Look inside the Edexcel A-Level Physics guide (9PH0)

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Edexcel · A-Level · 9PH0
Physics
Active Recall Guide
1,616 questions
PhysicsContents
Contents
13 topics, 158 subtopics
  1. Working as a Physicist82
  2. Mechanics180
  3. Electric Circuits148
  4. Materials101
  5. Waves and Particle Nature of Light288
  6. Further Mechanics98
  7. Electric and Magnetic Fields187
  8. Nuclear and Particle Physics120
  9. Thermodynamics96
  10. Space72
  11. Nuclear Radiation82
  12. Gravitational Fields62
  13. and 1 more
ii
MechanicsQuestions
Mechanics
The equations of uniformly accelerated motion
  1. Under what conditions may the equations of uniformly accelerated motion be used?
  2. State the equation of motion linking displacement, initial velocity, final velocity and time:
  3. State the equation of motion linking final velocity, initial velocity, acceleration and time:
  4. State the equation of motion giving displacement in terms of initial velocity, acceleration and time:
  5. State the equation of motion that does not contain time:
  6. Which equation of motion is most useful when time is neither given nor asked for?
  7. Why does s = (u + v)t/2 follow from the definition of average velocity when the acceleration is uniform?
  8. What sign convention should be used when analysing an object thrown vertically upwards?
  9. A car accelerates uniformly from rest at 2.5 m s⁻² for 8.0 s. What are its displacement and its final velocity?
12
MechanicsAnswers
Answers
The equations of uniformly accelerated motion
  1. They apply only when the acceleration is constant in both magnitude and direction and the motion is along a straight line.
  2. The equation is s = (u + v)t/2.
  3. The equation is v = u + at.
  4. The equation is s = ut + ½at².
  5. The equation is v² = u² + 2as.
  6. v² = u² + 2as, because it links the two velocities, the acceleration and the displacement without involving time.
  7. For constant acceleration the average velocity is the mean of the initial and final velocities, and displacement is average velocity multiplied by time.
  8. One direction is chosen as positive and used throughout, so if upwards is taken as positive then the acceleration due to gravity is entered as negative.
  9. The displacement is s = ½ × 2.5 × 8.0² = 80 m and the final velocity is v = 2.5 × 8.0 = 20 m s⁻¹.
13
Further MechanicsQuestions
Further Mechanics
Conservation of momentum in two dimensions
  1. A tennis ball of mass 0.058 kg leaves a racket at 45 m s⁻¹ from rest after a contact time of 5.0 ms. What average force acted on it?
  2. State the principle of conservation of linear momentum:
  3. How does conservation of momentum follow from Newton's third law?
  4. How is conservation of momentum applied to a collision that is not along a single line?
  5. Why can momenta in two dimensions not simply be added as magnitudes?
  6. A stationary object explodes into exactly two fragments. What can be said about their momenta?
  7. A snooker ball moving due east strikes an identical stationary ball and moves off at 30° north of east. What must be true of the second ball's motion?
  8. A 2.0 kg trolley moving north at 3.0 m s⁻¹ collides with a 1.0 kg trolley moving east at 6.0 m s⁻¹ and the two stick together. What is their speed afterwards?
34
Further MechanicsAnswers
Answers
Conservation of momentum in two dimensions
  1. The change of momentum is 0.058 × 45 = 2.61 kg m s⁻¹, so the average force is 2.61 ÷ (5.0 × 10⁻³) = 522 N, or 520 N to 2 significant figures.
  2. The total linear momentum of a system stays constant provided no resultant external force acts on it.
  3. Two interacting bodies exert equal and opposite forces on each other for the same length of time, so they receive equal and opposite impulses and their momentum changes cancel exactly.
  4. Momentum is conserved independently in two perpendicular directions, so the momenta are resolved into components and the totals before and after are equated in each direction separately.
  5. Momentum is a vector, so only components along a common axis may be added arithmetically; otherwise vector addition is required.
  6. The two fragments must carry equal magnitudes of momentum in exactly opposite directions, so that the total remains zero.
  7. The second ball must gain a southward momentum component equal in size to the first ball's northward component, and the two eastward components must add up to the original momentum.
  8. The northward momentum is 6.0 kg m s⁻¹ and the eastward momentum is 6.0 kg m s⁻¹, giving a resultant of 8.5 kg m s⁻¹, so dividing by the combined mass of 3.0 kg gives 2.8 m s⁻¹ at 45° east of north.
35
SpaceQuestions
Space
Luminosity, intensity and the inverse square law
  1. What is meant by the luminosity of a star?
  2. What is meant by the radiation intensity received from a star?
  3. State the equation relating the intensity received from a star to its luminosity and distance:
  4. Why does the distance appear as 4πd² in this relationship?
  5. What happens to the intensity received from a star if its distance from us were three times greater?
  6. A star of luminosity 3.9 × 10²⁶ W is 1.5 × 10¹¹ m away. What intensity is received?
  7. If the luminosity of a star and the intensity received from it are both known, how is its distance found?
  8. What two assumptions are made when using I = L/(4πd²) for a real star?
56
SpaceAnswers
Answers
Luminosity, intensity and the inverse square law
  1. It is the total radiant power the star emits in all directions, measured in watts.
  2. It is the radiant power arriving per unit area at the detector, measured in W m⁻².
  3. The equation is I = L/(4πd²), where d is the distance from the source.
  4. The emitted power spreads out uniformly over the surface of a sphere of radius d, and that sphere has area 4πd².
  5. The intensity would fall to one ninth of its value, because intensity is inversely proportional to the square of the distance.
  6. I = 3.9 × 10²⁶/(4π × (1.5 × 10¹¹)²) = 1.4 × 10³ W m⁻².
  7. Rearranging gives d = √(L/4πI).
  8. That the star radiates uniformly in all directions, and that no radiation is absorbed or scattered by dust and gas on the way to the observer.
57
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