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Raffles Junior College Last updated 26-Oct-07 1 H2 Physics Promotion Examination 2007 (Section A) Question Number Key 1 D 2 A 3 B 4 B 5 D 6 C 7 C 8 B 9 C 10 A 11 A 12 D 13 C 14 C 15 A 16 D 17 B 18 A Answers: 1. ( ) 2 2- 3 - 2 -1 -1 -1 2 9 (m )(kg m )(m s ) Units of kg m s (m s ) sf t rg v ρρ η = == 2. () 22 3 3 3 3 (25.15) Volume of the pipe, 150.80 74914.727 mm 44 2 0.05 0.02 2 25.15 150.80 307.8 mm 300 mm (to 1 sig. fig.) (74900 300) mm D VL VD L L VV V ππ ⎛⎞ = ⎜⎟ ⎝⎠ ΔΔ Δ =+ ∴Δ = + = = ∴= ±

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Raffles Junior College Last updated 26-Oct-07 2 3. -1 22 Consider the horizontal motion of the arrow. 45 m The time taken for it to hit the target, 1.8 s 25 m s 11 X (9.81)(1.8) 16 m (to 2 sig. fig.) t gt ⎛⎞ == ⎜⎟ ⎝⎠ = 4. Positive displacement (e.g. object moves to the right) starts from initial position to the time at point B , where the object is instantaneously at rest. The displacement is negative (e.g. object moves to the left) after the time at point B till the time at point D . Thus at point B the object has the largest displacemen t to the right of its initial position. The magnitude of this largest displacement corresponds to area under the v-t graph between 0 and time at point B . (Give a sketch of displacement-time graph to substantiate.) 5. Momentum of system after collision = momentum of system before collision [Conservation of linear momentum] = momentum of ball B p = Kinetic energy of system after collision = 2 2 2 2(7 ) 7 7 p m p E m 2 (momentum) [ Energy = ] 2(mass) Momentum of ball B after collision (common velocity after collision) velocity of ball B () 7 7 m m p = = = [Completely inelastic collision] Energy of ball B after collision = 2 2 2 7 24 9 4 9 p p m E m ⎜⎟ ⎜ ⎝⎠ ⎝
Raffles Junior College Last updated 26-Oct-07 3 6. The mast is in equilibrium. Let T F and T R to be the tension in the front and rear cable respectively. Resultant torque about the axis perpendicular to paper through point P is 0. 33 [(10)sin30 ] [(10)sin45 ] 0 sin30 sin30 (5.0x10 ) 3.5 x10 N sin45 oo FR RF TT −= == = sin30 = implies the horizontal component of the force exerted by the boat on the foot of mast is 0, i.e. the force exerted by the boat on the foot of mast is vertical. Let this force by F P . Resultant force on the mast along the vertical direction is zero. 3 cos30 cos45 0 6.8 x 10 N PF R R FT T T −−= =+= According to Newton’s 3 rd law, the force exerted by the foot of mast on the boat is 6.8 x 10 3 N, vertically downward . 7. The uniform rod is in equilibrium. Resultant torque about the axis perpendicular to paper through the base of rod is 0. ( sin35 ) 400 350 0 2 175 400 1002 N sin35 o o L TL L T ⎛⎞ ⎜⎟ ⎝⎠ + 8. Loss in kinetic energy of object = Gain in gravitational potential energy of object + Energy converted to heat due to friction, E 22 12 1 () 2 1 2 mv v mgh E Em v v m g h + =−

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Raffles Junior College Last updated 26-Oct-07 4 9. -2 power output power input 0.70 100 W (0.70)(100 W)(10x60 s) 428 kg (9.81 m s )(10 m) mgh t m η = = == 10.
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