[Music] hi everyone I hope you found this video helpful if so I'd really appreciate it if you could subscribe to my channel like this video share it with your friends and leave a positive comment your support and encouragement motivate me to create more great videos in this video I've covered all the outlines in the syllabus as shown in the figure linear momentum of the object is defined as the product of its mass and its velocity so we get the equation of linear momentum is p equal m UV where m is mass in kilog V is velocity in m/s P is momentum in kilog m/ second or Newton Second momentum is a vector quantity and its direction is the same as velocity Direction Newton's Laws of Motion Newton's first law state that a body will remain at rest or move with constant velocity unless acted on by a resultant Force so we can conclude that if no resultant Force so no acceleration causing the object to be rest or constant velocity Newton's first law is often called the law of inertia the inertia of an object is measured by its mass mass or inertia is the property of an object that resists change in motion Newton's Third Law of Motion state that if a body exerts a force on body B then body B will exert a force on body a of equal magnitude but in the opposite direction so we can conclude that a pair of forces in Newton's third law must be equal in magnitude opposite in direction act on different bodies and same type of forces Newton's second Laws of Motion state that a resultant force acting on a body will cause a change in momentum in the direction of the force so the resultant force is the rate of change in momentum so the formula of the resultant force is f equal Delta P / T where Delta p is change in momentum which is m v minus m u and vus U / T is the acceleration a so the resultant force can be written as F equal m a from this equation we move moves T to multiply with f like this the product of F and T is defined as the impulse I so the impulse is the change in momentum or the product of resultant force and time the impulse is a vector quantity and its unit is the same as momentum's unit we can conclude the Newton's second law of motion that if the resultant force is not zero causing the velocity to change and the object will be accelerated or decelerated if a ball of mass mass m kilg is hit a wall at at a speed U it bounces back with a speed V the Collision time t as shown we use a negative sign to represent the left Direction the change in momentum is m o v e minus m u substituting the negative V to indicate the left Direction like this so the change in momentum is negative m v + m u the negative sign indic Ates that the change in momentum of the mass to the left Direction during the hitting the mass m exerts a force on the wall with equal in size and opposite in direction as the wall exerts a force on the mass m this is because Newton's third law of motion about action and reaction the forces act on the mass and the wall for the same amount of during time T the graph of the average force F against the Collision time T of the collision between the mass m and wall is shown so the area undergr is the change in momentum or impulse the graph of momentum P of the mass m against during time T is shown the gradient of the graph is change in momentum over change in time which is the resultant Force so the magnitude of the resultant force is netive MV + m u / T this is the magnitude of resultant Force the negative sign indicates the left Direction therefore the gradient of the graph is the force of the mass to act on the wall or the force of the wall acting on the mass work example one a small pebble of mass 500 G is attached to the lower end of a light string a find the tension in the string when the pebble is moving upwards with an acceleration of 2 m/s squared we convert the mass in g into kilog by dividing 500 with 1,000 we get Mass equals 0 .5 kg the free body diagram of the pebble there are weight to act down and the tension acting upward from the equation F equal m a f is the resultant Force to act on the pebble upward due to it moves upward so F = tus w = m a substituting W = 0.5 * 9.81 m = 0.5 and a = 2 we solve the tension T to get t equal 5.9 Newtons for two significant figures B find the tension in the string when the pebble is moving downwards with an acceleration of 3 m/s squared from the equation FAL m a f is the resultant Force to act on the pebble downward due to it moves downward so f = wus t = ma a substituting W = 0.5 * 9.81 M = 0.5 and a = 3 we solve the tension T to get tals 3 point for Newtons for two significant figures C find the tension in the string when the pebble is moving downwards at a constant speed of 5 m/s from the equation F equal m a f is the resultant Force to act on the pebble it is zero due to constant speed and its acceleration is zero so W - t = 0 and then w = t = 0.5 * 9.81 we get the tension T equals for9 Newtons for two significant figures D find the tension in the string when the pebble is