Newton’s Second Law Archives

Newton’s first law (law of inertia) This law states that “A body continues in its state of rest or uniform motion unless an unbalanced force acts on it”. The mass of a body is a measure of its inertia. Inertia is the property that keeps an object in its state of motion and resists any efforts to change it. Newton’s second law (law of momentum) Momentum of a body is defined as the product of its mass and its velocity. Momentum ‘p’mv. The SI unit for momentum is kgm/s or Ns. The Newton’s second law states that “The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts” Change in momentum mv-mu Rate of change of momentum mv-mu/∆t Generally the second law gives rise to the equation of force Fma Hence Fmv-mu/∆t and F∆tmv-mu The quantity F∆t is called impulse and is equal to the change of momentum of the body. The SI unit for impulse is Ns. Examples A van of mass 3 metric tons is travelling at a velocity of 72 km/h. Calculate the momentum of the vehicle. Solution Momentummv72km/h(20m/s)×3×103 kg 6.0×104kgm/s A truck weighs 1.0×105 N and is free to move. What force willgiveit an acceleration of 1.5 m/s2? (take g10N/kg) Solution Mass of the truck (1.0×105)/106.0×104 Using Fma 1.5×10×104 1.5×104 N A car of mass 1,200 kg travelling at 45 m/s is brought to rest in 9 seconds. Calculate the average retardation of the car and the average force applied by the brakes. Solution Since the car comes to rest, v0, a(v-u)/t (0-45)/9-5m/s (retardation) Fma (1200×-5) N -6,000 N (braking force) A truck of mass 2,000 kg starts from rest on horizontal rails. Find the speed 3 seconds after starting if the tractive force by the engine is 1,000 N. Solution Impulse Ft1,000×3 3,000 Ns Let v be the velocity after 3 seconds. Since the truck was initially at rest then u0. Change in momentummv-mu (2,000×v) - (2,000×0) 2,000 v But impulsechange in momentum 2,000 v 3,000 v 3/21.5 m/s. Weight of a body in a lift or elevator When a body is in a lift at rest then the weight Wmg When the lift moves upwards with acceleration ‘a’ then the weight becomes W m (a+g) If the lift moves downwards with acceleration ‘a’ then the weight becomes W m (g-a) Example A girl of mass stands inside a lift which is accelerated upwards at a rate of 2 m/s2. Determine the reaction of the lift at the girls’ feet. Solution Let the reaction at the girls’ feet be ‘R’ and the weight ‘W’ The resultant force F R-W (R-500) N Using F ma, then R-500 50×2, R 100+500 600 N. Newton’s third law (law of interaction) This law states that “For every action or force there is an equal and opposite force or reaction” Example A girl of mass 50 Kg stands on roller skates near a wall. She pushes herself against the wall with a force of 30N. If the ground is horizontal and the friction on the roller skates is negligible, determine her acceleration from the wall. Solution Action reaction 30 N Force of acceleration from the wall 30 N F ma a F/m 30/50 0.6 m/s2 Linear collisions Linear collision occurs when two bodies collide head-on and move along the same straight line. There are two types of collisions; a) Inelastic collision:- this occurs when two bodies collide and stick together i.e. hitting putty on a wall. Momentum is conserved. b) Elastic collision:- occurs when bodies collide and bounce off each other after collision. Both momentum and kinetic energy are conserved. Collisions bring about a law derived from both Newton’s third law and conservation of momentum. This law is known as the law of conservation of linear momentum which states that “when no outside forces act on a system of moving objects, the total momentum of the system stays constant”. Examples A bullet of mass 0.005 kg is fired from a gun of mass 0.5 kg. If the muzzle velocity of the bullet is 300 m/s, determine the recoil velocity of the gun. Solution Initial momentum of the bullet and the gun is zero since they are at rest. Momentum of the bullet after firing (0.005×350) 1.75 kgm/s But momentum before firing momentum after firing hence 0 1.75 + 0.5 v where ‘v’ recoil velocity 0.5 v -1.75 v -1.75/0.5 - 3.5 m/s (recoil velocity) A resultant force of 12 N acts on a body of mass 2 kg for 10 seconds. What is the change in momentum