Newton's Laws of Motion PDF

NEWTON’S LAWS OF MOTION Motion can be defined as the relative change in the state of direction of a body with respect to force of reference. According to Newton the motion, of a body in a straight line can be summarized in three basic laws. Newton’s first law of motion: This states that a stationar...

NEWTON’S LAWS OF MOTION Motion can be defined as the relative change in the state of direction of a body with respect to force of reference. According to Newton the motion, of a body in a straight line can be summarized in three basic laws. Newton’s first law of motion: This states that a stationary body or a body moving will continue to maintain its state of rest or uniform motion unless acted upon by an external force. In other words, a body at rest will remain so if no force acts on it. The same is true for bodies in motion. It will be so continuously if no unbalanced force acts on it to slow it down to a stop. Such unbalances forces which tends to bring a moving body to a halt is the force of gravity or frictional force, otherwise a body once set in motion will continue with constant velocity. The tendency of a body to remain at rest or uniform motion on a straight line in the absence of applied forces is known as INERTIA. For Newton’s first law, we say that inertia is inherent in bodies at rest. That law also explains what force does but it does not suggest how it is measured. Newton’s first law is also known as the law of inertia. Newton’s second law of motion: This states that the rate of change of momentum is directly proportional to the impressed force and takes place in the direction of the force. MATHEMATICALLY, Force α rate of change of momentum F α ΔP 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 Fα 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑚𝑣 −𝑚𝑢 i.e. F α 𝑡 Note: momentum = mass x velocity = mv Where m = mass in kg; v = final velocity in m/s; u = initial velocity in m/s; t = time taken for the change in seconds; F = force in Newton. 𝑣−𝑢 But; Acceleration, a = 𝑡 𝑚(𝑣−𝑢) Thus; F = 𝑡 F = Kma (Where k = constant) The unit of force is chosen so that k = 1 :. F = ma The unit of force is Newton (N) which gives a mass of 1kg and an acceleration of 1 m/s2. 𝑚(𝑣−𝑢) F= 𝑡 Ft = m (v – u) The product “Ft” is known as the IMPULSE of the force. :. I = Ft = m(v-u) = change in momentum Impulse is measured in Newton-seconds (Ns). This is also the unit of momentum. Momentum is measured in kgm/s. from Newton’s second law, it can be seen that: i. force is the product of mass and acceleration. ii. force is the rate of change of momentum. Newton’s third law of motion: This states that to every action there is always an equal and opposite reaction. This implies that action and reaction are equal but opposite in direction. It also implies that the mutual actions of two bodies upon each other are always equal and directed to contrary parts. This implies that when a body, A exerts a force F A on a body B, the body B in turn exerts a force FB on body A. The forces FA and FB are equal in magnitude and opposite in direction. Since force is a vector, then FA = - FB FA gives the action while FB gives the direction. Example 1: A ball of mass 0.3kg moving at a velocity of 20m/s is suddenly hit by a force of 5N for a time of 0.03 seconds. Find its new velocity. SOLUTION: From Newton’s second law 𝑚(𝑣−𝑢) F= 𝑡 𝐹𝑡 V=u+ 𝑡 5 𝑥 0.03 = 20 + 0.3 = 20.5 m/s Example 2: When taking a penalty kick, a footballer applies a force of 30 N for a period of 0.05 seconds. If the mass of the ball is0.075kg, calculate the speed with which the ball moves off. SOLUTION: 𝑚(𝑣−𝑢) F= 𝑡  0.075 𝑥 𝑣 30 = 0.05 V = 20 m/s APPLICATIONS OF NEWTON’S THIRD LAW OF MOTION 1. Rocket propulsion: A rocket is propelled by the ejection of a portion of its mass to the rear. The forward force on the rocket is the reaction to the backward force on the ejected materials. 2. Recoil of a gun: When a bullet is shot from gun, the person firing experiences backward impact. 