moving downward with a deceleration of 4 m/s squared from the equation F equal m a f is the resultant Force to act on the pebble downward due to it moves downward so f = wus t = ma a substituting W = 0.5 * 9.81 m = 0.5 and a equal NE F due to deceleration we solve the tension T to get t equal 6.9 Newtons for two significant figures work example two p and Q of masses 2 kg and 3 3 Kg are attached to the ends of a light in extensible string that passes over a smooth pulley the system is released from rest a find the acceleration of each Mass the free body diagram of P there are weight two G to act down and the tension T acting upward G is the acceleration of Free Fall which is 9.81 m/s squared the free body diagram of Q There are weight Three G to act down and the T acting upward the tensions T act on P and Q are the same due to same tension in same string the mass Q moves downward and mass P moves upward with same acceleration a because mass of Q is more than mass of P this causes the acceleration of Q downward and the acceleration of P upward we write the equation of motion for p and Q from the equation FAL M A the resultant Force F on P is upward due to its acceleration upward the equation of motion of p is tus 2 g = 2 a the resultant Force F on Q is downward due to its acceleration downward the equation of motion of Q is 3 Gus T = 3 a we add two equations together so we get G = 5 a we solve the acceleration a by subtracting G = 9.81 we get the acceleration a = 2 2.0 m/s squared for two significant figures B find the tension in the string we substitute the acceleration a into the equation of motion of P or Q so I use the equation of motion of P substituting a = 1.96 2 and G = 9.81 we get the tension T equals 24 Newtons for two significant figures C find the force exer exted on the pulley two tension T act on the pulley downward like this so the resultant Force exerts on the pulley downward with magnitude 2 T substituting t equal 23.5 for 4 Newtons we get the resultant Force f equals 47 Newtons for two significant figures D find the distance moved by Q in the first for seconds we calculate the distance using the kinematic equations due to constant acceleration we use the positive sign to indicate downward Direction when Q moves downward the displacement s need to be find the initial velocity U equals z the final velocity a v is unknown the acceleration a = 1.96 2 and time T = 4 we use the equation s equal u t + half of a t^2 substituting U equal = 0 t = 4 a = 1.96 2 we get the distance equals 16 M for two significant figures exam style question one a box of mass 8 kg rests on a horizontal rough surface a string attached to box passes over a smooth pulley and supports a 2 kg mass at other end when the box is released a frictional force of 6 Newtons acts on it what is the acceleration of the Box the mass 2 kg moves downward to pull the Box 8 kg to the right with same acceleration this causes the acceleration of Mass 2 kg downward and the acceleration of box 8 kg to the right the free body diagram of the Box 8 kg there are the tension T to act rightward and the frictional force acts to the left like this the free body diagram of the mass 2 kg there are the tension T to act upward and the weight 2 G acts downward like this G is the acceleration of Free Fall which is 9.81 m/s squared we write the equation of motion for p and Q from the equation FAL M A the resultant Force F on box is point leftward due to its acceleration leftward the equation of motion of box 8 kg is t - 6 = 8 a the resultant Force F on Mass 2 kg is downward due to its acceleration downward the equation of motion of Mass 2 kg is 2 Gus T = 2 a we add two equations together so we get 2 g - 6 = 10 a substituting G = 9.81 we get the acceleration a = 1. for m/s squared for 2 significant figures exam style question two a golf ball of mass m is dropped onto a hard surface from a height H1 and rebounds to a height H2 the momentum of the ball just as it reaches the surface is different from its momentum just as it leaves the surface what is the total change in the momentum of the golf ball between these two instants ignore air resistance we calculate the velocity of mass m that hits the surface using the kinematic equations due to constant acceleration as gravity G we use the negative sign to indicate the downward Direction ction when the mass m drops from rest to the hard surface the displacement s equal H1 the initial velocity U equal 0 the final velocity V equal V1 the acceleration equals g and time T is unknown we use the equation V ^2 = U ^2 + 2 a s substituting V = V1 U = 0 a = G and S = H1 we get the velocity V1 = < TK 2 G H1 the negative sign indicates downward when the mass m moves upward the displacement s equals H2 the initial velocity U equal V2 the final velocity V equal 0 the acceleration a equal G and time T is unknown we use the equation V ^2 = U ^2 + 2 a s substituting V = 0 U = V2 equal G and S = H2 we get the velocity V2 = < TK 2 G H2 the change in