Tuesday, November 25, 2014

Physics 9702 Doubts | Help Page 25

  • 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 = ½mv2. 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 A2 cos2(60°) = 0.250A2
[Since A2 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 variation with distance x along the wire (the cross-sectional area of which varies with length) of the potential difference V relative to the reference zero (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.

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 = ρL/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 d2 (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 d2). [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 mv2
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 = 0.5v2.
So, kinetic energy is halved at the highest point.









Question 144: [Waves > Graph]
Speed v of waves in deep water is given by equation v2 = 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 f2 against λ
B f against λ2
C f against 1 / λ
D f2 against 1/ λ

Reference: Past Exam Paper – June 2014 Paper 11 Q24



Solution 144:
Answer: D.
The equation given is:             v2 = gλ / 2π
Speed of wave, v = fλ

Substituting v = fλ in the 1st equation gives (fλ)2 = gλ / 2π which is simplified into
f2 = g / 2πλ = (g / 2π) (1 / λ)
Equation of a straight line:      y = mx + c
If the straight line passes through the origin, the equation becomes: y = mx

Therefore, a plot of f2 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 10–15kg                    B 10–20kg                     C 10–25kg                     D 10–30kg

Reference: Past Exam Paper – June 2008 Paper 1 Q39



Solution 145:
Answer: C.
The unified atomic mass constant, u = 1.66x10-27kg

Uranium contains about 238 nucleons.
Therefore, mass of uranium nucleus = 238 x 1.66x10-27 ≈ 10-25kg



20 comments:

  1. The interference patterns from a diffraction grating and a double slit are compared.
    Using 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

    ReplyDelete
  2. Check question 787 at
    http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-159.html

    ReplyDelete
  3. please help me with these questions
    2008 o/n P1 Q36
    2006 o/n P1 Q27
    2011 m/j P1 Q33
    2004 m/j P1 Q25
    2003 m/j P1 Q30

    ReplyDelete
    Replies
    1. For 2008 o/n P1 Q36, see solution 801 at
      http://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

      Delete
  4. 2011 m/j P12 Q33
    thank u very much...

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/09/9702-june-2011-paper-12-worked.html

      Delete
  5. q141
    when it is reduced from 100 to 20mA doesn't the current stay at 20mA?

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

      Delete
  6. Thank you for posting all this, it's very helpful!

    ReplyDelete
  7. Thank you so much
    All this is extremely helpful!

    ReplyDelete
  8. Thank you so much
    All this is extremely helpful!

    ReplyDelete
  9. 2015 m/j, P12, Q32....... Thanks in advance :)

    ReplyDelete
  10. 2015 m/j, P12, Q32....... Thanks in advance :)

    ReplyDelete
    Replies
    1. See solution 1111 at
      http://physics-ref.blogspot.com/2016/05/physics-9702-doubts-help-page-238.html

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

    ReplyDelete
    Replies
    1. ok. you know that charge Q = It

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

      you should obtain the same answer.

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

    ReplyDelete
    Replies
    1. Good that it's helping others. let your friends know so that they can benefit too.

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

    ReplyDelete

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