JPJC 2026 Motion and Forces Tutorial Solutions

Kinematics Tutorial soln 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS/8867 H1 PHYSICS MOTION AND FORCES TUTORIAL SOLUTIONS Part 1: Motion (Kinematics) Self-Check Questions S1 Displacement: The change in position of an object in a particular direction. Velocity: The rate of change of displacement. Acceleration: The rate of change of velocity. It can be an increase or decrease in speed, or a change in direction. S2 Information that we can obtain from • Displacement-time graph: o Position of an object at a particular instance o Average velocity of an object over a time interval o Instantaneous velocity of an object via the gradient at that instance o How the velocity is changing by analysing the change in gradient for the graph trend • Velocity-time graph, we can obtain: o Velocity of an object at a particular instance o Average Acceleration of an object over a time interval o Acceleration of an object via the gradient at that instance o Change in d isplacement of an object over a time interval via the area under graph for that time interval o How the acceleration is changing by analysing the change in gradient for the graph trend • Acceleration-time graph, we can obtain: o Acceleration of an object at a particular instance o Change in velocity of an object over a time interval via the area under graph for that time interval S3 For kinematics graphs, (1) gradient of velocity-time graph = acceleration a and (2) area under velocity-time graph = displacement s For the velocity-time graph shown, gradient = a = vu t − − 0 area under graph = s = ( )1 2 v u t+ Hence, vua t −=  v u at=+ - - - - (1) ( )1 2s v u t=+ - - - - (2) Substituing (1) into (2) by replacing v, 21 2s ut at=+ - - - - (3) Substituting (1) into (2) by replacing t, 22 2 v u as=+ - - - - (4) A velocity-time graph illustrating uniform acceleration v u 0 t velocity time

Kinematics Tutorial soln 2 Self-Practice Questions 1 Velocity is determined by gradient of s-t graph. Gradient in first half of s-t graph is positive and decreasing in magnitude, second half is negative and increasing in magnitude. Gradient is zero in the middle. [ Ans: C ] 2 For the portion of the graph between 0 and point B, acceleration is positive, implying that acceleration is in the same direction as direction of motion of object. Hence, object’s speed keeps increasing until it reaches its maximum speed at point B. After point B, acceleration becomes negative. This implies that the acceleration is now directed opposite to the direction of motion. Hence, speed of the object will start to decrease after point B. [ Ans: B ] 3 Find T: −35.0 = (0.0 − 23.0) / (T − 0.250) T = 0.9071 s Minimum stopping distance = area under graph = 23.0 x 0.250 + ½(0.9071 − 0.250) x 23.0 = 13.3 m [ Ans: 13.3 m ] 4 5 (a) (i) Take downward as positive, given u = 0 m s−1, s = 125 m, a = g = 9.81 m s−2. Using s = ut + 1 2 at2 → t = 2 2(125) 9.81 s g = = 5.05 s (ii) Using v = u + at , final velocity is v = 0 + (9.81)(5.05) = 49.5 m s−1 v / m s −1 t / s 23.0 T 0.250 10 5 7 2 t / s 5 -5 a / ms−2 27.5 10 0 t / s 10 5 50 40 s / m 2 7