momentum is m v minus mu U substituting V2 and V1 like this we get the change in momentum like this exam style question three water is pumped through a horsep pipe at a rate of 90 kg per minute it emerges from the horsep pipe horizontally with a speed of 20 m/s what force is required from a person holding the Hors pipe to prevent it moving backwards when the water moves from the HSE pipe there is force to push the water out this causes the force to act the hand backward in equal size due to Newton's third law this force is equal to the force required to hold the Hors pipe we calculate this Force by the equation f equals m v minus mu U / T the mass per second is 90 / 60 it is 1.5 kg/s substituting m t equal 1.5 final velocity V equal 20 and initial velocity U equal Z we get the force equals 30 Newtons exam style question four two blocks X and Y of masses M and 3 m respectively are accelerated along a smooth horizontal Surface by a force F applied to block X as shown what is the magnitude of force exerted by block X on block y during this acceleration we calculate the acceleration of two blocks from the equation F equal m a f is the resultant force and M is the combined mass of X and Y we get the acceleration a equal f over for M the free body diagram of block X there are the force F and the force exerted by y the free body diagram of block y there is force exerted by X the force exerted by y on x equals the force exerted by X on y due to Newton's Third Law two blocks move with same acceleration like this we calculate the magnitude of force exerted by block X on block y by the equation FAL m a substituting FAL force on y m = 3 m and AAL F over 4 M we get the force on y equal 3F over 4 exam style question five a 1.2 kg's mass is supported by a person's hand and 2 Newton M as shown when a person's hand is removed what is the vertical acceleration of the mass we break the for Force for Newton's into vertical component is for CU 37° upward the force three Newton's break into vertical component is 3 CU 53° upward we calculate the acceleration by the equation FAL M A the resultant Force F equal 12 - 3 cuz 53° - 4 CU 37° substituting m equal 1.2 we get the acceleration a equal 6.0 m/s squared exam style question 6 the graph shows the variation with time of the momentum of a ball as it is kicked in a straight line initially the momentum is P1 at the time T1 at the time T2 the momentum is P2 what is the magnitude of average resultant force acting on the ball between T1 and T2 we calculate the resultant Force by the equation f equals change in momentum Delta p over time T the change in momentum Delta P equals P2 - P1 we get the resultant Force F equal P2 + P1 / T the negative sign indicates the direction the magnitude of the resultant force is this mass and weight mass is a measure of the amount of matter in an object at rest resisting its movement mass is a scalar quantity and its unit is kilogram mass is constant anywhere in the universe weight is a gravitational force on an object that has mass weight is a vector quantity and its unit is Newtons weight varies anywhere in the universe due to the gravitational field strength we calculate the weight of the mass by this equation exam style question 7 a man weighs 24 Newtons Mars where the acceleration of freeold G is for m/s squared on the moon G is 2 m/ second squared which statement is correct the mass of the man Remains the Same at any position in the universe we calculate the mass of the man by the equation m = w / G substituting W = 240 newtons on Mars and G of Mars equals 4 we get the mass of the man equals 60 kg the weight of the man depends on the acceleration of freeold G so the weight of the man on the moon calculate by the equation W = mg substituting M = 60 and g = 2 we get the weight of the man on the moon equals 120 Newtons a the man has a mass on Mars of 60 newtons this is incorrect B the man has a mass on the moon of 120 kg this this is incorrect C the man weighs 120 newtons on the moon this is correct d the man weighs 240 newtons on the moon this is [Music] incorrect hi everyone I hope you found this video helpful if so I I'd really appreciate it if you could subscribe to my channel like this video share it with your friends and leave a positive comment your support and encouragement motivate me to create more great videos in this video I've covered all the outlines in the syllabus as shown in the figure frictional forces friction is a force that opposes the movement of an object it acts in the direction opposite to the motion friction is present almost everywhere and it can slow down or even stop moving objects this happens because friction converts kinetic energy into thermal energy viscous or drag forces viscous or drag forces are the frictional force in a fluid this is the resistance and object experiences when moving through a fluid two main factors affecting fluid friction the surface area of the object in contact with the fluid increases drag also increases the