EJC 2023 GCE A Level H3 Physics 9814 Suggested Solutions

©EJC 2022 9814/H2 PHYSICS 2022 PHYSICS SUGGESTED SOLUTIONS 9814 October/November 2023 1(a) (i) torque = Iθ = − mglsinθ ≈ − mglθ, where moment of inertia I = ml2 ⇒ 2 2 , where angular frequency mg g g mθ θ θ ωθ ω=− =− ≡− = l lll 2 2T g π πω⇒= = l Note 1: The negative sign in the first line is because, while the positive direction for the angle θ (as defined in the diagram) is anticlockwise, the torque is negative (clockwise). Note 2: One can also consider the oscillation of the displacement of the particle to arrive at the same conclusion. 1(a)(ii) pen, max 2 min 1 1 9.81 0.910 Hz2 30.0 10f T π −= = =× 1(b)(i) Let the charge on the capacitor be q, p.d. across the capacitor q C= 2 2emf induced in the inductor d dqLLdt dt= = I By Kirchoff’s law, 2 2 2 2 2 0 1, 11 22 q dqLC dt dq q qLCdt LC f LC ωω ω ππ += =− ≡− = = = 1(b)(ii) 36 11 1 1 2.81 Hz22 82 10 39000 10 LCf LCππ −− = = = ×× × pen, max 2.81 3.08 30.910 LCf f = = ≈ 1(b)(iii) The frequency of the LC circuit needs to be lowered, hence a larger capacitance (by connecting a few capacitors in parallel) is needed. Assume the length of the pendulum remains (the minimum value) 30.0 cm. Let the number of capacitors needed to match the frequencies be n. 23 6 1 , or 30.0 10 9.81 82 10 39000 10 9.6 g gLC LC n n −− − = = × = × × ×× × = ll Thus at least 10 capacitors are needed in parallel. To match frequencies, the length of the pendulum must be 369.81 82 10 10 39000 10 0.3137 m 31.4 cmgLC −−= = ×× ×× × = =l θ mg mgsinθ l

2 ©EJC 2024 9749/H2 PHYSICS 2024 2(a) The magnitude of the dipole moment is the product of the magnitude of either charge and the separation between the charges. 2(b) Let the net charge be q. 12 3095.8 10 0.784 1.5 3.34 10q −−×× × = ×× 19 0.6670.667 10 0. 421.6q ee−= ×= = 2(c) net electric dipole moment = 2 × 1.50D × cos104 2  =1.847D ≈ 1.85D Note 1: Electric dipoles are vectors. We are adding up two vectors of equal length (1.50D), with an angle 104° between them. Note 2: As a show question, it is important to show intermediate value (1.847D) before rounding to the value that we are trying to show. 2(d)(i) Note: Oxygen atomic mass = 16, hydrogen atomic mass = 1. With two hydrogen atoms, total atomic mass = 2. Hence, the centre of mass is much closer to the oxygen atom (negative charge). Distance ratio should be roughly 1:8. 2(d)(ii) The two electric forces form a couple. torque sin sinFd E r pEδθ θ= = ×= 2(d)(iii) Maximum rotational kinetic energy is attained at θ = 0°. Method 1: max KE = potential energy at 63° − potential energy at 0° = −pEcos63° − (−pEcos0°) = pE(cos0° − cos63°) = 1.85 × 3.34 × 10−30 × 9500 × (1 − cos63°) = 3.21 × 10−26 J Method 2: 3 30 26 0 6 1.85 3.34 10 9500 1 cos63 3.21 10 J max KE sin (cos0 cos63 ) () pE d pEθθ − − = −= − = × × ×− ° = × × ∫    3(a) X N F Note 1: Since the object is moving at a high speed, a large centripetal force is required. Hence, the frictional force must have a leftward component to contribute to the centripetal force. Note 2: Normal force is normal to the slope and upwards. Its vertical component is equal in length to weight.

