RI 2025 Prelim (Answers)

2025 RI Prelims H3 Physics Solutions 1 (a) (i) 1 2 Taking up as positive, sphere starts out with initial velocity after the n th impact and hits the plate with a final velocity of just before the ( n + 1)th impact. Using (ii) Time taken from release to 1 st impact: From (a) (ii) Time taken between 1 st and 2 nd impact: From (a) (i) Hence, time taken between 1 st and 2 nd impact Total time 1

Since the sphere is dropped from height h initially, using Sub into total time (iii) 1 (b) (i) The vector sum can be found as follows: 2

Hence, v CM is at 36.9° clockwise below the horizontal. (ii) The combined meteorite of A and B must move off at the velocity v CM because in the absence of external forces, the total momentum of the system is conserved, and hence v CM is conserved before and after the collision. Therefore, once A and B are combined into a single mass, it must move at the velocity v CM . (iii) Using a vector diagram, one can see that if v CM ’, the velocity of the combined meteorite relative to the spaceship has to be vertically downwards, the velocity of the spaceship has to be angled as shown in the diagram. Using sine rule, 2 (a) 3

(b) Resolving horizontally: ---------- (1) Resolving vertically: ---------- (2) Since (c) To minimize F , the denominator must be maximized. Subst into (b) 4

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3 (a) Well-insulated ⇒ adiabatic process , , (b) Isothermal process ⇒ 6

4 (a) (b) Ray 2 has a phase shift of radians at Q. (c) (i) Sub (2) into (1) (ii) (iii) Testing of optical lenses. If the lens surface is ground to a perfectly symmetrical curvature, a circular diffraction pattern as shown in Fig. 4.2 will be obtained. 7

Variations from symmetry will result in fringes that are not smooth circular shape. These variations indicate that the lens must be reground and repolished to remove imperfections. 5 (a) (b) (i) The vertical component F y (is perpendicular to the velocity of the proton and this provides the centripetal force which) causes it to move in a circular path in the vertical plane. The horizontal component F x causes the proton to accelerate towards the right where the magnetic field is stronger. (ii) ● helical path around the magnetic field lines, and pitch of helical path is wider in the region of weaker magnetic field than in the region of stronger magnetic field. ● (when the proton crosses over the mid-point of the magnetic field to the stronger field region on the right, the direction of the horizontal component 8

of the magnetic force is reversed, causing it to) accelerate back towards the left. (This process repeats itself and the proton spirals back and forth between the two coils and is unable to leave the magnetic field). (iii) As the charged particles will be trapped (or confined) within the magnetic field region between the coils, this set-up can have further applications. [This setup is called a magnetic bottle, and the two coils are magnetic mirrors, reflecting the charged particles back and forth.] 6 (a) KE = loss of electric PE = q Δ V = (1.60 × 10 –19 ) (15 000) = 2.40 × 10 –15 (b) Δ p y Δ y ≳ h m Δ v y Δ y ≳ h Δ v y ≳ h / ( m Δ y ) = = 1.455 = 1.46 m s –1 (c) The uncertainty in position Δ r = (Δ v y )( t ) = 9

7(a) (b) Explanation : After a sufficiently long time, the nuclides in the mixture that have shorter half-lives would have decayed away, leaving the nuclide with the longest half-life. The tail end of the graph of ln A mix versus t will then be a straight line. This straight line is the graph of ln A longest-lived nuclide versus t for the nuclide with the longest half-life in the mixture. The gradient of this line will be equal to , where λ and τ are the decay constant and half-life of the longest-lived nuclide respectively. Calculations : From the straight line at the end of the graph drawn in Fig. 7.2, and using gradient co-ordinates (6.25, 8.0) and (17.5, 6.5), , 10