IBDP Physics- A.2 Forces and momentum- IB Style Questions For HL Paper 2 -FA 2025
▶️ Answer/Explanation
(a)
Initial kinetic energy:
\(KE_i = \frac{1}{2}(3.0)(6.0)^2 = 54\,J\).
Final velocities from the graph are \(v_X = -2.0\,m\,s^{-1}\) and \(v_Y = 4.0\,m\,s^{-1}\).
Final kinetic energy:
\(KE_f = \frac{1}{2}(3.0)(-2.0)^2 + \frac{1}{2}(6.0)(4.0)^2 = 6 + 48 = 54\,J\).
Since the total kinetic energy is unchanged, the collision is elastic.
(b)
(i) The average force on trolley Y is given by the rate of change of its momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{6.0(4.0 – 0)}{40 \times 10^{-3}} = 6.0 \times 10^{2}\,N\).
(ii) The average power delivered to Y is the rate of increase of its kinetic energy:
\(P = \frac{48}{40 \times 10^{-3}} = 1200\,W\).
(iii) At \(t = 30\,ms\), both trolleys have a speed of \(2.0\,m\,s^{-1}\).
Total kinetic energy at this instant:
\(KE_{30} = \frac{1}{2}(3.0 + 6.0)(2.0)^2 = 18\,J\).
Using conservation of energy, the elastic energy stored in the spring is:
\(E_{spring} = 54 – 18 = 36\,J\).