object speed increases drag also increases terminal velocity in air a badminton shuttlecock dropped from rest will reach terminal velocity as it falls through the air at the start the initial velocity is zero an initial drag forces is also zero this causes the resultant force equal weight acting downward causing the initial acceleration is equal to the acceleration of free fall when the badminton shuttlecock moves downward its speed increases from zero causing the drag Force to increase from zero this causes the resultant Force decreases to act downward and it is equal to weight minus drag Force so the acceleration is also decreased when the badminton shuttlecock moves downward its speed increases to maximum which is called terminal velocity causing the drag Force to increase to equal weight the resultant force is zero because weight equals drag Force so the acceleration is also zero and speed remains constant the graph of displacement varies with time T of this motion is shown like this recall that the gradient of displacement time graph is velocity this section shows that the velocity increases from zero to a maximum as its gradient increases to a maximum this section shows that maximum constant velocity occurs due to the maximum constant gradient the graph of velocity varies with time T of this motion is shown like this recall that the gradient of velocity time graph is the acceleration as the velocity increases but the acceleration decreases due to the gradient decreases until the velocity is constant and the acceleration is zero this constant velocity is terminal velocity the graph of acceleration varies with time T of this motion is shown like this the initial acceleration is equal the acceleration of free p and then it decreases until zero when reached terminal velocity assuming the blue ball has a larger mass than the green ball but same size both dropped from the same height at the same time the blue ball will likely reach the ground first and with a higher terminal velocity this is because the blue ball has bigger Mass so bigger weight too causing it accelerate for a longer time before drag Force to equal weight leading it reaches higher terminal velocity and reach reach ing the floor faster the green ball has smaller Mass so smaller weight too causing it accelerate for a shorter time before drag Force to equal weight leading it reaches lower terminal velocity and reaching the floor slower the graph of velocity varies with time t for both balls is shown like this the blue ball reaches the floor first with higher terminal velocity like this the green ball reaches the floor later with lower terminal velocity like this viscous or drag force of liquid is the resistance force when an object travels through a liquid it depends on the shape speed and temperature of liquid a metal sphere dropped from rest will reach terminal velocity as it falls through the liquid like as shown at the start the resultant Force acts on the metal is weight minus up thrust causing it accelerate downward and it velocity increases the drag Force increases upward as velocity increases so the result force is weight minus up thrust minus drag Force causing the resultant Force to decrease and acting downward leading to the acceleration also decreases and the velocity increases at slower rate the drag Force increases until the resultant force is zero meaning that weight is equal to up thrust plus drag Force causing the acceleration is also zero and the velocity remains constant this is called terminal velocity recall that the up thrust and weight are always constant while drag Force increases from zero to maximum when a ball reaches terminal velocity exam style question one a a sphere of radius are is moving through a fluid with constant speed V there is a frictional force F acting on the sphere which is given by the expression FAL 6i Dr R V where where D depends on the fluid one show that the SI base units of the quantity D are kilog per m/s the derived unit of force f is kilog m/s squared the SI base unit of radius R is meter the derived unit of speed V is m/s we rearrange the equation and making D is the subject like this we find the derived unit of d by substituting the derived units of f r and v like this we simplify the derived unit of D like this we get the derived of D is kilog per m/s two a raindrop of radius 1.5 mm Falls vertically in air at a velocity of 3.7 m/s the value of D for air is 6.6 * 10 power Nega for kg per m/ second the density of water is 1,000 kg per cubic m 1 calculate the magnitude of the frictional force F we substitute D = 6.6 * 10^4 R = 1.5 * 10^ -3 M and V = 3.7 m/s we get the force F equal 6.9 * 10^5 Newtons for two significant figures two calculate the acceleration of the raindrops when the raindrops moves downward like this there are the weight acting downward and the frictional force acting upward causing the Raindrop to accelerate downward we calculate the acceleration from the equation F equal m a f is the