61_BPGH_2021 4E Prelim PHY P1 ANSWERS

4 E PRELIM P1 ANSWERS 1 2 3 4 5 6 7 8 9 10 B B A B B D A B C C 11 12 13 14 15 16 17 18 19 20 D D D C D C A C D D 21 22 23 24 25 26 27 28 29 30 B B A A A B A B B C 31 32 33 34 35 36 37 38 39 40 D B B B C C C C C B Explanations: 1 B Unit of Angular momentum = unit of mass x unit of velocity x unit of radius = kg ms-1 m = kg m2s-1 2 B Method 1: Distance between two stones = Distance travelled by the first stone – distance travelled by the second stone = ½ gt2- ½ g(t-5)2 = ½ g(10t + 25) Since 10t+ 25 increases with t, the distance between the two stones will also increase with time. Method 2: Consider the v-t graph of the two stones above. The shaded area represents the distance between the two stones, which increases as time passes. 3 A Constance force ≡ constant acceleration. The velocity time graph has a constant gradient. 4 B By conservation of energy: Loss in GPE = Gain in KE mgh = ½ mv2 10 x 0.21 = ½ v2 v = 2.0 ms-1

5 B 6 D Since the car is moving with constant speed, the net force on the car is zero. Since there is no additional external forces acting on the car as the sand spills, the net force remains at zero. Hence, there is no change in the acceleration of the car. 7 A After force F is removed, the net force on the block will be equal to the frictional force acting against the motion of the block. Hence, there will be a negative acceleration which slows the block down. 8 B Force on hand by book = force on book by hand = 30 N Force on book by ceiling = Force on ceiling by book = F- Weight of book = 10 N 9 C W = mg 10 C Taking moments about P, By principle of moments, Sum of CW moments = Sum of ACW moments Wboard x 1.5 m = F x 2m F = 60×10×1.5 2 = 450 N 11 D Loss in GPE = Gain in KE + Work done against friction 50 J = Gain in KE + 20 J Gain in KE = 50 – 20 = 30 J ½ m(v22 – v12 )= 30 J v22 – v12 = 30×2 50 = 1.2 v22 = 1.2 + 3.02 = 10.2 v2 = 3.2 ms-1 12 D The time taken for each person is needed. 13 D 150 𝑁 5.0 = 𝐿𝑜𝑎𝑑 100 Load = 3000 N 14 C Let the pressure on the air column in the horizontal position and vertical position be phorizontal and pvertical respectively. At the same temperature, 𝑃ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 P𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 = V𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑉ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 = 18 20 = 9 10 Pvertical= 10 9 Phorizontal = 10 9 p

Pressure by the mercury column on the trapped air = pvertical - p= 10 9 p- p = 1 9 p 15 D 100℃−80℃ 80℃−20℃ = 28cm-24cm 24cm-h h = 12.0 cm 16 C KE of a gas represents the average speed of the gas molecules. 17 A As the temperature of the gas increases, the average KE of the gas molecules also increases. This causes gas molecules to collide more forcefully and more frequently against the walls o