Question 3.5: Additional Answer Options

2021 - May/June - Paper 1 - Question 3

Or

Option 2

Upwards as positive:

Comparing differences in heights:

Work done by the floor = change in Ek (ΔEp = 0)

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(-9,8)(1,1)^2

=-5,93\mathrm{~m}

\Delta y=5,93\mathrm{~m}(5,929)

\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{E}_{\mathrm{p}}+\Delta\mathrm{E}_{\mathrm{k}}

=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)

=(0,06)(9,8)(5,93-8,10)+0

=-1,28\mathrm{~J}

Or

Downwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(9,8)(1,1)^2

=-5,93\mathrm{~m}

\Delta y=5,93\mathrm{~m}

\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{E}_{\mathrm{p}}+\Delta\mathrm{E}_{\mathrm{k}}

=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)

=(0,06)(9,8)(5,93-8,10)+0

=-1,28\mathrm{~J}

Or

Option 3

Upwards as positive:

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

0=v_{i}+(-9,8)(1,1)

v_{i}=10,78\mathrm{~m}\cdot\mathrm{s}^{-1}

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

=\left(\frac{10,78+0}{2}\right)(1,1)

\Delta y=5,93m

Or

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

0=(10,78)^2+2(-9,8)\Delta y

\Delta\mathrm{y}=5,93\mathrm{~m}

 

\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{E}_{\mathrm{p}}+\Delta\mathrm{E}_{\mathrm{k}}

=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)

=(0,06)(9,8)(5,93-8,10)+0

=-1,28\mathrm{~J}

Or

Downwards as positive:

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

0=v_{i}+(9,8)(1,1)

v_{i}=-10,78

\mathrm{v}_{\mathrm{i}}=10,78\mathrm{~m}\cdot\mathrm{s}^{-1}

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

=\left(\frac{-10,78-0}{2}\right)(1,1)

\Delta y=-5,93m

Or

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

0=(-10,78)^2+2(9,8)\Delta y

\Delta y=-5,93m

\Delta y=5,93m

 

\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{E}_{\mathrm{p}}+\Delta\mathrm{E}_{\mathrm{k}}

=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)

=(0,06)(9,8)(5,93-8,10)+0

=-1,28\mathrm{~J}

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