Kinematics Tutorial soln 3 (b) Take downward as positive, given s = 125 m, a = g = 9.81 m s−2, t = 4.0 s. Using s = ut + 1 2 at2 , we have 125 = u(4.0) + 1 2 (9.81)(4.0)2 → the initial downward velocity, u = 11.6 m s−1 (c) (i) Take downward as positive, given u = −40.0 m s−1 , s = 125 m, a = g = 9.81 m s−2 Using s = ut + 1 2 at2 , we have 125 = (−40.0)t + 1 2 (9.81)t2 → 4.905t2 − 40t −125 = 0 → t = 10.6 s or t = −2.41 s (rejected since t > 0) (ii) Take upward as positive, given u = 40.0 m s−1 , a = −9.81 m s−2 Using v2 = u2 + 2as , let h be the maximum height from point of projection. 0 = 40.02 + 2(−9.81)h → h = 81.5 m Maximum height reached from bottom of cliff = 125 + 81.5 = 207 m 6 (a) sA = (40)(20) = 800 m (b) aB = 50 25 20 − = 1.25 m s−2 (c) sB = 1 2 (20)(25 + 50) = 750 m (d) For B to catch up with A, they must have the same displacement from t = 0. sA = sB 800 + 40t = 750 + 50t t = 5.0 s (e) sA = sB = 800 + 40(5.0) = 1000 m (f) sA = 40t OR Let t1 be the time when the two graphs sB = 25t + 1 2 (1.25)t2 = 25t + 0.625t2 intersect, when speed of B is 40 m s−1. sA−B = 40t − 25t − 0.625t2 So, aB = 1 40 25 t − = 1.25 (from (b)) = 15t − 0.625t2 → t1 = 40 25 1.25 − = 12 s For maximum sA−B , let ABds dt − = 0 Before t1, B is slower than A. After t1, B → 15 − 2(0.625)t = 0 → t = 12 s is faster than A. So the maximum Maximum sA−B = 15(12) − 0.625(12)2 distance between cars occurs at t1. = 90 m Refer left column for answer. Discussion Questions Rectilinear Motion 1 (a) Yes it is possible. An object is moving with constant velocity on a frictionless surface or an object falling with terminal velocity in air. (b) Yes it is possible. An object thrown upwards and resting momentarily at its maximum height experiences acceleration of free fall.

Kinematics Tutorial soln 4 (c) Yes it is possible. (1) An object being thrown upwards experiences acceleration of free fall. Taking upward to be positive, its velocity is positive while its acceleration is negative. (2) A car (moving to the right) that is slowing down. Taking right to be positive, the velocity is positive (in the direction of motion) while the acceleration is negative (in the opposite direction to that of motion). Refer notes p7: graph (e). 2 (a) (i) The height may be determined by the area under the graph prior to hitting the ground (i.e. height of drop = ( )( )11 1 2 vt ). (ii) t4 (b) 3 (a) (i) Let t1 be the time taken for the bottom edge to reach the light beam. Using s = ut + 1 2 at2 , where u = 0 m s−1, s = 1.00 m , g = 9.79 m s−2 ( )= = =st. g. 1 1 212 0 4529 79 s (ii) Let t2 be the time taken for the top edge to reach the light beam. Since the time taken for plate to pass through the beam is 0.052 s, t2 = t1 + 0.052 = 0.504 s Using s2 = ut2 + 1 2 a(t2)2 , where u = 0 m s−1, a = g = 9.79 m s−2 s2 = 0 + 1 2 (9.79)(0.504)2 → s2 = 1.24 m Length of metal plate = 1.24 − 1.00 = 0.24 m (b) 1. The effect of air resistance may cause the time taken to be of a higher value. 2. The metal plate does not fall vertically downwards with the base remaining horizontal throughout. It may tilt as it falls and cause the time to be different. 4 (a) (i) Taking upward to be positive, v u as=+22 2 0 = 20.02 + 2(−9.81)s Therefore, maximum height s = 20.4 m. (ii) 21 2=+s ut at 0 = 20.0 t + 1 2 (−9.81)t2 4.905 t2 – 20 t = 0 t = 0 or 4.08 s Therefore, the time required is 4.08 s. s t t1 t2 t3

Kinematics Tutorial soln 5 (b) (i) (ii) Part 2: Forces (Dynamics) Self-Check Questions S1 1st law: a body at rest will stay at rest, and a body in motion will continue to move at constant velocity, unless acted on by a resultant external force; 2nd law: 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; 3rd law: the force exerted by one body on a second body is equal in magnitude and opposite in direction to the force simultaneously exerted by the second body on the first body. S2 The mass of an object is a measure of its inertia. The weight of an object refers to the gravitational force exerted on it by the Earth. The mass of an object remains the same regardless of its location but the weight of the object depends on the strength of the gravitational field it is placed in. S3 The tw