Solutions: Option C When a body is released from rest, the only force acting on the body is its weight due to gravity . Both bodies experience acceleration due to gravity . Hence for both bodies (regardless of mass), will have the same initial acceleration of 10 m/s2. As the two bodies are of the same size and shape , they will  experience the same air resistance for any particular speed . As speed increases, air resistance increases. For terminal velocity to be reached, air resistance has to be equal to the weight . Since weight is greater for the ball with larger mass, the air resistance has to be bigger. Thus, the ball has to accelerate more (air resistance increases with speed) for the larger air resistance to be equal to the weight. Hence the ball with larger mass will have larger terminal velocity . [NOTE] Do not confuse ‘speed of the body is independent of the mass’ as learned in Work Done, Energy and Power. This concept is based on the assumption that there is no air resistance. So not applicable in this question as for terminal velocity to occur, air resistance must be present.

When comes to filament lamp, the non-ohmic characteristic properties comes to mind. But for this quesitons, you have to understand that the p.d. across the lamp remains constant. Refer to the video for explanation. Answer: Option: A

For these type of questions, it is important to identify which tyre/wheel is connected/powered by the engine or the person. On each tyre or wheel, there are actually two types of frictional force. Frictional force on wheel Frictional force on road/ground You can view the explanation plus the 4 examples from Olevel p1 in the video below. Examples used in the video. 2007PP P1Q5 Answer: Option D 2004PP P1Q5 Answer: Option D 2020PPp1q6 Answer: Option B 2021PP P1Q7 Answer: Option C

Instead of P = IV to explain, it might be easier to explain using P = I^2R when the components are connected in series where current I is constant. Hence power P is directly proportional to R. The bigger the R, the more power it uses. Likewise, if the components are connected in parallel, it will be easier to use P = V^2/R, as the potential difference is constant for the components connected in parallel. Hence, power P is inversely proportional to R. The smaller the R, the more power it uses. SP2021Q8 PP2017Q11

Resistors may be connected in series or parallel. When asked which resistor has higher power (hence higher energy), in general we will think of using P = IV. But when using P = IV, it cannot relate to resistance R of the resistors directly. Hence it will be more appropriate for us to P = I^2R or P = V^2/R. For resistors in series , the current is the same. Hence it will be more direct to compare P and R using P = I^2R , where P is directly proportional to R . For resistors in parallel , the p.d. is the same. Hence it will be more direct to compare P and R using P = V^2/R , where P is inversely proportional to R . Refer to the video for explanation

When two components are connected in series, where one is a fixed resistor and the resistance of the other can vary (e.g. thermistor, light dependent diode, variable resistor etc), you always make use of the fixed resistor to explain the its p.d. as resistance is constant. Then use the concept of sum of p.d. across both components is equal to the e.m.f of the circuit to explain how the p.d. across the other component varies . Though thermistor is not in SciPhy syllabus, it can still be tested as long as the information on thermistor is given. Refer to the videos below where 2 are from Sciphy and the other is from Pure. SP2020P2Q9 SP2013P2Q11 PP2008P2Q11OR

Using the concepts that POM, we know 1) Ma = Mc 2) net force on the stick is 0N. Most of us will most likely to take pivot about P as default, and find F, then P using the concept of net force = 0N. But a body is in equilibrium, you can actually take pivot about any point in the body to do calculation based on Principle of Moments (POM). This method is especially using to find the answer for P straight away. The video below will guide you the normal and the short-cut. Answer: Option D

Instead of imaging how the particles will move and guess their motion, there is a technique which can help you to determine that. In addition, when particles are in phase, it means that both particles on the wave have the same velocity and same displacement. It means both have the same speed in the same direction and same distance and direction away from the rest position. Particles that are out of phase means both are having the same speed but in opposite direction. Refer to the video below to find out more. Question 01: 2005/2008 PP P1 Q20/Q18 Question 02: 2018 PP P1 Q24

These 2 questions are actually the same. Q23 is from 2010 Pure Physics P1 while Q11 is from 2014 Sci Physics P1. Take a look at these 2 questions. If you are not sure, view the video below for the explanation. Answer to Q23: Option A Answer to Q11: Option D If you do not know how to answer these 2 questions, view this video and also refer to the lens summary below.

2016PPP1Q40 Solutions: Option C [There is another similar question in 2013 Nov Pure Physics P1Q40. The answer is Option C too. Refer to the last section of this post] Refer to the 2 videos below. The 1st one is faster if you know that the bigger the resistance, by proportion, the bigger the potential difference of the component as it will take a larger portion of the emf wrt to the other components in series. This method is especially useful for MCQ. If you are still unsure, you may put in values to find the I, V and P across the components. This will be more tedious and time-consuming. Another similar question is 2013 Nov Pure Physics P1 Q40.

Below are some examples of questions that involve finding the resultant force due to two or three forces. Below are the questions in the video. Any question you are unsure, you can just watch the respective section of the video based on the time stated. (1) 2010PPp1q2 similar to 2002PPp1q7 (0s to 3.02 min) Ans: Option D and D respectively (2) 2013SPp1q2 (3.02min to 4.25 min) Ans: Option A (3) 2005PPp1q10, 2012SPp1q1, 2018PPp1q2 (4.25 min to 6.40 min) Ans: Option A (3) 2014PPp1q8 (6.40 min to 7.48 min) Ans: Option B (4) 2019PPp1q2 (7.48 min to 8.22 min) Ans: Option B ( 5) 2020PPq1q3 (8.22 min to end) Ans: Option B

(1) 2003PPq1q7 (0 min to 2.42 min) Ans: Option A (2) 2023PPp1q4 (2.47 min 3.54 min) Ans: Option C (3) 2018PPp1q2 (3.45 min to 6.22 min) Ans: Option D (4) 2011PPp1q6, 2016PPp1q6 (6.22 min) Ans: Option A