9814 H3 Physics - HCI 2024 Complete H3 Topical Examples and Tutorial Solutions

Hwa Chong Institution (College) MOE H3 Physics 2024 A1. Inertial Frames of Reference 1 Suggested Solutions for Worked Examples 1 a) Car,Truck Car,Earth Earth,Truck -1 Car,Truck 25 30 5 m s V V V V =+ = − =− The observer in the truck observes that the car moves in the opposite direction at a constant speed of 5 m s-1. b) Truck,Car Truck,Earth Earth,Car -1 Truck,Car 30 25 5 m s V V V V =+ = − =+ The observer in the car observes that the truck moves in the forward direction at a constant speed of 5 m s-1. c) Car,Truck Car,Truck ( 5)(60) 300 md V t = = − =− 2 spider,Earth spider,passenger passenger,tr ain train,Earth -1 spider,Earth 0.5 1.2 3.1 3.8 m s V V V V V = + + =− + + = 3 a) Yes. In both Earth and reference frame M, the speeds of the carts are constant. Hence, their kinetic energies are constant. b) No net external force acting on the system. Hence, no change in momentum and energy. Therefore, the isolated system containing only cart 1 is closed. c) No net external force acting on the system. Hence, no change in momentum and energy. Therefore, the isolated system containing only cart 2 is closed. 4 a) Merry-go-round: Its velocity changes, there is an acceleration towards the centre of the circle. It is a non-inertial reference frame. b) Airplane taking off: The velocity must increase. There is an acceleration associated with it. It is a non-inertial reference frame. c) Train at constant speed: The velocity is constant and no acceleration. It is an inertial frame of reference.

Hwa Chong Institution (College) MOE H3 Physics 2024 A1. Inertial Frames of Reference 2 5 a) Head-on collision in the Earth frame, By RSOA = RSOS, 2 1 1 2 1 2 1 20.80 0 0.80 u u v v v v v v − = − − = − → = + By PCOLM, 12 22 -1 2 -1 1 (0.36)(0) (0.12)(0.80) (0.36) (0.12) (0.12)(0.80) (0.36)(0.80 ) (0.12) 0.40 m s (moves rightwards) 0.40 m s (moves leftwards) vv vv v v + = + = + + =− = The relative velocity is 0.80 m s-1. -1 1 1 1 -1 2 2 2 0.40 0 0.40 m s (leftwards) 0.40 ( 0.80) 1.20 m s (rightwards) V v u V v u  = − = − =  = − =− − + =− b) -1 1 1 1 -1 2 2 2 (0.36)(0.40) 0.144 kg m s (leftwards) (0.12)( 1.20) 0.144 kg m s (rightwards) p m V p m V  =  = =  =  = − =− Please note that the total change in momentum in the cart 1 and 2 system is zero. This indicates that there is no net external force acting on the system. c) 22 ,1 22 ,2 1 (0.36)(0.40 0 ) 0.0288 J2 1(0.12)(( 0.40) 0.80 ) 0.0288 J2 k k E E  = − =  = − − =− Please note that the total change in the kinetic energy of the system is zero. This indicates that there is no energy lost in this elastic collision. d) Reference frame M moves at a constant speed of 0.20 m s -1 towards right (towards cart 2) -1 1 , 1 , , -1 2, 2, , 0 0.20 0.20 m s (leftwards) 0.80 0.20 1.00 m s (leftwards) M Earth Earth M M Earth Earth M u u u u u u = + = + = = + = + = In reference frame M, Head-on collision in the Earth frame, by RSOA = RSOS,

Hwa Chong Institution (College) MOE H3 Physics 2024 A1. Inertial Frames of Reference 3 2, 1 , 1 , 2, 1 , 2, 1 , 2,1.00 0.20 0.80 M M M M M M M M u u v v v v v v − = − − = − → = + By PCOLM, 1 , 2, -1 2, -1 1, (0.36)(0.20) (0.12)(1.00) (0.36) (0.12) 0.20 m s (moves rightwards) 0.60 m s (moves leftwards) MM M M vv v v + = + =− = Please check that the relative velocity is 0.80 m s -1 in this reference. It was the same in the Earth frame as well. This shows that the relative velocity is the same in all inertial frames of reference. Change in velocity in reference frame M, -1 1 , 1 , 1 , -1 2, 2, 2, 0.60 0.20 0.40 m s (leftwards) 0.20 (1.00) 1.20 m s (rightwards) M M M M M M V v u V v u  = − = − =  = − =− − =− Change in momentum in reference frame M, -1 1 , 1 1 , -1 2, 2 2, (0.36)(0.40) 0.144 kg m s (leftwards) (0.12)( 1.20) 0.144 kg m s (rightwards) MM MM p m V p m V  =  = =  =  = − =− 22 ,1 22 ,2 1 (0.36)(0.60 0.20 ) 0.0576 J2 1 (0.12)(( 0.20) 1.00 ) 0.0576 J2 k k E E  = − =  = − − =− Please take note that the individual change in kinetic energy in cart 1 or cart 2 is different in this frame compared to the Earth frame. However, the total change in kinetic energy is still zero . This reinforces the fact that the total change in kinetic energy is the same in all inertial frames of reference regardless of the type of collision. e) Now, the collision is elastic and the velocity of cart 1 is +0.30 m s -1 after the collision (moves towards left). In the Earth frame, by PCOLM, 2 -1 2 (0.36)(0) (0.12)(0.80) (0.36)(0.30) (0.12) 0.10 m s (moves rightwards) v v + = + =− 22 ,1 22 ,2 1 (0.36)(0.30 0 ) 0.0162 J2 1(0.12)(( 0.10) 0.80 ) 0.0378 J2 0.0162 ( 0.0378) 0.0216 J k k k E E E  = − =  = − − =−  = + − =−

