Question 5.4: Additional Answer Options

2021 - November - Paper 1 - Question 5

Or

Option 2

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

-14=2\left(v_{f}-4\right)

v_{f}=-3m\cdot s^{-1}

\Delta\mathrm{E}_{\mathrm{K}}=1/2mv_{f}^2-1/2mv_{i}^2

=1/2(2)\left[(0)^2-(-3)^2\right]

=-9\mathrm{~J}

Or

Option 3

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

14=2\left(v_{f}-(-4)\right)

v_{f}=3\mathrm{~m}\cdot\mathrm{s}^{-1}

\Delta\mathrm{E}_{\mathrm{K}}=1/2mv_{f}^2-1/2mv_{\mathrm{i}}^2

=1/2(2)\left[(3)^2-(-4)^2\right]

=-7\mathrm{~J}

Or

Option 4

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

14=2\left(v_{f}-(-4)\right)

v_{f}=3\mathrm{~m}\cdot\mathrm{s}^{-1}

\Delta\mathrm{E}_{\mathrm{K}}=1/2mv_{\mathrm{f}}^2-1/2mv_{\mathrm{i}}^2

=1/2(2)\left[(0)^2-(-3)^2\right]

=-9\mathrm{~J}

Related subjects & topics
Physical Sciences
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