IBDP Physics 2.4 – Momentum and impulse: IB style Question Bank SL Paper 2

a. only non-renewable is depleted/cannot re-generate whereas renewable can / consumption rate of non-renewables is greater than formation rate and consumption rate of renewables is less than formation rate;

Do not allow “cannot be used again”.

(i) volume released = \((22 \times {10^6} \times 6 = ){\text{ }}1.32 \times {10^8}{\text{ (}}{{\text{m}}^3}{\text{)}}\);

volume per second \(\frac{{1.32 \times {{10}^8}}}{{6 \times 3600}}{\text{ }}( = 6111{\text{ }}{{\text{m}}^3})\);

(ii) use of average depth for calculation (3 m);

gpe lost \(6100 \times 1000 \times 9.81 \times 3\);

0.18 (GW);

Accept g = 10 m\(\,\)s^–2.

Award [1 max] if 6 m is used and an “average” is used at end of solution without mention of average depth.

(iii) converts/states output with units; } (allow values quoted from question without unit)

converts/states input with units; } (allow values quoted from question without unit)

calculates efficiency from \(\frac{{{\text{output}}}}{{{\text{input}}}}\) as 0.31;

Award [3] for bald correct answer.

eg:

power output \(\frac{{5.4 \times {{10}^8}}}{{365 \times 24 \times 3600}}{\text{ }}\left( { = 17{\text{ kW}}\,{\text{h}}\,{{\text{s}}^{ – 1}}} \right)\);

\( = 17 \times 3600000 = 6.16 \times {10^7}{\text{ (W)}}\);

efficiency = \(\left( {\frac{{6.16 \times {{10}^7}}}{{2.0 \times {{10}^8}}} = } \right){\text{ }}31\% \)\(\,\,\,\)or\(\,\,\,\)0.31;

or

0.2 GW is \(1.752 \times {10^9}{\text{ (kW}}\,{\text{h}}\,{\text{yea}}{{\text{r}}^{ – 1}}{\text{)}}\);

\(\frac{{5.4 \times {{10}^8}}}{{1.752 \times {{10}^9}}}\);

efficiency \( = 0.31\);

(iv) cloud cover / weather conditions;

latitude;

time of year / season;

nature/colour of surface;

(i) (total) momentum unchanged before and after collision / momentum of a system is constant; } (allow symbols if explained)

no external forces / isolated system / closed system;

Do not accept “conserved”.

(ii) final momentum of neutron \( = {\text{neutron mass}} \times 9.8 \times {10^6} – 1{\text{u}} \times 12 \times 1.5 \times {10^6}\); } (allow any appropriate and consistent mass unit)

final speed of neutron \( = 8.0\)\(\,\,\,\)or\(\,\,\,\)\(8.2 \times {10^6}{\text{ (m}}\,{{\text{s}}^{ – 1}}{\text{)}}\);

\(\left( { \approx {\text{8.0}} \times {\text{1}}{{\text{0}}^6}{\text{ (m}}\,{{\text{s}}^{ – 1}}{\text{)}}} \right)\)

Allow use of 1 u for both masses giving an answer of 8.2 \( \times \) 10⁶ (m\(\,\)s^–1).

(iii) initial energy of neutron \( = 8.04 \times {10^{ – 14}}{\text{ (J)}}\) and final energy of neutron \( = 5.36 \times {10^{ – 14}}{\text{ (J)}}\); } (both needed)

fractional change in energy \( = \left( {\frac{{(8.04 – 5.36)}}{{8.04}} = } \right){\text{ }}0.33\);

or

fractional change \( = \left( {\frac{{\frac{1}{2}mv_{\text{i}}^2 – \frac{1}{2}mu_{\text{f}}^2}}{{\frac{1}{2}mv_{\text{i}}^2}}} \right)\); }(allow any algebra that shows a subtraction of initial term from final term divided by initial value)

\(\left( { = \frac{{{{(9.8 \times {{10}^6})}^2} – {{(8.0 \times {{10}^6})}^2}}}{{{{(9.8 \times {{10}^6})}^2}}}} \right)\) (allow omission of 10⁶)

\( = 0.33\); (allow 0.30 if 8.2 used)

Do not allow ECF if there is no subtraction of energies in first marking point.

(iv) \({(0.33)^n} = {10^{ – 6}}\);

\(n = 13\); (allow n = 12 if 0.3 is used)

(v) neutrons produced in fission have large energies;

greatest probability of (further) fission/absorption (when incident neutrons have thermal energy or low energy);

Do not accept “reaction” for “fission reaction”.