Question 3.4: Additional Answer Options

2019 - November - Paper 1 - Question 3

3.1

(Motion during which) the only force acting is the force of gravity.

Tip: free fall is the motion of an object under only its own weight.

2 marks

(1 x 2 marks)

3.2

How to approach this question:

The velocity of stone A at the highest point in the motion is 0 m·s^-1.
Remember, vectors have directions and the initial velocity and displacement are in the opposite direction to the gravitational force.

Option 1:

Upward as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$ ⁽¹⁾

$0=(10)^2+(2)(-9,8)\Delta y$ ⁽²⁾

$\Delta y=5,10m\quad(5,102)$

The question asks for the height above the ground, therefore, upward displacement must be added to the original height stone A was thrown from.

$\text{ Height }=40+{5.10}$ ⁽³⁾

$=45,10\mathrm{~m}$ ⁽⁴⁾

Downward as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$ ⁽¹⁾

$0=(-10)^2+(2)(9,8)\Delta y$ ⁽²⁾

$\Delta y=-5,10m\quad(5,102)$

$\text{ Height }=40+5.10$ ⁽³⁾

$={45,10}\mathrm{~m}$ ⁽⁴⁾

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

4 marks

(4 x 1 mark)

Marking criteria:

⁽¹⁾Any appropriate formula for Δy
⁽²⁾Whole substitution to calculate 5,1 m
⁽³⁾40+ answer from calculation
⁽⁴⁾Final answer: 45,10 m

3.3

9,8 m·s^-2 downwards.

Tips:

The object experiences a constant acceleration of 9.8 m/s² throughout its motion.
A common error among students is that the acceleration becomes zero at the highest point (where the velocity is zero). This is not true. If the acceleration were truly zero at that point, the object would remain stationary, not continue moving downwards.

2 marks

(2 x 1 mark)

3.4

How to approach this question:

Calculate the time taken for stone A to reach a height which is 29,74m above the ground where the two stones pass each other, this is a displacement of 10,26 m below the projection point of stone A.

Option 1:

Upward as positive:

Displacement from roof to meeting point: -40 + 29,74 = -10,26m.

Stone A:

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

$-10.26=10\Delta t+1/2(-9.8)\Delta t^2$ ^{(2 + 3)}

$\Delta\mathrm{t}=2,79\mathrm{~s}$

Stone B:

Calculate the time taken for stone B to fall to a height which is 29,74m above the ground where the two stones pass each other, this is also a displacement of 10,26m below the position that stone B was dropped from.

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

$-10,26=0+1/2(-9,8)\Delta t^2$ ⁽⁴⁾

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

The difference in time between when stone A is projected and stone B is dropped can be determined by subtracting the two values.

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

$\left[-10,26=0(2,79-x)+1/2(-9,8)(2,79-x)^2\right]$ ^{(4 + 5)}

x=1,34(s) ⁽⁶⁾

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

=0^2+2(-9,8)(-10,26)

$v_{f}=-14,18\mathrm{~m}\cdot\mathrm{s}$

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

$-14,18=0+(-9,8)\Delta t$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Downward as positive:

Displacement from roof to meeting point: 40 - 29,74 = 10,26m.

Stone A:

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

$10.26=-10\Delta t+1/2(9,8)\Delta t^2$ ^{(2 + 3)}

$\Delta\mathrm{t}=2,79\mathrm{~s}$

Stone B:

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

$10,26=0+1/2(9,8)\Delta t^2$ ⁽⁴⁾

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

$\left[-10,26=0(2,79-x)+1/2(-9,8)(2,79-x)^2\right]$ ^{(4 + 5)}

x=1,34(s) ⁽⁶⁾

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

=0^2+2(9.8)(10.26)

$v_{f}=14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$14,18=0+(9,8)\Delta t$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1,34(\mathrm{~s})$ ⁽⁶⁾

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

6 marks

(6 x 1 mark)

Marking criteria:

⁽¹⁾Calculation/use of 10,26 m
⁽²⁾Appropriate formula to calculate Δt
⁽³⁾Substitution for stone A
⁽⁴⁾Substitution for stone B
⁽⁵⁾Calculating the time difference between two stones
⁽⁶⁾Final answer 1,34 (s)

3.5.1

Tips:

The upward direction is taken as positive.
Stone A has a positive velocity when the curve's gradient is positive or when its position increases between 0 and e. However, the only time value between these points is at d.

1 mark

3.5.2

Tip: the maximum height reached must be at a time of e and a height value of a.

1 mark

3.5.3

Tips:

Stone B was dropped after Stone A was thrown upwards.
Stone B was thrown upwards from a height of b at a time of f.

1 mark

3.5.4

Tip: the stones pass each other where the position is the same for both of the position graphs. This happens at a time of g but a height of b.

