DHS 2022 JC1 MYE Physics (Solutions)

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 1 Answers to Mid-Year Exam Section A 1 2 3 4 5 6 7 8 9 10 C D C B C B D D A A 11 12 13 14 15 C A D B A 1 C 4 2 23 23 24 34 P AT kg m s mUnits of P s kg m s kg m sUnits of mK kg s K σ σ − − − −− = ×= = = = 2 D 22 1 1 () 5.0 12.0 13m s 5sin ( )13 23 Change invelocity final velocity initial velocity vvuv u Magnitudeof v θ − − = − ∆ = − = +− ∆= + = = = ° -1Thechangeinvelocityis13ms at adirection23° West of South v = 5.0 m s-1 -u = 12.0 m s-1 ∆v

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 2 3 C During fall in air, weight of ball W - viscous force F = ma The ball viscous force increases with speed, hence acceleration decreases. At terminal speed, net force is zero, hence W = F If mass m is less, terminal speed is less. At sufficiently large t, the s – t graph is a straight line as the gradient of s – t graph gives the velocity. Answer is C. 4 B ( ) 1 22 22 1 ms spe 81 ed 6 58 4 7 7 m 62 9 68 s 5 yy yx v . v u at . . . v − − = + = + = = + = −+ = 5 C Considering the free body diagram of the measuring device, 598 1 53 64 N T (. ) () T −= = 6 B The component of the weight along the slope = (12) (9.81) sin 20o = 40 N Hence resultant force = 200 – 40 = 160 N 7 D Normal reaction acts perpendicularly (normally) to the surface. W F

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 3 8 D Consider free body diagram of the mass the beaker of water T + U = M -----(1) N = W + U ------(2) From (1), U = M – T Hence, N = W + M - T 9 A 10 A ( )( )( ) 32 2 3 pressure 600 kg m 9.81 m s 20.0 10 m 1.18 10 Pa p ghρ −− − = = × = × F’ F’ a a F’’ F = ma = (2.0 + 1.0) a a = F / 3 F’ = kx1 = F / 3 x1 = F / 3k F’’ =F/3 x 2 kx2 = 2F/3 x2 = 2F / 3k = 2 x1 ⇒ x1 : x2 = 1 : 2 T M U U W N

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 4 11 C ( )( ) θ= = = o work done, cos 2.1 N 4.0 m cos30 7.3 J W Fs 12 A ( ) −= ⇒ + −= ⇒= − = = = = 1 o o constant 30 m s zero resultant force equation of motion down the slope: sin6 2000 0 2000 1500 9.81 sin6 461.9 N power output of the car's engine 13856 W 14 kW engine engine engine v F mg F Fv 13 D ( ) 50 6025 3022 5 5 2050 2 actual indicated indicated actual actual v ,v vv v v %v ωω ωω ω  = = = =   ∆= − = + ∆+ = = + 14 B ( )( ) 1 222 1 4 0.250 s4 4 4 0.250 = 4.00 s 22 1.571 rad s4.00 0.500 1.571 1.23 m s T T T ar ππω ω − − = × =×× = = = = = = 15 A Like an external resultant force, the centripetal force is NOT an additional force acting on the object. It should not be drawn on any free-body diagrams.

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 5 Section B 1 (a) Density is defined as mass per unit volume. [B1] Comments: “ma ss over volume” not accepted. (b) vernier calipers [B1] (c) percentage uncertainty = 0.0004100% 100% 0.95%0.0420 r r ∆ × = ×= [A1] Comments: As ab solute uncertainty is given to 1 s.f., the percentage uncertainty should be recorded to 1 s.f. or at most 2 s.f. (d) (i) nMr kLρ = units of LHS = kg m-3 units of RHS = kg (mn ) (m-1) [M1] Comparing indices, -3 = n – 1 [M1] Thus n = -2 [A0] Comments: Some students were very poor in their presentation, and some wrongly equated physical quantities with units e.g. “M = kg” (ii) 100% ( 2 ) 100% nMr kL M rL M rL ρ ρ ρ = ∆ ∆ ∆∆× = + +× [C1] percentage uncertainty = [(0.001)/(1.072) + (2)(0.0004)/(0.0420) + (0.0001)/(0.1242)] x 100% = 2.1% [A1] (iii) ρ = (1.072) (0.0420)-2 / [(2.094) (0.1242)] = 2336.7 kg m-3 [C1] ∆ρ = (0.021) (2336.7) = 50 kg m-3 (1.s.f) [C1] ρ = (2340 ± 50) kg m-3 [A1] = (234 ± 5) x 101 kg m-3 Comments: Some students calculated this by finding the maximum or minimum possible value of ρ. As long as the working is clear, this method is also given full credit.

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 6 2 (a) Speed is a physical quantity and cannot be defined by a unit (seconds). [B1] S peed is distance travelled per unit time. [B1] Comment: Very few stated that speed is a physical quantity. Instantaneous speed or average speed were quoted as having the definition of “ distance travelled per second”. (b) (i) Method 1: Using equations of motion Taking downwards to be positive: [M1] ( )( ) 22 2 1 2 1.5 2 9.81 0.65 3.9 m s v u as v − = += + = ± Final speed = 3.9 m s-1 [A1] OR Method 2: Using conservation of energy Loss in GPE = Gain in KE [M1] ( ) 22 22 1 1 2 2 3.9 m s mgh m v u v gh u v − = − = + = Final speed = 3.9 m s-1 [A1] Comment: Method 1 is the common method used for this part which were well done.

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 7 (ii) Straight line from X to t1 [B1] Straight line from X to - 3.9 m s-1 [B1] Gradient straight line from t1 to t2 must be parallel to line from 0 to t1. [B1] Comment: A few drew curves for this part. The straight line drawn from t1 to t2 is not parallel to the line drawn from 0 to t1. (iii) The speed of the ball after rebound is less than the speed just before impact. The ground/Earth is assumed to be stationary, hence there is a loss in the kinetic energy of the system during the bounce. [B1] Hence the bounce is not elastic. [B1] Comment: The system of Earth and ball was not stated. Many only mentioned kinetic energy was not conserved but did not specify whether it was the kinetic energy of ball or system. Some mentioned bo unce was elastic but the explanation contradicted with the answer.

DUNMAN HIGH SCHOOL 2022 PHYSICS H2 (YEAR 5) 8 3 (a)(i) [B1] Comments : Care must be given to ensure that the graph is symmetrical about the horizontal axis. At the very least, the amplitude must be correct. This question was marked leniently and many answers were given BOD even though the graph drawn wasn’t symmetrical. (a)(ii) The principle of conservation states that the total momentum in a system is constant if no net force acts on the system. [B1] The negative and positive areas represent the changes in momentum (or impulses) of the two trucks. [B1] By Newton’s third law, they are equal and opposite in direction, hence the total change in momentum for the two-trucks system is zero. [B1] Hence total momentum of system is constant. [A0] Comments: Thi s question requires you to make reference to (a)(i). (b)(i) take original direction of big truck to be +ve By Conservation of Momentum, m 1u1 + m2u2 = m1v1 + m2v2 [C1] 3mv + m(–v) = 3m(0.60v) + mv2 v 2 = 0.20v = 0.20 x 70 = 14 km h–1 = 3.9 m s-1