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