resultant Force which is m g minus F equal m a we rearrange the equation to isolate the acceleration like this where m is mass it calculates by multiplying density row time volume v the Raindrop is sphere so the volume V is for over 3 piun R cubed substituting g = 9.81 f = 6.9 * 10^-5 row = 1,000 R = 1.5 * 10^ 3 we get the acceleration equals for9 m/s squared for two significant figures B the variation with time T of the speed V of the raindrops in a is shown in the graph below one state the variation with time of the acceleration of the Raindrop the gradient of velocity time graph is the acceleration at t equal 0 second the initial acceleration equals g and then the acceleration decreases as time increases due to the gradient decreases until the acceleration is zero when the gradient of graph is zero two a second raindrop has a radius that is smaller than that given in a on the graph sketch the variation of speed with time for a second raindrop smaller radius causes the weight and the frictional force decreases but weight decrease greater that frictional force so the Raindrop will be reached the terminal velocity faster and lower we draw the graph like this exam style question two a sky diver jumps from a high altitude balloon a 1 explain briefly why the acceleration of the sky diver decreases with time drag Force increases as speed increases causing the resultant Force decreases two explain briefly why the acceleration of the sky diver is 9.8 m/s squared at the start of the jump the resultant force is weight acting downward and no drag Force B the variation with time T of the vertical speed V of the sky diver is shown in the graph below use the graph to determine the magnitude of the acceleration of the sky diver at time t equal 6 seconds we calculate the the acceleration by the gradient of velocity time graph we draw the tangent at t equal 6 seconds like this we draw the large triangle like this Delta y = 30 - 12 is equal to 18 like this Delta X = 11.2 - 1.6 is equal to 9.6 like this we calculate the gradient equal 18 / 9.6 we get the acceleration equals 1 9 m/s squared for two significant figures you got marks if the acceleration is in range 1.7 to 2.1 m/s squared C the sky diver and his equipment have total mass 90 kg 1 calculate for the sky driver and his equipment the total weight we calculate the weight by the equation W equal m g substituting m equal 90 kg and G = 9.8 1 m/s squared we get the weight equals 880 Newtons for two significant figures two calculate for the sky driver and his equipment the acceleration force at time t equal 6 seconds we calculate the resultant Force by the equation f = m a substituting m equal 90 kg and a equal 1.875 m/s squared we get the weight equals 70 Newtons for two significant figures three use your answer in 1 and two to determine the total resistance force acting on the sky diver at time tal 6 seconds we calculate the resistance force by the equation FAL ma a f is the resultant Force which is wus resistance force equal m a so resistance force equals wus ma a substituting w = 8829 Newtons and Ma equal 168.75 Newtons we get the resistance force equals 710 Newtons for two significant [Music] figures hi everyone I hope you found this video helpful if so I'd really appreciate it if you could subscribe to my channel like this video share it with your friends and leave a positive comment your support and encouragement motivate me to create more great videos in this video I've covered all the outlines in the syllabus as shown in the figure principle of conservation of momentum states that the total momentum of a closed system before an interaction is equal equal to the total momentum of that system after the interaction a closed system that is a system on which no external forces act momentum in collisions the total momentum of objects that Collide Remains the Same this is called conservation of momentum this means that the total momentum before Collision to equal the total momentum after Collision when a ball a with mass ma a collides to a ball B with mass m b like this during the each ball exerts a force on the other with equal in size and opposite in direction this is because Newton's three law about action and reaction so negative FAL FB these forces act each ball for the same amount of during time T this means the f * T being equal in magnitude but opposite in direction for each ball like this F A * T is the change in momentum of ball a FB * T is the change in moment of Ball B the negative shows that both change in momentum are in opposite direction before the Collision the initial velocity of the ball a is u a and the initial velocity of the ball B is u b after the Collision the final velocity of the ball a is VT a and the final velocity of the ball B is v b so the change in momentum of ball a is equal in magnitude but opposite in direction to the change in momentum of Ball B we rearrange range the equation like this this ensures that the total momentum before the Collision remains