3. Jet engines: A fired engine burns by smoothing gas downward from the tail of the rocket at high. 4. why walking is possible: when a person pushes his or her foot against the ground, the ground in turn also pushes or exerts a force which I equal and opposite to the push on the person. CONSERVATION OF LINEAR MOMENTUM Momentum is defined as the product of the mass of a body or particle and the velocity in a given direction. Momentum is a vector quantity, measured in Kgm/s. PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The principle of conservation of linear momentum states that if two or more bodies collide in a closed system, the total momentum before and after collision remains unchanged. Assume a body of mass M1 and acceleration a1 moving towards another body M2 and acceleration a2; and their forces are equal and opposite. Then from Newton’s third law: F1 = - F2 From Newton’s second law F = ma, hence M1a1 = M2a2 Recall, acceleration, a, is given by: 𝑽−𝑼 a= 𝒕 Therefore, 𝑉1− 𝑈1 𝑉2 − 𝑈2 M1 = - M2 𝑡 𝑡 M1 (V1 – U1) = - M2 (V2 – U2) :. M1V1 – M1U1 = -M2V2 +M2U2 M1V1 + M2V2 = M1U1 + M2U2 Where U = Initial velocity before collision V = Final velocity after collision t = constant time of the body during collision TYPES OF COLLISON 1. ELASTIC COLLISION: If after two bodies had collided, their kinetic energy K.E both before collision and after collision is the same, then, such collision is perfectly elastic. In such occasion both momentum and energy are conserved. In such situation the bodies moves with different velocities after collision. Consider two bodies of masses M1 and M2 moving with initial velocities u1 and u2 after collision, we arrive at the following equations if collision is perfectly elastic: M1U1 + M2U2 = M1V1 + M2V2 For their kinetic energy ½ M1U12 + ½ M2U22 = ½ M1V12 + ½ M2V22 2. INELASTIC COLLISION: Conversely, in inelastic collision, the kinetic energy decreases after collision, but the momentum is conserved all the same. In such case, the colliding bodies stoic together and move in one direction (direction of the body with the greatest momentum). Meaning: V = V1 = V2 Hence, M1U1 + M2U2 = (M1 + M2) V The kinetic energy of the system before collision and after collision is given by E.K1 = ½ M1U12 + ½ M2U22 After collision E.K2 = ½ M1V12 + M2V22 = ½ (M1 + M2) V2 Since for a completely inelastic collision, the kinetic energy (E.K) before collision is greater than the kinetic energy (E.K) after collision. Example 1: A ball of mass 200g travelling with a velocity of 100ms-1 collides with another ball of mass 800g moving at 50ms-1 in the same direction. If they stick together, what will be their common velocity? Solution M1U1+ M2U2 = (M1 + M2) V V = M1 U1 + M2 U2 M1 + M2 = (0.2)100 + (0.8)50 0.2 + 0.8 = 20 + 40 1 V = 60ms-1 E.K before collision = ½M1U12 + ½ M2U22 = ½ (0.2)(100)2 + ½(0.8)(50)2 = 1000 + 1000 = 2000J E.K After collision = ½(M1 + M2) V2 = ½(0.2 + 0.8)(60)2 = ½(1) 3600 = 1800J Loss in energy = E.K before – E.K after = 2000 - 1800 = 200J (This however, is for inelastic collision). Example 2: A body of mass 4.2kg moving with velocity of 10ms-1 due east, hits a stationary body of mass 208kg. If they stick together after collision and move with a velocity V due east, Calculate the value of V. Solution M1V1 + M2 U = (M1 + M2) V (4.2)10 + 2.8(0) = (4.2 + 2.8) V 42 = 7V V = 6ms-1 E.k before collision = ½M1V12 = ½ (4.2) (10)2 = ½ (4.2)(100) = ½ (4.2)(100) = 420/2 = 210 E.k after collision = ½ (M1 + M2) V2 = ½ (4.2 + 2.8) (6)2 = ½ (7) 36 = 126J E.k loss = ½ M1V12 - ½ (M1 + M2) = 210 – 126 = 84J Example 3: A ball of mass 6.0kg moving with a velocity of 10ms-1 collides with a 2.0kg ball moving in the opposite direction with a velocity of 5.0ms-1. After the collision the two balls coalesce and move in the same direction. Calculate the velocity of the composite body. Solution M1V1 – M2V2 = (M1V2) V 6(10) – 2(5) = (6 + 2) V 60 – 10 = 8V 50 = 8V V = 50 8 = 6.25ms-1 E.K before collision = ½ M2V12 + ½ M2V22 = ½ (6) (10)2 + ½ (2) (5)2 = ½ [600 + 50] = ½ (650) = 325J E.K after collision = ½ (V1 + M2) V2 = ½ (6 + 8) (6.25)2 = 273.4J E.K loss = 325 – 273.4 = 51.6J