2024 RI Prelim H1 Phy Paper 2 - Solutions

Raffles Institution Year 5-6 Physics Department 1 2024 H1 Physics Paper 2 Preliminary Examination Solution (JC Sharing) 1 (a) (i) Take direction up the slope as positive.   2 2 0 2 0 0 2 0 7.0 2 9.81sin30 4.9949 4.99 m v u as s s         OR By the principle of conservation of energy, increase in G.P.E. = decrease in K.E.   2 2 2 0 2 1 0, where is the max. vertical height fr om ground2 2 7.0 2(9.81) sin30 7.0 sin302(9.81) 4.9949 4.99 m mg h mv h vh g hs               (ii) 1.      2 2 2 1 1 2 sin2 2 1 sin2 K K E mv m u g s E mu mg s         Graph of EK-s is a straight line with negative gradient and vertical intercept 21 2 mu . 2.       sin sin P P E mg h mg s E mg s       Graph of EP-s is a straight line through the origin with positive gradient. energy EP EK s / m 0 0

Raffles Institution Year 5-6 Physics Department 2 (b) (i)   2 2 2 2 1 2 4.014.0 2 9.81sin30 sin30 10.841 10.8 m s (shown) v u as v v               OR decrease in K.E. = increase in G.P.E.        2 2 2 2 2 1 1 1 2 2 2 14.0 2 9.81 4.0 10.841 10.8 m s (shown) mu mv mg h v u g h v            *Must show non-rounded off value before stating the final answer. (ii) Take directions to the right and downwards as positive. Time of flight after the ball leaves the top of the slope to the ground:     21 2 214.0 10.8sin30 9.81 2 1.6080 s or 0.50713 s (NA) y y ys u t a t t t t          Horizontal distance travelled:    21 2 10.8cos30 1.6080 0 15.040 15.0 m x x xs u t a t       *Correct method and substitution must be shown. * Wrong root must be rejected.

Raffles Institution Year 5-6 Physics Department 3 2 (a) The initial total momentum of both balls is not zero. Since there is no net external force acting on the balls as a system, by the principle of conservation of momentum, the total momentum of both balls must remain unchanged and cannot be zero. Hence, the balls could not be stationary at the same time. (b) By the principle of conservation of momentum, 4.0 ( 1.0) 3.0 ----- (1) A A B B A A B B A B A B A B A B m u m u m v m v u u v v v v v v             Since collision is elastic, ( 1.0) 4.0 5.0 ----- (2) B A A B A B A B u u v v v v v v          1 (1) (2) 2 3.0 ( 5.0) 4.0 m s (shown) B B v v       * Both equations must be solved. (c) By Newton’s second law, the average force on ball B by ball A is    , 0.50 4.0 1.0 0.25 10 N B net B pF t       By Newton’s third law, the average force on ball A by ball B has the same magnitude of 10 N. OR From equation (1) or (2) in part (b), vA = 1.0 m s1. By Newton’s second law, the average force on ball A by ball B is    , 0.50 1.0 4.0 0.25 10 N A net A pF t       *Signs for direction of vectors must be coherent and shown.

Raffles Institution Year 5-6 Physics Department 4 (d) Balls A and B will exchange velocities and momenta as both balls have the same mass. From (b),   , 0.50 4.0 2.0 N sB fp   . Since duration of collision is 0.25 s, c onstant final momenta to start from 0.75 s to 1.5 s, with lines joining 0.50 s to 0.75 s during the collision. -3 -2 -1 0 1 2 3 momentum / N s t / s pA pB 0 1.0 0.5 1.5

Raffles Institution Year 5-6 Physics Department 5 3 (a) When the disc is just about to rotate, the contact force by the ground just becomes zero. Perpendicular distance from corner of box to line-of-action of F is 2 R . Perpendicular distance from corner of box to line -of-action of W is 2 2 3 2 2 RR R      . Applying the principle of moments about the corner of box, 3 2 2 3 1.7321 1.73 RF W R F W       *Need to state or show understanding that normal contact force is zero. (b) F acting at O needs to be inclined upwards such that it is at an angle above the horizontal to produce a clockwise moment about the corner to overcome the anticlockwise moment due to the weight. OR F acting at O needs to be inclined upwards so that there is a n upward vertical component to produce a clockwise moment about the corner to overcome the anticlockwise moment due the weight. OR F needs to be shifted upwards above O so that there is a moment arm from the corner of the box to the line -of-action of F, to produce a clockwise moment about the corner to overcome the anticlockwise moment due to the weight. Note: Increasing the magnitude of the horizontal force F acting at O will not cause any rotation as F has no moment about the corner of the box because the perpendicular distance from the corner to the line-of-action of F is zero. *Position of F must be such that there is a moment arm from the corner of the box to the line-of-action of F. *Direction of F must be such as to provide a clockwise moment about the corner of the box to overcome the anticlockwise moment due to the weight. F O R box ground disc force by ground force by box W

Raffles Institution Year 5-6 Physics Department 6 4 (a) (i) At the top of the circle, 2 T T mvT mg L  For the ball to just complete the vertical circle, the tension TT at the top of the circle is zero. 2 2 T T T mvmg L v gL v gL    *Explains or states clearly that tension at the top is zero. (ii) By the principle of conservation of energy, as the ball moves from the top to the bottom of the circle, it s gravitational potential energy decreases and its kinetic energy increases. This means the speed at the bottom is greater than the speed at the top. Hence 1B T v v  . (iii) 3 3 B T B T v v v v   As the ball moves from the top to the bottom, increase in K.E. = decrease in G.P.E.       2 2 2 2 2 2 2 1 1 22 2 1 1 3 22 2 9 1 22 2 4 2 1 2 B T T T T T T T mv mv mg L m v mv mg L mv mv mgL mv mgL v gL         As 1 2 gL gL , where gL is the value of vT at which the string just goes slack, the ball will not be able to complete a full circle if 3B T v v  . Hence, 3B T v v  is not possible to achieve. (b) Considering forces along the radial direction, 2sinT mr     2 2 Since sin , sin sin r L T m L T mL         Since m and L are constants, 2T  . Hence when the angular velocity is doubled, the tension in the string is 4T.

Raffles Institution Year 5-6 Physics Department 7 5 (a) Newton’s law of gravitation states that two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. (b) (i) Gravitational force provides the centripetal force on the satellite. 2 2 , where is the mass of the satellit e (shown) E E GM m mv mrr GMv r   (ii)       2 10 10 11 24 10 6 7 10 10 10 1 0 7.67 102 1 7.67 102 6.67 10 6.0 101 1600 7.67 102 6.4 10 2.1 10 1.17 10 7.67 10 8.84 10 J total K P E E E E mv GMm r                             

Raffles Institution Year 5-6 Physics Department 8 6 (a) (i) Effective resistance of Q and LDR, 1 1 1 1 1 1 6.0 8.0 3.4286 k eff Q LDR R R R                  1 3 3 3 9.0 3.4286 4.0 10 1.2115 10 1.21 10 A A T EI R           (ii) Potential difference across Q and LDR, 3.4286 9.03.4286 4.0 4.1539 V eff eff eff P RV E R R          2 3 4 4 4.1539 8.0 10 5.1924 10 5.19 10 A LDR A LDR VI R          (b) (i) Resistance of the LDR increases when light intensity is lowered. The effective resistance of Q and the LDR increases. Since the potential difference across P and Q is the same at 9.0 V,