IB DP Physics Mock Exam HL Paper 2 Set 1 - 2025 Syllabus
IB DP Physics Mock Exam HL Paper 1B Set 1
Prepare for the IB DP Physics Mock Exam HL Paper 2 Set 1 with our comprehensive mock exam set 1. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam
Question
A glider is an aircraft with no engine. To be launched, a glider is uniformly accelerated from rest by a cable pulled by a motor that exerts a horizontal force on the glider throughout the launch.
a. The glider reaches its launch speed of 27.0 m s–1 after accelerating for 11.0 s. Assume that the glider moves horizontally until it leaves the ground. Calculate the total distance travelled by the glider before it leaves the ground. [2]
Draw the forces acting on the glider to complete the free-body diagram. The dotted lines show the horizontal and vertical directions. [2]
Answer/Explanation
Markscheme
a.correct use of kinematic equation/equations
148.5 or 149 or 150 «m»
Substitution(s) must be correct.
a = \(\frac{{27}}{{11}}\) or 2.45 «m s–2»
F – 160 = 492 × 2.45
1370 «N»
Could be seen in part (a).
Award [0] for solution that uses a = 9.81 m s–2
ALTERNATIVE 1
«work done to launch glider» = 1370 x 149 «= 204 kJ»
«work done by motor» \( = \frac{{204 \times 100}}{{23}}\)
«power input to motor» \( = \frac{{204 \times 100}}{{23}} \times \frac{1}{{11}} = 80\) or 80.4 or 81 k«W»
ALTERNATIVE 2
use of average speed 13.5 m s–1
«useful power output» = force x average speed «= 1370 x 13.5»
power input = «\(1370 \times 13.5 \times \frac{{100}}{{23}} = \)» 80 or 80.4 or 81 k«W»
ALTERNATIVE 3
work required from motor = KE + work done against friction «\( = 0.5 \times 492 \times {27^2} + \left( {160 \times 148.5} \right)\)» = 204 «kJ»
«energy input» \( = \frac{{{\text{work required from motor}} \times 100}}{{23}}\)
power input \( = \frac{{883000}}{{11}} = 80.3\) k«W»
Award [2 max] for an answer of 160 k«W».
\(\omega = \) «\(\frac{v}{r} = \)» \(\frac{{27}}{{0.6}} = 45\)
rad s–1
Do not accept Hz.
Award [1 max] if unit is missing.
drag correctly labelled and in correct direction
weight correctly labelled and in correct direction AND no other incorrect force shown Award [1 max] if forces do not touch the dot, but are otherwise OK.
vertical/all forces are in equilibrium/balanced/add to zero
OR
vertical component of lift mentioned
as equal to weight
quotes their answer(s) to 3 significant figures
speed = 12.7 m s–1 or direction = 9.46º or 0.165 rad «below the horizontal» or gradient of \( – \frac{1}{6}\)