3 ©EJC 2022 9749/H2 PHYSICS 2022 3(b) The centripetal force is the sum of the horizontal components of F and N: c cos sinFF N θθ= + The maximum frictional force is Nµ , hence the maximum centripetal force is 2 max c, max cos sin vF N Nm rµθ θ= += Consider the vertical direction: sin cosN N mgµθ θ− += Taking ratio between the above two equations and simplify, max ( tan ) 1 tan rgv µθ µθ += − . 3(c)(i) Using the expression for vmax above, for 12° ≤ θ ≤ 45°: Since tanθ increases monotonically with θ, tan 1 tan µθ µθ + − increases monotonically with θ , since the numerator increases, and denominator decreases, as θ increases. Hence, the optimal value for θ is 45°. Since vmax ∝ r , the optimal value for r is the maximum value: optimal 20 (0.20 0.70)cos 45 20.6 mr = + + °= . Note: The above arguments are a little sketchy, since r is also a function of θ: 20 (0.20 0.70)cosr θ= ++ tan 1 tan µθ µθ + − increases with θ, but 20 (0.20 0.70)cosr θ= ++ decreases with θ. Ideally, one needs to replace r by the above function of θ in the expression for v max, differentiate the resulting expression w.r.t. θ and set the derivative to 0, to solve for the optimal value for θ at the stationary point. But the resulting equation is too tedious to solve. Alternatively, one can plot the resulting function using a graphing calculator, and discover that it monotonically increases with θ. In other words, our conclusion in our solution is correct. 3(c)(ii) max ( tan ) 20.6 9.81 (0.70 tan45 ) 1 tan 1 0.70 tan45 rgv µθ µθ + ×× + °= =− −× ° 120.6 9.81 (0.70 1) 33.8 m s1 0.70 −×× += =− 4(a) By conservation of momentum, 01 2 2 12 2 2 2 cos , being the angle between and hori zontal 2. 4 mv mv mv v xmv mv L x θθ= + = +  + Hence, 01 2 2 2 ( ) , where ( ) 2 4 xv v fx v fx L x = + =  + .

4 ©EJC 2024 9749/H2 PHYSICS 2024 4(b) time taken for the two balls to reach middle pockets = time taken for the cue ball to reach the right side: 2 2 12 2 21 2 4 22 4 L LLx xx vvvv Lx +  ++ = ⇒= +  2 2 0 12 1122 22 01 1 422 244 261. 2 2 Lx xxv vv vv LLL xxx x Lxvv v L Lxx + =+ = +× ×  +++     += +=  + + 4(c) Kinetic energy remains constant before and after an elastic collision: 2 2 2 22 2 0 1 2 01 2 11 1 2 or 222 2mv mv mv v v v= +× = + Using equations in parts (a) and (b) to express v0 and v2 in terms of v1, simplify and rearrange: 2 224 8 0 2 Lx xL+ −= Solving the quadratic equation for x, ( 7 2)12 Lx = − . Hence, b = 7. 4(d) Yes. The frictions on the balls are of equal magnitude. Since the cue ball moves a longer distance, friction does more negative work on it. The cue ball slows down more than the other two balls, hence will hit the right-hand cushion after the two balls have been potted. 5(a)(i) 3 228360 3.0 10 (12.0 10 ) 0.361 kgM −−= ×× × × = Using parallel axis theorem, 22 22 22 22 2 moment of inertia 12 12 (12.0 10 )0.361 ( 25.0 10 ) 0.02299 0.023 kg m12 Ma aMl M l − − = += +    ×=× +× = =  5(a)(ii) mass of one cylindrical rod = 23 26.0 108360 25. 0 10 0.0591 kg2π − − ×× ××=  moment of inertia due to a single cylindrical rod about one end ( ) 22 2211 0.0591 25.0 10 0.00123 kg m33ML −= = × ×× = total moment of inertia = (0.023 + 0.00123) × 8 = 0.193 = 0.19 kg m2.

5 ©EJC 2022 9749/H2 PHYSICS 2022 5(a)(iii) By conservation of energy, loss in GPE = gain in rotational kinetic energy 2 1 12 2 4 4 1.81 9.81 11.0 64.1 rad s0.19 mgh mgh ω ω − = ×××= = = I I 5(a)(iv) Assume the angular acceleration α of the paddles to be constant. The angular acceleration α of the paddles and the linear acceleration a of the masses are related by a = rα, r being the radius of the drum. Let the time for the mass to move 11.0 m be t. 2 22 64.1 1 1 1 8.6 10 64.1 11.0 7.98 8.0 s.22 22 t at r t t t α α − = ×= = × × ×= ⇒= = 5(a)(v) no. of rounds one paddle turns 2 11.0 11.0 40.78.6 10dπ π −= = =×× volume of air displaced by one paddle 2 2 22 3no. of rounds 2 40.7 2 25.0 10 (12.0 10 ) 0.92 mlaππ −−= × ×= ×× × × × = 5(b)(i) mass of water mwater = 3999 93.2 10 93.1 kg−××= Assuming all GPE lost is converted into internal energy of water, and there is no heat loss. water water 2 2 2 1.81 9.81 11.0 0.001 K4190 93.1 mgh cm T mghT cm = ∆ ×××∆= = =× 5(b)(ii) 2 22 12 0.3052 0.000432 2 2 9.81 11.0 mv v mgh gh × = = =×× 1 0.07613 = Since the heat capacity of the apparatus is a larger fraction of the total heat capacity than the kinetic energy of the mass is of the total GPE loss, the former will have a lar