Question 2.1.4: Additional Answer Options

2019 - May/June - Paper 1 - Question 2

2.1.1

 

Fcheck F/ 90N / F90
wcheck Fg / Fw / weight/mg / Gravitational force
fcheck (Kinetic) Friction / Ff / / fk
Ncheck FNormal / Normal / FN

 

light bulb Tips:

  • As of 2024 no marks will be awarded for free-body diagrams if the weight has been resolved into two components.
  • Since the question is out of four marks, it indicates that four forces are acting on the lawn mower.

4 marks

(4 x 1 mark)

2.1.2

 

It is moving at constant speed in a straight line, the acceleration is zero/ the net force (resultant) acting on it is zero/it is moving at constant velocity. check

light bulb Tips:

  • When an object is in equilibrium, there is no resultant force acting on the system so the object moves at a constant velocity or remains at rest.
  • In this case, the person is pushing the lawnmower at a constant speed so it must be in equilibrium.

1 mark

 

2.1.3

 

Option 1

\left.\begin{array}{l}\mathrm{F}_{\text{net }}=\mathrm{ma}\\ \mathrm{F}_{\text{net }}=0\\ \mathrm{F}_{x}=f\\ \mathrm{F}_{\mathrm{x}}-f=0\\ \mathrm{F}\cos40^{\circ}-f=0\end{array}\right\}Any one (1)check

90\cos40^{\circ}-f=0 (2)check

f=68,94\mathrm{~N} (3)check

Or

Option 2

\left.\begin{array}{l}\mathrm{F}_{\text{net }}=\mathrm{ma}\\ \mathrm{F}_{\text{net }}=0\\ \mathrm{F}_{\mathrm{x}}=f\\ \mathrm{F}_{\mathrm{x}}-f=0\\ \mathrm{F}_{\cos320^{\circ}-f=0}\end{array}\right\}Any one (1)check

90\cos320^{\circ}-f=0 (2)check

f=68,94\mathrm{~N} (3)check

Or

Option 3

\left.\begin{array}{l}\mathrm{F}_{\text{net }}=\mathrm{ma}\\ \mathrm{F}_{\text{net }}=0\\ F_{x=f}\\ \mathrm{F}_{x}-f=0\\ \mathrm{F}_{\sin50^{\circ}-f=0}\end{array}\right\}Any one (1) check

90\sin50^{\circ}-f=0 (2) check

f=68,94\mathrm{~N} (3)check

light bulb  Tip: the horizontal component of the applied force must be equal to the friction for the lawnmower to move at a constant speed.

3 marks

(3 x 1 mark)

Marking criteria:

  • (1) Formula check
  • (2) Substitution check
  • (3) Answer check

 

 

2.1.4

 

light bulb How to approach this question:

  • You need to calculate the acceleration using the given information.

Option 1

Part 1:

v_{f}=v_{i}+a\Delta t

2=0+a(3)checkcheck

\mathrm{a}=0,67\mathrm{~m}\cdot\mathrm{s}^{-2}

Part 2:

  • Use Newton's second law to calculate the force.

\mathrm{F}_{\text{net }}=\mathrm{ma}check

F\cos40^{\circ}-68,94=15(0,67)checkcheck

\mathrm{F}=103,11\mathrm{N}(103,05\mathrm{N}-103,11\mathrm{N})check

  • Remember, you are asked for the force and not just the horizontal component of the force.

Or

F_{net}=macheck

\mathrm{F}\cos320^{\circ}-f=15(0,67)

F\cos320^{\circ}-68,94=15(0,67)checkcheck

\mathrm{F}=103,11\mathrm{~N}check

There are more ways to answer this question. To view other options, click here.

6 marks

(6 x 1 mark)

 

2.2

 

Option 1

F=G\frac{m_1m_2}{r^2} check

20^{}=\left(6,67\times10^{-11}\right)\frac{\mathrm{m}_{\text{planet }}(10)}{\left(6\times10^5\right)^2} checkcheck

m_{\text{planet }}=1,08\times10^{22}\mathrm{~kg} check

Or

Option 2

\mathrm{w}=\mathrm{mg} check

20=(10)(g) check

\mathrm{g}=2\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}

2=\frac{\left(6,67\times10^{-11}\right)\mathrm{M}}{\left(6\times10^5\right)^2} check

M=1,08\times10^{22}\mathrm{~kg}check

light bulb Tip: the universal gravitational constant, G, is given on the formula sheet.

4 marks

(4 x 1 mark)


Or

Option 2

\mathrm{F}_{\mathrm{net}}\cdot\Delta\mathrm{t}=\Delta\mathrm{p}

F\cos40^{\circ}-(68,94)(3)=15(2-0)

F=103,11\mathrm{~N}


Or

Option 3

\left.\begin{array}{l}F_{\text{net }}=ma\\ F_{x}-f_{k}=ma\end{array}\right\}Any one

F_{x}-68,94=15\frac{(2-0)}{3^{}}

F_{x}=78,94N

\tan\theta=\frac{F_{y}}{F_{x}}

\tan40^{\circ}=\frac{F_{y}}{78,94}

F_{y}=66,24N

F^2=F_{x}^2+F_{y}^2

F^2=(78,94)^2+(66,24)^2

F=103,05\mathrm{~N}


Or

Option 4

\Delta x=\frac{v_{i}+v_{f}}{2}\Delta t

=\frac{(2+0)}{2}^{}(3)

\Delta x=3m

\mathrm{W}_{\mathrm{net}}=\Delta\mathrm{K}

\mathrm{W}_{\mathrm{F}}+\mathrm{W}_{f}=\Delta\mathrm{K}

\mathrm{F}\Delta\mathrm{x}\cos\theta+f\Delta\mathrm{x}\cos\theta=\Delta\mathrm{K}

F(3)\cos40^{\circ}+68,94(3)\cos180^{\circ}=1/2(15)\left(2^2\right)-1/2(15)(0)^2

\mathrm{F}=103.06\mathrm{~N}

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