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.

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

Teacher tip

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⁾

-15,2=(0)+1 / 2(-9.8) \Delta t^2 ⁽²⁾

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

Downwards as positive:

\Delta y=v_i \Delta t+1 / 2 a \Delta t^2 ⁽¹⁾

15,2=(0)+1 / 2(9,8) \Delta t^2 ⁽²⁾

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

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Teacher tip

3 marks

(3 x 1 mark)

Marking criteria:

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

3.2.2

Option 1:

Upwards as positive:

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

-3,2=(0)+1 / 2(-9,8)(\Delta t)^2 ⁽²⁾

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

\Delta t(B)=1,76-0,81 ⁽³⁾

\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)=v_i(0.95)+1 / 2(-9.8)(0.95)^2 ⁽⁴⁾

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

Downwards as positive:

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

3,2=(0)+1 / 2(9,8)(\Delta t)^2 ⁽²⁾

\Delta t=0.81 \mathrm{~s}

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

\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⁾

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

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Teacher tip

5 marks

(5 x 1 mark)

Marking criteria:

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

3.3

Upwards as positive:

Downwards as positive:

Teacher tip

5 marks

(5 x 1 mark)

Marking criteria:

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

Part 1 (No marks awarded)

Option 2

Upwards as positive:

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

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

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

Upwards as positive:

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

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

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

Option 3

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

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

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

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

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

Option 4

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

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

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

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

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

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⁾

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⁾

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⁾

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⁾

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⁾

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⁾