Question 4.2: Additional Answer Options

2022 - November - Paper 1 - Question 4

Or

Option 2

East as positive:

For Y:

\left.\begin{array}{l}\mathrm{F}_{\text{net }}\Delta\mathrm{t}=\Delta\mathrm{p}\\ \mathrm{F}_{\text{net }}\Delta\mathrm{t}=\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}\right)\end{array}\right\}\text{ Any one}

F_{\text{net }}(0,1)=0,5(-9,6-0)

F_{\text{net }}=-48\mathrm{~N}

F_{\text{net }}=48\mathrm{~N}

Or

West as positive:

For Y:

\left.\begin{array}{l}\mathrm{F}_{\text{net }}\Delta t=\Delta\mathrm{p}\\ \mathrm{F}_{\text{net }}\Delta\mathrm{t}=\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}\right)\end{array}\right\}Any one

F_{\text{net }}(0,1)=0,5(9,6-0)

F_{\text{net }}=48\mathrm{~N}

Or

Option 3

East as positive:

For X:

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

-4=-8+a(0,1)

a=-40\mathrm{~m}\cdot\mathrm{s}^{-2}

F_{\text{net }}=\mathrm{ma}

F_{\text{net }}=(1,2)(-40)

F_{\text{net }}=-48\mathrm{~N}

F_{\text{net }}=48\mathrm{~N}

Or

West as positive:

For X:

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

4=8+a(0,1)

a=40\mathrm{~m}\cdot\mathrm{s}^{-2}

F_{\text{net }}=\mathrm{ma}

F_{\text{net }}=(1,2)(40)

F_{\text{net }}=48\mathrm{~N}

Or

Option 4

East as positive:

For X:

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

\Delta x=\left(\frac{8+4}{2}\right)(0,1)

\Delta\mathrm{x}=0,6\mathrm{~m}

F_{\text{net }}\Delta x\cos\theta=1/2mv_{f}^2-1/2mv_{i}^2

F_{\text{net }}(0,6)\cos180^{\circ}=1/2(1,2)(4)^2-1/2(1,2)(8)^2

F_{net}=48\mathrm{~N}

Or

West as positive:

For X:

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

\Delta x=\left(\frac{-8-4}{2}\right)(0,1)

\Delta x=-0,6m

F_{\text{net }}\Delta x\cos\theta=1/2mv_{f}^2-1/2mv_{i}^2

F_{net}(0,6)\cos0^{\circ}=1/2(1,2)(-4)^2-1/2(1,2)(-8)^2

F_{net}=-48\mathrm{~N}

F_{net}=48\mathrm{~N}

Or

Option 5

\text{ Gradient }=\frac{\Delta y}{\Delta x}

=\frac{\Delta\mathrm{v}}{\Delta\mathrm{t}}

=\frac{4-8}{0,1}

=-40\mathrm{~m}\cdot\mathrm{s}^{-2}

F_{\text{net }}=\mathrm{ma}

F_{\text{net }}=(1,2)(-40)

F_{net}=-48\mathrm{~N}

F_{\text{net }}=48\mathrm{~N}

Related subjects & topics
Physical Sciences
Explore similar posts in our community