Oxford Revise AQA GCSE Physics | Chapter P13 answers

P13: Braking and momentum

Answers

Extra information

Mark

AO / Specification reference

01.1

Factor

Affects thinking distance

Affects braking distance

road conditions

distractions in the car

speed

one mark for each correct column

2

AO1

AO2

4.5.6.3.1

4.5.6.3.2

4.5.6.3.3

01.2

appropriate example e.g., tiredness

if you are tired your reaction time is greater

the thinking distance will increase

accept other examples e.g., drugs/alcohol

1

1

1

AO1

4.5.6.3.2

01.3

appropriate example e.g.,

icy road conditions will increase the braking distance

the stopping distance = the thinking distance + the braking distance

the stopping distance will increase

1

1

1

AO1

4.5.6.3.1

4.5.6.3.3

02.1

thinking distance = speed × reaction time

= 13.4 × 0.67 = 8.978 m

= 9.0 m

answer given to two significant figures

1

1

AO2

4.5.6.3.2

02.2

stopping distance = 13.9 m + 8.978 m = 22.878 m

= 23 m

answer given to two significant figures

1

1

AO2

4.5.6.3.1

02.3

\(
\begin{array}{lr}
\rm{thinking\ distance\ at\ 50\ mph} = \rm{\ thinking\ distance\ at\ 30\ mph} \times \frac{50}{30} = 9.00 \times \frac{50}{30}\\
= 15\ \rm{m}\\
\end{array}
\)

the distance is proportional to the speed

if the reaction time is constant

accept working out thinking distance using equation

1

1

1

AO2

AO3

4.5.6.3.1

4.5.6.3.2

4.5.6.3.3

02.4

braking distance = stopping distance – thinking distance

(= 53.0 – 15) = 38 m

1

1

AO2

4.5.6.3.1

4.5.6.3.2

4.5.6.3.3

03.1

independent – type of surface

dependent – distance it travels on the surface before stopping

control – any two from:

  • height of ramp
  • position of release of trolley
  • type of trolley/mass of trolley

one mark for each correct point up to a maximum of two marks

1

1

2

AO2

03.2

method e.g.,

  • raise one end of the ramp by a height h
  • cover the floor at the other end of the ramp with a type of surface
  • place a trolley at the top of a ramp and releases it
  • measure the distance the trolley travels from the bottom of the ramp to the place that it stops
  • repeat the experiment twice more with the same surface
  • replace the surface with a different material and repeat the experiment with ramp at same height h
  • identify outliers; do not include them in the calculation of mean

max 5

AO1

03.3

appropriate example with improvement e.g.,

releasing the trolley from exactly the same place each time

improvement: draw a line on the ramp and line up the back of the trolley with the line each time

or leaving ramp – make transition as smooth as possible by making sure the surface is at the same height as the bottom of the ramp

1

1

AO3

03.4

this is a good model because different surfaces will affect the stopping distance

difference surfaces produce different frictional forces on the trolley, so do different amounts of work on it.

1

1

1

AO2

AO3

4.5.6.3.4

03.5

the investigation does not involve braking

because the trolley does not have brakes

or no thinking distance can be included as no ‘brain’ in the trolley

1

1

AO3

04.1

as distance increases, the reaction time increases at a decreasing rate / they are not directly proportional

if you have a longer reaction time the ruler will fall further

1

1

AO2

AO3

4.5.6.3.2

04.2

0.23 s

accept 0.22 – 0.24

1

AO2

4.5.6.3.2

04.3

yes

typical reaction times range from 0.2 – 0.9 s

1

1

AO1

AO2

4.5.6.3.2

04.4

appropriate example e.g., the partner holds the ruler a big distance above the student’s hand

the distance it falls will be larger than it should be

the reaction time will be smaller than it actually is as they are no longer proportional

1

1

1

AO3

4.5.6.3.2

05.1

the data is checked by other scientists

peer review

1

1

AO1

05.2

force is measured in newtons/kilograms or tonnes are both a unit of mass, not force

1

AO1

4.5.1.3

05.3

weight = mass × gravitational field strength

allow W = mg

1

AO1

4.5.1.3

05.4

W = 1000 × 9.8

= 9800 N

1

1

AO2

4.5.1.3

05.5

people (without a science background) reading the article may be able to relate better to tonnes than to newtons.

