IB Diploma Physics

1. Signed graph areas are +8, +4, −4 and −8 m, so their sum is zero displacement.
2. Distance adds the magnitudes, giving 24 m. The final position is the initial 2 m plus zero displacement.
3. Average speed = 24/8 = 3 m s⁻¹, whereas average velocity = 0/8 = 0. Distance and speed are scalars; displacement and velocity require direction.
4. The gradient is (−4−4)/(6−2) = −2 m s⁻² throughout the interval.
5. Before 4 s acceleration opposes velocity, so speed decreases. After 4 s acceleration and velocity both point west, so speed increases.
6. A negative acceleration does not alone mean slowing. Compare its direction with the instantaneous velocity.

Answer (a) Displacement = 0 m; distance = 24 m; final position = 2 m east; average speed = 3 m s⁻¹; average velocity = 0. (b) Acceleration = −2 m s⁻²; it slows until 4 s and then speeds up westward.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Net heating power is 100−20 = 80 W, since the sample does not receive the full heater input.
2. Melting needs mL = 0.10(3.36×10⁵) = 33600 J; duration is 33600/80 = 420 s.
3. Warming liquid by 20 K needs mcΔT = 0.10(4200)(20) = 8400 J; duration is 8400/80 = 105 s. Both match the supplied curve.
4. During melting, transferred energy changes intermolecular potential energy rather than increasing mean kinetic energy, so temperature need not rise.
5. The internal energy still increases through the phase change; temperature is not a measure of the total internal energy.
6. Ignoring losses gives 33600/100 = 336 s, 84 s too short. Insulation and a lid can reduce unwanted transfers; repeats alone do not remove systematic heat loss.

Answer (a) Melting takes 420 s; subsequent warming takes 105 s. (b) Temperature stays constant while internal energy increases through a change in particle arrangement; ignoring losses underestimates the required time.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. The graph gives F = −20x, so a = F/m = −40x: acceleration is proportional to displacement and directed toward equilibrium.
2. Comparing a = −ω²x gives ω = √40 rad s⁻¹ and T = 2π/ω ≈0.993 s.
3. The amplitude is 0.10 m. From energy conservation or vmax = ωA, the maximum speed is 0.632 m s⁻¹ at equilibrium.
4. At x = 0.060 m, acceleration is −40(0.060) = −2.4 m s⁻², toward equilibrium.
5. Energy gives v² = ω²(A²−x²) = 40(0.0100−0.0036) = 0.256; speed is √0.256 ≈0.506 m s⁻¹, with direction depending on the motion phase.
6. A driving frequency near the natural frequency can produce a large response. Damping transfers mechanical energy to the surroundings and reduces and broadens the resonance peak, useful when large vibrations would cause damage.

Answer (a) ω = √40 ≈6.32 rad s⁻¹; T ≈0.993 s; vmax ≈0.632 m s⁻¹. (b) Speed magnitude ≈0.506 m s⁻¹; acceleration = −2.4 m s⁻².

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. At 2R the field is g₀/4 = 2.5 N kg⁻¹, so the gravitational force is mg = 250 N. The satellite's mass remains 100 kg.
2. Gravity alone supplies mv²/r: v = √(gr) = √(2.5×1.2×10⁷)≈5477 m s⁻¹.
3. Period is circumference/speed, T = 2πr/v≈1.38×10⁴ s. 'Centripetal force' names the inward resultant; it is not an additional interaction alongside gravity.
4. For a circular orbit g∝r⁻², v∝r⁻¹/² and T∝r³/². Doubling r therefore gives the stated ratios.
5. Astronauts and spacecraft accelerate together in free fall. Little support force acts on the astronauts, so their apparent weight can be near zero.
6. The gravitational field is not zero in orbit, and constant speed still requires acceleration because velocity direction changes.

Answer (a) Weight = 250 N; orbital speed ≈5.48 km s⁻¹; period ≈1.38×10⁴ s. (b) Field becomes 1/4, speed 1/√2 and period 2√2 times the initial orbital values.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.