Thursday, July 16, 2015

Physics 9702 Doubts | Help Page 178

  • Physics 9702 Doubts | Help Page 178



Question 879: [Measurement]
Uncertainty in the value of the momentum of a trolley passing between two points X and Y varies with the choice of measuring devices.
Measurements for the same trolley made by different instruments were recorded.
1 distance between X and Y using a metre rule with cm divisions = 0.55 m
2 distance between X and Y using a metre rule with mm divisions = 0.547 m
3 timings using a wristwatch measuring to the nearest 0.5 s at X = 0.0 s and at Y = 4.5 s
4 timings using light gates measuring to the nearest 0.1 s at X = 0.0 s and at Y = 4.3 s
5 mass of trolley using a balance measuring to the nearest g = 6.4 × 10–2 kg
6 mass of trolley using a balance measuring to the nearest 10 g = 6 × 10–2 kg
Which measurements, one for each quantity measured, lead to the least uncertainty in the value of the momentum of the trolley?
A 1, 3 and 6                B 1, 4 and 6                C 2, 3 and 6                D 2, 4 and 5

Reference: Past Exam Paper – June 2011 Paper 12 Q4



Solution 879:
Answer: D.
We need to identify which of the measurements would give the least uncertainty in the value of the momentum of the trolley.

Momentum p = mv = ms / t
where s is the distance between X and Y

Δp / p = (Δm / m) + (Δs / s) + (Δt / t)

For the uncertainty in the momentum, Δp, to have to least value, the uncertainties in the quantities measured should be smallest. That is Δm, Δs and Δt should be small. This is observed to be the case for measurements 2, 4 and 5, compared to the corresponding quantity measured.









Question 880: [Simple harmonic motion]
(a)
(i) Define simple harmonic motion.
(ii) On the axes of Fig.1, sketch variation with displacement x of the acceleration a of a particle undergoing simple harmonic motion.

(b) A strip of metal is clamped to the edge of a bench and a mass is hung from its free end as shown in Fig.2.

The end of the strip is pulled downwards and then released. Fig.3 shows variation with time t of the displacement y of the end of the strip.


On Fig.4, show corresponding variation with time t of the potential energy Ep of the vibrating system.

(c) The string supporting the mass breaks when the end of the strip is at its lowest point in an oscillation. Suggest what change, if any, will occur in period and amplitude of the subsequent motion of the end of the strip.

Reference: Past Exam Paper – June 2002 Paper 4 Q4



Solution 880:
(a)
(i) In simple harmonic motion, the acceleration is proportional to the displacement / distance (from fixed points) and it is directed towards a fixed point.

(ii) The graph is a straight line through the origin, in the quadrants 2 and 4.




(b) The graph is a sinusoidal curve, all about the t-axis. It should have the correct period and a correct ‘phase’.
{EP = mgh where h is the magnitude of the displacement y. Energy is a scalar and cannot have a negative value. At the equilibrium position (y = 0), h = 0 and so, Ep = 0.}



(c) The period is shorter. The amplitude is larger.









Question 881: [Matter > Deformation of Solids]
Two steel wires P and Q have lengths l and 2l respectively, and cross-sectional areas A and A / 2 respectively. Both wires obey Hooke’s law.
What is the ratio (tension in P / tension in Q) when both wires are stretched to the same extension?
A 1 / 4                         B 1 / 2                         C 2 / 1                         D 4 / 1

Reference: Past Exam Paper – June 2005 Paper 1 Q22 & June 2009 Paper 1 Q20



Solution 881:
Answer: D.
Since both wires are made of steel, they have the same Young modulus E.

Young modulus E = stress / strain
Stress = Tension / Area = F / A
Strain = extension / original length = e / L

Young modulus E = (F/A) / (e/L) = FL / Ae
Tension F = EAe / L

For wire P: FP = EAe / l
For wire Q: FQ = E (A/2) e / (2l) = EAe / 4l = FP / 4

Ratio = FP / FQ = FP / (FP/4) = 4 / 1
 


4 comments:

  1. This comment has been removed by the author.

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    Replies
    1. For Q14, see solution 884 at
      http://physics-ref.blogspot.com/2015/07/physics-9702-doubts-help-page-179.html

      Delete
  2. Explain June 2002 Paper 4 Q4c? Thanks

    ReplyDelete
    Replies
    1. The mass is no longer attached to the strip. So, as the strip moves up towards its highest point, the weight of the mass is no longer acting downwards against this motion. So, the amplitude is larger. Similarly, the strip moves quicker and thus, the period is shorter.}

      Delete

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