Additional Answer Options: 4.2.2

2023 - May/June - Paper 1 - Question 4

4.1

 

In an isolated system the total (linear) momentum is conserved / remains
constant.   

Teacher tip

2 marks

(2 x 1 mark)

4.2.1

 

Option 1:

Right as positive:

\left.\begin{array}{rl}\Sigma p_i & =\sum p_f \\ m_Av_{Ai}+m_BV_{Bi} & =m_Av_{Af}+m_Bv_{Bf} \\ m_Av_{Ai}+m_Bv_{Bi} & =\left(m_A+m_B\right)v_f\end{array}\right\}Anyone  

(7,2)(0,4)+(0)=(7,2+5,3) v_f  

v_{\mathrm{f}}=0,23 \mathrm{~m} \cdot \mathrm{s}^{-1}  

OR

Left as positive:

\left.\begin{array}{rl}\sum p_i & =\sum p_f \\ m_Av_{Ai}+m_Bv_{Bi} & =m_Av_{Af}+m_Bv_{Bf} \\ m_Av_{Ai}+m_Bv_{Bi} & =\left(m_A+m_B\right)v_f\end{array}\right\}Anyone  

(7,2)(-0,4)+(0)=(7,2+5,3) v_f  

v_f=-0,23 

\therefore \mathrm{v}_{\mathrm{f}}=0,23 \mathrm{~m} \cdot \mathrm{s}^{-1}  

OR

Option 2:

Right as positive:

\left.\begin{array}{rl}\Delta\mathrm{p}_{\text{trolley }} & =-\Delta\mathrm{p}_{\text{trolley }}\mathrm{B} \\ \mathrm{m}_{\mathrm{A}}\left(\mathrm{v}_{\mathrm{Af}}-\mathrm{v}_{\mathrm{Ai}}\right) & =-\mathrm{m}_{\mathrm{B}}\left(\mathrm{v}_{\mathrm{Bf}}-\mathrm{v}_{\mathrm{Bi}}\right)\end{array}\right\}Anyone  

(7,2)\left(v_f-0,4\right)=-(5,3)\left(v_f-0\right)  

v_f=0,23 \mathrm{~m} \cdot \mathrm{s}^{-1}  

OR

Left as positive:

\left.\begin{array}{rl}\Delta\mathrm{p}_{\text{trolley }} & =-\Delta\mathrm{p}_{\text{trolley }\mathrm{B}} \\ \mathrm{m}_{\mathrm{A}}\left(\mathrm{v}_{\mathrm{Af}}-\mathrm{v}_{\mathrm{Ai}}\right) & =-\mathrm{m}_{\mathrm{B}}\left(\mathrm{v}_{\mathrm{Bf}}-\mathrm{v}_{\mathrm{Bi}}\right)\end{array}\right\}Anyone  

(7,2)\left(v_f+0,4\right)=-(5,3)\left(v_f-0\right)  

v_f=-0,23 

\therefore \mathrm{v}_{\mathrm{f}}=0,23 \mathrm{~m} \cdot \mathrm{s}^{-1}  

Teacher tip

3 marks

(3 x 1 mark)

4.2.2

 

Option 1:

Right as positive:

\left.\begin{array}{l}F_{net}\Delta t=\Delta p \\ F_{net}\Delta t=m\left(v_f-v_i\right)\end{array}\right\}Anyone  

F_{\text {net }}(0.02)=7.2(0.23-0.4)  

F_{\text {net }}=-61,2 

\therefore \mathrm{F}_{\text {net }}=61,2 \mathrm{~N}  

(60,95\mathrm{N}\text{ to }61,2\mathrm{N}) 

OR

Left as positive:

\left.\begin{array}{l}F_{net}\Delta t=\Delta p \\ F_{net}\Delta t=m\left(v_f-v_i\right)\end{array}\right\}Anyone  

F_{\text {net }}(0,02)=7,2(-0,23+0,4)  

F_{n e t}=61,2 \mathrm{~N}  

(60,95\mathrm{N}\text{ to }61,2\mathrm{N}) 

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

Teacher tip

3 marks

(3 x 1 mark)

Or

Option 2

Right as positive:

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

F_{\text{net }}(0,02)=5,3(0,23-0) (2)

\mathrm{F}_{\mathrm{net}}=60,95\mathrm{~N} (3)

Or

Left as positive:

\left.\begin{array}{l}\mathrm{F}_{\mathrm{net}}\Delta\mathrm{t}=\Delta\mathrm{p}\\ \mathrm{F}_{\mathrm{net}}\Delta\mathrm{t}=\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}\right)\end{array}\right\}Anyone (1)

F_{\text{net }}(0,02)=5,3(-0,23-0) (2)

F_{\text{net }}=-60,95

\therefore F_{\text{net }}=60,95\mathrm{~N} (3)

Or

Option 3

Right as positive:

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

0,23=0,4+a(0,02)

a=-8,5\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{F}_{\text{net }}=\mathrm{ma} (1)

=(7,2)(-8,5) (2)

=-61,20

F_{\text{net }}=61,20\mathrm{~N} (3)

Or

Left as positive:

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

-0,23=-0,4+a(0,02)

a=8,5\mathrm{~m}\cdot\mathrm{s}^{-2}

F_{\text{net }}=ma (1)

=(7,2)(8,5) (2)

=61,20\mathrm{~N}

\mathrm{F}_{\text{net }}=61,20\mathrm{~N} (3)

Or

Option 4

Right as positive:

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

0,23=0+a(0,02)

\mathrm{a}=11,5\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{F}_{\text{net }}=\mathrm{ma} (1)

=(5,3)(11,5) (2)

=60,95\mathrm{~N}

F_{net}=60,95\mathrm{~N} (3)

Or

Left as positive:

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

-0,23=0+a(0,02)

\mathrm{a}=-11,5\mathrm{~m}\cdot\mathrm{s}^{-2}

\mathrm{F}_{\text{net }}=\mathrm{ma} (1)

=(5,3)(-11,5) (2)

=-60,95\mathrm{~N}

F_{\text{net }}=60,95\mathrm{~N} (3)

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