Oxford Revise AQA GCSE Physics | Chapter P9 answers

P9: Forces

Answers

Extra information

Mark

AO / Specification reference

01.1

force: newtonmeter or amount of masses/weights on end of spring

extension: ruler

1

1

AO2

4.5.3

01.2

measure the length of the spring with the ruler

apply a known force/ (hang up the spring and) hang a known weight on it

measure the length again

find the extension by subtracting the original length from the stretched length.

1

1

1

1

AO2

AO3

4.5.3

01.3

to get more accurate/precise measurements

1

AO2

AO3

4.5.3

01.4

repeat it/ignore it when they are calculating the mean

1

AO2

AO3

4.5.3

01.5

line graph

the data are continuous/all numbers and no words/names

1

1

AO2

4.5.3

02.1

non-contact – weight/force of the Earth on the wood

contact force – upthrust/upwards force of the water on the wood

1

1

AO1

AO2

4.5.1.2

02.2

the forces are equal in magnitude

and opposite in direction

1

1

AO1

AO2

4.5.1.1

4.5.1.4

02.3

water resistance

contact force

1

1

AO1

AO2

4.5.1.2

03.1

work done = force × distance

accept W = Fs

1

AO1

4.5.2

03.2

work done = 20 × 30

= 600 (Nm or J)

accept 600 (Nm or J) with no working for two marks

1

1

AO2

4.5.2

03.3

newton metres (Nm)

joules (J)

1

AO1

4.5.2

03.4

friction

1

AO1

4.5.2

03.5

chemical energy store will decrease (food/oxygen)

thermal energy store (of the surrounding) will increase

do not accept answers involving changes in kinetic energy store (at constant speed, the kinetic energy store will stay at constant level)

1

1

AO2

4.5.2

04.1

any sensible suggestion, e.g., difficult to see the undeflected position of the ruler to measure from, difficult to see the extension

1

AO3

4.5.3

04.2

ignore the outlier 17

\(
\begin{array}{lr}
\rm{average\ of\ the\ other\ two\ readings}\ = \frac{10 + 12}{2}\\
= 11
\end{array}
\)

1

AO2

4.5.3

04.3

one mark for correct plotting of four points

two marks for correct plotting of all points

one mark for curved line of best fit

one mark for appropriate y-axis label and scale

4

AO2

AO3

4.5.3

04.4

no

the line is not straight/linear (through origin)

1

1

AO3

05.1

the first column is should be labelled mass in grams and not weight which would be in newtons (N)

they should convert g to kg and then to N using weight = mass (in kg) × g

1

1

AO2

4.5.1.3

05.2

one mark for correct values of force converted from g

one mark for correct plotting of at least four points

one mark for labelled axes

one mark for appropriate line of best fit

4

AO2

4.5.1.3

4.5.3

05.3

original length = intercept on x axis/when force on sample is zero

= 3.0 cm

allow answers between 2.5 cm and 3.5 cm

1

1

AO3

4.5.3

05.4

as the force increases the material becomes less stiff/easier to stretch

or the same increase in force produces a bigger increase in length

1

AO2

AO3

4.5.3

05.5

it would not be suitable

the extension if not proportional to the force

1

1

AO3

4.5.3

06.1

the force of the hand on the bag

or words to that effect

1

AO2

4.5.3

06.2

inelastic deformation is deformation where the object does not return to its original size and shape when the force is removed

1

AO1

4.5.3

06.3

Statement

Correct

the graph for the plastic bag shows a non-linear relationship between force and extension

the graph for the plastic bag shows that is proportional to extension

a graph that is a straight line is likely to be for a spring

the material that produced a linear graph has been inelastically deformed

one mark for each correct row

2

AO3

4.5.3

06.4

extension = stretched length – unstretched length

= 3 cm – 2 cm/0.03 – 0.02

= 1 cm/0.01 m

1

1

AO2

4.5.3

07.2

force = spring constant × extension

allow F = ke

1

AO1

4.5.3

07.3

\(
\begin{array}{lr}
2 = \rm{k}\ \times 0.01\\
\rm{k} = \frac{2}{0.01} = 200\ \rm{N/m}\\
\end{array}
\)

1

1

AO2

4.5.3

07.4

energy = 0.5 × spring constant × extension2

= 0.5 × 200 × 0.012

= 0.01 J

allow E = 0.5 ke2

accept 0.01 (J) with no working for two calculation marks

1

1

AO2

4.5.3

07.5

0.01 J

the work done on the spring is equal to the elastic energy stored in the spring

1

1

AO2

4.5.3

08.1

up arrow: force of the workbench on the tub

down arrow: force of the Earth on the tub

down arrow should be larger than the up arrow

accept ‘normal’ or ‘reaction’

accept ‘weight’

do not accept ‘gravity’

one mark for two equal length arrows pointing in opposite directions

3

AO2

4.5.1.4

08.2

the weight can be resolved into two components, one down the ramp and one at 90 degrees to the ramp

there is a force of friction opposing the component of weight down the ramp

which is smaller than the component of the weight (so there is a resultant force down the ramp and the tub accelerates)

1

1

1

AO2

4.5.1.4

08.3

one mark for correct x and y labels

one mark for horizontal line

2

AO3

08.4

any sensible suggestion, e.g.,

  • as the mass increases, the frictional force increases
  • as the mass increases the component of the weight down the slope also increases, so the two effects cancel out

