NJC Motion and Forces Exercise Solutions 2025

National Junior College Science Department | Physics 3. Motion and Forces Exercises Solution E1 (a) displacement (direction: SW; magnitude: 200 km) (b) speed (c) velocity (direction: along straight edge of table; magnitude: 2 mm s–1) Note: The description of the direction may be unclear (not sure exactly which direction along the straight edge ) but there is undoubtedly a mention of the direction. (d) distance E2 Speedometer shows the speed your car is moving at. It does not provide information on the direction of motion. E3 Draw a vector diagram (a) distance = 3.0 + 4.0 = 7.0 km (b) magnitude of displacement |𝑠| = √32 + 42 = 5.0 km tan 𝜃 = 4 3 ⇒ 𝜃 = 53° displacement is 5.0 km 53o east of north E4 average acceleration 𝑎̅ = Δ𝑣 Δ𝑡 = 𝑣𝑓−𝑣𝑖 𝑡 = 9−0 1.5 = 6.0 m s−2 E5 average acceleration 𝑎̅ = Δ𝑣 Δ𝑡 = 𝑣𝑓−𝑣𝑖 𝑡 = 0−115×103 60×60 1.5×60 = −0.35 m s−2

National Junior College Science Department | Physics E6 speed = distance travelled time taken = 2𝜋×1.5×1011 365×24×60×60 = 3.0 × 104 m s−1 = 30 km s−1 In the course of one year, its displacement is zero, so its average velocity is zero. Note: As the Earth orbits the Sun, its direction of motion keeps changing. Hence its instantaneous velocity keeps changing. E7 average speed = distance travelled time taken = 20×2+40×2+60×6 2+2+6 = 48 m s−1 E8 s-t graph is a straight line through the origin. velocity = gradient of s-t graph = 340−0 4−0 = 85 m s−1 E9 (a) (b) acceleration for first 10 s = gradient of straight line from 0 to 10 s = 30−0 10−0 = 3.0 m s−2 (c) acceleration for last 15 s = 0−30 30−15 = −2.0 m s−2 (d) total distance = area under graph = 1 2 × (5 + 30) × 30 = 525 m

National Junior College Science Department | Physics E10 Draw a tangent to the curve at point P acceleration = 320−60 12−0 = 21.7 m s−2 E11 (a) We know u, a and t and we want to know v, so we use the equation 𝑣 = 𝑢 + 𝑎𝑡. Velocity 𝑣 = 0 + 2.0 × 10 = 20 m s−1 (b) We know u, a and t and we want to know s, so we use the equation 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2. Distance 𝑠 = 0 × 10 + 1 2 × 2.0 × 102 = 100 m (c) We know u, v and a and we want to know t, so we rearrange the equation 𝑣 = 𝑢 + 𝑎𝑡. Time 𝑡 = 24−0 2.0 = 12 s E12 (a) We know u, v and t and we want to know a, so we use the equation 𝑣 = 𝑢 + 𝑎𝑡. Acceleration 𝑎 = 20−4.0 100 = 0.16 m s−2 (b) Average velocity 𝑣𝑎𝑣𝑔 = 𝑣+𝑢 2 = 20+4.0 2 = 12 m s−1 (c) [Method 1] Distance = average speed × time = 12 × 100 = 1200 m [Method 2] We know u, v and t and we want to know a, so we use the equation 𝑠 = 1 2 (𝑢 + 𝑣)𝑡. Distance 𝑠 = 1 2 (4.0 + 20) × 100 = 1200 m (12,320)× (0,60)×

