Question 3.2.3: Additional Answer Options

2020 - November - Paper 1 - Question 3

3.1

 

Motion of an object under the influence of the gravitational force only. checkcheck  

Or

Motion in which the only force acting on the object is gravitational force. checkcheck 

light bulb  Tips: 

  • Free fall and vertical projectile motion both describe the motion of an object when the only force acting on the object is the gravitational force.
  • If your response focuses on the object rather than its motion, you will lose the mark.
  • As of 2024 you will not get marks for using the term weight or gravity, you need to refer to gravitational force.

2 marks

(1 x 2 marks)

3.2.1

 

Δt = 0,67 – 0,64 = 0,03 s checkcheck

light bulb Tip: the time that the ball is in contact with the floor, is the time when the position is zero as shown in the graph below:

2 marks

(1 x 2 marks)

3.2.2

 

Option 1

\Delta t=\frac{(1,90-0,67)}{2}check

=0,62\mathrm{s}(0,615\mathrm{s})check

light bulb Tips: 

  • The time that it takes for the ball to reach the maximum height from its projection is equal to the time it takes for the ball to return from that height back to the point of projection. Therefore, divide the time by 2.

 

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

2 marks

(1 x 2 marks)

3.2.3

 

Option 1

Upwards as positive:

v_{f}=v_{i}+a\Delta t (1)check

0=v_{i}+(-9,8)(0,62) (2)check

v_{i}=6,08\mathrm{~m}\cdot\mathrm{s}^{-1}\left(6,076\mathrm{~m}\cdot\mathrm{s}^{-1}\right) (3)check

Or

Downwards as positive:

v_{f}=v_{i}+a\Delta t (1)check

0=v_{j}+(9,8)(0,62) (2)check

v_{i}=-6,08

\therefore6,08\mathrm{~m}\cdot\mathrm{s}^{-1}\left(6,076\mathrm{~m}\cdot\mathrm{s}^{-1}\right) (1)check

light bulb Tip: the ball must move in the opposite direction to the direction of gravity. One value must be positive and the other must be negative.

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3 marks

(3 x 1 mark)

Marking criteria:

  • (1) Any appropriate formula check
  • (2) Correct substitutioncheck
  • (3) Final answer: 5,94 to 6,08 m·s-1 check 

3.2.4

 

light bulb How to approach this question: 

  • This question provides multiple parts with multiple alternative options.

Part 1:

Option 1

Downwards as positive:

\mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2\mathrm{a}\Delta\mathrm{y} (1) check

0=v_{i}^2+2(9,8)(-1,2)(2)check

\mathrm{v}_{\mathrm{i}}=-4,85\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Upwards as positive:

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

0=v_{i}^2+2(-9,8)(1,2) (2) check

\mathrm{v}_{\mathrm{i}}=4,85\mathrm{~m}\cdot\mathrm{s}^{-1}

Part 2:

Option 1

Upwards as positive:

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

1.2=(4.85)\Delta t+1/2(-9.8)\Delta t^2 (4) check

\Delta t=0,4898\mathrm{~s}/0,5\mathrm{~s}

t=1,97+2(0,4898) (5) check

=2,95\mathrm{~s}/2,97\mathrm{~s} (6) check

Or

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

0=(4,85)\Delta t+1/2(-9,8)\Delta t^2 (4)check

\Delta\mathrm{t}=0,9898\mathrm{~s}(\text{ or }\Delta\mathrm{t}=0)

t=1,97+0,9898=2,96s (5 + 6) checkcheck

Or

Downwards as positive:

\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2 (3) check

1,2=(-4,85)\Delta t+1/2(9,8)\Delta t^2 (4) check

\Delta\mathrm{t}=0,4898\mathrm{~s}/0,5\mathrm{~s}

t=1,97+2(0,4898) (5) check

=2,95\mathrm{~s}/2,97\mathrm{~s} (6) check

  • Each mark can only be awarded once.

