\n<\/h2>\n
- \n
- \nNewton’s laws of motion\n<\/div>\n<\/li>\n
- \n
Conservation of linear momentum\n<\/div>\n<\/li>\n- \n
Collision\n<\/div>\n<\/li>\n<\/ul>\nNEWTON’S LAWS OF MOTION
\n<\/strong>Newton’s first law of motion<\/em> states that everybody continues in its state of rest or of uniform motion in straight line unless it is acted upon by a force. The tendency of a body to remain at rest or, if moving, to continue its motion in a straight line is called the inertia<\/em>. That is why Newton’s first law is otherwise referred to as the law of inertia.
\n Newton’s second law of motion<\/em> states that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in of the force.
\nF \u03b1 mv \u2013mu
\n\t\t\u00a0\u00a0\u00a0\u00a0t
\nF \u03b1 m (v \u2013u)
\n\t\tt
\nF \u03b1 ma
\n\t\tF = kma
\nWhere k =1
\nF =ma
\nMOMENTUM
\nMomentum<\/em> of a body is the product of the mass and velocity of the body. The S.I. unit of momentum is kgm\/s.\u00a0
\nIMPULSE
\nImpulse<\/em> is the product of a force and time. It is also defined as the change in momentum. Thus both momentum and impulse have ‘Ns’ as unit
\nF = m (v-u)\/t
\nFt = mv \u2013 mu (where ‘mv-mu’ is the change of momentum)
\nF x t = I (Ns)
\nNewton’s third law of motion<\/em> states that to every action, there is an equal but opposite reaction. A practical demonstration of this law can be observed when a bullet is fired from a gun, the person holding it experiences the backward recoil force of the gun (reaction) which is equal to the propulsive force (action) acting on the bullet.
\nAccording to Newton second law of motion, force is proportional to change in momentum
\nTherefore the momentum of the bullet is equal and opposite to the momentum of the gun i.e.
\nMass of bullet x muzzle velocity = mass of gun x recoil velocity
\nHence, if: m= mass of bullet, v= velocity of bullet, M=mass of gun, V= velocity of the recoil of the gun.
\nThen, the velocity, V, of the recoil of the gun is given by:
\nMV = mv
\nV = mv\/M
\nCONSERVATION OF LINEAR MOMENTUM<\/strong>
\n\t\tThe principle of conservation of linear momentum states that when two or more bodies collide, their momentum remain constant provided there is no external force acting on the system<\/em>. This implies that in a closed or isolated system where there is no external force, the total momentum after collision remains constant. The principle is true for both elastic and inelastic collision.
\nCOLLISION<\/strong>
\n\t\tThere are two types of collision- elastic and inelastic.
\nIn elastic collision<\/em> the two bodies collide and then move with different velocities. Both momentum and kinetic energy are conserved<\/em> e.g. collision between gaseous particles, a ball which rebounds to its original height etc.
\nIf the two colliding bodies have masses m1<\/sub>and m2<\/sub> initial velocities u1<\/sub> and u2<\/sub> and final velocities v1<\/sub> and v2<\/sub>. The conservation principle can be mathematically expressed as:
\n\u00a0m1<\/sub>u1<\/sub> + m2<\/sub>u2<\/sub> = m1<\/sub>v1<\/sub> + m2<\/sub>v2<\/sub>\u00a0
\nIn an inelastic collision<\/em>, the two bodies join together after the collision and with the same velocity. Here, momentum is conserved but kinetic energy is not conserved <\/em>because part of it has been converted to heat or sound energy, leading to deformation.
\nThus, the conversation principle can be re-written as:
\nm1<\/sub>u1<\/sub> + m2<\/sub>u2<\/sub> = v (m1<\/sub> +m2<\/sub>)
\nSince momentum is a vector quantity, all the velocities must be measured in the same direction, assigning positive signs to the forward velocities and negative signs to the backward or opposite velocities
\nTWO BODIES MOVING IN THE SAME DIRECTION BEFORE COLLISION
\n![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi1.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi2.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi3.png\")
VA<\/sub> VB<\/sub>
\n\t\t\u00a0\u00a0
\n\u00a0MA<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0MB<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0MA<\/sub> MB<\/sub>
\n\t\tBEFORE COLLISION AFTER COLLISION
\nMA<\/sub>VA<\/sub> + MB<\/sub>VB<\/sub> = V (MA<\/sub> + MB<\/sub>) \u00a0
\nV= common velocity
\nV= MA<\/sub>VA<\/sub> + MB<\/sub>VB<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a0(MA<\/sub> + MB<\/sub>)<\/p>\n\u00a0TWO BODIES TRAVELLING IN OPPOSITE DIRECTION
\n\u00a0![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi4.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi5.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi6.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi7.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi8.png\")
\u00a0
\n![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi9.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi10.png\")
=
\nMA<\/sub>\u00a0\u00a0\u00a0\u00a0MB<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0MA<\/sub> MB<\/sub>
\n\t\tMA<\/sub>VA<\/sub> – MB<\/sub>VB<\/sub> = V (MA<\/sub> + MB<\/sub>)
\nV= MA<\/sub>VA<\/sub> + MB<\/sub>VB<\/sub>
\n\t\tMA<\/sub> + MB<\/sub>
\n\t\t\u00a0
\n\u00a0COLLISION BETWEEN A STATIONARY AND MOVING BODY
\n![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi11.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi12.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi13.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi14.png\")
\u00a0
\n![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi15.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi16.png\")
VA<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V
\n MA<\/sub> MB<\/sub> MA<\/sub>MB<\/sub>
\n\t\t\u00a0
