2024 RI Prelim H1 Phy Paper 1 - Solutions

Raffles Institution Year 5-6 Physics Department 1 2024 H1 Physics Paper 1 Preliminary Examination Solution and Mark Scheme Qn Ans Solution 1 A A basketball has a mass of about 0.6 kg, or a weight of about 6 N. 2 A Unit of 2 1 1 2 kg m s kg m smP     Unit of 3kg m        1 2 2 2 3 kg m s unit of m s kg m nn n n P              2 2 1m s m s nn n    Comparing indices of m: 2n  1  n  ½ 3 D 1 2 1 2 0.02 0.03 100 1.4 %16.24 12.78 d ddiff diff d d         4 C At times t1 and t2 after release, ball falls through a distance of s1 and s2, respectively. 2 2 1 1 2 2 2 2 2 1 2 1 2 2 2 1 1 1 and 2 2 1 ( )2 2 ( ) s gt s gt h s s g t t hg t t         5 D The ball hits the floor at A and gets compressed from A to B. The ball uncompresses from B to C and leaves the floor at C. From B to C to D, the ball is moving away from the floor , before coming to an instantaneous rest at D. Hence, D is the time the ball has reached maximum height after bouncing from the floor. 6 C From 0t1, since at graph is constant, gradient of vt graph is constant. From t1t2, since at graph is decreasing, gradient of vt graph is decreasing. After t1, since at graph is zero, the velocity is constant. 7 D Option A: Possible, if lift is decelerating / decreasing in speed on its way up. Option B: Possible, if lift is moving upwards at a constant speed. Option C: Possible, if lift is accelerating / increasing in speed on its way up. Option D: Hence, all options above is possible, depending on the lift’s acceleration.

Raffles Institution Year 5-6 Physics Department 2 8 C Applying N2L on a system of both crates: , 2 100 5.0 9.81 5.0 50.95 5.0 10.19 m s net both bothF m a a a a          Applying N2L on 2.0 kg crate, OR Applying N2L on 3.0 kg crate, ,2 2 100 2.0 9.81 2.0 10.19 60 N net kg kgF m a T T        ,3 3 3.0 9.81 3.0 10.19 60 N net kg kgF m a T T       9 B Impulse or the change in momentum is area under the force-time graph,         1 1 2 2 1 400 4.5 1.0 400 2.0 800 6004.5 1.5 400 4.5 1.5 3.0 m s p Fdt v v v               10 D Resolving the normal forces horizontally, sin 45 sin30 sin 45 1.4sin30 o o Y X o X o Y N N N N    11 C The resultant force of the road on the wheel is due to the upward normal force on the wheel and the leftward frictional force on the wheel. 12 A Take moments about the lower hinge, (so that we can ignore FY and force exerted by lower hinge) FX (1.8) + (40)(9.81)(1.5) = (200 sin30o)(3.0) + (200 cos30o)(1.8) FX = 13 N to the right 13 A Work done by force is area under Fs graph, not sF graph. Energy P is returned or released upon removal of force, and energy Q is used to permanently stretch to spring, i.e., energy used to separate the particles of the spring further apart. W Y NX X NY 30 45 30° 1.8 m 3.0 m FX W FY 200 N

Raffles Institution Year 5-6 Physics Department 3 14 D 3 3 72 1012 10 60 60 600 N P Fv F F                  from fuel tocar 6 30 30 40 10 12 10 1 60 60 3 6 kg E E . m m .        15 B Since speed of block remains constant, net force on block is zero. Therefore, magnitude of frictional force is equal to magnitude of component of weight along slope. Hence, rate of work done by frictional force = Fv = (mg sin 25)(2.0) = 13.3 W 16 C 2 60 60 2 12 60 60 12 60 601.5 1860 60 m h m m m h h h v r v r               17 B Applying Newton’s 2nd Law along the vertical and horizontal directions:     2 2 -1 vertically: sin15 (1) horizontally: cos15 0 (2) (1) (2) : tan15 150 9.81 tan15 20 ms mvN r N mg v rg v         18 A The gravitational force of attraction between the two stars provides the centripetal force for the stars to undergo uniform circular motion about the same centre with the same angular speed , where r is the radius of orbit of star of mass m and R is the radius of orbit of star of mass M: 2 2 2 2 2 GMm mr MRd mr MR mr MR         Since X has a larger mass M, its radius R will be smaller. In addition, b oth stars must orbit diametrically opposite each other and in the same direction to maintain the same centre of orbit.

Raffles Institution Year 5-6 Physics Department 4 19 A 2 2 E E d EI LR L A dI         2 2 1 1 2 14 1 16 1 8 2 1 9 X Y X total dI I d I I             20 B       3 16 19 19 16 1 12.2 10 2.58 10 1.60 10 ' 3.20 10 ' 2.52 10 s electrons ions N N                 I I I 21 B From the graph, E = 3.0 V  3.0 0.75 0.8 3.0 0.8 2.930.75 E Ir V r r        22 D There is no current flowing from one circuit to the other. Hence, the p.d. across the 100  and 200  resistors are the same. 100 200 100 200 12 2450 100 200 400 V V R R                   23 B Assume that the lower potential terminal of the cell is 0 V. Hence, potential at A is equal to p.d. across the 11 resistor and potential at B is equal to p.d. across the 1.5  resistor. 11 1.5 11 1.56 6 3.5 V11 1.0 1.5 3.0 AB A BV V V V V                     

Raffles Institution Year 5-6 Physics Department 5 24 A 1 1 1 Assume each resistor has resistance . 1 1Across PR: 2 2 1 1 1 1Across QS: 2 2 2 1 3 5Across PS & QR: (see below) 5 8 eff eff eff r R r r r R r r r r R r r r                       25 D The electron experiences a downward force within the electric field which causes it to gain speed in the vertical direction. Upon exiting the field, the electron’s speed has increased. 26 B Gain in K.E. of electron is due to work done on it by the electric field:     19 5 15 1 60 10 3 0 10 0 080 3 8 10 J W Fd qEd . . . .          27 C The current in wire X produces a magnetic field along the circumference of coil Y in the clockwise direction. This magnetic field is parallel to the current in coil Y . Hence. no magnetic force acts on any part of coil Y. 28 C Energy released = (89 × 8.6169 + 144 × 8.2656) – (235 × 7.5909) = 173.289 = 173 MeV 29 C     * 4 4 0 1 2 1 4 4 0 2 1 8 4 0 3 2 1 Option A: (same element) Option B: (different element) Option C: 2 (different isotope) Option D: 2 (different element) A A Z Z A A Z Z A A Z Z A A Z Z X X X X X X X X                           30 B For X, 3 1 1 1 2 8 2 X t T                lg 1 8 3lg 1 2X t T   . For Y, 2 1 1 1 2 4 2 Y t T                lg 1 4 2lg 1 2Y t T   . 2 0.673 X Y T T   R Q r 5/3 r R Q r r 2/3 r R Q r r r 2r

Raffles Institution Year 5-6 Physics Department 6 7A 8B 8C 7D