RI 2023 Y5 H2 Phy Promo Sect BC Solution

Raffles Institution Year 5-6 Physics Department 1 2023 Year 5 H2 Physics Promotional Examination Solution Section B 1 (a) xx x x 1 20.0 1.40 14.286 14.3 m s s ut su t − = = = = = *Final answer to 3 s.f. (b) y x tan32.0 u u°= yx 1 tan32.0 14.286 tan32.0 8.9269 8.93 m s uu − = ° = ° = = *Final answer to 3 s.f. (c) Consider vertical motion of the ball from starting point to point of maximum height. ( ) yy 0 8.9269 9.81 8.9269 0.90998 0.910 s9.81 v u at t t = + = +− = = = *Final answer to 3 s.f. (d) (i) vertical velocity when the ball hits the wall: ( )( ) 1 yy 8.9269 9.81 1.40 4.8071 4.81 m sv u at −= + = +− = − = − OR gradient, y 1 y 0 9.81 4. 8071 4.81 m s1.40 0.90998 v v −− = − ⇒= − = −− *Straight line with negative gradient, positive and negative vy *Correct label values vy / t / s 0 1.40 0 8.93 – 4.81 0.910

Raffles Institution Year 5-6 Physics Department 2 (ii) Height of the wall is the area under the graph from 0 st = to 1.40 st = . OR Height of the wall is the difference in the absolute areas from 0 st = to 0.910 st = and 0.910 st = to 1.40 st = . (e) (i) With air resistance, there is now a resultant force in the horizontal direction that causes the horizontal component of the ball’s velocity to decrease continuously / gradually / at a decreasing rate. (ii) Preferred answer (comparing the rate at which vertical component of velocity changes with and without air resistance) : With air resistance, when t he ball is moving upwards towards its highest point, the vertical component of its velocity decreases to zero at a faster rate compared to the rate when there is no air resistance. When the ball is movi ng downwards from its highest point, the vertical component of its velocity increases at a slower rate compared to the rate when there is no air resistance. Al ternative answer (describing the effect of air resistance on the variation of vertical component of velocity): With air resistance, the acceleration in the vertical direction is no longer constant. When the ball is mo ving upwards, the vertical component of its velocity decreases at a decreasing rate. When the ball is moving downwards, the vertical component of its velocity increases at a decreasing rate. Comments (a) (b) (c) • A number of students stated the answer in terms of the initial velocity u. The question stated ‘calculate’ which means numerical answers are expected. • Students should present their worked solutions clearly. Defining equations with appropriate symbols should be stated first before substituting numerical values. For example in part (c), students could have written yyv u at= + first, where subscript y represents the vertical components of the vectorial physical quantities in the equation. (d) (i) • The area under the graph above the horizontal axis is larger than that below the horizontal axis. This is because the former represents the vertical distance of the ball from the ground to the highest point and the latter is the vertical distance of the ball from the highest point to the top of the wall. • A number of students did not indicate the final vertical velocity of the ball. (ii) • It is important to state the time interval which the area under the graph should be considered, or state ‘TOTAL area under the graph’. • Phrasing needs to be improved. The area under the graph from 0 s to 0.910 s is positive while the area under the graph from 0.910 s to 1.40 s is negative, since the area under the graph is displacement, a vector quantity. So the sum of these two areas would actually give the height of the wall. Answers that suggest subtracting the areas should rightfully state ‘subtract the absolute areas’ • Some students labelled the two areas with ‘A’ and ‘B’ and used these symbols to explain how to determine the height of the wall. This was clear and accepted. (e) (i) (ii) • Many students misread the question and described the effect of air resistance on the components of displacement instead of velocity for both parts. • For (ii), some gave long answers that described acceleration and made no reference to velocity. Such answers were not given credit. • It should be stated clearly which direction the ball is moving, and to consider both upward and downward motions.

Raffles Institution Year 5-6 Physics Department 3 2 (a) The normal contact force / resultant force always points towards the centre of the circular track and is thus always perpendicular to the instantaneous velocity of the ball. This only changes the direction of the velocity of the ball while the magnitude of its velocity remains constant. Hence the speed and kinetic energy of the ball remain constant. OR The normal contact force / resultant force always points towards the centre of the circular track and is thus always perpendicular to the instantaneous velocity of the ball. Since there is no displacement in the direction of the resultant force, work done by the resultant force is zero. This does not cause a change to the kinetic energy of the ball and kinetic energy remains constant. (b) (i) 1. At point P, normal contact force on ball and weight of ball provides the centripetal force. 2 2 vN mg m r vN m mgr += = − For ball to complete the vertical circle, it cannot lose contact w ith the track. 2 0 0 N vm mgr v rg ≥ −≥ ≥ minimum speed at P, ( ) min 1 0.60 9.812 1.7155 1.72 m s v rg − = =  = = OR explain The normal contact force on the ball decreases (since speed decreases) as it moves from the bottom of the track to point P. Hence for the ball to just complete the vertical circle without losing contact, the minimum speed at point P is when 0N = at P. This means that the ball will not lose contact with the track at any other points along the vertical circle. 2. By conservation of energy from just before spring is released to point P, decrease in EPE = increase in GPE + increase in KE ( ) ( ) ( ) i f f i fi 22 f i min EPE EPE GPE PE KE KE 11 0022 G kx mg h h mv − = − +− −= − + −  ( ) ( ) ( )( ) ( ) 2 f i min 3 2 21 2 2 45 10 19.81 0.60 1.715580 2 0.090979 0.0910 m mx gh h vk − = −+  × = +  = =

Raffles Institution Year 5-6 Physics Department 4 (ii) 1. At minimum speed, the centripetal acceleration at P is always the acceleration of free fall and is independent of mass . Hence the minimum speed remains the same. OR The minimum speed is only dependent on the diameter of the circular track and the acceleration of free fall. Since it is independent of mass, the minimum speed at P remains the same. 2. With a greater mass, the increase in gravitational potential energy and kinetic energy needed to reach P is greater . Hence the initial compression of the spring must be greater. OR With a greater mass, the total energy at P is greater . Hence the initial compression of the spring must be greater. OR With a greater mass, a greater force is needed to project the ball to the same minimum initial speed and consequently the same minimum speed at P. Hence the initial compression of the spring must be greater. (iii) *Parabolic path in the correct direction, initial part of the path tangential to track at point Q, path within circumference of track Comments For all problems on circular motion, students have to distinguish if they are analysing a body moving in a horizontal or vertical circular motion and if its speed is uniform or not. Is this question, (a) is on horizontal circular motion with uniform speed while (b) is on vertical circular motion with non-uniform speed (since there is no constant input of energy to keep the speed constant). (a) Students have to distinguish the different quantities and terms and use them precisely in the explanations. For