Question 4.3: Additional Answer Options

2021 - November - Paper 1 - Question 4

Or

Option 2

East as positive:

\text{ For X:}

\left.\begin{array}{l}F_{net}\Delta t=\Delta p\\ F_{net}\Delta t=m\left(v_{f}-v_{i}\right)\end{array}\right\}Any one

F_{\text{net }}(0.1)=10(0-2)

F_{\text{net }}=-200\mathrm{~N}

F_{\text{net }}=200\mathrm{~N}

Or

West as positive:

\text{ For X:}

\left.\begin{array}{l}F_{net}\Delta t=\Delta p\\ F_{net}\Delta t=m\left(v_{f}-v_{i}\right)\end{array}\right\}Any one

F_{\text{net }}(0,1)=10\{0-(-2)

F_{\text{net }}=200\mathrm{~N}

Or

Option 3

East as positive:

\text{ For Y:}

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

6=-4+a(0,1)

\mathrm{a}=100\mathrm{~m}\cdot\mathrm{s}^{-2}

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

=2(100)

=200\mathrm{~N}

Or

West as positive:

\text{ For Y:}

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

-6=4+a(0,1)

a=-100\mathrm{~m}\cdot\mathrm{s}^{-2}

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

=2(-100)

=-200\mathrm{~N}

F_{\text{net }}=200\mathrm{~N}

Or

Option 4

East as positive:

\text{ For X:}

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

0=2+a(0,1)

a=-20\mathrm{~m}\cdot\mathrm{s}^{-2}

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

=10(-20)

=-200\mathrm{~N}

\mathrm{F}_{\text{net }}=200\mathrm{~N}

Or

West as positive:

\text{ For X:}

\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

0=-2+a(0,1)

\mathrm{a}=20\mathrm{~m}\cdot\mathrm{s}^{-2}

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

=10(20)

F_{\text{net }}=200\mathrm{~N}

Or

Option 5

East as positive:

\text{ For X:}

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

0=2+a(0,1)

a=-20\mathrm{~m}\cdot\mathrm{s}^{-2}

v_{f}^2=v_{i}^2+2a\Delta x

0=(2)^2+2(-20)\Delta x

\Delta\mathrm{x}=0,10\mathrm{~m}

\Delta\mathrm{x}=\left(\frac{\mathrm{v}_{\mathrm{f}}+\mathrm{v}_{\mathrm{i}}}{2}\right)\Delta\mathrm{t}

=\left(\frac{0+2}{2}\right)(0,1)

=0,10\mathrm{~m}

\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{k}}

\mathrm{F}_{\mathrm{net}}\Delta\mathrm{x}\cos\theta=1/2\mathrm{~m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right)

F_{\text{nett }}(0.1)\cos180^{\circ}=1/2(10)\left(0^2-2^2\right)

F_{net}=200\mathrm{~N}

Or

West as positive:

\text{ For X:}

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

0=-2+a(0,1)

a=20\mathrm{~m}\cdot\mathrm{s}^{-2}

v_{f}^2=v_{i}^2+2a\Delta x

0=(-2)^2+2(20)\Delta x

\Delta x=-0,10m

\Delta\mathrm{x}=\left(\frac{\mathrm{v}_{\mathrm{f}}+\mathrm{v}_{\mathrm{i}}}{2}\right)\Delta\mathrm{t}

=\left(\frac{0+(-2)}{2}\right)(0,1)

=-0,10\mathrm{~m}

\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{k}}

F_{net}\Delta x\cos\theta=1/2m\left(v_{f}^2-v_{i}^2\right)

F_{\text{net }}(0.1)\cos180^{\circ}=1/2(10)\left(0^2-2^2\right)

F_{\text{net }}=200\mathrm{~N}

Or

Option 6

East as positive:

\text{ For Y:}

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

6=-4+a(0,1)

a=100\mathrm{~m}\cdot\mathrm{s}^{-2}

v_{f}^2=v_{i}^2+2a\Delta x

(6)^2=(-4)^2+2(100)\Delta x

\Delta\mathrm{x}=0,10\mathrm{~m}

\Delta x=\left(\frac{v_{f}+v_{i}}{2}\right)\Delta t

=\left(\frac{6-4}{2}\right)(0,1)

=0,10\mathrm{~m}

W_{\text{net }}=\Delta E_{k}

\mathrm{F}_{\mathrm{net}}\Delta\mathrm{x}\cos\theta=1/2\mathrm{~m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right)

F_{net}(0.1)\cos0^{\circ}=1/2(2)\left(6^2-(-4)^2\right)

\mathrm{F}_{\text{net }}=200\mathrm{~N}

Or

West as positive:

\text{ For Y:}

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

-6=4+a(0,1)

a=-100\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2\mathrm{a}\Delta\mathrm{x}

(-6)^2=(4)^2+2(-100)\Delta x

\Delta x=-0,10m

\Delta\mathrm{x}=\left(\frac{\mathrm{v}_{\mathrm{f}}+\mathrm{v}_{\mathrm{i}}}{2}\right)\Delta\mathrm{t}

=\left(\frac{-6+4}{2}\right)(0,1)

=-0,10\mathrm{~m}

\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{k}}

F_{net}\Delta x\cos\theta=1/2m\left(v_{f}^2-v_{i}^2\right)

F_{\text{net }}(0,1)\cos0^{\circ}=1/2(2)\left((-6)^2-(4)^2\right)

F_{\text{net }}=200\mathrm{~N}

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