Physics 1st Ch 04 Newtonian Mechanics

Physics - First Paper Chapter 4: Newtonian Mechanics Selected Questions and Answers (HSC English Version) Chapter 4: Newtonian Mechanics B. Creative Questions 1. Mr. Anis is an efficient hunter. He went to the Sundarban for hunting deer and during firing a bullet weighing 10 g with speed 10 m/s from his gun weighing 10 kg, there was a recoil in his gun. As a result Mr Anis was hurt slightly and moved backward. (a) What is friction? (b) When a bullet is fired from a gun, why does it give thrust backward? (c) Calculate the recoil velocity of gun of Mr Anis. Solution : Given: mbullet = 10 g = 0.01 kg, vbullet = 10 m/s, Mgun = 10 kg. By conservation of momentum (initial total = 0): 0 = MgunVgun + mbulletvbullet Vgun = −mbullet · vbullet Mgun = −0.01 × 10 10 Vgun = −0.01 m/s (backward) (d) Does the event of the stimulus obey the principle of conservation of momentum? Give your opinion with mathematical analysis. Answer: (a) Friction is the resistive force that opposes relative motion (or tendency of motion) between two surfaces in contact. It acts tangentially along the surface and arises due to microscopic irregularities and intermolecular adhesion at the contact surfaces. (b) This is

Total final momentum = total initial momentum = 0. The principle of conservation of momentum is fully obeyed: in the absence of any external horizontal force on the (gun + bullet) system, total momentum is conserved. 2. Sajib Kazi, a fruit businessman, loaded the truck by purchasing fruits. The mass of the truck along with fruits is 1600 kg. The truck is moving from Dhaka to Chittagong at night with a velocity of 20 km/h. Suddenly the truck pushes another truck of mass of 1400 kg standing by the side of the road. Both the trucks are moving combinedly. (a) What is the principle of conservation of momentum? (b) Keeping the mass constant if the force is increased or decreased then what change of accel- eration will occur? (c) What will be the velocity of the trucks combined? Solution : m1 = 1600 kg, u1 = 20 km/h = 5.556 m/s; m2 = 1400 kg, u2 = 0. Perfectly inelastic collision (move together): v = m1u1 + m2u2 m1 + m2 = 1600 × 5.556 + 0 3000 = 8888.9 3000 v ≈2.963 m/s ≈10.67 km/h (d) “Although the momentum of the two trucks is conserved after collision but the kinetic energy is not conserved” - justify this with mathematical analysis. Answer: (a) The total linear momentum of a system of

(b) A body will remain in equilibrium position if its acceleration is zero - explain. (c) If the mass of a body is 10 kg, initial velocity v0 = 5 m/s, final velocity v = 15 m/s and time of action of force is 5 s, then find the magnitude of the applied force. Solution : a = (v −v0)/t = (15 −5)/5 = 2 m/s2. F = ma = 10 × 2 F = 20 N (d) Derive Newton’s first law from Newton’s second law of motion - show by mathematical analysis. Answer: (a) A fundamental force is one of the four basic interactions in nature that cannot be explained in terms of other forces: (1) Gravitational force, (2) Electromagnetic force, (3) Strong nuclear force, (4) Weak nuclear force. (b) By Newton’s second law,⃗Fnet = m⃗a. If⃗a = 0, then⃗Fnet = 0, meaning all forces on the body are balanced. A body with⃗Fnet = 0 either remains at rest or continues with uniform velocity (Newton’s first law) - both are states of equilibrium (static or dynamic). Hence, zero acceleration ⇔mechanical equilibrium. (c) (d) Newton’s second law:⃗F = md⃗v dt . If no net external force acts:⃗F = 0: md⃗v dt = 0 =⇒d⃗v dt = 0 (since m̸ = 0) =⇒⃗v = constant This means if initially at rest (⃗v = 0), it remains at rest; if in motion (⃗v̸ = 0), i