1 mark

Option 2

Upwards as positive:

Displacement from roof to meeting point: -40 + 29,74 = -10,26m.

Displacement of stone A from max height to meeting point: -15,36 m.

Stone A

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

$0=10+(-9,8)\Delta t$

$\Delta\mathrm{t}=1,02\mathrm{~s}$

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

$-15,36=0+1/2(-9,8)\Delta t^2$ ⁽²⁾

$\Delta\mathrm{t}=1,77\mathrm{~s}$

$\Delta\mathrm{t}_{\text{total }}=1,77+1,02=2,79\mathrm{s}$

Stone B

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

$-10,26=0+1/2(-9,8)\Delta t^2$ ^{(3 + 4)}

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

$x=2,79-1,45=1,34(\mathrm{s})$ ^{(5 + 6)}

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

=0^2+2(-9,8)(-10,26)

$v_{f}=-14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$-14,18=0+(-9,8)\Delta t$ ^{(3 + 4)}

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1.34(\mathrm{~s})$ ⁽⁶⁾

Downwards as positive:

Displacement from roof to meeting point: 40 - 29,74 = 10,26m.

Displacement of stone A from max height to meeting point: 15,36 m.

Stone A

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

$0=-10+(9,8)\Delta t$

$\Delta\mathrm{t}=1,02\mathrm{~s}$

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

$15,36=0+1/2(9,8)\Delta t^2$ ⁽²⁾

$\Delta\mathrm{t}=1,77\mathrm{~s}$

$\Delta\mathrm{t}_{\mathrm{total}}=1,77+1,02=2,79\mathrm{s}$

Stone B

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

$10,26=0+1/2(9,8)\Delta t^2$ ⁽³⁾

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(4 + 5)}

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

=0^2+2(9,8)(10,26)

$\mathrm{v}_{\mathrm{f}}=14.18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$14.18=0+(9.8)\Delta t$ ⁽³⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ^{(4 + 5)}

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Option 3

Upwards as positive:

Displacement from roof to meeting point: -40 + 29,74 = -10,26m.

Stone A

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

$0=10+(-9,8)\Delta t$

$\Delta\mathrm{t}=1.02\mathrm{~s}$

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

$-10,26=-10+1/2(-9,8)\Delta t^2$ ^{(2 + 3)}

$\Delta\mathrm{t}=0,75\mathrm{~s}$

$\Delta\mathrm{t}_{\mathrm{total}}=1,02+1,02+0,75=2,79\mathrm{s}$

Stone B

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

$-10,26=0+1/2(-9,8)\Delta t^2$ ⁽⁴⁾

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

=0^2+2(-9,8)(-10,26)

$v_{f}=-14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$-14.18=0+(-9.8)\Delta t$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Downwards as positive:

Displacement from roof to meeting point: 40 - 29,74 = 10,26m.

Stone A

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

$0=-10+(9,8)\Delta t$

$\Delta\mathrm{t}=1,02\mathrm{~s}$

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

$10,26=10+1/2(9,8)\Delta t^2$ ^{(2 + 3)}

$\Delta\mathrm{t}=0,75\mathrm{~s}$

$\Delta t_{\text{total }}=1,02+1,02+0,75=2,79\mathrm{s}$

Stone B

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

$10,26=0+1/2(9,8)\Delta t^2$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

=0^2+2(9,8)(10,26)

$v_{f}=14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$14,18=0+(9,8)\Delta t$ ⁽⁴⁾

$\Delta t=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Option 4

Upwards as positive:

Displacement from roof to meeting point: -40 + 29,74 = -10,26m.

Stone A

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

$-5,10=0+1/2(-9,8)\Delta t^2$

$\Delta\mathrm{t}=1,02\mathrm{~s}$

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

$-10,26=-10+1/2(-9,8)\Delta t^2$ ^{(2 + 3)}

$\Delta t=0,75s$

$\Delta\mathrm{t}_{\mathrm{total}}=1,02+1,02+0,75=2,79\mathrm{s}$

Stone B

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

$-10,26=0+1/2(-9,8)\Delta t^2$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

=0^2+2(-9,8)(-10,26)

$v_{f}=-14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$-14,18=0+(-9,8)\Delta t$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Downwards as positive:

Displacement from roof to meeting point: 40 - 29,74 = 10,26m.

Stone A

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

$5,10=0+1/2(9,8)\Delta t^2$

$\Delta\mathrm{t}=1,02\mathrm{~s}$

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

$10,26=10+1/2(9,8)\Delta t^2$ ^{(2 + 3)}

$\Delta t=0,75s$

$\Delta\mathrm{t}_{\mathrm{total}}=1,02+1,02+0,75=2,79\mathrm{s}$

Stone B

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

$10.26=0+1/2(9.8)\Delta t^2$ ⁽⁴⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

=0^2+2(9,8)(10,26)

$v_{f}=14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$14,18=0+(9,8)\Delta t$ ⁽⁴⁾

$\Delta t=1,45\mathrm{~s}$

x=2,79-1,45 ⁽⁵⁾

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Option 5

Upwards as positive:

Displacement from roof to meeting point: -40 + 29,74 = -10,26m.

