Question 3.2.3: Additional Answer Options

2021 - November - Paper 1 - Question 3

3.1

 

No check

Gravitational force is not the only force acting on the balloon/other forces are acting on the balloon. check

Or
Its acceleration is not 9,8 m∙s-2/is zero. check 

Or

It has constant velocity/no acceleration. check

light bulb Tip: when an object is in free fall, the only force acting on it is its weight. This would cause it to accelerate downwards at 9,8 m·s-2. In this case, the balloon is moving upwards at a constant velocity so it is not in free-fall.

2 marks

(2 x 1 mark)

 

3.2.1

 

light bulb Tips:

  • It is important to realise that the final velocity, acceleration and displacement are all downwards and therefore, must be given the same sign.

Option 1

Upwards as positive:

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

(-62,68)^2=v_{i}^2+2(-9,8)(-200) check

v_{i}=2,96\mathrm{~m}\cdot\mathrm{s}^{-1} check

Or

Downwards as positive:

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

(62,68)^2=v_{i}^2+2(9,8)(200) check

v_{i}=-2,96\mathrm{~m}\cdot\mathrm{s}^{-1}

=2,96\mathrm{~m}\cdot\mathrm{s}^{-1}check

  • There is not enough information about the hot air balloon to be able to calculate an answer from the information provided.
  • However, it is not known that the stone and balloon had the same initial velocity. Therefore, if you calculate the initial speed of the stone, it will also be the initial speed of the balloon.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

3.2.2

 

Option 1

Upwards as positive:

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

-62,68=2,96+(-9,8)\Delta tcheck

\Delta t=6,70\mathrm{s}\quad(6,698) check

Or

Downwards as positive:

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

62,68=-2,96+9,8\Delta tcheck

\Delta t=6,70\mathrm{s}(6,698)check

light bulb Tips:

  • This question is marked positively, which means that if you got the incorrect answer in question 3.2.1 but used it in any of these equations, you can still be awarded all 3 marks for this question.
  • The answer given in brackets (6,698) is the unrounded answer. All answers should always be given to a minimum of 2 decimal places.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

3.2.3

 

light bulb How to approach this question:

  • Stone A hits the ground in a time of 6,70s (calculated in question 3.2.2), since stone B was dropped 5s later, it has fallen downwards starting from the same initial velocity of the hot air balloon (2,96m·s-1 upwards) for a time of 1,70s (6,70s-5s).

Option 1

Upwards as positive:

Stone B:

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

=2,96(6,7-5)+1/2(-9,8)(6,7-5)^2 (2 + 3)checkcheck

=-9,13\mathrm{m}(-9,09\text{ to }-9,13)

Distance travelled by stone B: 9,13 m

Hot-air balloon:

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

=2.96(6.7-5)+0 (4)check

=5,03\mathrm{~m}

Stone\text{ }{B}=9.13+5.03 (5)check

=14,16\mathrm{m}(14,11-14,16) (6)check

Or

Upwards as positive:

Stone B:

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

=-2.96(6.7-5)+1/2(9.8)(6.7-5)^2 (2 + 3)checkcheck

=9,13\mathrm{m}(9,09\text{ to }9,13)

Distance travelled by stone B: 9,13 m

Hot-air balloon:

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

=-2.96(6.7-5)+0 (4)check

=-5,03\mathrm{~m}

Stone\text{ }{B}=9.13+5.03 (5)check

=14,16\mathrm{m}(14,11-14,16) (6)check

  • It is essential that the initial velocity of stone B is in the opposite direction to the acceleration and final velocity of stone B.

There are more ways to answer this question. To view other options, click here.