Hwa Chong Institution (College) MOE H3 Physics 2024 A1. Inertial Frames of Reference 4 The loss in the kinetic energy is expected as this is an inelastic collision. The loss in kinetic energy goes to increase the internal energy of the system. Reference frame M, -1 1 , 1 , , -1 2, 2, , 0 0.20 0.20 m s (leftwards) 0.80 0.20 1.00 m s (leftwards) M Earth Earth M M Earth Earth M u u u u u u = + = + = = + = + = -1 1 , 1 , , -1 2, 2, , 0.30 0.20 0.50 m s (leftwards) 0.10 0.20 0.10 m s (leftwards) M Earth Earth M M Earth Earth M v v v v v v = + = + = = + =− + = 22 ,1 22 ,2 1 (0.36)(0.50 0.20 ) 0.0378 J2 1 (0.12)(( 0.10) 1.00 ) 0.0594 J2 0.0378 ( 0.0594) 0.0216 J k k k E E E  = − =  = − − =−  = + − =− Please take note that the energy loss in reference frame M is the same as that of in Earth frame. This indicates that the gain in internal energy is the same in all inertial frames of reference. 6 a) -1(0.36)(0) (0.12)(0.80) 0.20 m s (leftwards)0.36 0.12 CMV +== + b) -1 1 , 1 , , -1 2, 2, , -1 1, -1 2, 0 0.20 0.20 m s (rightwards) 0.80 0.20 0.60 m s (leftwards) 0.40 0.20 0.20 m s (leftwards) 0.40 0.20 0.60 m s (rightwards) CM Earth Earth CM CM Earth Earth CM CM CM u u u u u u v v = + = − =− = + = − = =−= =− − =− Please take note that the velocities simply change sign after the elastic collision in this centre of mass frame (zero-momentum frame). This always happens for elastic collision in CM frames and it simplifies tedious calculations greatly. -1 1, -1 2, -1 1 , 1 1 , -1 2, 2 2, 0.20 ( 0.20) 0.40 m s (leftwards) 0.60 ( 0.60) 1.20 m s (rightwards) (0.36)(0.40) 0.144 kg m s (0.12)( 1.20) 0.144 kg m s CM CM CM CM CM CM V V p m V p m V  = − − =  =− − + =−  =  = =  =  = − =− 22 ,1 , 22 ,2, 1(0.36)(0.20 ( 0.20) ) 02 1 (0.12)(( 0.60) 0.60 ) 02 k CM k CM E E  = − − =  = − − =

Hwa Chong Institution (College) MOE H3 Physics 2024 A1. Inertial Frames of Reference 5 7 Since it is a uniform piece of sheet metal, let the mass of each square be m. 3 (15) (5) 2 (10) 11.7 cm32 CM m m mx m m m ++== ++ 3 (5) (15) 2 (25) 13.3 cm32 CM m m my m m m ++== ++ The coordinate of the CM is (11.7 cm, 13.3 cm) which is outside of the sheet metal. 8 a) Since the rod is thin, we assume that it has negligible thickness. The infinitesimal mass of the rod: Mdm dx dx L== The centre of mass along the x-direction: ( ) 2 0 2 11 () 2 / 22 L cm cm Lx xdm x dx xdxM M M M ML LLx M = = = = ==    b) It is a non-uniform rod which has mass per unit length varying with x. The infinitesimal mass of the rod: dm dx xdx== Integrating it to find the total mass of the non-uniform rod gives 2 2 LM = 2 33 2 1 1 1 ( ) ( ) 2 33 32 cm cm x xdm x dx x xdx x dxM M M M L L Lx M L    = = = = = = =       9 a) Neglecting air resistance, the only external force acting on the projectile is the gravitational force. Thus, if the projectile did not explode, it would continue to move along the parabolic path indicated by the dashed line.

Hwa Chong Institution (College) MOE H3 Physics 2024 A1. Inertial F