PHY2: MOTION

MOTION A body is said to be motion when its relative position changes with time

• Speed

This is the rate of change of the distance moved e.g If a car travels 30m in 5 seconds. The speed will be 6m/s.  During motion, the distance a body covers may not be the same at every interval of time that may be chosen. Therefore, we consider average speed.  📷

• Velocity

This is the rate of change of displacement.  SI unit is ms-1 📷

• Acceleration

This is the rate of change of velocity.  The SI unit for acceleration is ms-2

• Uniform acceleration.

If the rate change of velocity is constant then uniform acceleration is said to be taking place.  Acceleration is regarded as positive if it is increasing and negative if it is decreasing. Equations of uniform acceleration V is final velocity, U is initial velocity, a is acceleration, t is time, S is distance . 1st Equation. Consider a body moving at initial velocity u, accelerates with uniform acceleration a to the final velocity v in time t. Then acceleration a =v-u/t V = u +at ...........1st equation. 2nd equation  A body moving with uniform acceleration has an average velocity equal to half of the sum of its initial velocity u and final velocity v. And substituting for v in equation 2.  Total distance s = average velocity x time  S = (u+v)/2 x t but from equation 1, V = u + at  S = (u +u +at)/2 x t  S = (2u +at)/2 x t  S = ut+ ½at² ....second equation. 3rd equation This is obtained by eliminating time t from equation 1and 2. S =ut + ½at² as t = (v - u)/a V²=u²+2aS ......3rd equation. The three equations of motion are; V= u +at S = u t+½at² V² = u² +2as Examples. A car starts from rest and is accelerated uniformly at a rate of 1m/s² in 20 seconds. Find its final velocity and distance covered. Solution: V = u +at = 0+1 x 20 = 20m/s S = u t +½at² = 0 x20 +½ x1x20² = 200m A car accelerates uniformly at a speed of 20m/s for 4 seconds .find final velocity if acceleration is 2m/s² and the distance traveled. Solution: V = u +at V = 20 +2 x4 V = 28m/s S = u t+½at² S =20 x4 + ½ x2 x4 S = 96m. A body moving with velocity of 20 m/s accelerates to a velocity of 40m/s in 5 seconds.Find Acceleration hence the distance traveled in 5s. a=(v- u)/2  =(40-20)/2  = 4m/s² S = u t + ½at² S =20 x 5+ ½ x4 x5²  =150m A body at rest at height of 20m falls freely to the ground. Calculate; a) The velocity with which it hits the ground b) The time before striking the ground. Solution a = g =10 V²=u² +2as V² =0²+2 x10 x20 V =20m/s V = u+at 20 = 0+10t t =2s. UNIFORM UNITS Given the table below;- Displacement 10  20  30  40  60  70 Time 2    4    6     8   12  41 Plot a graph of displacement against time 📷 If the displacement covered is increasing by the same amount in every time interval considered, the body is said to be moving with uniform velocity. Considering the graph above for every 2 second the displacement increases by 10m.  Therefore velocity is 5ms-1. In the case the body is moving with uniform velocity and gradient/slope of the displacement time graph gives the velocity. From the above information, a velocity time graph can be between. Velocity   5    5    5    5    5     5 Time (s).   2    4    6    8   10   12    A sketch below is of a velocity – time graph a body with uniform acceleration. The gradient of a velocity time graph gives acceleration . NB : The area under a velocity time graph of a body represents the total distance covered.    Example A car starts from rest and accelerates uniformly for 4ms-² for 7 seconds. It is to maintain the acquired velocity for further 10 seconds. Brakes are then applied uniformly until the car comes to rest in 20 seconds.

1. Sketch a velocity time graph for the whole journey
2. Distance covered when car was accelerating
3. Find the distance covered when car was moving at constant velocity
4. Find the distance travelled from time when the brakes were applied till it stopped
5. Find the total distance covered

Velocity against time  Distance covered when accelerating  1/2 x b x h  1/2 x 7 x 28  = 98m Distance for constant velocity  L x w  10 x 28 = 280m Brakes applied  = 1/2  x 20 x 28  = 280m  PROJECTILES A projectile is a body that is made to move initially with non – zero velocity. Such a body must be given an initial push. Case 1:  A body projected vertically up wards.  When a body is projected vertically upwards, gravity acts on it and as a result, its velocity decreases as it ascends. In other wards it has an acceleration of -10ms-2. Time to descend is the same as the time to ascend. At the maximum point the velocity is zero. Exmple: A stone is thrown vertically upwards with a velocity of 200ms-1

1. Find maximum height reached.
2. Time taken to reach the maximum height
3. The time taken for the stone to return to the point of projection.

