2023 HCI H3 Prelim P1 SS

2023 H3 Prelim Suggested Solutions 1 (a) It is a frame, which moves at a constant velocity and Newton’s laws are applicable in such frames. B1 B1 Comments: Several students forgot to mention about Newton’s laws in their answers. (b) Let the identical mass be m. The kinetic energy of the system in this frame is ( ) ( ) ( ) 22 21 1 12 3 132 2 2K m u m u mu= + = A frame that moves towards cart A with a speed of u. In this frame, the speed of cart A becomes 3u and the speed of cart B becomes 2u. Hence, the same kinetic energy K is obtained in this inertial frame. B1 A1 Comments: Many students managed to express the correct frame of reference. (c) Let the speed of the mass m be u. The collision must be a perfectly inelastic one. By the principle of conservation of linear momentum, 0.5(0) 1.5 0.667 common common mu mv vu += = The initial kinetic energy of the system before the collision is 21 2 mu . The kinetic energy after the collision is ( )( ) 2 21 1.5 0.667 0.3332 m u mu = . The increase in internal energy of the system is ( ) 220.5 0.333 0.167mu mu−= The fraction of the initial KE that is convertible to internal energy of the system would be 2 2 0.167 1 30.5 mu mu = B1 B1 A1 Comments: This part was a distinguishing factor. Weaker students could not properly set their thought process on the inelastic nature of collision, only in which, energy losses take place.

2 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 2 (a)(i) ( ) 22 2 22 M R d R d   = + A1 (a)(ii) Moment of inertia of a circular disc about an axis through its center is given by 𝐼𝑑𝑖𝑠𝑐 = ∫ ( 𝑚 𝜋𝑅2) (2𝜋𝑟𝑑𝑟) 𝑅 0 𝑟2 = 𝑚𝑅2 2 ( ) ( )( ) ( ) ( ) ( )( )( )( ) ( ) ( )( ) 222 2222 22 2 2 2 2 2 2 2 22 2 2 2 2 22 2 2 3218 2 2 33 11 2 9 3 inner outerinner outer reel reel reel V R V RM M r M R d R R d R R d R d M R d R RRd MR MR     ++==   + +=  + += == I I I B1 B1 B1 Comments: The proof for the moment of inertia of a disc should be shown in the working as the expression is not given in the formula list.

3 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 (b) Let the tension in the string be T. Consider the box falling with acceleration a, 𝑀𝑔 − 𝑇 = 𝑀𝑎 Consider the translational motion of reel, 𝑇 − 2𝑀𝑔𝑠𝑖𝑛30° − 𝑓 = 2𝑀(2𝑅𝛼) Consider the rotational motion of reel, 𝑇(𝑅) + 𝑓(2𝑅) = (11𝑀𝑅2 3 ) 𝛼 Solving, 3𝑇 2 − 𝑀𝑔 = 4𝑀𝑅𝛼 + 11𝑀𝑅𝛼 6 = 35𝑀𝑅𝛼 6 Further, as the reel rotates, 𝑣𝑀 = 𝑣𝑂𝑌 + 𝑣𝑂 where vM is velocity of box wrt earth VOY is velocity of Y wrt reel center of mass VO is velocity of reel wrt earth differentiate the velocities with respect to time, 𝑑𝑣𝑀 𝑑𝑡 = 𝑑𝑣𝑂𝑌 𝑑𝑡 + 𝑑𝑣𝑂 𝑑𝑡 𝑎 = 𝑅𝛼 + 2𝑅𝛼 = 3𝑅𝛼 3𝑇 2 − 𝑀𝑔 = 35𝑀𝑎 18 Solving, 1 2 𝑀𝑔 − 3 2 𝑀𝑎 = 35𝑀𝑎 18 𝑎 = 9𝑔 62 M1 M1 M1 M1 A1 Comments: Quite a number of candidates have difficulty forming the equations of motion. As the system is going through acceleration, the tension in the string is not equal to the weight of the box. Also, many students often neglect friction when writing down the equation of motion for the reel, for both its translational as well as its rotational motion. X O M Y