Displacement of stone A from max height to meeting point: -15,36 m.

Stone A

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

$v_{f}^2=(0)^2+(2)(-9,8)(-15,36)$

$v_{f}=-17,35\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$-17,35=0+(-9,8)\Delta t$ ⁽²⁾

$\Delta\mathrm{t}=1,77\mathrm{~s}$

$\Delta\mathrm{t}_{\mathrm{total}}=1,02+1,77=2,79(\mathrm{s})$

Stone B

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

$-10,26=0+1/2(-9,8)\Delta t^2$ ^{(3 + 4)}

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

$x=2,79-1,45=1,34\text{ \lparen s\rparen}$ ^{(5 + 6)}

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

=0^2+2(-9,8)(-10,26)

$v_{f}=-14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$-14,18=0+(-9,8)\Delta t$ ⁽³⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ^{(4 + 5)}

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Downwards as positive:

Displacement from roof to meeting point: 40 - 29,74 = 10,26m.

Displacement of stone A from max height to meeting point: 15,36 m.

Stone A

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

$v_{f}^2=(0)^2+(2)(-9,8)(-15,36)$

$v_{f}=-17,35\mathrm{~m}\cdot\mathrm{s}^{-1}$

$v_{f}=v_{i}+a\Delta t$ ⁽¹⁾

$17.35=0+(9.8)\Delta t$ ⁽²⁾

$\Delta\mathrm{t}=1,77\mathrm{~s}$

$\Delta\mathrm{t}_{\mathrm{total}}=1,02+1,77=2,79(\mathrm{s})$

Stone B

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

$10,26=0+1/2(9,8)\Delta t^2$ ^{(3 + 4)}

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

=0^2+2(9,8)(10,26)

$v_{f}=14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$14,18=0+(9,8)\Delta t$ ⁽³⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ^{(4 + 5)}

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Option 6

Upwards as positive:

Displacement from roof to meeting point: -40 + 29,74 = -10,26m.

Stone A

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

$v_{f}^2=(-10)^2+(2)(-9,8)(-10,26)$

$v_{f}=-17,35\mathrm{~m}\cdot\mathrm{s}^{-1}$

$v_{f}=v_{i}+a\Delta t$ ⁽¹⁾

$-17,35=-10+(-9,8)\Delta t$ ⁽²⁾

$\Delta\mathrm{t}=0,75\mathrm{~s}$

$\text{ Ball A: }\Delta\mathrm{t}=1,02+1,02+0,75=2,79(\mathrm{~s})$

Stone B

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

$-10,26=0+1/2(-9,8)\Delta t^2$ ^{(3 + 4)}

$\Delta\mathrm{t}=1,45\mathrm{~s}(1,447\mathrm{~s})$

x=2,79-1,45=1,34(s) ^{(5 + 6)}

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

=0^2+2(-9,8)(-10,26)

$v_{f}=-14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$-14,18=0+(-9,8)\Delta t$ ⁽³⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ^{(4 + 5)}

$=1,34(\mathrm{~s})$ ⁽⁶⁾

Downwards as positive:

Displacement from roof to meeting point: 40 - 29,74 = 10,26m.

Stone A

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

$v_{f}^2=(10)^2+(2)(9,8)(10,26)$

$\mathrm{v}_{\mathrm{f}}=17,35\mathrm{~m}\cdot\mathrm{s}^{-1}$

$v_{f}=v_{i}+a\Delta t$ ⁽¹⁾

$17,35=10+(9,8)\Delta t$ ⁽²⁾

$\Delta\mathrm{t}=0,75\mathrm{~s}$

$\text{ Ball A: }\Delta\mathrm{t}=1,02+1,02+0,75=2,79(\mathrm{~s})$

Stone B

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

$10,26=0+1/2(9,8)\Delta t^2$ ^{(3 + 4)}

$\Delta t=1,45\mathrm{~s}(1,447\mathrm{~s})$

$x=2,79-1,45=1,34\text{ \lparen s\rparen}$ ^{(5 + 6)}

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

=0^2+2(9,8)(10,26)

$v_{f}=14,18\mathrm{~m}\cdot\mathrm{s}^{-1}$

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

$14,18=0+(9,8)\Delta t$ ⁽³⁾

$\Delta\mathrm{t}=1,45\mathrm{~s}$

x=2,79-1,45 ^{(4 + 5)}

$=1,34(\mathrm{~s})$ ⁽⁶⁾