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

6 marks

(6 x 1 mark)

Option 1/2/3/4 marking criteria:

Calculating initial velocity:

  • (1) Appropriate formula check
  • (2) Substitutioncheck

Calculating Δt:

  • (3) Appropriate formula check
  • (4) Substitutioncheck
  • (5) 1,97 s + Δt check
  • (6) Final answer: 2,95 – 2,97 s check

Marking criteria:

  • (1) Any appropriate formula.
  • (2) Correct substitution.
  • (3) Final answer: 5,94 to 6,08 m·s-1.

Or

Option 2

Upwards positive:

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

1,85=v_{i}(0,62)+1/2(-9,8)(0,62)^2 (2)

\mathrm{v}_{\mathrm{i}}=6,02\mathrm{~m}\cdot\mathrm{s}^{-1}\left(6,022\mathrm{~m}\cdot\mathrm{s}^{-1}\right) (3)


Or

Downwards positive:

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

1.85=v_{i}(0.62)+1/2(9.8)(0.62)^2 (2)

v_{i}=-6,02

\therefore\mathrm{v}_{\mathrm{i}}=6,02\mathrm{~m}\cdot\mathrm{s}^{-1}\left(6,022\mathrm{~m}\cdot\mathrm{s}^{-1}\right) (3)


Or

Option 3

Motion from top to bottom:

Downwards positive:

v_{f}^2=v_{j}^2+2a\Delta y (1)

v_{f}^2=0+2(9,8)(1,85) (2)

v_{f}=6,02\mathrm{~m}\cdot\mathrm{s}^{-1}

Initial velocity = 6,02 m·s-1. (3)


Or

Upwards positive:

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

v_{f}^2=0+2(-9,8)(-1,85)(2)

v_{f}=6,02m\cdot s^{-1}

Initial velocity = 6,02 m·s-1. (3)


Or

Motion from bottom to top:

Downwards positive:

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

0^2=v_{i}^2+2(9,8)(-1,85) (2)

v_{i}=6,02\mathrm{~m}\cdot\mathrm{s}^{-1} (3)


Or

Upwards positive:

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

0=v_{i}^2+2(-9,8)(1,85) (2)

\mathrm{v}_{\mathrm{i}}=6,02\mathrm{~m}\cdot\mathrm{s}^{-1} (3)


Or

Option 4

Upwards positive:

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

0=v_{i}(1,23)+1/2(-9.8)(1,23)^2 (2)

\mathrm{v}_{\mathrm{i}}=6,03\mathrm{~m}\cdot\mathrm{s}^{-1} (3)


Or

Downwards positive:

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

0=v_{i}(1,23)+1/2(9,8)(1,23)^2 (2)

v_{i}=-6,03m\cdot s^{-1}

\text{ speed }=6,03\mathrm{m}\cdot\mathrm{s}^{-1} (3)


Or

Option 5

\Delta y=\left(\frac{v_{f}+v_{i}}{2}\right)\Delta t (1)

1,85=\left(\frac{0+v_{i}}{2}\right)(0,62) (2)

\mathrm{v}_{\mathrm{i}}=5,97\mathrm{~m}\cdot\mathrm{s}^{-1} (3)


Or

Option 6

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

m(9,8)(0,62)=m\left(0-v_{i}\right) (2)

v_{i}=6,08\mathrm{~m}\cdot\mathrm{s}^{-1} (3)


Or

Option 7

\left(E_{p}+E_{k}\right)_{\text{floor}}=\left(E_{p}+E_{k}\right)_{\text{top}} (1)

\left(mgh+1/2\mathrm{mv}^2\right)_{\text{floor}}=\left(\mathrm{mgh}+1/2\mathrm{mv}^2\right)_{\text{top}}

0+1/2v^2=(9.8)(1.85)+0 (2)

\mathrm{v}=6,02\mathrm{~m}\cdot\mathrm{s}^{-1} (3)

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