IB DP Physics Mock Exam SL Paper 2 Set 1 - 2025 Syllabus
IB DP Physics Mock Exam SL Paper 1B Set 1
Prepare for the IB DP Physics Mock Exam SL 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
(i) the magnitude of the average force acting on Y.
(ii) the average power transferred to Y.
(iii) the elastic potential energy stored in the spring at \( t = 30\,\text{ms} \).
▶️ Answer/Explanation
(a)
ALT 1 – Kinetic energy check
\( \text{KE before} = \frac12 \times 3.0 \times 6.0^2 = 54.0 \ \text{J} \quad \checkmark \)
\( \text{KE after} = \frac12 \times 3.0 \times (-2.0)^2 + \frac12 \times 6.0 \times 4.0^2 = 6.0 + 48.0 = 54.0 \ \text{J} \quad \checkmark \)
Since kinetic energy is conserved, the collision is elastic.
ALT 2 – Relative speed rule (for elastic collisions)
\( |u_X – u_Y| = |v_X – v_Y| \)
\( |6.0 – 0| = |-2.0 – 4.0| \ \Rightarrow \ 6.0 = 6.0 \quad \checkmark \)
(b)(i)
Average force = change in momentum / time interval
\( F_{\text{av}} = \frac{\Delta p_Y}{\Delta t} = \frac{6.0 \times 4.0 – 0}{40 \times 10^{-3}} = \frac{24.0}{0.040} = 6.0 \times 10^2 \ \text{N} \quad \checkmark \)
(b)(ii)
Average power = change in kinetic energy of Y / time interval
\( \Delta E_{K,Y} = \frac12 \times 6.0 \times 4.0^2 = 48.0 \ \text{J} \)
\( P = \frac{48.0}{40 \times 10^{-3}} = 1200 \ \text{W} \quad \checkmark \)
(b)(iii)
At t = 30 ms, from the graph: both carts have the same velocity v = 2.0 m/s
Total kinetic energy at this moment:
\( KE_{\text{total}} = \frac12 (3.0 + 6.0) \times 2.0^2 = 18.0 \ \text{J} \)
Initial kinetic energy = 54.0 J
Elastic energy stored in spring = 54.0 – 18.0 = 36.0 J
\( \boxed{36 \ \text{J}} \quad \checkmark \)