Tuesday, November 17, 2015

Physics 9702 Doubts | Help Page 225

  • Physics 9702 Doubts | Help Page 225



Question 1069: [Matter > Deformation]
The diagram represents a steel tube with wall thickness w which is small in comparison with the diameter of the tube.


The tube is under tension, caused by a force T, parallel to the axis of the tube. To reduce the stress in the material of the tube, it is proposed to thicken the wall.
The tube diameter and the tension being constant, which wall thickness gives half the stress?
A w / 2                        B 2 w                        C 2w                           D 4w

Reference: Past Exam Paper – June 2015 Paper 12 Q23



Solution 1069:
Answer: C.
This is a difficult question.
Stress = Force / Area

The steel tube is hollow in the centre with its wall having thickness w. When we calculate stress, we need to consider the area of the steel ‘material’, not the hollow space.

Because the wall is thin in comparison with the diameter of the tube, the area of the annulus (representing the wall thickness) is proportional to w.

Suppose the length of the tube is L and the diameter is d. Image that we have cut the steel tube so that it is now in the form of a rectangular sheet.
Area of annulus = (circumference of tube) × w
Area of annulus = 2π(d/2) × w = πdw
Thus, the area of the annulus is proportion to w.

But the diameter d and the force T are kept constant. To halve the stress, w must be doubled (since stress = force / area).










Question 1070: [Dynamics > Collision]
Two equal masses X and Y are moving towards each other on a frictionless air track as shown. The masses make an elastic collision.

Which row gives possible velocities for the two masses after the collision?

velocity of X                           velocity of Y
A         zero                                         20 cm s–1 to the right
B         10 cm s–1 to the right               10 cm s–1 to the right
C         20 cm s–1 to the left                 zero
D         30 cm s–1 to the left                 50 cm s–1 to the right

Reference: Past Exam Paper – June 2008 Paper 1 Q10



Solution 1070:
Answer: D.
For an elastic collision,
Relative speed of approach = Relative speed of separation

Since the masses are moving towards each other initially, they are ‘approaching’.
Relative speed of approach = 50 + 30 = 80 cm s–1

The relative speed of separation should also be 80 cm s–1.
A and C are incorrect since their relative speed is 20 cm s–1.
B is incorrect since the relative speed is zero.

For choice D, mass X is moving to the left while mass Y is moving to the right – so they are separating.
Relative speed of separation = 30 + 50 = 80 cm s–1











Question 1071: [Pressure]
A rectangular metal bar exerts a pressure of 15 200 Pa on the horizontal surface on which it rests.
If the height of the metal bar is 80 cm, what is the density of the metal?
A 190 kg m–3               B 1900 kg m–3             C 19 000 kg m–3          D 190 000 kg m–3

Reference: Past Exam Paper – November 2009 Paper 11 Q19 & Paper 12 Q18



Solution 1071:
Answer: B.
Pressure = Force / Area
Here, the force is the weight of the metal bar. [Weight = mg]

Mass m = Density ρ × Volume V       [m = ρV]
Volume V = base area A × height h of metal bar       [V = Ah]
Weight = mg = (ρV) × g = ρAhg

Pressure P = ρAhg / A = hρg

{OR simply start with P = hρg}

{g may be approximated to 10 ms–2 here.}
Density ρ = P / hg = 15200 / (0.80 × 10) = 1900kg m–3



4 comments:

  1. please consider answering Q20 9702/12/M/J/13

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/07/9702-june-2013-paper-12-worked.html

      Delete
  2. 9702/12/M/J/15 q36 thanks

    ReplyDelete
    Replies
    1. See solution 1118 at
      http://physics-ref.blogspot.com/2016/08/physics-9702-doubts-help-page-241.html

      Delete

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