EJC 2026 JC2 H2 PHYSICS PRELIM P2 MS

2026 EJC H2 Physics P2 Solutions - Review Copy 2026 EJC H2 Physics P2 Solutions Qns Answer Marks 1(a)(1) At t = 0, the gradient of the graph is 0. Hence the speed of the ball is 0. Marker’s comment: This question was generally well done. However, students should be mindful that this is a distance-time graph and thus, speed should be used instead of velocity when justifying what the gradient represents. A handful of students mentioned that when t = 0, d = 0 and thus, speed is 0. This is untrue as an object can go at some speed at its initial position. A small minority mentioned that speed = d t = 0 0 = 0 which is untrue. A1 1(a)(2) The gradient of the graph eventually becomes constant after the t = 0.80 s mark, indicating that the ball has reached terminal velocity. The acceleration, and thus the resultant force is zero. Since Fnet = mg – Fair = 0, this shows that air resistance is significant and balances the weight when terminal velocity is reached. Marker’s comment: This question was not answered well as students did not refer to Fig. 1.1 to support their explanation. These students were not awarded any marks. A handful of students mentioned that the speed is increasing at an increasing rate which is incorrect. Even though the speed is increasing (as indicated by the gradient of the distance - time graph), it increases at a decreasing rate since the acceleration is getting smaller due to a larger air resistance (that increases with speed). M1 A1 1(b) Draw a tangent to Fig. 1.1 at t = 0.40 s and determine its gradient. Speed = (2.8 ± 0.3) m s⁻¹. Marker’s comment: This question was generally well done though students are reminded to draw the tangent at a point in solid line. The tangent should also be long enough for students to easily identify suitable coordinates to find the gradient. Students who did not arrive at the answer either used a ratio � 𝑑𝑑 𝑡𝑡 � or attempted to use the kinematics formula (e.g. v2 = u2 + 2as) provided in the formula sheet. M1 A1

2026 EJC H2 Physics P2 Solutions - Review Copy Qns Answer Marks 1(c) Marker’s comment: Performance on this question was weak, with most students drawing a straight line that omitted the effect of air resistance. Students who deduced the general shape of the graph often failed to show a curve that becomes progressively less steep, maintaining a gradient at t2 that was incorrectly steep. *Note: At t2 the gradient should not be zero because Fig. 1.1 shows that at d = 2.0 m the object has not quite reached terminal speed. Students were not penalized for missing this intricacy, but they are reminded to check for such details in future questions. M1 A1 A1 TOTAL FOR QUESTION 1 8 2(a) The rate of change of momentum of a body is (directly) proportional to the resultant force acting on the body and is in the same direction as the resultant force. Marker’s comment: This question was generally well done. A small minority did not mention that the force is a resultant force , which is crucial when stating Newton’s second law of motion. In addition, some students substituted "in the direction of the resultant force" with "in the direction of the change in momentum" . Credit was not awarded for this, as explicitly stating the formal is a strict requirement when defining Newton’s second law of motion. B1 2(b)(i) The sand initially has zero horizontal momentum but accelerates to the speed of the belt. Therefore, the belt exerts a forward horizontal force on the sand to produce this change in momentum. By Newton’s third law, the sand exerts an equal and opposite horizontal force on the belt. Marker’s comment: This question was not well answered. Students who were not credited often stated that the belt exerts a horizontal force on the sand without explaining how that force came about. Additionally, many wrote that the initial momentum was zero, failing to recognize that the falling sand has vertical momentum. The term horizontal momentum must be specified. Lastly, partial-credit responses identified the force acting on the sand but stopped short of applying Newton’s third law to explicitly state the corresponding force acting on the belt. M1 A1

2026 EJC H2 Physics P2 Solutions - Review Copy Qns Answer Marks 2(b)(ii) ( )( )= ∆= = 0 100 0 050 0 0050 N mF v. . t . Marker’s comment: This question was not well answered. Students who were not credited often provided values without showing the equation. This is incorrect as a show question requires proper workings that lead to the answer. This is a strict requirement by Cambridge. Students are also reminded to leave the final answer in the correct significant figures as required by the question. Marks were not penalized for this mistake this time round. M1 A0 2(b)(iii)1 P = Fv = 0.0050 × 0.050 = 2.5 × 10−4 W Marker’s comment: This question was well answered with full credit given. B1 2(b)(iii)2 In one second, rate at which sand gains kinetic energy, K: ( ) ( ) − ∆ = = −   = −       = × 2 22 2 4 1 12 2 1 100 5 0 02 1000 100 1 25 10 W fi mv mK vvtt . . Marker’s comment: This question was not well answered. Students who lost marks often applied the standard formula for kinetic energy ( Ek= 1 2 mv2) without accounting for the fact that the question asked for the rate at which the sand gains kinetic energy. Consequently, the working required an explicit rate term per unit time, for example, using m t or Ek t . Additionally, some candidates incorrectly assumed that the rate of kinetic energy gain is equal to the rate at which the motor does work. This is incorrect because half of the work done by the motor is dissipated to the internal store of the surroundings. B1 2(b)(iv) Perfectly inelastic collision. Energy transferred mechanically via work done by the motor is to the (horizontal)l kinetic store of the sand and to the internal store of the system (heating/friction). Marker’s comment: This question was not well answered. For the first mark, students were penalized for failing to explicitly state that the collision was perfectly inelastic. Including the word "perfectly" is crucial, as the sand and belt coalesce upon contact. For the second mark, credit was lost when candidates incorrectly claimed that energy was transferred from the belt's kinetic store to the sand. Because the belt moves at a constant horizontal velocity, its kinetic energy remains constant. Thus, the energy gained by the sand originates directly from the work done by the motor. Additionally, some candidates irrelevantly discussed the sand's energy in the gravitational potential store or its kinetic store from falling vertically. This failed to address the core requirement of the question, which strictly concerned accelerating the sand horizontally up to the speed of the belt. B1 B1

2026 EJC H2 Physics P2 Solutions - Review Copy Qns Answer Marks TOTAL FOR QUESTION 2 8 3(a) Electric potential at a point is the work done per unit charge by an external force in bringing a small positive test charge from infinity to that point. Marker’s comment: This question was not well answered. Most common mistakes include the exclusion of “by an external force”. B1 3(b)(i) ( ) 2 1 gradient of graph 2 50 0 [M1]0 2 50 10 100N C [ A1] dVE dr E . . − − =− = −= −× = Acceptable range: 88 <=E<= 1105 Direction: in positive x-direction / towards right / from left to right / → [B1] Marker’s comment: This question was generally well answered. Common mistakes include failure to convert cm to m in the denominator. A handful of students also applied the isolated point charge formula of E = Q/4πε r2 which is not applicable for this case. M1 A1 B1 3(b)(ii) The graph is symmetrical. Marker’s