2021 RI Yr CT Sect B Soln

Raffles Institution Year 5-6 Physics Department 1 2021 Year 5 H2 Physics Term 3 Common Test Solution Section B 1 (a) 2 2 12 base unit of forcebase unit of pressure base unit of area kg m s m kg m s (shown) − −− = = = (b) ( )( ) ( ) ( ) 13 12 1 2 22 1 1 1 2 2 2 2 21 1 1 m kg m s kg molbase unit of m kg m s K mol K skg m s m kg m s s m kg s Z −− − −− − − −− − =   =   = = (c) (i) ( ) ( ) 12 12 12 4 2.5 100 2.5 800101100 20003 2 10 Pa (1 s.f.) pp pp pp ∆− =− ∆− = × = = × (ii) ( ) ( ) ( ) 6 12 0.80 0.02 10 Pa 0.80 0.02 MPa pp−= ± × = ± (d) (i) Systematic errors result in all the measurements being either above or below the true value by a fixed amount. Random errors result in the measurements being scattered about a mean value. These measurements have equal probability of being above or below the mean value. (ii) ( ) ( ) 12 12 113 22 0.03 2.5 0.001 1 1.7 1 2.8 13 si nce 3.34 100 0.153 2 100 2 100 2 0.08098 8.1% (2 s.f.) ppZr Lm T Z r pp L m T rdrd rd ∆−∆∆ ∆∆ ∆    = + ++ +    −     ∆∆   = + + + + = ⇒=       = = Comments (a) Students should not use square brackets [ ] as they represent dimensions and not units. Students must write “units of” of a physical quantity.

Raffles Institution Year 5-6 Physics Department 2 (b) The most common mistake is missing the power of ½ for the term in square root. Students who are weak in manipulating algebraic expressions tend to get this wrong. (c) (i) The pressure gauge measures the pressure difference directly, and the percentage uncertainty of 2.5% refers to the pressure difference. Weaker students erroneously double the percentage uncertainty to 5%. (ii) Those who got this wrong tend to miss out the “mega” prefix in “MPa”, or they presented inconsistent decimal places for the numerical values. (d) (i) Many were careless in their use of terminologies in their statements. Marks were not awarded for inaccurate descriptions. (ii) Common mistakes include: using r = d/2, but not ∆r = ∆d/2; missing out factor of 3 for percentage error in r; missing out factor of ½ for percentage error in m and T; not using the given 2.5% for percentage error in ∆(p1 − p2). 2 (a) (i) Take direction upwards and to the right as positive. Consider the trajectory of the arrow from the archer to the cauldron. Vertically: ( )( ) 22 2 1 2 0 2 9.81 24 470.88 21.70 m s yyyy y y v u as u u − = + = +− = = ( )0 21.70 9.81 2.21 2.2 s (shown) yyyv u at t t = + = +− = = (ii) Since there is no air resistance, horizontal component of the velocity of the arrow remains constant throughout and acceleration in the horizontal direction is zero. 1 60 2.2 27.27 27.3 m s xx x x s ut su t − = = = = = (iii) 1 tan 21.70tan 27.27 38.51 38.5 y x u uθ θ − = =  = = °

Raffles Institution Year 5-6 Physics Department 3 (b) (i) (ii) cart’s acceleration along plane sing φ= (iii) Yes, the ball will land back on the cart. Assuming no air resistance, the ball undergoes free -fall acceleration g in air. The component of this acceleration parallel to the plane, //a is sing φ , the same as the cart’s acceleration. Furthermore, the component of the initial velocity that is parallel to the plane, //u , is also the same for the cart and the ball. Hence, at any point in time t, their displacements parallel to the plane will be the same as shown by 2 // // // 1 2s ut at= + . So the ball will land on the cart when it falls back to the plane. Comments (a) (i) Students who used symbols with subscripts y (e.g. uy) had no problem substituting uy into second kinematics equation. In comparison, for some who used usinθ , it was less obvious to them they had to substitute a value for usinθ in the second equation. Some went to find θ value which was unnecessary. (ii)- (iii) Generally well-done. (b) (i) Some students drew curves that were obviously not parabolic. Many also erroneously drew paths with symmetry about a line perpendicular to the slope. This is wrong as the symmetry should be about the vertical through the peak (as is typical of projectile motion with ball projected at an angle to the horizontal). The final part of trajectory should not be vertical as there is a non- zero horizontal component of velocity throughout the flight and thus ball should not land vertically. Note the initial part of trajectory should not be perpendi cular to the motion of the cart, the question states the ball is launched such that its initial velocity with respect to the cart is perpendicular to the motion of the cart. This means the initial velocity is perpendicular when viewed from within the moving cart, but this is not true for the observer outside. The initial velocity must factor in the velocity of the cart (and ball before launch) that is along the plane. (ii) Most answered correctly. Parabolic path, symmetry about the vertical through the maximum point Cannot land vertically

Raffles Institution Year 5-6 Physics Department 4 (iii) Some students loosely said aball is gsinφ (not true as aball is g), without clarifying this is only true for the // component of a ball. There is a serious misconception that the acceleration of the ball remains the same as the cart after launch (just because the ball is one system with the accelerating cart at first). Some erroneously reasoned that since the launching angle is 90 o, the // component of a ball is unaffected by the launch (not true as aball in air is due to free-fall g, regardless of launching force/angle). These conceptual errors had to be penalised. One needs to know right after launch, there is only gravitational force at work and hence a ball is g, which we then resolve into // and perpendicular components as required by the question. Very few students recognised the importance of mentioning the component of the initial velocity parallel to the plane, //u is also the same for the cart and ball. If this condition is not met, the //s will not be the same for both and the ball will not land back in the cart. One can justify since the launching angle is 90o, //u is unaffected by the launch. 3 (a) Since the balls are moving at the same speed, they will meet at the midpoint. Distance travelled by each ball 0.950 0.030 0.475 0.030 0.445 m2= −=−= distancetime taken speed 0.445 0.20 2.225 2.23 s = = = = (b) (i) The collision is inelastic and since the system loses kinetic energy, total kinetic energy is not conserved. (ii) final initalKE 0.80 KE= 22 22 A A BB A A BB 11 11 + 0.8022 22mv mv mu mu = + 1 A BA B A BSince , 0.20 m s and m mu u v v v−= = = = = ( ) ( ) 222211 1 1+ 0.80 0.20 0.2022 2 2vv = + 21speed after collision, 0.80(0.20) 0.1789 0.179 m s v −= = = (iii) Taking direction to the right as positive, AFt p∆= ∆ () 0.22( 0.179 0.20) 0.070 1.191 1.19 N (force on A by B is to the left) ApF t mv u t ∆= ∆ −= ∆ −−= = −= − magnitude of the average force on A by B = 1.19 N

Raffles Institution Year 5-6 Physics Department 5 (iv) 1. 2. By Newton’s third law, the force on ball A by ball B and the force on ball B by bal l A are equal in magnitude and opposite in direction. By Newton’s second law, the accelerations of balls A and B are inversely proportional to their masses for the same force. Since the mass of ball A is greater than the mass of ball B, the acceleration of ball A will be smaller than the acceleration of ball B. Comments (a) Many students make the mistake of neglecting the radius of the ball when calculating the distance travelled by each ball. a / m s −2 t / s −amax,A 2.23 2.30 B A amax,B 0 a / m s −2 t / s 2.23 2.30 B’ A