1

AO3

05.6

\(
\begin{array}{lr}
\rm{momentum\ } = \rm{\ mass} \times \rm{velocity}\\
\rm{acceleration\ } = \frac{\rm{change\ in\ velocity}}{\rm{time}}\\
\rm{velocity\ } = \rm{acceleration\ } \times \rm{\ time\ } = 9.8 \times 2.5\\
= 24.5\ \rm{m/s}\\
\rm{momentum} = 2.5 \times 24.5\\
= 61.3\ \rm{kg}\ \rm{m/s}
\end{array}
\)

or Ft = mv-mu

2.5 × 9.8 × 2.5 = mv – 0

mv = 61.25 kg m/s

1

1

1

1

AO1

AO2

06.1

momentum = mass × velocity

= 65 × 5

= 325 kg m/s

1

1

1

AO1

AO2

4.5.7.1

06.2

momentum is conserved

the mass of the two skaters together after the collision is bigger than the mass of the single skater before the collision

so the velocity of the two people together will be less than 5.0 m/s

assuming the system is closed/there are no external forces acting

1

1

1

1

AO1

AO2

4.5.7.1

06.3

momentum before = momentum after

\(
\begin{array}{lr}
325 = (65 + 85) \times \rm{velocity\ afterwards}\\
\rm{velocity\ afterwards\ } = \frac{325}{65 + 85}\\
= 2.17/2.2\ \rm{m/s}
\end{array}
\)

1

1

1

AO2

4.5.7.1

06.4

the barrier exerts an external force

the system is not closed

1

1

AO1

AO2

4.5.7.1

07.1

\(
\begin{array}{lr}
70\ \rm{miles\ } = 70 \times 1609\ \rm{m} = 112\ 630\rm{\ m}\\
1\ \rm{hour\ } = 3600\rm{\ s}\\
\rm{speed\ } = \frac{112\;630}{3600}\\
= 31(.3)\ \rm{m/s}\\
\end{array}
\)

1

1

1

AO1

AO2

4.5.6.1.2

07.2

keep the cars far enough apart/a big distance apart

so that the driver can brake in time/the distance is bigger than the stopping distance

1

1

AO3

4.5.6.3.1

4.5.6.3.2

07.3

thinking distance = stopping distance – braking distance

\(
\begin{array}{l}
= 96 – 75 = 21\ \rm{m}\\
\rm{distance} = \textrm{speed} \times \textrm{time}\\
21 = 31.3 \times \textrm{time}\\
\textrm{time} = \frac{21}{31.3}\\
= 0.67\textrm{ s}
\end{array}
\)

1

1

1

1

AO1

AO2

4.5.6.3.1

4.5.6.3.2

07.4

the distance is half the stopping distance

1

AO3

4.5.6.3.1

07.5

correct suggestion e.g.,

the driver will start to brake when they see the brake lights of the car in front

during that time the car in front is also coming to a stop

or

to see 2 chevrons, the cars need to be 3 chevrons apart

so 120 m, not 80 m

1

1

or

1

1

AO3

4.5.6.3.1

08.1

graph B

in an emergency stop the driver presses the brake pedal harder/uses a bigger force

producing a bigger deceleration

at every speed the braking distance is shorter

1

1

1

1

AO2

AO3

4.5.6.3.3

4.5.6.3.4

08.2

appropriate suggestion e.g.,

internal organs can be damaged/the organs of your body continue to move

or extra friction on/damage to the tyres

1

AO2

4.5.6.3.4

08.3

no difference/it is the same at a particular speed

thinking distance is related to reaction time, which is the same in both situations at the same speed

1

1

AO1

AO2

4.5.6.3.1

4.5.6.3.2

09.1

texting and conversation increase your reaction time

conversation by up to 1.9 and texting by up to three times as long

which increases your thinking distance/stopping distance

causing more accidents

1

1

1

AO3

4.5.6.3.2

09.2

thinking distance = speed × reaction time

with no phone: = 13.4 × 1.00 = 13.4 m

complex conversation: = 13.4 × 1.80 = 24.12 m

difference = 24.12 – 13.4 = 10.7 m

1

1

1

AO2

4.5.6.3.2

09.3

\(
\begin{array}{lr}
24.12\ \rm{m} = \rm{\ speed\ } \times 1.00\ \rm{s}\\
\rm{speed} = \frac{24.12}{1.00}\\
= 24.12\ \rm{m/s}\\
24.12 \times \frac{30}{13.4} = 54\ \rm{mph}
\end{array}
\)

having a conversation has the effect of nearly doubling the effective speed of the car in terms of thinking distance.

