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Question 22N.2.HL.TZ0.8

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Date November 2022 Marks available [Maximum mark: 6] Reference code 22N.2.HL.TZ0.8
Level HL Paper 2 Time zone TZ0
Command term Determine, Outline, Show that Question number 8 Adapted from N/A
8.
[Maximum mark: 6]
22N.2.HL.TZ0.8
(a)

The diagram shows field lines for an electrostatic field. X and Y are two points on the same field line.

Outline which of the two points has the larger electric potential.

[2]

Markscheme

potential greater at Y ✓

 

«from E=-ΔVeΔr» the potential increases in the direction opposite to field strength «so from X to Y»

OR

opposite to the direction of the field lines, «so from X to Y»

OR

«from W=qVe» work done to move a positive charge from X to Y is positive «so the potential increases from X to Y» ✓

Examiners report

A significant majority guessed at X, probably because the field lines are closer together. Those that identified Y were generally successful in their explanation.

A satellite is launched from the surface of Earth into a circular orbit.

The following data are given.

                                        Mass of the satellite = 8.0 × 102 kg

Height of the orbit above the surface of Earth = 5.0 × 105 m

                                                 Mass of Earth = 6.0 × 1024 kg

                                              Radius of Earth = 6.4 × 106 m

(b.i)

Show that the kinetic energy of the satellite in orbit is about 2 × 1010 J.

[2]

Markscheme

orbital radius =6.4×106+5.0×105 «=6.9×106 m» ✓

KE=12×8.0×102×6.67×10-11×6.0×10246.9×106  OR  2.3×1010 «J» ✓

 

Award [1] max for answers ignoring orbital height (KE = 2.5 × 1010 J).

Examiners report

This question was well done, with only a few missing the height of the satellite.

(b.ii)

Determine the minimum energy required to launch the satellite. Ignore the original kinetic energy of the satellite due to Earth’s rotation.

[2]

Markscheme

change in PE =6.67×10-11×6.0×1024×8.0×10216.4×106-16.9×106= «3.6×109 J» ✓

energy needed = KE + ΔPE = 2.7×1010 «J» ✓

 

Allow ECF from 8(b)(i).

Examiners report

Generally, this question was not well done. Most carried out a calculation based on the formula for escape velocity. An opportunity to remind candidates of reading back the stem for the sub-question when answering a second or any subsequent part of it.