4 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 3 (a) Kepler’s First Law states the planets move in elliptical orbits with the Sun at one focus of the ellipse. Alternative answer: Every planet (in the solar system) orbits the Sun in an elliptical orbit, with the Sun at one of the ellipse’s foci. B1 B1 Comments: Kepler’s Laws are empirical laws about the planets in the solar system. It was only later found that they follow naturally from Newton’s law of universal gravitation – and that other systems, such the moon’s of Jupiter and extrasolar planets – follow similar laws. Students who did not get the mark often missed the second part, that the Sun is at one focus of the ellipse. There were also quite a few answers that were written so badly that the meaning of the answer distorted. (b) (i) Elliptical orbit [B1] that is completely outside Venus’ orbit and completely inside Earth’s orbit [B1] Comments: Apparently, it was difficult for students to draw elliptical orbits. As such, it is advisable to annotate the sketch as being elliptical. (ii) Let vE be the speed and rE be the distance from the Sun of the rocket at launch point, and vV the speed and rV the distance from the Sun when it reaches Venus’ orbit. Then, by the principle of conservation of energy, 𝐸 = 1 2 𝑚𝑣𝐸 2 − 𝐺𝑀𝑚 𝑟𝐸 = 1 2 𝑚𝑣𝑉 2 − 𝐺𝑀𝑚 𝑟𝑉 [M1] Given that, for an elliptical orbit, the total mechanical energy is 𝐸 = − 𝐺𝑀𝑚 2𝑎 where a is the semi-major axis of the elliptical orbit. Combining, we get 1 2 𝑚𝑣𝐸 2 − 𝐺𝑀𝑚 𝑟𝐸 = − 𝐺𝑀𝑚 2𝑎 [M1] Rearranging, and using that 2a = rE + rV, [M1] we can solve for vE, transfer orbit (elliptical) v Sun 2a rE rV

5 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 𝑣E = √ 2𝐺𝑀 𝑟E − 2𝐺𝑀 𝑟E+ 𝑟V = √ 2(6.67×10−11)(1.99×1030) (1.50×1011) − 2(6.67×10−11)(1.99×1030) (1.50×1011)+ (1.08×1011) [M1] 𝑣E = 2.72 × 104 m s−1 = 27.2 km s−1 (shown) [A0] Alternative answer: The semi-major axis of the Hohmann transfer orbit would be 𝑎 = 𝑟E+ 𝑟V 2 = 1.50×1011+1.08×1011 2 = 1.29 × 1011 m [M1] At the Earth’s orbit, the gravitational potential energy is GPE = − 𝐺𝑀𝑚 𝑟E = − (6.67×10−11) (1.99×1030) ( 𝑚 kg) (1.50×1011) = −8.85 × 108 ( 𝑚 kg) J [M1] The total energy in the Hohmann transfer orbit should be 𝐸 = − 𝐺𝑀𝑚 2𝑎 = (6.67×10−11) (1.99×1030) ( 𝑚 kg) 2𝑎 = −5.14 × 108 ( 𝑚 kg) J [M1] Hence, the kinetic energy upon entering the Hohmann transfer orbit should be KE = 𝐸 − GPE = (−5.14 × 108 ( 𝑚 kg)) − (−8.85 × 108 ( 𝑚 kg)) = 3.71 × 108 ( 𝑚 kg) J [M1] KE = 1 2 𝑚𝑣2 ⇒ 𝑣 = √2 × (3.71 × 108) = 27222 m s−1 = 27.2 km s−1 (shown) [A0] Comments: Several students could do the required calculations. However, the presentation left a lot to be desired. Students should explain what they are doing, not just include random statements. For instance, several of them mentioned the principle of conservation of energy. As this question is only about one pint in the rocket’s orbit, the principle of conservation of energy is not relevant. (iii) By the principle of conservation of angular momentum, 𝑚𝑟E 𝑣E = 𝑚𝑟V 𝑣V [M1] Solving for vV, 𝑣V = 𝑟E 𝑟V 𝑣E = 1.50×1011 1.08×1011 (27.2 km s−1) 𝑣V = 37.8 km s−1 = 3.78 × 104 m s−1 [A1] Alternative answer: At Venus’ orbit, GPE = − (6.67×10−11) (1.99×1030) ( 𝑚 kg) (1.08×1011) = −1.229 × 109 ( 𝑚 kg) J [M1] 𝐸 = −5.14 × 108 (𝑚 kg) J KE = 7.15 × 108 (𝑚/kg) J 𝑣 = 3.78 × 104 m s−1 [A1]

6 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 Comments: The principle of conservation of angular momentum should have made this one of the easiest questions in the paper. Still, several students left this part blank. Part (ii) was a show question. In an exam, “show” questions suggest that the quoted value can be used in subsequent parts. (iv) By Kepler’s Third Law, 𝐺𝑀 𝑎3 = 4𝜋2 𝑇2 . Solving for the