Additional Answer Options: 2.5

2023 - May/June - Paper 1 - Question 2

2.1

Option 1:

Downwards as positive:

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

(3,41)^2=(0)^2+(2)a(1,5)

$\mathrm{a}=3,88\mathrm{~m}\cdot\mathrm{s}^{-2}$

Upwards as positive:

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

(-3,41)^2=(0)^2+(2)a(-1,5)

a=-3,88

$a=3.88\mathrm{~m}\cdot\mathrm{s}^{-2}$

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

Teacher tip

3 marks

(3 x 1 mark)

2.2

w.	F_g / F_w / weight / mg / gravitational force / F_{Earth on block} / 73,5N
T.	Tension / F_Tension / F_Rope / F_T / F

Teacher tip

2 marks

(2 x 1 mark)

2.3

When a resultant / net force acts on an object, the object will accelerate in the direction of the force with an acceleration that is directly proportional to the force and inversely proportional to the mass of the object.

The resultant / net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant / net force.

Teacher tip

2 marks

(2 x 1 mark)

2.4

Calculation of tension (Block B):

Downwards as positive:

$\left.\begin{array}{rl}\mathrm{F}_{\mathrm{net}} & =\mathrm{ma}\\ \mathrm{F}_{\mathrm{g}}+\mathrm{T} & =\mathrm{ma}\\ \mathrm{mg}-\mathrm{T} & =\mathrm{ma}\end{array}\right\}Anyone$ ⁽¹⁾

7,5(9,8)-T=7,5(3,88) ^{(2 + 3)}

$\mathrm{T}=44,40\mathrm{~N}$

Upwards as positive:

$\left.\begin{array}{rl}\mathrm{F}_{\mathrm{net}} & =\mathrm{ma}\\ \mathrm{T}-\mathrm{F}_{\mathrm{g}} & =\mathrm{ma}\\ \mathrm{T}-\mathrm{mg} & =\mathrm{ma}\end{array}\right\}Anyone$ ⁽¹⁾

T-7,5(9,8)=7,5(-3,88) ^{(2 + 3)}

$\mathrm{T}=44,40\mathrm{~N}$

Calculation of mass (Block A):

Upwards as positive:

$\mathrm{F}_{\mathrm{net}}=\mathrm{ma}$

$\mathrm{T}-\mathrm{F}_{\mathrm{g}}=\mathrm{ma}$

$\mathrm{T}-\mathrm{mg}=\mathrm{ma}$

44,40-m(9,8)=m(3,88) ⁽⁴⁾

$\mathrm{m}=3,25\mathrm{~kg}$ ⁽⁵⁾

Downwards as positive:

$\mathrm{F}_{\mathrm{net}}=\mathrm{ma}$

$\mathrm{F}_{\mathrm{g}}-\mathrm{T}=\mathrm{ma}$

$\mathrm{mg}-\mathrm{T}=\mathrm{ma}$

$\mathrm{m}(9,8)-44,40=\mathrm{m}(-3,88)$ ⁽⁴⁾

$\mathrm{m}=3,25\mathrm{~kg}$ ⁽⁵⁾

Teacher tip

5 marks

(5 x 1 mark)

Marking criteria:

⁽¹⁾Any correct formula.
^{(2 + 3)}Correct substitution to calculate tension.
⁽⁴⁾Correct substitution to calculate mass of block A.
⁽⁵⁾Correct final answer: 3,25 kg.

2.5

Option 1:

Upwards as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$ ⁽¹⁾

$\left(0^2\right)=(3,41)^2+(2)(-9,8)\Delta y$ ^{(2 + 3)}

$\Delta y=0,59\mathrm{~m}$

$\text{ Maximum height }=0,59+1,5$ ⁽⁴⁾

$=2,09\mathrm{~m}$ ⁽⁵⁾

Downwards as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$ ⁽¹⁾

$\left(0^2\right)=(-3.41)^2+(2)(9.8)\Delta y$ ^{(2 + 3)}

$\Delta y=-0,59$

$\Delta y=0,59m$

$\text{ Maximum height }=0,59+1,5$ ⁽⁴⁾

$=2,09\mathrm{~m}$ ⁽⁵⁾

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

Teacher tip

5 marks

(5 x 1 mark)

Marking criteria:

⁽¹⁾Any correct formula.
⁽²⁾Correct substitution of v_i and v_f.
⁽³⁾Correct substitution of 9,8 m·s^-2.
⁽⁴⁾Adding 1,5 m to calculated Δy.
⁽⁵⁾Correct final answer: 2,09 m.

Option 2

Upwards as positive:

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

$0=3,41+(-9,8)\Delta t$

$\Delta\mathrm{t}=0,35\mathrm{~s}$

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

=(3,41)(0,35)+y/2(-9,8)(0,35)^2

$=0,59\mathrm{~m}$

$\text{ Maximum height }=0,59+1,5$

$=2,09\mathrm{~m}$

Downwards as positive:

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

$0=-3,41+(9,8)\Delta t$

$\Delta\mathrm{t}=0,35\mathrm{~s}$

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

=(-3,41)(0,35)+1/2(9,8)(0,35)^2

=-0,59

$\Delta y=0,59\mathrm{~m}$

$\text{ Maximum height }=0,59+1,5$

$=2,09\mathrm{~m}$

Option 3

Upwards as positive:

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

$0=3,41+(-9,8)\Delta t$

$\Delta\mathrm{t}=0,35\mathrm{~s}$

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

$\Delta y=\left(\frac{3,41+0}{2}\right)(0,35)$

$=0,59\mathrm{~m}$

$\text{ Maximum height }=0,59+1,5$

$=2,09\mathrm{~m}$

Downwards as positive:

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

$0=-3,41+(9,8)\Delta t$

$\Delta\mathrm{t}=0,35\mathrm{~s}$

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

$\Delta y=\left(\frac{-3,41+0}{2}\right)(0,35)$

$=-0,59\mathrm{~m}$

$\text{ Maximum heiaht }=0.59+1.5$

$=2,09\mathrm{~m}$

Option 4

$\left.\begin{array}{rl}\left(E_{\text{mech }}\right)_{\text{top }} & =\left(E_{\text{mech }}\right)_{\text{bottom }}\\ \left(E_{p}+E_{k}\right)_{\text{top }} & =\left(E_{p}+E_{k}\right)_{\text{bottom }}\\ \left(mgh+1/2mv_{i}^2\right)_{\text{top }} & =\left(\mathrm{mgh}+1/2mv_{f}^2\right)_{\text{bottom }}\end{array}\right\}Anyone$

(9.8)(h)+(0)=(0)+1/2(3,41)^2

$\mathrm{h}=0,59\mathrm{~m}$

$\text{ Maximum height }=0,59+1,5$

$=2,09\mathrm{~m}$

Option 5

$\left.\begin{array}{rl}W_{nc} & =\Delta K+\Delta U\\ W_{nc} & =\Delta K+mg\left(h_{f}-h_{i}\right)\\ 0 & =1/2mv_{f}^2-1/2mv_{i}^2+mgh_{f}-mgh_{i}\end{array}\right\}Anyone$