Question 3.3: Additional Answer Options

2022 - May/June - Paper 1 - Question 3

Option 4 marking criteria:

  • (1) Formula ΔyC = viΔt + 1⁄2aΔt2.
  • (2) Substitute to calculate ΔyB.
  • (3) Substitute to calculate ΔyC.
  • (4) Substitute Δt = 0,5.
  • (5) Equating yB and yC.
  • (6) Final answer: 1,71 s.

Or

Option 2

Upwards as positive:

\Delta y_{c}=v_{i}\Delta t+1/2a\Delta t^2 (1)

=15\Delta t+1/2(-9,8)\Delta t^2 (2)

=15\Delta\mathrm{t}-4,9\Delta\mathrm{t}^2\ldots\ldots(Equation1)

 

\Delta y_{B}=v_{i}\Delta t+1/2a\Delta t^2 (3)

={40}\left(\Delta t-0,5^{}\right)+1/2(-9,8)(\Delta t-0,5)^2 (4)

 

ΔyC in terms of:

30+\Delta y_{c}=40(\Delta t-0,5)+1/2(-9,8)(\Delta t-0,5)^2

\Delta\mathrm{y}_{\mathrm{C}}=-4,9\Delta\mathrm{t}^2+44,9\Delta\mathrm{t}-51,225..\ldots\ldots\left(Equation2\right)

Equate the 2 equations:

15\Delta t=44,9\Delta t-51,225 (5)

\Delta t=1,71\mathrm{~s}(6)


Or

Downwards as positive:

\Delta\mathrm{y}_{\mathrm{c}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2 (1)

=-15\Delta t+1/2(9,8)\Delta t^2 (2)

=4,9\Delta\mathrm{t}^2-15\Delta\mathrm{t}

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

=-40(\Delta t-0,5)+1/2(9,8)(\Delta t-0,5)^2 (3 + 4)

ΔyC in terms of ΔyB:

30+\Delta y_{c}=(-40)(t-0,5)+1/2(9,8)(\Delta t-0,5)^2

\Delta\mathrm{y}_{c}=4,9\Delta\mathrm{t}^2-44,9\Delta\mathrm{t}+51,225

Equate both equations:

-15\Delta t=-44,9\Delta t+51,225 (5)

\Delta\mathrm{t}=1,71\mathrm{~s} (6)


Or

Option 3

Upwards as positive:

\Delta\mathrm{y}_{\mathrm{c}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2 (1)

=15(\Delta t+0,5)+1/2(-9,8)(\Delta t+0,5)^2 (2 + 3)

=-4,9t^2+10,1\Delta t+6,275

\Delta\mathrm{y}_{\mathrm{B}}=\mathrm{v}_{\mathrm{j}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

=40\Delta t+1/2(-9,8)\Delta t^2 (4)

ΔyC in terms of ΔyB:

30+\Delta y=40\Delta t+1/2(-9,8)t^2

\Delta y=-4,9\Delta t^2+40t-30

Equate the 2 equations:

10.1\Delta t+6.275=40\Delta t-30 (5)

\Delta\mathrm{t}=1,21\mathrm{~s}

\Delta\mathrm{t}_{\mathrm{TOT}}=1,21+0,5

=1,71\mathrm{~s} (6)


Or

Downwards as positive:

\Delta y_{c}=v_{i}\Delta t+1/2a\Delta t^2 (1)

=-15(\Delta t+0,5)+1/2(9,8)(\Delta t+0,5)^2 (2 + 3)

=4,9\Delta\mathrm{t}^2-10,1\Delta\mathrm{t}+6,275

\Delta\mathrm{y}_{\mathrm{B}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

=-40\Delta t+1/2(9,8)\Delta t^2 (4)

ΔyC in terms of ΔyB:

30+\Delta y=-40\Delta t+1/2(9,8)t^2

\Delta\mathrm{y}=-4,9\Delta\mathrm{t}^2-40\Delta\mathrm{t}-30

Equate the 2 equations:

-10,1\Delta t+6,275=-40\Delta t-30 (5)

\Delta\mathrm{t}=1,21\mathrm{~s}

\Delta\mathrm{t}_{\mathrm{TOT}}=1,21+0,5

=1,71\mathrm{~s} (6)


Or

Option 4

Upwards as positive:

Displacement of C after 0,5s:

\Delta\mathrm{y}_{\mathrm{c}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2 (1)

=15(0,5)+1/2(-9,8)(0,5)^2 (2)

\Delta y_{c}=6,28m

Velocity of C after 0,5 s:

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

=15+(-9,8)(0,5)

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

Displacement of C at meeting point:

\Delta\mathrm{y}_{\mathrm{c}}=\mathrm{v}_{i}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

=10.1\Delta\mathrm{t}+1/2(-9.8)\Delta\mathrm{t}^2 (3)

=10,1\Delta t-4,9\Delta t^2

Displacement of B at meeting point:

\Delta\mathrm{y}_{\mathrm{B}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

=40\Delta\mathrm{t}+1/2(-9,8)\Delta\mathrm{t}^2 (4)

=40\mathrm{t}-4,9\mathrm{t}^2

At meeting point:

\Delta v_{C}+\Delta v_{B}=-\left[10.1\Delta t-4.9\Delta t^2\right]+40\Delta t-4.9\Delta t^2 (5)

36,28=-10,10\Delta t+40\Delta t

\Delta\mathrm{t}=1,21\mathrm{~s}

\Delta\mathrm{t}_{\mathrm{tot}}=1,21+0,5=1,71\mathrm{~s} (6)


Or

Downwards as positive:

Displacement of C after 0,5 s:

\Delta\mathrm{y}_{\mathrm{c}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2 (1)

=-15(0,5)+1/2(9,8)(0,5)^2 (2)

\Delta y_{c}=-6,28m

Velocity of C after 0,5 s:

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

=-15+(9,8)(0,5)

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

Displacement of C at meeting point:

\Delta\mathrm{y}_{\mathrm{c}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

=-10.1\Delta t+1/2(9.8)\Delta t^2 (3)

=-10,1\Delta t+4,9\Delta t^2

Displacement of B at meeting point:

\Delta\mathrm{y}_{\mathrm{B}}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

=-40\Delta t+1/2(9,8)\Delta t^2 (4)

=-40t+4,9t^2

At meeting point:

\Delta v_{c}+\Delta v_{B}=-\left[-10.1\Delta t+4.9\Delta t^2\right]-40\Delta t+4.9\Delta t (5)

-36,28=10,10\Delta t-40\Delta t

\Delta t=1,21\mathrm{~s}

\Delta\mathrm{t}_{\mathrm{tot}}=1,21+0,5=1,71\mathrm{~s} (6)

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