Sunday, June 14, 2015

Physics 9702 Doubts | Help Page 173

  • Physics 9702 Doubts | Help Page 173

Question 863: [Dynamics > Newton’s laws of motion]
A body experiences a varying resultant force that causes its momentum to vary, as shown in graph.
At which point does the resultant force have largest value?

Reference: Past Exam Paper – November 2014 Paper 13 Q10

Solution 863:
Answer: B.
From Newton’s 2nd law, the resultant force is the rate of change of momentum.
Resultant force F = Δp / t

In a momentum-time graph, the resultant force is given by the gradient of the graph.

The resultant force will have the largest value at the point where the gradient has the largest value. This is indicate by the steepness of the tangent at that point. The steeper the tangent, the greater is the gradient and hence the resultant force.

Of the 4 points, the tangent is steepest at point B.

Question 864: [Dynamics > Equilibrium]
(a) Define centre of gravity.

(b) A uniform rod AB is attached to vertical wall at A. Rod is held horizontally by a string attached at B and to point C, as shown in Fig.1.

Angle between the rod and the string at B is 50°. The rod has length 1.2 m and weight 8.5 N. Object O of mass M is hung from the rod at B. Tension T in the string is 30 N.
(i) Use resolution of forces to calculate the vertical component of T.
(ii) State principle of moments.
(iii) Use principle of moments and take moments about A to show that weight of object O is 19N.
(iv) Hence determine mass M of the object O.

(c) Use concept of equilibrium to explain why a force must act on the rod at A

Reference: Past Exam Paper – June 2013 Paper 22 Q3

Solution 864:
(a) Centre of gravity is defined as the point where (all) the weight (of the body) is considered / seems to act.

(i) Vertical component of T = 30cos(40) = 22.98 ≈ 23N

(ii) The principle of moments states that the sum of the clockwise moments about a point equals the sum of the of the anticlockwise moments (about the same point)

Let W be the weight of object O
Sum of clockwise moments about A = Sum of anti-clockwise moments about A
(8.5 × 0.60) + (1.2 × W) = 23 × 1.2
1.2W = 27.6 – 5.1
W = 18.75 19N

{Here is a suggested solution for Q3(b)(iii):
The previous part (vertical component of T) is equal to 23 N
Now my method of solving it is:
Tan 50 = AC / 1.2
AC = 1.43
23 * 1.43 = (8.5+W) * 1.2
W = 18.9 = 19.0 N
Now the method in the marking scheme is a bit different, but my final answer ( 19.0 ) is still the same. Is my method acceptable or not?

Why this method is not appropriate, even if the correct answer is obtained?
1st mistake:
Moment is the product of force and the perpendicular distance of the force from the pivot. Here BOTH the distance AC and the force are VERTICAL. This does not have any turning effect at A.
2nd mistake:
The centre of gravity does not act at 1.2m from the pivot BUT at the CENTRE.}

W = Mg
So, M = W / g = 19 / 9.81 = 1.9 kg

(c) For equilibrium, the resultant force and moment must be zero. A force must act on the rod at A since the upward force does not equal downward force OR the horizontal component of T is not balanced by the forces shown.
{The force would start at point A and its direction would be about north east (the angle with the horizontal is not necessary 45° - it needs to be determined)}

Question 865: [Measurement > Uncertainty]
In an experiment to determine acceleration of free fall g, the period of oscillation T and length l of a simple pendulum were measured. Uncertainty in the measurement of l is estimated to be 4%, and the uncertainty in the measurement of T is estimated to be 1%.
Value of g is determined using the formula
g = 4π2 l / T2
What is the uncertainty in the calculated value for g?
A 2%                           B 3%                           C 5%                           D 6%

Reference: Past Exam Paper – June 2013 Paper 11 Q5

Solution 865:
Answer: D.
g = 4π2 l / T2
Δg / g = (Δl / l) + 2(ΔT / T)

Percentage uncertainty (Δl / l) = 4%
Percentage uncertainty (ΔT / T) = 1%

Percentage uncertainty in g = 4 + 2(1) = 4 + 2 = 6%

Question 866: [Matter > Deformation]
Metal cube of side l is placed in a vice and compressed elastically by two opposing forces F.

How will Δl, the amount of compression, relate to l?
A Δl α 1 / l2                 B Δl α 1 / l                   C Δl α l                        D Δl α l2

Reference: Past Exam Paper – November 2011 Paper 11 Q26 & Paper 13 Q25

Solution 866:
Answer: B.
Young modulus E = stress / strain = (Force/Area) / (extension/original length)
Young modulus E = (F/A) / (e/l) = Fl / Ae

The Young modulus E is proportional to l / Ae.
Area = l2
Compression, e = Δl

E α l / (l 2 × Δl)
E α 1 / (l × Δl)
So, Δl α 1 / (El) giving Δl α 1 / l since the Young modulus E is a constant.

1 comment:

  1. awesome, thank you for all the help , this work is well appreciated


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