Question 3.2.4: Additional Answer Options

2020 - November - Paper 1 - Question 3

3.1

(Motion of an object) under the influence of gravity (weight) only.

(Motion in which) the only force acting on the object is gravity (weight).

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 your mark.

2 marks

(1 x 2 marks)

3.2.1

Δt = 0,67 – 0,64 = 0,03 s

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

(2 x 1 mark)

3.2.2

Option 1:

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

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

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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, we can divide the time by 2.

2 marks

(2 x 1 mark)

3.2.3

Option 1:

Upwards as positive:

$v_{f}=v_{i}+a\Delta t$ ⁽¹⁾

$0=v_{i}+(-9,8)(0,62)$ ⁽²⁾

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

Downwards as positive:

$v_{f}=v_{i}+a\Delta t$ ⁽¹⁾

$0=v_{j}+(9,8)(0,62)$ ⁽²⁾

$v_{i}=-6,08$

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

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The ball must move in the opposite direction to the direction of gravity. One value must be positive and the other must be negative.

3 marks

(3 x 1 mark)

Marking criteria:

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

3.2.4

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⁾

$0=v_{i}^2+2(9,8)(-1,2)$ ⁽²⁾

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

Upwards as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$ ⁽¹⁾

$0=v_{i}^2+2(-9,8)(1,2)$ ⁽²⁾

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

Part 2:

Option 1:

Upwards as positive:

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

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

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

t=1,97+2(0,4898) ⁽⁵⁾

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

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

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

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

t=1,97+0,9898=2,96s ^{(5 + 6)}

Downwards as positive:

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

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

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

t=1,97+2(0,4898) ⁽⁵⁾

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

Each mark can only be awarded once.

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6 marks

(6 x 1 mark)

Option 1 / 2 / 3 / 4 marking criteria:

Calculating initial velocity:

⁽¹⁾Appropriate formula.
⁽²⁾Substitution.

Calculating Δt:

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

Marking criteria for option 1/2/3/4:

Calculating initial velocity:

⁽¹⁾Appropriate formula.
⁽²⁾Substitution.

Calculating Δt:

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

Option 5 marking criteria:

⁽¹⁾Formula.
⁽²⁾Substitution: Δy = 1,2.
⁽³⁾Substitution: 0 + 1⁄2(9,8) Δt².
⁽⁴⁾1,97 s +.
⁽⁵⁾2 Δt.
⁽⁶⁾Final answer: 2,95 - 2,97 s.

Part 1

Option 2

$\left.\begin{array}{l}\left(E_{\text{mechanical }}\right)_{\text{top }}=\left(E_{\text{mechanical }}\right)_{\text{bottom}}\\ \left(E_{p}+E_{k}\right)_{\text{top }}=\left(E_{p}+E_{k}\right)_{\text{bottom}}\end{array}\right\}Anyone$ ⁽¹⁾

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

(9,8)(1,2)+0=0+(1/2)v^2 ⁽²⁾

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

Option 3

$\left.\begin{array}{l}\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{E}_{\mathrm{p}}+\Delta\mathrm{E}_{\mathrm{k}}\\ 0=(0-\mathrm{mgh})+1/2\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right)\end{array}\right\}Anyone$ ⁽¹⁾