6 marks

(6 x 1 mark)

Marking criteria:

  • (1) Formula to calculate Δy of stone B check
  • (2) Substitution of t = 1,7s (tA – 5) check
  • (3) Substitution to calculate Δy of stone B check
  • (4) Substitution to calculate Δy of the balloon check
  • (5) Calculating the distance between balloon and stone B check
  • (6) Final answer: 14,16 m. (14,11 to 14,16) check

 

3.3

 

Upwards as positive:

checkcheckcheckcheck  

Or

Downwards as positive:

checkcheckcheckcheck

light bulb Tips:

  • The line for the hot air balloon must be a straight line since it continues to travel at a constant velocity for the entire duration of the graph.
  • The graph for stone A must be convex since the stone accelerates in the opposite direction to its original motion. Stone A should finish at a height of 0m when it strikes the ground.

4 marks

(4 x 1 mark)

Marking criteria:

  • Correct shape for stone A not starting from 0 m
  • Correct shape and initial position for a hot-air balloon
  • The gradient for a hot-air balloon is higher than that of stone A until stone A reaches the maximum height
  • Both graphs start at the same position and end at the same time

Option 2

Upwards as positive:

Stone B:

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

=2,96+(-9,8)(6,70-5) (1)

=-13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

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

(-13,7)^2=(2,96)^2+2(-9,8)\Delta y (3)

\Delta y=-9,13m

Distance travelled by stone B: 9,13 m

Hot-air balloon:

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

=-2,96(6,70-5)+0 (4)

=-5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ stone }{B}=9,13+5,03 (5)

=14,16\mathrm{m}(14,11-14,16) (6)


Or

Downwards as positive:

Stone B:

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

=-2,96+(9,8)(6,70-5) (1)

=13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

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

(13,7)^2=(-2,96)^2+2(9,8)\Delta y (3)

\Delta y=9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Hot-air balloon:

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

=-2,96(6,70-5)+0 (4)

=-5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone }{B}=9,13+5,03 (5)

=14,16\mathrm{m}(14,11-14,16) (6)


Or

Option 3

Upwards as positive:

Stone B:

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

=2,96+(-9,8)(6,70-5) (1)

=-13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

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

=\left(\frac{+2,96+(-13,7)}{2}\right){(6,70-5)}(3)

=-9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Hot-air balloon:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2

=2.96(6.70-5)+0 (4)

=5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone }B=9,13+5,03 (5)

=14,16\mathrm{m}\quad(14,11-14,16) (6)


Or

Downwards as positive:

Stone B:

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

=-2,96+(9,8)(6,70-5) (1)

=13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

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

=\left(\frac{-2,96+(13,7)}{2}\right){(6,70-5)} (3)

=9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Hot-air balloon:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2=-13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

=-2,96(6,70-5)+0 (4)

=-5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone }{B}=9,13+5,03 (5)

=14,16\mathrm{m}\quad(14,11-14,16) (6)


Or

Option 4

Upwards as positive:

Stone B:

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

=2,96+(-9,8)(6,70-5) (1)

=-13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

Balloon's height after 5 s: 214,8 m

\left(\mathrm{E}_{\mathrm{mech}}\right)_{214,8\mathrm{m}}=\left(\mathrm{E}_{\mathrm{mech}}\right)_{1,7\mathrm{s}}

\left(E_{P}+E_{K}\right)_{214,8m}=\left(E_{P}+E_{K}\right)_{1,7\mathrm{~s}}

\left(mgh+1/2mv^2\right)=\left(mgh+1/2m^2\right)_{1,7s} (2)

(9,8)(214,9)+1/2(2,96)^2=

(9,8)h+1/2(13,7)^2 (3)

\therefore\mathrm{h}=205,67\mathrm{~m}

Distance travelled by stone B: 214,8 – 205,67 = 9,13 m

Hot-air balloon:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2

=2,96(6,70-5)+0 (4)

=5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone B: }9,13+5,03=14,16\mathrm{m} (5 + 6)

(14.11\text{ to }14.16)


Or

Downwards as positive:

Stone B:

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

=-2,96+(9,8)(6,70-5) (1)