1

AO3

09.1

free body diagram e.g.,

up arrow: normal/reaction

left arrow: driving force/18 000N

right arrow: resistive force/12 000 N

down arrow: weight/15 kN

right arrow should be longer than all the other arrows

one mark for arrow left labelled driving force/18 000 N

one mark for arrow right labelled resistive force/12 000 N

one mark for arrow downwards labelled weight/15 kN

one mark for arrow upwards labelled normal force

weight and normal arrows the same length, driving force arrow longer than resistive force arrow

4

AO2

4.5.1.1

4.5.1.2

4.5.1.4

09.2

horizontally: resultant = 18 000 – 12 000

= 6000 N

to the left

vertically: resultant = 15 000 – 15 000

= 0 N

1

1

1

AO2

4.5.1.4

09.3

\(
\begin{array}{lr}
\rm{weight} = 15 000\ \rm{N}\\
\rm{weight} = \rm{mass} \times \rm{gravitational\ field\ strength}\\
15\ 000 = \rm{mass} \times 9.8\\
\rm{mass} = \frac{15 000}{9.8}\\
= 1531\rm{kg}
\end{array}
\)

1

1

1

1

AO2

4.5.1.3

09.4

both vertical arrows would change slightly in length

but still cancel out / be the same size

the horizontal arrows would not change

1

1

1

AO3

4.5.1.2

4.5.1.3

4.5.1.4

10.1

appropriate scale diagram e.g., 1 cm = 10 N

answer = 153 N

accept answers from 148 N to 158 N

one mark for clear scale

one mark for parallelogram drawn

one mark for correct answer

3

AO3

4.5.1.4

10.2

(if angle increases) cos (angle) decreases so that tension increases

so that the component of the tension stays the same/so that the resultant of the 2 tension forces stays the same

1

1

AO3

4.5.1.4

10.3

the angle between the vertical component of tension and the string = 60° but the vertical component is T sin60

2T sin60 = W =153

T = 88N

which is less than in arrangement A

1

1

1

1

1

AO3

4.5.1.4

11.1

moment = force × distance

accept M = Fd

1

AO1

4.5.4

11.2

moment = 8 × 1000

= 8000 Nm

accept 8000 (Nm) with no working

1

1

AO2

4.5.4

11.3

\(
\begin{array}{lr}
\rm{clockwise\ moments} = \rm{anticlockwise\ moments\ if\ the\ crane\ is\ balanced}\\
\rm{counterbalance\ weight\ \times\ distance\ from\ pivot\ =\ moment\ of\ load}\\
\rm{counterbalance\ weight}\ \times\ 4 = 8000\\
\rm{counterbalance\ weight} = \frac{8000}{4}\\
= 2000\ \rm{N}
\end{array}
\)

one mark for evidence of using law of moments

1

1

1

1

AO1

AO2

4.5.4

11.4

low, high

both answers required for the mark

1

AO1

4.5.4

12.1

force for spring 1 = 30 N, force for spring 2 = 20 N

\(\rm{the\ force\ for\ spring\ 1\ is}\ \frac{30}{20} = 1.5\ \rm{times\ bigger}\)

1

1

AO2

4.5.3

12.2

spring 1 would feel stiffer than spring 2

it takes a bigger force to extend it by the same length

1

1

AO3

4.5.3

12.3

force (applied to a spring) = spring constant × extension

accept F = ke

1

AO1

4.5.3

12.4

spring 1:

48 = k1 × 0.08

k1 = 600 N/m

spring 2:

32 = k2 × 0.08

k2 = 400 N/m

accept any correct pair of values from the graph

1

2

1

2

AO2

4.5.3

13.1

force = spring constant × extension

accept F = ke

1

AO1

13.2

extension = final length – original length

= 20.14 – 20.00 = 0.14 m

1

AO2

13.3

\(
\begin{array}{lr}
\rm{weight\ of\ climber} = \rm{mass\ \times\ g}\\
= 80 \times 9.8\\
= 784\ \rm{N}\\
\rm{using\ force} = \rm{spring\ constant\ \times\ extension}\\
784 = \rm{spring\ constant\ \times\ 0.14}\\
\rm{spring\ constant} = \frac{784}{0.14}\\
= 5600\ \rm{N/m}
\end{array}
\)

answer is given to two significant figures

1

1

1

1

1

AO1

AO2

4.5.1.3

4.5.2

13.4

the weight stays the same

so the spring constant will be smaller

accept units of k are N/m so if weight (N) remains same and more metres then k is smaller

1

1

AO3

4.5.1.4

14.1

weight = mass × gravitational field strength

accept W = mg

1

AO1

4.5.1.3

14.2

weight = 140 × 9.8

= 1372 N

accept 1372 (N) with no working for the two calculation marks

1

1

AO2

4.5.1.3

14.3

the centre of mass

accept centre of gravity

1

AO1

4.5.1.3

14.4

(weight = mass × gravitational field strength)

= 140 × 3.8

= 532 (N)

difference = 1372 – 532

= 840 N

or

\(\frac{1372}{532} = 2.6\)

the difference in weight is 840 N or the weight on Earth is 2.6 times bigger

accept 532 (N) with no working for two calculation marks

evidence of calculating difference or ratio with relevant comment for three marks

1

1

1

1

AO2

4.5.1.3

14.5

non-contact

the objects do not ned to be in contact for the force to act

1

1

AO1

4.5.1.3

Loading...