Material Science

Question 1

Crystalline Material

Answer 1

Atoms are arranged in structured lattices with long-range order.

Question 2

Amorphous Material

Answer 2

Atoms are arranged in random structures with short-range order.

Question 3

Polymeric Materials

Answer 3

Atoms are arranged in chain-like molecules.

Question 4

Ductile

Answer 4

Material property describing its ability to be drawn out into wire.

Question 5

Brittle

Answer 5

Material property describing its high stiffness, but low toughness. Brittle materials have a very small plastic region.

Question 6

Strong

Answer 6

Material property describing high maximum stress that can be endured before fracture.

Question 7

Tough

Answer 7

Material property describing its ability to absorb a lot of energy before fracture.

Question 8

Necking

Answer 8

The cross-sectional area at the weakest surface point due to imperfections decreases as the material deforms plastically.

Question 9

Hysteresis

Answer 9

The different behaviour of polymeric materials in a stress-strain relationship during loading and unloading.

Question 10

Dislocations

Answer 10

Gaps in the ordered atomic lattice structure of crystalline materials which allows adjacent atomic planes to break their bonds, slip and reform at the terminating edge of the dislocated plane under stress. This causes the dislocation to move through the material, decreasing its strength.

Question 11

Grains / Grain Boundaries

Answer 11

Crystallites, small areas of atomic planes with similar orientation, within polycrystalline or crystalline materials (no material is a perfect crystal).

The interfaces between them are difficult for dislocations to cross due to differences in atomic planes, meaning introducing more boundaries increases the strength of the material

Question 12

Crack Propagation

Answer 12

Due to surface imperfections, surface bonds will break, but most stress is concentrated on inner bonds which break, causing crack propagation downwards.

Question 13

Brittle Materials under Compression

Answer 13

Putting brittle materials under compression decreases the rate of crack propagation as more energy is required to separate each side of the crack. E.g. steel under strain in concrete.

Question 14

Cross-Links

Answer 14

Placing extra molecules between polymer chains increases stiffness.

Question 15

Proportional Limit

Answer 15

Point where Hooke’s Law stops being obeyed, stress is not longer proportional to strain.

Question 16

Elastic Limit

Answer 16

Maximum stress a material can endure before plastic deformation.

Question 17

Yield Point

Answer 17

Point where a material begins to permanently plastically deform.

Question 18

Experiment details

Answer 18

Procedure:
Use longer wire for lower uncertainty in length
Measure diameter at several points on the wire
Ensure there are no kinks in the wire
Add load in incremental steps and acquire varying extensions
More loads
Repeat measurements
Constant temperature

Analysis:
Plot a graph of stress-strain to find gradient (Young’s Modulus)

Question 19

Negligible uncertainty

Answer 19

Very low percentage uncertainty may be negligible in calculations
as it is small relative to more impactful uncertainties.

Question 20

3rd Law

Answer 20

Every force has an equal and opposite reaction force.

Not 3rd law if forces not same type
Not 3rd law if forces acting on the same body

Question 21

2nd Law

Answer 21

Momentum is the rate of change of force or F = ma

momentum = change in force / change in time

Question 22

1st Law

Answer 22

An object will remain in inertia (stationary or constant velocity) unless an external force acts on it.

Question 23

Compare star properties

Answer 23

Surface temperature
Colour (based on peak intensity wavelength)
Brightness (light intensity)

Question 24

Black Body

Answer 24

Object that absorbs all EM radiation that falls on it.

Question 25

Drop time depends on…

Answer 25

Drop height
Vertical acceleration component / gravitational force
Vertical velocity component

Question 26

Conditions for equillibrium

Answer 26

Net force = 0
Sum of clockwise moments = sum of anticlockwise moments

Question 27

Experiment

Answer 27

[diagram]
1. Set up equipment as shown.
2. Measure unstretched length of wire by holding it taut, but not extended.
3. Measure diameter of the wire using a micrometer, making sure to not compress the wire.
3. Set up ruler with 0 mark at the end of the hung wire and keep it still.
4. Add incrementing loads to the wire, recording the respective extension of the wire for each by reading from the 0 mark on the meter ruler to the new end.
5. Plot a graph of force-extension/stress-strain and find gradient / Young’s Modulus.

Question 28

Momentum conservation

Answer 28

Momentum in a closed system remains constant, providing there are no external forces acting.

Question 29

Center of Mass

Answer 29

Point where weight of an object seems to act.