| Date | May 2017 | Marks available | 2 | Reference code | 17M.3.SL.TZ2.6 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Determine | Question number | 6 | Adapted from | N/A |
Question
A cylindrical space probe of mass 8.00 x 102 kg and diameter 12.0 m is at rest in outer space.
Rockets at opposite points on the probe are fired so that the probe rotates about its axis. Each rocket produces a force F = 9.60 x 103 N. The moment of inertia of the probe about its axis is 1.44 x 104 kg\(\,\)m2.
The diagram shows a satellite approaching the rotating probe with negligibly small speed. The satellite is not rotating initially, but after linking to the probe they both rotate together.
The moment of inertia of the satellite about its axis is 4.80 x 103 kg\(\,\)m2. The axes of the probe and of the satellite are the same.
Deduce the linear acceleration of the centre of mass of the probe.
Calculate the resultant torque about the axis of the probe.
The forces act for 2.00 s. Show that the final angular speed of the probe is about 16 rad\(\,\)s–1.
Determine the final angular speed of the probe–satellite system.
Calculate the loss of rotational kinetic energy due to the linking of the probe with the satellite.
Markscheme
zero
[1 mark]
the torque of each force is 9.60 x 103 x 6.0 = 5.76 x 104 «Nm»
so the net torque is 2 x 5.76 x 104 = 1.15 x 105 «Nm»
Allow a one-step solution.
[2 marks]
the angular acceleration is given by \(\frac{{1.15 \times {{10}^5}}}{{1.44 \times {{10}^4}}}\) «= 8.0 s–2»
ω = «\(\alpha \)t = 8.0 x 2.00 =» 16 «s–1»
[2 marks]
1.44 x 104 x 16.0 = (1.44 x 104 + 4.80 x 103) x ω
ω = 12.0 «s–1»
Allow ECF from (b).
[2 marks]
initial KE \(\frac{1}{2} \times 1.44 \times {10^4} \times {16.0^2} = 1.843 \times {10^6}\) «J»
final KE \(\frac{1}{2} \times \left( {1.44 \times {{10}^4} + 4.80 \times {{10}^3}} \right) \times {12.0^2} = 1.382 \times {10^6}\) «J»
loss of KE = 4.6 x 105 «J»
Allow ECF from part (c)(i).
[3 marks]