1

1

1

1

1

AO2

AO3

4.5.6.3.2

10.1

force = spring constant × extension

1

AO1

10.2

\(
\begin{array}{lr}
\textrm{extension\ } = 1.25 – 1.00 = 0.25 \textrm{ m}\\
100 = \textrm{spring constant\ } \times 0.25\\
\textrm{spring constant\ } = \frac{100}{0.25}\\
= 400 \textrm{ N/m}
\end{array}
\)

1

1

1

1

AO2

10.3

H = L2 + d

= 1.25 + 0.5

= 1.75 m

1

1

AO3

11.1

\(
\begin{array}{lr}
\rm{maximum\ force\ with\ airbag\ } = 800\ \rm{N}\\
\rm{maximum\ force\ without\ airbag\ } = 4750\ \rm{N}\\
\rm{ratio\ of\ forces\ } = \frac{4750}{800} = 5.9
\end{array}
\)

the force without an airbag is nearly 6 times the force with an air bag

readings from graph

1

1

1

AO2

AO3

4.5.7.3

11.2

estimate between 500 N and 800 N

1

AO3

4.5.7.3

11.3

\(
\begin{array}{lr}
\rm{force\ } = \frac{\rm{change\ in\ momentum}}{\rm{time}}\\
\rm{momentum\ } = \rm{\ mass\ \times\ velocity}\\
\rm{force} = \frac{\rm{\ mass\ \times\ change\ in\ velocity}}{\rm{time}}\\
750\ \rm{ms} = 750 \times 10^{-3}\ \rm{s}\\
750 = \frac{60 \times \rm{change\ in\ velocity}}{750 \times 10^{-3}}\\
\rm{change\ in\ velocity\ } = \frac{750}{60} \times 750 \times 10^{-3}\\
= 9.4\ \rm{m/s}\\
\textrm{final velocity = 0 m/s, so initial velocity = 9.4 m/s}
\end{array}
\)

1

1

1

1

1

AO1

AO2

4.5.7.3

11.4

the acceleration is less/change in velocity happens over a longer time

the force required is much smaller for the same change in momentum

1

1

AO1

AO2

4.5.7.3

12.1

work is done by friction/energy transferred mechanically

1

AO1

4.5.6.3.4

12.2

the brakes/the surroundings

1

AO1

4.5.6.3.4

12.3

\(
\begin{array}{lr}
\rm{deceleration\ } = \frac{\rm{change\ in\ speed}}{\rm{time\ taken}}\\
= \frac{20}{4.3}\\
= 4.7\ \rm{m/s}^2\\
\end{array}
\)

assuming the acceleration is constant

the acceleration is probably not constant

because the brakes do not exert a constant force

1

1

1

1

1

AO1

AO2

AO3

4.5.6.3.4

12.4

braking force = mass × deceleration

= 1250 × 4.7

= 5875 (= 5900 N)

this is similar to the forces exerted by car engines

accept F = ma

1

1

1

AO2

4.5.6.3.4

13.1

weight = mass × gravitational field strength

= 1 × 9.8

= 9.8 N

1

1

1

AO1

AO2

4.5.1.3

13.2

work = force × distance

= 9.8 N × 1 m

= 9.8 J

1

1

1

AO1

AO2

4.5.2

13.3

9.8 N

allow error carried forward

1

AO2

4.5.1.3

14.1

kinetic energy = 0.5 × mass × speed2

= 0.5 × 1500 × 13.42

= 134 670

= 1.35 × 105 J

1

1

1

AO1

AO2

4.1.1.2

14.2

\(
\begin{array}{lr}
\rm{work\ done\ } = \rm{\ force\ \times\ distance}\\
1.35 \times 10^5 = 10000 \times \rm{\ distance}\\
\rm{distance} = \frac{1.35\ \times\ 10^5}{10\;000}\\
= 13.5\ \rm{m}\\
\rm{difference\ } = 13.5 – 9.4 = 4.1\ \rm{m}
\end{array}
\)

1

1

1

1

1

AO1

AO2

4.5.2

4.1.1.2

14.3

\(
\begin{array}{lr}
(\rm{final\ velocity})^2 – (\rm{initial\ velocity})^2 = 2 \times \rm{\ acceleration\ \times\ distance}\\
\rm{final\ velocity\ } = \sqrt{\left(\rm{2\ \times\ acceleration \times distance}\right) + \left(\rm{initial\ velocity}\right)^2}\\
\rm{final\ velocity\ } = \sqrt{\left(2 \times \left( -6.7 \right) \times 9.4 \right) + 13.4^2}\\
\rm{final\ velocity\ } = 7.32\ \rm{m/s}\\
= 7.32 \times \frac{3600}{1609}\\
= 16.4\ \rm{mph}
\end{array}
\)

1

1

1

1

1

AO2

4.5.6.1.5

14.4

sensible suggestion with reason e.g.,

approach two

because people can relate to the fact that the car will still be moving, so will cause injury to a person that they hit at that speed.

1

1

AO3

Loading...