\n\u00a0The momentum of a stationary body is zero because velocity is zero
\nMA<\/sub>VA<\/sub> + 0 = V (MA<\/sub> + MB<\/sub>)
\nV =\u00a0\u00a0\u00a0\u00a0 MA<\/sub>VA
\n<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0MA<\/sub> + MB
\n<\/sub>EXAMPLE<\/strong><\/p>\n- \n
- \nTwo moving toys of masses 50kg and 30kg are traveling on the same plane with speeds of 5 m\/s and 3 m\/s respectively in the same direction. If they collide and stick together, calculate their common velocity.\n<\/div>\n<\/li>\n<\/ol>\n
MA<\/sub>VA<\/sub> + MB<\/sub>VB<\/sub> = V (MA<\/sub> + MB<\/sub>)
\nV= MA<\/sub>VA<\/sub> + MB<\/sub>VB<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a0(MA<\/sub> + MB<\/sub>)
\nV = (50 x 5) + (30 x 3)
\n\t\t50 + 30
\nV = 250 + 90
\n\t\t80
\nV = 340
\n\t\t80
\nV =<\/strong> 4.05 m\/s<\/p>\n- \n
- \nTwo balls of masses 0.5 kg and 0.3kg move towards each other in the same line at speeds of 3 m\/s and 4 m\/s respectively. After the collision, the first ball has a speed of 1m\/s in the opposite direction. What is the speed of the second ball after collision\n<\/div>\n
3m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V\n<\/li>\n<\/ol>\n
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi17.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi18.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi19.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi20.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi21.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi22.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi23.png\")
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100423_0859_Week5SS2Fi24.png\")
<\/p>\n\u00a00.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.3
\nBefore\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0After
\n3×0.5 + (0.3 x-4) = 0.5 (-1) + 0.3v <\/p>\n- \n
- \n– 1.2 = -0.5 + 0.3v\n<\/div>\n<\/li>\n<\/ol>\n
0.3v = 2.0 – 1.2
\nV = 0.8 \/ 0.3
\n\t\t\tV = 2.7m\/s<\/p>\n- \n
- \nA gun of mass 100kg fires a bullet of mass 20g at a speed of 400m\/s. What is the recoil velocity of the gun?\n<\/div>\n
Solution
\n<\/strong>Momentum gun = momentum of bullet
\n\t\t\t\t\t<\/strong>MV = m v
\n\t\t\t\t\t<\/strong><\/li>\n<\/ol>\n10 x V = 0.002 x 400
\nV = 0.002 x 400
\n\t\t10
\nV= 0.8 m\/s
\nCLASSWORK
\n<\/strong><\/p>\n- \n
- \nDerive from Newton’s law the relationship between Force, mass and acceleration\n<\/div>\n<\/li>\n
- \n
State Newton laws of motion and explain the consequences of each law\n<\/div>\n<\/li>\n- \n
State the principle of conservation of linear momentum.\n<\/div>\n<\/li>\n- \n
A 15kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200N and breaks as the elevator accelerates. What was the elevators minimum acceleration (g=10m\/s2<\/sup>).\n<\/div>\n<\/li>\n<\/ol>\nASSIGNMENT
\n<\/strong>SECTION A<\/strong><\/p>\n- \n
- \nA force acts on a body for 0.5s changing its momentum from 16kgms-1<\/sup> to 21kgms-1<\/sup>. Calculate the magnitude of the force (a) 42N (b) 37N (c) 32N (d) 10N\n<\/div>\n<\/li>\n
- \n
A ball of mass 6kg moving with a velocity of 10m\/s collides with a 2kg ball moving in the opposite direction with a velocity of 5m\/s. After the collision the two balls coalesce and move in the same direction. Calculate the velocity of the composite body (a) 5m\/s (b) 6.25m\/s (c) 8.75m\/s (d) 12m\/s\n<\/div>\n<\/li>\n- \n
A machine gun with mass of kg fires a 50g bullet at a speed of 100m\/s. The recoil speed of the machine gun is (a) 0.5m\/s (b) 1.0m\/s (c) 1.5m\/s (d) 2.0m\/s\n<\/div>\n<\/li>\n- \n
When taking a penalty kick, a footballer applies a force of 30N for a period of 0.05S. If the mass of the ball is 0.075kg, calculate the speed with which the ball moves off (a) 4.5m\/s (b) 11.25m\/s (c) 20m\/s (d) 45m\/s\n<\/div>\n<\/li>\n- \n
A jet engine develops a thrust of 270N when the velocity of the exhaust gases relative to the engine is 300m\/s, what is the mass of the gas ejected per second? (a) 81.00kg (b) 9.00kg (c) 0.90kg (d) 0.09kg\n<\/div>\n<\/li>\n<\/ol>\nSECTION B
\n<\/h2>\n- \n
- \nAn object of mass 5kg slides down a smooth plane at an angle of 250<\/sup>. If the object starts from rest, find; (i) its velocity after 3m (ii) its momentum 3m from the starting point (iii) the force causing it moves (g=m\/s2<\/sup>)\n<\/div>\n<\/li>\n
- \n
State the law of conservation of linear momentum. A 3kg rifle lies on a smooth table when it suddenly discharges, firing a bullet of 0.02kg with a speed of 500m\/s. Calculate the recoil speed of the gun.\n<\/div>\n<\/li>\n- \n
A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20m\/s. The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the: (i) magnitude of the resistance (ii) the distance moved by the bullet in the wood\n<\/div>\n<\/li>\n<\/ol>\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\t\t\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"WEEK 5 NEWTON’S LAW OF MOTION \u00a0 CONTENTS Newton’s laws of motion Conservation of linear…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,242],"tags":[],"class_list":["post-2996","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-physics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2996","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2996"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2996\/revisions"}],"predecessor-version":[{"id":2997,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2996\/revisions\/2997"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2996"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2996"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2996"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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