A wire of length 1.70 m hangs vertically from a fixed point, as shown in Fig. 4.1.

Question 11

(a) Define, for a wire,

(i) stress, [1]

(ii) strain. [1]

(b) A wire of length 1.70 m hangs vertically from a fixed point, as shown in Fig. 4.1.

Fig. 4.1

The wire has cross-sectional area 5.74 × 10^-8 m² and is made of a material that has a

Young modulus of 1.60 × 10¹¹ Pa. A load of 25.0 N is hung from the wire.

(i) Calculate the extension of the wire. [3]

(ii) The same load is hung from a second wire of the same material. This wire is

twice the length but the same volume as the first wire. State and explain how the

extension of the second wire compares with that of the first wire. [3]

Reference: Past Exam Paper – June 2011 Paper 21 Q4

Solution:

(a)

(i) Stress is defined as the force acting per unit (cross-sectional) area.

(ii) Strain is defined as the ratio of the extension of the wire to its original length.

(b)

(i)

Young modulus E = Stress / Strain

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

Extension e (= FL / AE) = (25 × 1.70) / (5.74×10^-8 × 1.6×10¹¹)

Extension e = 4.6×10^-3 m

(ii)

{Since the wire is of the same material the Young modulus is the same as the previous wire.

The force F is also constant as the same load is used.}

The area A becomes A/2       OR the stress is doubled.

{Volume = Area × length = AL

Since the length is now twice (= 2L), the area must be halved (= A/2) for the volume to remain the same (2L × A/2 = AL = V).

Stress = Force / Area = F / (A/2) = 2 × F/A

The stress is doubled.}

The extension e is proportional to (L / A)   OR Substitute into the full formula

{Extension e = FL / AE

As force F and the Young modulus E are constants, the extension is proportional to L/A

e ∝ L / A

The length is doubled while the area is halved. So,

New extension = 2L / (A/2) = 4 × L/A = 4 × extension of first wire}.

So, the total extension increase (for the second wire) is 4e.

Two balls X and Y are moving towards each other with speeds of 5 m s-1 and 15 m s-1 respectively.

Question 20

Two balls X and Y are moving towards each other with speeds of 5 m s^-1 and 15 m s^-1

respectively.

They make a perfectly elastic head-on collision and ball Y moves to the right with a speed of 7 m s^-1.

What is the speed and direction of ball X after the collision?

A 3 m s^-1 to the left

B 13 m s^-1 to the left

C 3 m s^-1 to the right

D 13 m s^-1 to the right

Reference: Past Exam Paper – March 2019 Paper 12 Q8

Solution:

Answer: B.

For a perfectly elastic head-on collision,

Relative velocity of separation = Relative velocity of approach

Initially, the two balls are moving towards each other.

Relative velocity of approach = 5 + 15 = 20 m s^-1

After the collision, ball Y moves to the right at a speed of 7 m s^-1.

The relative velocity of separation should be equal to the relative velocity of approach (= 20 m s^-1). 

Ball Y is moving to the right at 7 m s^-1. So, ball X should move in such a way to increase the increase the relative velocity of separation.

Ball X should move in the opposite direction (i.e. to the left).

v + 7 = 20

Velocity v = 20 – 7 = 13 m s^-1

An isolated solid metal sphere is positively charged. The variation of the potential V with distance x from the centre of the sphere is shown in Fig. 5.1.

Question 24

(a) Define electric potential at a point. [2]

(b) An isolated solid metal sphere is positively charged.

The variation of the potential V with distance x from the centre of the sphere is shown in

Fig. 5.1.

Fig. 5.1

Use Fig. 5.1 to suggest

(i) why the radius of the sphere cannot be greater than 1.0 cm, [1]

(ii) that the charge on the sphere behaves as if it were a point charge. [3]

(c) Assuming that the charge on the sphere does behave as a point charge, use data from Fig. 5.1 to determine the charge on the sphere. [2]

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q5

Solution:

(a) The electric potential at a point is defined as the work done in moving unit positive charge from infinity (to the point).

(b)

(i) The electric potential is constant inside the sphere

{Electric potential decreases from the surface of the sphere as we go further away. BUT inside the sphere the electric potential should be constant (as the electric field strength is zero inside the sphere).

Since the graph starts at x = 1.0 cm it can be deduced that any distance smaller than x = 1.0 cm is inside the sphere.

If the radius was greater than 1.0 cm, the electric potential should have been constant for some values of x greater than 1.0 cm. However, we can observe that the potential decreases as from x = 1.0 cm. So the radius cannot be greater than 1.0 cm.}

(ii)

{V = Q / 4πε₀x

Since 4πε₀ is constant,

V ∝ Q / x

Vx ∝ Q

The product of Vx is proportional to the charge Q.

For a point charge, the value of Q should be constant – that is, the charge cannot be changing.}

For a point charge, the product Vx is constant.

{From the graph, a point represents (x, V). So for any point on the graph, the product of Vx should be given the same (constant) value.

We need to use points from the graph to show that the products for these points are constant.}

The co-ordinates should be clear and we need to determine two values of Vx at least 4 cm apart.

{Consider the points (6, 30) and (2, 90) from the graph. (the points cannot be too close)

The product of both of them give a value of 180 (6×30 = 180 and 2×90 = 180). So, the product obtained is constant at different points.}

Since the product is constant, the charge on the sphere behaves as if it were a point charge.

(c)

{As found above, the product is 180 Vcm

Vx = 180 Vcm = 180 × (1.0×10^–2) Vm

V = Q / 4πε₀x

Q = V×4πε₀x = 4πε₀[Vx] }

Charge Q = 4πε₀[Vx] = 4π × (8.85×10^–12) × [180 × (1.0×10^–2)] = 2.0×10^–10 C