National Junior College Science Department | Physics E13 (a) We know s = 0.80 m (↓) and a = 9.81 m s–2 (↓), and that u = 0, and we need to find t. Use equation 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 and take downwards as positive. 0.80 = 0 × 𝑡 + 1 2 × 9.81 × 𝑡2 Time 𝑡 = √2×0.80 9.81 = 0.40 s. (b) We know s = 0.80 m (↓) and a = 9.81 m s–2 (↓), and that u = 0, and we need to find v. Use equation 𝑣2 = 𝑢2 + 2𝑎𝑠 and take downwards as positive. Impact velocity 𝑣 = √02 + 2 × 9.81 × 0.8 = 4.0 m s−1. E14 [Method 1] Drop an object towards the sensor, but take care not to break it (e.g. use a softer material such as plasticine ball) [Method 2] A better method is to use a sloping ramp with a low friction trolley. Resultant force = mg sin θ Acceleration = mg sin θ / m = g sin θ Gradually increase the angle of slope. Deduce the value of the acceleration when as the ramp approaches vertical (θ → 90o). E15 Define initial velocity direction as positive. change in momentum = 0 − 4.5 × 10−3 × 0.12 = −5.4 × 10−4 𝑘𝑔 𝑚𝑠−1 The negative implies that the average force is opposite to the initial velocity.

National Junior College Science Department | Physics E16 Answer: D constant velocity implies zero resultant force E17 Answer: B uniform acceleration means acceleration is constant resultant force = mass × acceleration = constant E18 Answer: B constant velocity implies zero resultant force E19 Answer: B resultant force = 12 – f ma = 12 – f f = 12 – 0.60 × 4.0 = 9.6 N E20 Answer: D Resolve forces along the incline plane, resultant force = mg sin θ – friction ma = mg sin θ – R a = g sin θ – R / m frictional force f θ R friction θ mg a

National Junior College Science Department | Physics E21 (a) acceleration is downwards for this question, so gravitational force > force by cable resultant force = mass × acceleration 500 × 9.81 – F = 500 × 3 F = 3400 N (2 s.f.) (b) even though lift is moving downwards, acceleration is upwards (since slowing down), so gravitational force < force by cable resultant force = mass × acceleration F – 500 × 9.81 = 500 × 2 F = 5900 N (2 s.f.) force exerted by cable, F gravitational force = 500g acceleration = 3.0 m s–2 force exerted by cable, F gravitational force = 500g acceleration = 2.0 m s–2

National Junior College Science Department | Physics E22 Answer: D force diagrams of both masses together and individually, (leftmost diagram) acceleration of X and Y, a = resultant force / mass = F / 4m Consider horizontal forces on block Y, 𝐹X on Y = resultant force = 3𝑚𝑎 = 3𝐹/4 E23 (a) resultant force = 30 – (6 + 8) = 16 N (right) (b) acceleration = 16 / (2 + 4) = 2.7 m s–2 (2 s.f.) (right) (c) The ends of the string (assumed massless) exert force of equal magnitude F on the blocks. The free body diagrams of the two blocks are shown. Note: The two forces F are not action-reaction pair. Method 1 (forces acting on the 4 kg block) resultant force = mass × acceleration 30 – F – 8 = 4 × 2.6667 F = 11 N (2 s.f.) Force exerted by the string is 11N to the left. Method 2 (forces acting on the 2 kg block) F – 6 = 2 × 2.6667 F = 11 N (2 s.f.) Magnitude of force acting by the string on the 4 kg block equals F, so the force is 11N to the left. X YF a X Y F a FY on X FX on Y a 2.0 kg 4.0 kg 30 N 6.0 N 8.0 N F acceleration acceleration F

National Junior College Science Department | Physics E24 Take left as positive, force on stream 𝐹 = Δ𝑚 Δ𝑡 (𝑣𝑓 − 𝑣𝑖) = 2.0 × [5 − (−5)] = 20 N By Newton’s third law, force exerted by water is 20 N and directed to the right. E25 Mass flow rate = Δ𝑚 Δ𝑡 = 𝜌𝐴𝑣 Force exerted on the hose by water = force exerted on water by hose = Δ𝑚 Δ𝑡 (𝑣𝑓 − 𝑣𝑖) = 𝜌𝐴𝑣(𝑣 − 0) = 1000 × 𝜋 ( 0.01 2 ) 2 × 0.502 = 0.020 N wall 5.0 m s–1 pipe 5.0 m s–1 force on wall by water force on water by wall