Module 3 Standard Answers

Question 1

Explain how Galileo disproved the idea that heavier objects fall faster than lighter ones

Answer 1

Two canon balls of different mass (same size) are dropped from the same height and at the same time.

They hit the ground at the same time and so both have the same acceleration.

Question 2

Describe and explain factors that increase the braking distance of the car

Answer 2

Greater speed: braking distance is directly proportional to speed²
Greater mass: braking distance is directly proportional to mass
Worn tires/brakes: friction acting between the car and road deceases
Road conditions:friction acting between the car and road decreases
Downhill slope: the component of the acceleration parallel to the slope increases

Question 3

Describe factors that increase the thinking distance of the car

Answer 3

Consumption of drugs/alcohol
Driver fatigue / distractions
A greater speed: greater distance traveled during the reaction time

Question 4

Describe how a seatbelt/airbag can protect a driver from injury during a collision

Answer 4

A seatbelt/airbag increases the time taken for the driver to stop
By a= (v -u)/ t, the deceleration of the driver decreases
By F=ma, the resultant force exerted on the driver decreases

Question 5

A spaceship engine provides a constant driving force (D). Explain why the acceleration of the spaceship increases after liftoff

Answer 5

The mass, and weight (W), of the spaceship decreases as it burns fuel.
The resultant force F=D-W increases.
By F=ma, as the resultant force increases and mass decreases, the acceleration increases.

Question 6

Why does F=ma not apply to an object travelling at very high speeds

Answer 6

F=ma only applies to an object with constant mass.
However, the mass of an object increases at its speed approaches the speed of light.

Question 7

State the factors that affect the magnitude of the drag force acting on a moving object

Answer 7

Cross-sectional area
Speed: drag is directly proportional to speed²
Roughness of surface / aero shape / streamlining of object
Air viscosity/ temperature/ density

Question 8

Describe and explain the motion of an object in free fall…………as it begins to fall

Answer 8

Speed of object≈0
Drag is directly proportional to speed², the drag (R) acting on the object ≈ 0
The resultant F=W=mg
By F=ma, the object’s acceleration a=g (the acceleration due to g)

Question 9

Describe and explain the motion of an object in free fall………before terminal velocity is reached

Answer 9

Speed of the object increases
As drag is directly proportional to speed², the drag (R) acting on an object increases
The resultant force F=W-R decreases
By F=ma, the object’s acceleration a<g

Question 10

Describe and explain how the motion of an object in free fall……..at terminal velocity

Answer 10

Speed of the object increases to a constant maximum value.
As drag is directly proportional to speed², the drag (R) acting on the object increases to a constant maximum value.
The resultant force F=0 because W=R
By F=ma, the object’s acceleration a=0

Question 11

Consider the motion of the same object in free fall which reaches terminal velocity. Suggest changes to this objects motion if it’s mass increases (same size/shape)

Answer 11

Magnitude of terminal velocity increases (as weight = drag and drag is direct proportional to speed²)
Time taken for object to reach terminal velocity increases
Initial acceleration of the object is unchanged p: a=g

Question 12

Explain why the efficiency of a mechanical device can never be 100%

Answer 12

There is some work done by the device to overcome resistive/frictional forces
So useful energy output<total energy output
By efficiency = useful energy output/ total energy input,efficiency < 100%

Question 13

Explain why the work done in stretching/compressing a material is not given by w=fd

Answer 13

The tensile force in the material is not constant; it’s continually becoming larger in magnitude in order to keep extending the material
The area under a forre vs extension graph is equal to the work done or elastic potential energy

Question 14

Explain why the work done in stretching/ compressing a material might not be equal to the elastic potential energy stored in the material in reality

Answer 14

There may be some work done in raising the temperature of the material

Question 15

Describe the features of a brittle material ( cast iron )

Answer 15

A brittle only behaves elastically
No plastic behaviour/permanent deformation
Fractures at its elastic limit
The uts is equal to the breaking strength
Material obeysHooke’s law throughout the graph is a straight line through the origin; tensile stress is directly proportional to tensile strain.

Question 16

Describe the features of a ductile material ( gold )

Answer 16

A ductile material behaves elastically and plastically
It has become permanently deformed when it is stretched beyond its elastic limit
Before the limit of proportionality is reached, the material obeys Hooke’s law as the graph is a straight line through the origin here; it’s directly proportional to tensile strain
The UTS is reached before the breaking point

Question 17

Describe the features of a polymer ( rubber )

Answer 17

A polymer only behaves elastically
It will not experience plastic behaviour or permanent deformation
The graph is not a straight line through the origin so Hooke’s law is not obeyed so there is no directly proportional relationship between tensile stress and tensile strain
More work is done on the polymer during load than the polymer does during unloading
A hysteresis loop is formed by the loading and unloading curve

Question 18

Describe the points associated with a material’s deformation

Answer 18

The limit of proportionality at P is the point up until which the material obeys Hooke’s law
The Elastic limit at E is where the material behaves elastically up to this point but behaves plastically beyond causing permanent deformation
Yield point at Y is the point where the material extends rapidly
The UTS is the maximum tensile stress the material can withstand
The breaking strength at B is where the tensile stress applied to an object at the point of fracture

Question 19

State the necessary conditions for Newtons Third Law to apply to a pair of forces

Answer 19

The must be the same type of force
They must act on different objects

Question 20

Explain why linear momentum is a vector quantity

Answer 20

Velocity is a vector quantity and mass is a scalar quantity
The product of vector*scalar=vector

Question 21

Two objects A and B collide. Use Newton’s Third Law to explain the relationship between the impulse acting on A and the impulse acting on B during the collision

Answer 21

By N3L, the force F exerted on B due to A is equal in magnitude and opposite in direction to the force exerted on A due to B
The time of contact during the collision is the same for both objects
By impulse =FΔt, the impulse on A is equal in magnitude and opposite in direction to the impulse on B

Question 22

Experiment to estimate the acceleration of free fall

Answer 22

Measure…..Distance ball falls with metre ruler
Time the ball is falling using a stopwatch
How……S=ut+1/2at² and uv0
Plot a graph of s vs t² so the grad=g/2
Estimate….. because drag acting on the ball is ignored
Reaction time is needed for a stopwatch
Parallax errors with the time the ball hits the ground

Question 23

Describe an experiment to determine the centre of gravity of a metal plate

Answer 23

Suspend the plate for a point
Use a plumb line to mark a vertical line on the planet form the point
Suspend the plate from another point
Use the plumb line to mark a vertical line on the plate from the second point
The centre of gravity is the point of intersection

Question 24

Young modulus of a metal wire

Answer 24

Measure…….OG length of wire using a metre ruler
Extension of the wire using a vernier scale
Mass of hanging masses using a top pan balance
Diameter with usual method
How….σ=F/A
W=mg
A=πr²
ε=x/L₀
Ε=σ/ε