📷 Case 2;  When the body is projected horizontally  📷 When a body is projected horizontally, it travels both vertically and horizontally  Gravity has effect only on the vertical motion and no effect on the horizontal motion. NEWTON’S LAWS OF MOTION Newton’s first law of motion (law of inertia) It states that everybody continues in its state of rest or uniform motion in a straight line unless compelled by some external force to act otherwise. This law is also known as the law of inertia and may be evidenced by the following observations.  There is an innate reluctance for a body initially at rest to move when a force is applied on it. Our bodies jerk forward when vehicles in which we are travelling are suddenly made to stop; this is so because the body tends to maintain its state of motion.  MOMENTUM The product of the mass of the body and its velocity is called momentum.  Momentum = mass x velocity  = mv The SI unit of momentum is kgms¯¹  Examples A body of mass 1.5 tons is initially moving at 3m/s. several seconds later the body is said to be moving at 5m/ s.  a)Find the initial momentum. b)The final momentum c)The change in momentum Solution:

• 1 ton = 1000kg

1.5 = 1.5 x 1000  = 1500kg a)Initial Momentum = mu  = 1500 x 3  = 4500kgm/s b)Final Momentum = mv  = 1500 x 5  = 7500kgm/s c)Change = mv – mu  = 7500 – 4500  = 3000kgm/s Newton’s second law of motion The rate of change of momentum is directly proportional to the applied force and it takes place in the direction in which force acts.  📷 📷 Where k is a constant By definition newton is the force that accelerates a body of mass 1kg at a rate of 1m/s2. If F = IN, a = 1m/s², m = kg. F = kma  1 = k x 1 x 1= 1 F = ma Newton’s third law of motion This law states that to every action there is an equal and opposite reaction. When a bullet is fired from a gun, equal and opposite forces are exerted on bullet and the gun during the time the bullet passes through the barrel. Conservation of linear momentum Since both the bullet and the gun are acted by equal forces for the same time, both will in accordance with the second law of motion hence acquiring equal and opposite momentum. mass of the bullet x Muzzle velocity = mass of gun x recoil velocity. The momentum of a bullet and gun are equal but opposite in opposite direction. Consequently the sum (total) of the momentum is zero.  This leads to an important principle arising out of Newton’s second law and third law which is known as the law of conservation of momentum.  It states that when two or more bodies act upon one another, their total momentum remains constant provided no external force act on them. If we have two bodies one with mass m1 and initial velocity u1 and final velocity v1 and the other body with mass m2 and initial velocity u2 and final velocity v2 .  From the law of conservation of momentum; M1 u1 + m2 u2 = m1 v1 + m2 v2 📷 📷 THIS VIDEO EXPLAINS THE NEWTON'S THREE LAWS OF MOTION ROCKED PROPULSION If you release an inflated balloon, it flies in the opposite direction to that of the escaping air.  📷 Escaping air : This is the principle of rockets and jets. In both a high velocity stream of hot gas is produced by burning fuel and leaves the exhaust pipes with large momentum. Space rocks carry their own oxygen supply and jet engines use the air around. The rockets contain solid chemicals which burn to produce a high velocity blust of hot gases.  Space rockets designed to travel long distances have tanks of liquid fuel together with a supply of liquid oxygen or some other liquid which produces oxygen to enable the fuel to burn.  In either case, a chemical reaction takes place inside the rocket and reacts on a large force which propels the gaseous products out through the tail nozzle with trichinous velocity. An equal momentum is imported to the rocket in the opposite direction so that inspite of its large mass, the rocket builds up a high velocity.  Inelastic collision Total inelastic collision is that in which the two objects collide and remain together and move with a common velocity. Elastic collision This is the collision where momentum is always conserved and energy also conserved. A total elastic collision is one in which the kinetic energy is conserved. Initial kinetic energy = Final kinetic energy TICKER TIME A ticker timer has a timing device which consists of a rapidly vibrating hammer which prints dots on a length of paper tape.  A ticker timer might print 50 dots per second,which gives the frequency in Hz.  📷 A ticker timer can be used to find velocity. If it makes 50dots per second,then each dot is made in 1/50 which is equal to 0.02 seconds.  📷 Total time = 0.02 x number of spaces  = 0.02 x 8  = 0.16 seconds. Example.  📷 If the frequency of the ticker timer is measured as 50Hz; find

1. The total time
2. Velocity of the trolley

Total time = 0.02 x number of spaces  = 0.02 x 10  = 0.2 seconds  📷