of the body? Solution Change in momentum ∆P mv – mu Ft 12×10 12 Ns A minibus of mass 1,500 kg travelling at a constant velocity of 72 km/h collides head-on with a stationary car of mass 900 kg. The impact takes 2 seconds before the two move together at a constant velocity for 20 seconds. Calculate a) The common velocity b) The distance moved after the impact c) The impulsive force d) The change in kinetic energy Solution a) Let the common velocity be ‘v’ Momentum before collision momentum after collision (1500×20) + (900×0) (1500 +900)v 30,000 2,400v v 30,000/2,400 12.5 m/s (common velocity) b) After impact, the two bodies move together as one with a velocity of 12.5 m/s Distance velocity × time 12.5×20 250m c) Impulse change in momentum 1500 (20-12.5) for minibus or 900 (12.5 – 0) for the car 11,250 Ns Impulse force F impulse/time 11,250/2 5,625 N d) K.E before collision ½ × 1,500 × 202 3 × 105 J K.E after collision ½ × 2400 × 12.52 1.875×105 J Therefore, change in K.E (3.00 – 1.875) × 105 1.25× 105 J Some of the applications of the law of conservation of momentum Rocket and jet propulsion:- rocket propels itself forward by forcing out its exhaust gases. The hot gases are pushed through exhaust nozzle at high velocity therefore gaining momentum to move forward. The garden sprinkler:- as water passes through the nozzle at high pressure it forces the sprinkler to rotate. Solid friction Friction is a force which opposes or tends to oppose the relative motion of two surfaces in contact with each other. Measuring frictional forces We can relate weight of bodies in contact and the force between them. This relationship is called coefficient of friction. Coefficient of friction is defined as the ratio of the force needed to overcome friction Ff to the perpendicular force between the surfaces Fn. Hence µ Ff/ Fn Examples A box of mass 50 kg is dragged on a horizontal floor by means of a rope tied to its front. If the coefficient of kinetic friction between the floor and the box is 0.30, what is the force required to move the box at uniform speed? Solution Ff µFn Fn weight 50×10 500 N Ff 0.30 × 500 150 N A block of metal with a mass of 20 kg requires a horizontal force of 50 N to pull it with uniform velocity along a horizontal surface. Calculate the coefficient of friction between the surface and the block. (take g 10 m/s) Solution Since motion is uniform, the applied force is equal to the frictional force Fn normal reaction weight 20 ×10 200 N Therefore, µ Ff/ Fn 50/ 200 0.25. Laws of friction It is difficult to perform experiments involving friction and thus the following statements should therefore be taken merely as approximate descriptions: Friction is always parallel to the contact surface and in the opposite direction to the force tending to produce or producing motion. Friction depends on the nature of the surfaces and materials in contact with each other. Sliding (kinetic) friction is less than static friction (friction before the body starts to slide). Kinetic friction is independent of speed. Friction is independent of the area of contact. Friction is proportional to the force pressing the two surfaces together. Applications of friction Match stick Chewing food Brakes Motion of motor vehicles Walking Methods of reducing friction Rollers Ball bearings in vehicles and machines Lubrication / oiling Air cushioning in hovercrafts Example A wooden box of mass 30 kg rests on a rough floor. The coefficient of friction between the floor and the box is 0.6. Calculate a) The force required to just move the box b) If a force of 200 N is applied the box with what acceleration will it move? Solution a) Frictional force Ff µFn µ(mg) 0.6×30×10 180 N b) The resultant force 200 – 180 20 N From F ma, then 20 30 a a 20 / 30 0.67 m/s2 Viscosity This is the internal friction of a fluid. Viscosity of a liquid decreases as temperature increases. When a body is released in a viscous fluid it accelerates at first then soon attains a steady velocity called terminal velocity. Terminal velocity is attained when F + U mg where F is viscous force, U is upthrust and mg is weight. See also: REFRACTION OF LIGHT LINEAR MOTION FLUID FLOW SOUND WAVES

NEWTON LAWS OF MOTION