=13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

Balloon's height after 5 s: 214,8 m

\left(\mathrm{E}_{\mathrm{mech}}\right)_{214,8\mathrm{m}}=\left(\mathrm{E}_{\mathrm{mech}}\right)_{1,7\mathrm{s}}

\left(E_{P}+E_{K}\right)_{214,8m}=\left(E_{P}+E_{K}\right)_{1,7s} (2)

\left(mgh+1/2mv^2\right)=\left(mgh+1/2mv^2\right)_{1,7}

(9,8)(214,8)+1/2(2,96)^2=

(9,8)h+1/2(13,7)^2 (3)

\therefore\mathrm{h}=205,67\mathrm{~m}

Distance travelled by stone B: 214,8 – 205,67 = 9,13 m

Hot-air balloon:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2

=-2.96(6.70-5)+0 (4)

=-5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone B: }9,13+5,03=14,16\mathrm{m} (5 + 6)

(14,11\text{ to }14,16)


Or

Option 5

Upwards as positive:

Stone B:

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

=2,96+(-9,8)(6,70-5) (1)

=-13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{K}} (2)

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

=1/2m\left(v_{f}^2-v_{j}^2\right)

(9.8)\Delta h\cos0^{\circ}=1/2\left(13.7^2-2.96^2\right) (3)

\Delta\mathrm{h}=9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Hot-air balloon:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2

=2.96(6.70-5)+0 (4)

=5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone B: }9,13+5,03=14,16\mathrm{m} (5 + 6)

(14,11\text{ to }14,16)


Or

Downwards as positive:

Stone B:

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

=-2,96+(9,8)(6,70-5) (1)

=13,7\mathrm{~m}\cdot\mathrm{s}^{-1}

\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{K}} (2)

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

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

(9,8)\Delta h\cos0^{\circ}=1/2\left(13,7^2-2,96^2\right) (3)

\Delta\mathrm{h}=9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Hot-air balloon:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2

=-2,96(6,70-5)+0 (4)

=-5,03\mathrm{~m}

Distance travelled by hot-air balloon: 5,03m

Distance between hot-air balloon and
stone B:

\text{ Stone B: 9,13+5,03 }=14,16\mathrm{m} (5 + 6)

(14,11\text{ to }14,16)


Or

Option 6

Upwards as positive:

Using relative velocities:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2 (1)

=(2,96-2,96)(1,7)+\frac12(-9,8)(1,7)^2 (2 + 3 + 4 + 5)

=-14,16\mathrm{~m}

Distance between hot-air balloon and
stone B: 14,16m (6)


Or

Downwards as positive:

Using relative velocities:

\Delta y=v_{i}\Delta t+\frac12a\Delta t^2 (1)

=(2,96-2,96)(1,7)+\frac12(9,8)(1,7)^2 (2 + 3 + 4 + 5)

=14,16\mathrm{~m} (6)

Distance between hot-air balloon and
stone B: 14,16m


Or

Option 7

Upwards as positive:

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

=(2,96)(1,7)+1/2(-9,8)(1,7)^2 (2 + 3)

=-9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Height of stone B from the ground = 200 + 14,8 – 9,13 = 205,63 m

Height of balloon from the ground = 200 + (6,7)(2,96) = 219,83 m (4)

Distance between B and the balloon = 219,83 – 205,63 = 14,16 m (5 + 6)


Or

Downwards as positive:

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

=(-2,96)(1,7)+1/2(9,8)(1,7)^2 (2 + 3)

=9,13\mathrm{~m}

Distance travelled by stone B: 9,13 m

Height of stone B from the ground = 200 + 14,8 – 9,13 = 205,63 m

Height of balloon from the ground = 200 + (6,7)(2,96) = 219,83 m (4)

Distance between B and the balloon = 219,83 – 205,63 = 14,16 m (5 + 6)

Related subjects & topics
Explore similar posts in our community