Additional Answer Options: 3.2.1

2023 - May/June - Paper 1 - Question 3

3.1

Motion under the influence of gravity/weight/gravitational force only. checkcheck 

Or

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

light bulb Tips: 

  • Make sure that the focus of your definition is on a motion and not an object.
  • If you define a projectile, you are not defining the correct thing.
  • As of 2024, you will no longer be awarded marks if you refer to gravity or weight instead of gravitational force.

2 marks

(1 x 2 marks)

3.2.1

 

Option 1

Upwards as positive:

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

-15,2=(0)+1/2(-9.8)\Delta t^2 (2)check

\Delta\mathrm{t}=1,76\mathrm{~s} (3)check

Or

Downwards as positive:

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

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

\Delta\mathrm{t}=1,76\mathrm{~s} (3)check

light bulb Tips: 

  • When the ball is dropped, its initial velocity is 0 m·s-1.
  • Remember, velocity, displacement, and acceleration are all vector quantities; therefore, they must have a direction.
  • This results in the displacement and acceleration having the same sign because they both act downwards.

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

3 marks

(3 x 1 mark)

Marking criteria:

  • (1) Formula to calculate Δt check
  • (2) Correct substitution to calculate Δt check
  • (3) Final answer: 1,76 s check

3.2.2

 

light bulb How to approach this question: 

  • You need to start by determining how long it takes for ball A to fall 3,2m starting from rest.
  • Velocity, displacement and acceleration are all vector quantities; therefore, do require a direction.

Option 1

Upwards as positive:

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

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

\Delta t=0,81\mathrm{~s}

  • Since ball B is projected upward when ball A has fallen downward, it strikes the ground at the same time ball A strikes the ground.
  • You need to calculate the time of motion for ball B.

\Delta t(B)=1,76-0,81 (3)check

\Delta\mathrm{t}(\mathrm{B})=0,95\mathrm{~s}

  • Because ball B returns to the starting position and it is known that the displacement is 0m and the initial and final velocities have the same magnitudes but opposite directions, you can determine the velocity with which ball B was projected upward:

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

(0)=v_{i}(0.95)+1/2(-9.8)(0.95)^2 (4)check

\mathrm{v}_{\mathrm{i}}=4,66\mathrm{~m}\cdot\mathrm{s}^{-1} (5)check

Or

Downwards as positive:

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

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

\Delta t=0.81\mathrm{~s}

\Delta\mathrm{t}(\mathrm{B})=1.76-{0}.81 (3)check

\Delta\mathrm{t}(\mathrm{B})=0,95\mathrm{~s}

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

0=-\mathrm{V}_{\mathrm{i}}(0,95)+1/2(9,8)(0,95)^2 (4)check

v_{i}=4,66\mathrm{~m}\cdot\mathrm{s}^{-1} (5)check

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

5 marks

(5 x 1 mark)

Marking criteria:

  • (1) Correct substitution to calculate Δt for ball A check
  • (2) Subtraction 1,76 0,81 check
  • (3) Correct formula to calculate vi for ball B check
  • (4) Correct substitution to calculate vi for ball B check
  • (5) Final answer: 4,66 m·s-1 (4,66 to 4,7) check

 

 

3.3

 

Upwards as positive:

checkcheckcheckcheckcheck 

Or

Downwards as positive:

checkcheckcheckcheckcheck 

light bulb Tips: 

  • The bullet points in the question specify what must be indicated on the graph. The question will always tell you what must be in the graph.
  • Remember, since the ball is accelerating, the shape of the graph will be parabolic.

5 marks

(5 x 1 mark)

Marking criteria:

  • Initial position of ball A = 15,2 m and B = 0 m check
  • Starting times for A = 0 s and B = 0,81 s check
  • Both balls strike the ground at t = 1,76 s check 
  • Shape of graph for ball A check
  • Shape of graph for ball B check

Or

Part 1 (No marks awarded)

Option 2

Upwards as positive:

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

v_{f}^2=(0)^2+(2)(-9,8)(-15,2)

v_{f}=-17,26m\cdot\mathrm{s}^{-1}

Or

Upwards as positive:

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

\mathrm{v}_{\mathrm{f}}^2=(0)^2+(2)(9,8)(15,2)

v_{f}=17,26\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Option 3

\left(\mathrm{E}_{\text{mech }}\right)_{\text{top }}=\left(\mathrm{E}_{\text{mech }}\right)_{\text{bottom }}

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

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

(9,8)(15,2)+0=0+(1/2)\left(v_{f}\right)^2

\mathrm{v}_{\mathrm{f}}=17,26\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Option 4

\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{K}+\Delta\mathrm{U}

W_{nc}=\Delta K+mg\left(h_{f}-h_{i}\right)

0=1/2mv_{f}^2-1/2mv_{i}^2+mgh_{f}-mgh_{i}

0=1/2\left(v_{f}^2-0\right)+(9,8)(15,2)

v_{f}=17,26\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Option 5

\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{k}}

w\Delta y\cos\theta=1/2mv_{f}^2-1/2mv_{i}^2

(9,8)(15,2)\cos180^{\circ}=0-1/2\left(v_{f}\right)^2

\mathrm{v}_{\mathrm{f}}=17,26\mathrm{~m}\cdot\mathrm{s}^{-1}

Part 2

Option 1

Upwards as positive:

\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{i}+\mathrm{a}\Delta\mathrm{t} (1)

-17,26=(0)+(-9,8)\Delta t (2)

\Delta\mathrm{t}=1,76\mathrm{~s} (3)

Or

Downwards as positive:

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

17,26=(0)+(9,8)\Delta t (2)

\Delta\mathrm{t}=1,76\mathrm{~s} (3)

Or

Option 2

Upwards as positive:

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

-15,2=\left(\frac{0-17,26}{2}\right)\Delta t (2)

\Delta\mathrm{t}=1,76\mathrm{~s} (3)

Or

Downwards as positive:

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

15,2=\left(\frac{0+17,26}{2}\right)\Delta t (2)

\Delta\mathrm{t}=1,76\mathrm{~s} (3)

Or

Option 6

Upwards as positive:

\left.\begin{array}{rl}F_{net}\Delta t & =\Delta p=m\left(v_{f}-v_{i}\right)\\ mg\Delta t & =m\left(v_{f}-v_{i}\right)\end{array}\right\}Anyone (1)

-9,8\Delta t=-17,26-(0) (2)

\Delta t=1,76\mathrm{~s}(3)

Or

Downwards as positive:

\left.\begin{array}{l}F_{net}\Delta t=\Delta p=m\left(v_{f}-v_{i}\right)\\ mg\Delta t=m\left(v_{f}-v_{i}\right)\end{array}\right\}Anyone (1)

9,8\Delta t=17,26-(0) (2)

\Delta\mathrm{t}=1,76\mathrm{~s} (3)

Related subjects & topics
Physical Sciences
Explore similar posts in our community