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Question 304: [Forces > Hooke’s law]
A sample is placed in a tensile testing machine. It is extended by known amounts and the tension is measured.

What is the work done on the sample when it is given a total extension of 9mm?
A 0.31J                        B 0.36J                        C 0.43J                        D 0.72J

Reference: Past Exam Paper – N83 / II / 40 & J99 / I / 23



Solution 304:
Answer: C.
Work done on sample is given by the area under the graph. In this case, the area can be divided by that of a triangle (from extension – 0 to 5mm) and that of a parallelogram (from extension = 5 to 9 mm)

Area of triangle = ½ (60) (0.005) = 0.150J
Area of parallelogram = ½ (sum of parallel sides) (height)
Area of parallelogram = 0.5 (60 + 80) (0.009 – 0.005) = 0.280J

Total area = 0.150 + 0.280 = 0.430J











Question 305: [Matter > Young modulus]
Wire consists of a 3.0 m length of metal X joined to a 1.0 m length of metal Y.
Cross-sectional area of the wire is uniform.

A load hung from wire causes metal X to stretch by 1.5 mm and metal Y to stretch by 1.0 mm.
Same load is then hung from a second wire of the same cross-section, consisting of 1.0 m of metal X and 3.0 m of metal Y.
What is total extension of this second wire?
A 2.5 mm                    B 3.5 mm                    C 4.8 mm                    D 5.0 mm

Reference: Past Exam Paper – November 2010 Paper 11 Q22 & 13 Q21



Solution 305:

Answer: B.

For a material, the Young modulus is constant. Since the load and the cross-sectional area are the same, the stress (= Force / Area) is also the same.

Young modulus, E = stress / strain = stress / (e/L) = L (stress) / e
where L is the original length and e is the extension

Consider metal X.
For 1^st wire: E = 3 (stress) / 1.5
For 2^nd wire: E = 1 (stress) / e

Since Young modulus for a particular material is the same,
3 (stress) / 1.5 = stress / e
Extension, e of metal X = 1.5 / 3 = 0.5 mm


Consider metal Y.
For 1^st wire: E = 1 (stress) / 1
For 2^nd wire: E = 3 (stress) / e

Since Young modulus for a particular material is the same,
Stress = 3 (stress) / e
Extension, e of metal Y = 3.0 mm

Total extension = 0.5 + 3.0 = 3.5 mm









Question 306: [Matter > Polymer]
Which graph represents force-extension relationship of a rubber band that is stretched almost to its breaking point?

Reference: Past Exam Paper – June 2010 Paper 12 Q20 & Paper 13 Q19 & June 2013 Paper 13 Q20



Solution 306:
Answer: A.
Rubber is a polymer. When the elastic band is stretched, the long chain molecules are fully aligned so making any further stretching is almost impossible.

This is represented in the graph A: after the section with a straight line, to extend the rubber band by even a small amount, the force required is much higher [when compared with the ‘straight line’ section]. That is, the gradient at the later part is much higher than the gradient at the earlier part.

Graph B suggest that after the rubber band is stretched [after the straight line], less force is required to cause a small extension.

Graph C is that of a metal obeying Hooke’s law.

Graph D suggests that after the rubber is stretched, no additional force is required to extend the rubber.
[B, C and D are incorrect]









Question 307: [Circular motion]
An object of mass 0.20 kg tied to a string is made to move in a vertical circle. When the object is at the highest point, the tension in the string is zero. Determine the tension in the string when the object is at the lowest point.



Solution 307:
Since the object is moving in circular motion, there is a resultant centripetal acceleration.

The tension is in the string is always towards the centre of the circle and the force of gravity acts downwards (towards the surface of the Earth). The centripetal force is also always towards the centre of the circle.

Let the tension in the string be T and let the centripetal force be F_C.
Weight of object, W = mg (= 0.20 x 9.81 = 1.962 N)

At the highest point, the tension and the centripetal force also act downwards.
T + mg = F_C
Since the tension is stated to be zero at the highest point, centripetal force, F_C = mg

Centripetal force, F_C = mv² / r = mg
Speed, v_h of object at highest position = √(gr)

Note that the speed of the object will not be constant. This is because at the highest position, the object also has gravitational potential energy (this varies depending on the height of the object).

At the lowest position, the difference in height of the object when compared to the highest position is 2r where r is the radius of the circle formed. Therefore, the gravitational potential energy of the object at the highest position is greater than that at the lowest position by mg(2r). This additional potential energy will contribute to the kinetic energy of the object at the lowest position.

Let the velocity of the object at the lowest position be v_l.

From the law of conservation of energy,
KE at lowest position = KE at highest position + GPE at highest position
½ mv_l² = ½ mv_h² + mg(2r)                  [but v_h = √(gr)]
0.5 v_l² = 0.5(gr) + 2gr
Speed, v_l at lowest position = √(5gr)


Now, consider the forces at the lowest position
T – mg = mv_l² / r
Tension T = mg + m (5gr) / r = mg + 5mg = 6mg
Tension T = 6 (0.2) (9.81) = 11.8 N









Question 308: [Forces > Hooke’s law]
A catapult consists of 2 strands of rubber, each of unstretched length 0.20 m and each of which stretches by 0.1 m under a tension of 50 N. A stone of mass 0.060 kg is projected vertically upwards from the catapult after each strand has been extended to a total of 0.35 m. What is the energy stored in the stretched catapult? Find the maximum height attained by the stone. (Assume that the rubber obeys Hooke’s law and that air resistance is negligible.) [Acceleration of free fall, g = 10 ms^-2.]

Reference: Past Exam Paper – J76 / I / 11

Solution 308:

Hooke’s law: F = kx
For each strand, spring constant k = F / x = 50 / 0.1 = 500 Nm^-1

Extension of each strand due to stone = 0.35 – 0.20 = 0.15 m
Energy stored in 1 rubber strand = ½ kx² = 0.5 (500) (0.15)² = 5.625 J
Total energy stored = 2 (5.625) = 11.25 J

At the maximum height, the energy of the stone is entirely gravitational potential energy.
mgh = 11.25
0.060 (10) h = 11.25
Maximum height, h = 18.75 = 18.8 m