Additional Answer Options: 3.2.2

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.

Option 2

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⁾

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

$\Delta\mathrm{t}(\mathrm{B})=1,76-0,81$ ⁽³⁾

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

$\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}$

$-v_{i}=v_{i}+(-9,8)(0,95)$ ⁽⁴⁾

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

Downwards as positive:

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

$3,2=(0)+1/2(9,8)(\Delta t)^2$ ⁽²⁾

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

$\Delta\mathrm{t}(\mathrm{B})=1,76-0,81$ ⁽³⁾

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

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

$\mathrm{v}_{\mathrm{i}}=-\mathrm{v}_{\mathrm{i}}+(9,8)(0,95)$ ⁽⁴⁾

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

Part 1

Option 1

Upwards as positive:

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

$v_{f}^2=(0)^2+(2)(-9,8)(-3,2)$

$v_{f}=-7,92\mathrm{~m}\cdot\mathrm{s}^{-1}$

Downwards as positive:

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

$v_{f}^2=(0)^2+(2)(9,8)(3,2)$

$\mathrm{v}_{\mathrm{f}}=7,92\mathrm{~m}\cdot\mathrm{s}^{-1}$

Option 2

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

$\left(\mathrm{E}_{\mathrm{p}}+\mathrm{E}_{\mathrm{k}}\right)_{\mathrm{top}}=\left(\mathrm{E}_{\mathrm{p}}+\mathrm{E}_{\mathrm{k}}\right)_{\text{bot }}$

$\left(mgh+1/2mv_1^2\right)_{\text{too }}=\left(mgh+1/2mv_{f}^2\right)_{\text{bot }}$

$(9,8)(3,2)+0=0+(1/2)\left(v_{f}\right)^2$

Option 3

$\mathrm{v}_{\mathrm{f}}=7,92\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

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

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

$0=1/2\left(v_{f}^2-0\right)+(9,8)(3,2)$

$v_{f}=7,92\mathrm{~m}\cdot\mathrm{s}^{-1}$

Option 4

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

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

$(9,8)(3,2)\cos0^{\circ}=1/2v_{f}^2-0$

$\mathrm{v}_{\mathrm{f}}=7.92\mathrm{~m}\cdot\mathrm{s}^{-1}$

Part 2

Option 1

Upwards as positive:

$v_{f}=v_{j}+a\Delta t$

$-7,92=(0)+(-9,8)\Delta t$ ⁽¹⁾

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

$\Delta\mathrm{t}(\mathrm{B})=1,76-0,81$ ⁽²⁾

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

Downwards as positive:

$\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}$

$7,92=(0)+(9,8)\Delta t$ ⁽¹⁾

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

$\Delta\mathrm{t}(\mathrm{B})=1,76-0,81$ ⁽²⁾

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

Option 2

Upwards as positive:

$\Delta\mathrm{y}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t}$

$-3,2=\left(\frac{0-7,92}{2}\right)\Delta t$ ⁽¹⁾

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

$\Delta\mathrm{t}(\mathrm{B})=1,76-0,81$ ⁽²⁾

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

Downwards as positive:

$\Delta\mathrm{y}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t}$

$3,2=\left(\frac{0+7,92}{2}\right)\Delta t$ ⁽¹⁾

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

$\Delta\mathrm{t}(\mathrm{B})=1,76-0,81$ ⁽²⁾

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

Option 3

Upwards as positive:

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

$-12=-7,92\Delta t+1/2(-9,8)\Delta t^2$ ^{(1 + 2)}

$\Delta\mathrm{t}=0,95\mathrm{~s}$

Downwards as positive:

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

$12=7,92\Delta t+1/2(9,8)\Delta t^2$ ^{(1 + 2)}

$\Delta\mathrm{t}=0,95\mathrm{~s}$

Part 3

Option 1

Upwards as positive:

$\left(\Delta\mathrm{t}_{\mathrm{up}}\text{ and down }=0,95\mathrm{~s}\right)$

$v_{f}=v_{i}+a\Delta t$ ⁽³⁾

$-v_{i}=v_{i}+(-9,8)(0,95)$ ⁽⁴⁾

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

Downwards as positive:

$v_{f}=v_{i}+a\Delta t$ ⁽³⁾

$v_{i}=v_{i}+(9,8)(0,95)$ ⁽⁴⁾

=-4,66

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

Option 2

Upwards as positive:

$\left(\Delta\mathrm{t}_{\mathrm{up}}=0,475\mathrm{~s}\right)$

$v_{f}=v_{i}+a\Delta t$ ⁽³⁾

$0=v_{i}+(-9,8)(0,475)$ ⁽⁴⁾

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

Downwards as positive:

$\left(\Delta\mathrm{t}_{\mathrm{up}}=0,475\mathrm{~s}\right)$

$\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}$ ⁽³⁾

$0=v_{i}+(9,8)(0,475)$ ⁽⁴⁾

=-4,66

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

Option 3

Upwards as positive:

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

$0=v_{i}(0,95)+1/2(-9,8)(0,95)^2$ ⁽⁴⁾

$\mathrm{v}_{\mathrm{i}}=4,66\mathrm{~m}\cdot\mathrm{s}$

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

Downwards as positive:

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

$0=-v_{i}(0,95)+1/2(9,8)(0,95)^2$ ⁽⁴⁾

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

Option 7

Upwards as positive:

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

$(-9,8)(0,95)=-2v_{i}$ ^{(2 + 3 + 4)}

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

Downwards as positive:

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

$(9.8)(0.95)=2v_{i}$ ^{(2 + 3 + 4)}

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