Markscheme RC & the bullet

Answers to principle of moments 2

3. (a) Principle of moments

In equilibrium

sum of clockwise moment (about any point) is equal to sum of

anticlockwise moment (about that point)

2

(b)(i) Weight

Use of “width x thickness x length”

Use of “density = mass”

volume

Correct value 3

V = 1.2 × 0.6 × 200 (cm3) = 144 (cm3)

Using ρ = m/v

m = 8 (g cm-3) × 144 (cm-3) = 1152 g

Weight = mg = 1152 × 10-3 (kg) × 9.81 ( m s-2) = 11.3 (N) / 12

(N)

(ii) Force F

Correct substitution into correct formula

Correct value with correct unit

F × 60 (cm) = 11.3 (N) × 40 (cm) / 12 (N) × 40 (cm) / 11(N) × 40

(cm)

= 7.5 N / 8 N / 7.3 N

2

(iii) Force R

18.3 N / 18.8 N / 20 N

1

(iv) Sketch graph

Any line upwards

Correct shape for F [concave shaped curve]

2

Y12 Young's Modulus

Stress, strain and the Young modulus

1.         A long strip of rubber whose cross section measures 12 mm by 0.25 mm is pulled with a force of 3.0 N. What is the tensile stress in the rubber?

2.         Another strip of rubber originally 90 mm long is stretched until it is 120 mm long.  What is the tensile strain?

3.         The marble column in a temple has dimensions 140 mm by 180 mm.

a.     What is its cross-sectional area in mm2?

b.    Now change each of the initial dimensions to metres – what is the cross-sectional area in m2?

c.     If the temple column supports a load of 10 kN, what is the compressive stress, in N m–2?

d.    The column is 5.0 m tall, and is compressed by 0.1 mm. What is the compressive strain when this happens?

e.     Use your answers to parts 5 and 6 to calculate Young’s modulus for marble.

4.   A 3.0 m length of copper wire of diameter 0.4 mm is suspended from the ceiling. When a 0.5 kg mass is suspended from the bottom of the wire it extends by 0.9 mm.

a.     Calculate the strain of the wire.

b.    Calculate the stress in the wire.

c.     Calculate the value of the Young modulus for copper.

Cpacitance