| Date | May 2015 | Marks available | 1 | Reference code | 15M.2.SL.TZ1.6 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Calculate, Estimate, Show that, Sketch, and State | Question number | 6 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.
Part 1 Kinematics and gravitation
A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.
Part 2 Radioactivity
Two isotopes of calcium are calcium-40 \(\left( {\frac{{40}}{{20}}{\rm{Ca}}} \right)\) and calcium-47 \(\left( {\frac{{47}}{{20}}{\rm{Ca}}} \right)\). Calcium-40 is stable and calcium-47 is radioactive with a half-life of 4.5 days.
State the significance of the negative values of s.
Use the graph to
(i) estimate the velocity of the ball at t \( = \) 0.80 s.
(ii) calculate a value for the acceleration of free fall close to the surface of the Moon.
The following data are available.
Mass of the ball = 0.20 kg
Mean radius of the Moon = 1.74 \( \times \) 106 m
Mean orbital radius of the Moon about the centre of Earth = 3.84 \( \times \) 108 m
Mass of Earth = 5.97 \( \times \) 1024 kg
Show that Earth has no significant effect on the acceleration of the ball.
Calculate the speed of an identical ball when it falls 3.0 m from rest close to the surface of Earth. Ignore air resistance.
Sketch, on the graph, the variation with time t of the displacement s from the point of release of the ball when the ball is dropped close to the surface of Earth. (For this sketch take the direction towards the Earth as being negative.)
Explain, in terms of the number of nucleons and the forces between them, why calcium-40 is stable and calcium-47 is radioactive.
Calculate the percentage of a sample of calcium-47 that decays in 27 days.
The nuclear equation for the decay of calcium-47 into scandium-47 \(\left( {{}_{21}^{47}{\rm{Sc}}} \right)\) is given by
\[{}_{20}^{47}{\rm{Ca}} \to {}_{21}^{47}{\rm{Sc + }}{}_{ - 1}^0{\rm{e + X}}\]
(i) Identify X.
(ii) The following data are available.
Mass of calcium-47 nucleus = 46.95455 u
Mass of scandium-47 nucleus = 46.95241 u
Using the data, determine the maximum kinetic energy, in MeV, of the products in the decay of calcium-47.
(iii) State why the kinetic energy will be less than your value in (h)(ii).
Markscheme
upwards (or away from the Moon) is taken as positive / downwards (or towards the Moon) is taken as negative / towards the Earth is positive;
(i) tangent drawn to curve at 0.80s;
correct calculation of gradient of tangent drawn;
−1.3 ±0.1m s–1 or 1.3 ±0.1m s–1 downwards;
or
correct coordinates used from the graph; substitution into a correct equation;
−1.3 ±0.1m s–1 or 1.3 ±0.1m s–1 downwards;
(ii) any correct method used;
correct reading from graph;
1.6 to 1.7 m s–2;
values for masses, distance and correct G substituted into Newton’s law;
see subtraction (ie r value = 3.84 × 108 − 1.74 × 106 =3.82 × 108 m);
F=5.4 to 5.5 × 10–4 N / a=2.7 × 10–3 m s–2;
comment that it’s insignificant compared with (0.2 × 1.63 =) 0.32 to 0.33 N / 1.63 m s–2;
7.7 m s\(^{ - 1}\);
curve permanently below Moon curve;
smooth parabola; (judge by eye)
line passing through s = −3.00 m, t = 0.78 s or s = −3.50 m, t = 0.84 s (±1mm);
Ca-40 has 20 protons and 20 neutrons, Ca-47 has 20 protons and 27 neutrons / Ca-47 has 7 additional neutrons;
mention of strong/nuclear and coulomb/electrostatic/electromagnetic forces;
excess neutrons/too high a neutron-to-proton ration leads to the coulomb/electrostatic’ electromagnetic force being greater than the strong/nuclear force (so the nucleus is unstable);
Award [1 max] for an answer stating that Ca-47 has more neutrons so is bigger and less stable.
six half-lives occurred;
\(\left( {{{\left( {\frac{1}{2}} \right)}^6} = } \right)1.6\% \) remaining;
98.4 / 98% decayed;
(i)(electron) anti-neutrino / \(\overline v \) ;
(ii) 46.95455 u − (46.95241 u + 0.00055 u) = 0.00159 u;
1.48 MeV;
(iii) does not account for energy of (anti) neutrino/gamma ray photons;
