

2011 m/j P12 Q33

ReplyDeletethank u very much...

Go to

Deletehttp://physics-ref.blogspot.com/2014/09/9702-june-2011-paper-12-worked.html

q141

ReplyDeletewhen it is reduced from 100 to 20mA doesn't the current stay at 20mA?

It took 8.0s to reduce the current from 100mA to 20mA. We are calculate the charge flow during this amount of time.

DeleteThank you for posting all this, it's very helpful!

ReplyDeleteThank you so much

ReplyDeleteAll this is extremely helpful!

No problem.

DeleteThank you so much

ReplyDeleteAll this is extremely helpful!

2015 m/j, P12, Q32....... Thanks in advance :)

ReplyDelete2015 m/j, P12, Q32....... Thanks in advance :)

ReplyDeleteSee solution 1111 at

Deletehttp://physics-ref.blogspot.com/2016/05/physics-9702-doubts-help-page-238.html

I don't get the question 141 that we need to calculate avg current.

ReplyDeleteI have drawn the negative gradient graph, and I still don't get any clue that we needed to calculate average current. Can you please explain?

ok. you know that charge Q = It

Deletefrom the current-time graph, the charge is given by the area under graph.

you should obtain the same answer.

Idk what would I ever do w/o this blog <3

ReplyDeleteGood that it's helping others. let your friends know so that they can benefit too.

DeleteThis is the best site. Helped me so much through out A levels! Thankyou so muchhh

ReplyDelete