Additional Answer Options: 4.2.2

2023 - May/June - Paper 1 - Question 4

No memo is currently available

Or

Option 2

Right as positive:

\left.\begin{array}{l}F_{net}\Delta t=\Delta p\\ F_{net}\Delta t=m\left(v_{f}-v_{i}\right)\end{array}\right\}Anyone (1)

F_{\text{net }}(0,02)=5,3(0,23-0) (2)

\mathrm{F}_{\mathrm{net}}=60,95\mathrm{~N} (3)

Or

Left as positive:

\left.\begin{array}{l}\mathrm{F}_{\mathrm{net}}\Delta\mathrm{t}=\Delta\mathrm{p}\\ \mathrm{F}_{\mathrm{net}}\Delta\mathrm{t}=\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}\right)\end{array}\right\}Anyone (1)

F_{\text{net }}(0,02)=5,3(-0,23-0) (2)

F_{\text{net }}=-60,95

\therefore F_{\text{net }}=60,95\mathrm{~N} (3)

Or

Option 3

Right as positive:

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

0,23=0,4+a(0,02)

a=-8,5\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{F}_{\text{net }}=\mathrm{ma} (1)

=(7,2)(-8,5) (2)

=-61,20

F_{\text{net }}=61,20\mathrm{~N} (3)

Or

Left as positive:

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

-0,23=-0,4+a(0,02)

a=8,5\mathrm{~m}\cdot\mathrm{s}^{-2}

F_{\text{net }}=ma (1)

=(7,2)(8,5) (2)

=61,20\mathrm{~N}

\mathrm{F}_{\text{net }}=61,20\mathrm{~N} (3)

Or

Option 4

Right as positive:

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

0,23=0+a(0,02)

\mathrm{a}=11,5\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{F}_{\text{net }}=\mathrm{ma} (1)

=(5,3)(11,5) (2)

=60,95\mathrm{~N}

F_{net}=60,95\mathrm{~N} (3)

Or

Left as positive:

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

-0,23=0+a(0,02)

\mathrm{a}=-11,5\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{F}_{\text{net }}=\mathrm{ma} (1)

=(5,3)(-11,5) (2)

=-60,95\mathrm{~N}

F_{\text{net }}=60,95\mathrm{~N} (3)

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