equal to the total momentum after the Collision exam style question one a wooden block is freely supported on brackets at a height of 4 m above the ground as shown a bullet of mass 5 G is shot vertically upwards into the wooden block of mass 95 G it embeds itself in the block the impact causes the block to rise above its supporting brackets the bullet hits the block with a velocity of 200 m/s how far above the ground will the block be at the maximum height of its path when the bullet hits the block with a velocity of 200 m/s they stick together with a speed v as shown we calculate the speed V of the bullet and the block by the total momentum before Collision equals total momentum after Collision first we convert the masses to kill K bullet Mass 5 G equal 0.005 kg block Mass 95 G equal 0.095 kg so the total mass of the bullet and wooden block is 0.1 kg the momentum of the bullet before the Collision equals 0.005 * 200 the momentum of the block after the Collision equals 0.095 * 0 due to to the block rest the momentum of the bullet and block after the Collision equals 0.1 * V due to they stick together we get the speed V equals 10 m/s when both bullet and block rise above its supporting bracket to reach the maximum height at their speed is zero as shown during this moment the displacement s need to find the initial velocity U equal 10 the final velocity V equals 0 the acceleration equals 9 8 1 due to downward Direction and time T is unknown we calculate the displacement s by i v ^2 = u 2 + 2 a s substituting V = 0 u = 10 a = 9.81 we get the displacement s equals 5.09 68 M so the total distance from the ground equals 9.1 M for two significant figures Collision in two dimensions a ball a of mass ma with a velocity U moving towards a stationary Ball B of mass MB the balls Collide after the Collision ball a moves off with velocity VA at an angle Theta to its original velocity and ball B moves with a velocity VB at an angle of beta to the original velocity of a we break VA into horizontal component as VA cos Theta and vertical component as VA sin Theta like this we break VB into horizontal component as VB e cos beta and vertical component as VB sin beta like this horizontally total momentum before Collision equals ma U this is because momentum of Ball B is zero horizontally total momentum after Collision equals ma VA C the + MB VB e cos beta vertically total momentum before Collision equals zero this is because vertical momentum of both balls is zero as both balls move in horizontally along x-axis vertically total momentum after Collision equals ma a VA sin theta plus netive MB VB sin beta the negative sign indicates downward Direction the momentum must be conserved during the Collision the total momentum before Collision is shown by the momentum Vector of ma Au U like this the total momentum after Collision is shown by the sum of the momentum Vector of m a VA and MB VB like this this shows that m u equal m a VA a plus MB VB we can resolve the momentum as follows horizontally the total momentum is conserved so m m u equal m a v a COS theta plus MB VB cos beta vertically the total momentum is conserved so 0 equal m a VA sin Theta MB VB sin beta exam style question two a ball X moves along a horizontal frictionless surface and collides with another ball y as Illustrated in figure the balls stick together during the collision and then travel along line AB as Illustrated in figure A1 calculate to three significant figures the component of the initial momentum of ball y that is perpendicular to line AB we break a velocity of ball Y into horizontal and vertical components as follows 6 cos 60° and 6 sin 60° we calculate the momentum by P equal m v so the component of the initial momentum of ball y that is perpendicular to line AB equals 0.20 * 6. sin 60° we get the answer is 1.0 for kg m/ second for three significant figures two by considering the component of the initial momentum of each ball perpendicular to line a calculate to three significant figures VX we break a velocity of ball X into horizontal and vertical components as follows VX cos 60° and VX sin 60° downward vertically the total momentum is conserved vertical momentum of ball X before the Collision equals 0.30 * VX sin 60° the negative sign indicates the downward Direction vertical momentum of ball y before the Collision equals 0.20 * 6 0 sin 60° vertical momentum of both balls after the Collision equals z because they don't move in vertical we get the speed v x equals for. m/s for three significant figures three show that the speed of the two balls after the Collision is 2. for m/s we give the speed of two balls after the Collision equals V horizontally the total momentum is conserved horizontal momentum of ball X before the Collision equals 0.30 * VX cos 60° horizontal momentum of ball y before the Collision equals 0.20 * 6. cos 60° horizontal momentum of two balls after the Collision equals 0.50 * V we get the speed V equals 2.4 m/s this is shown exam style question three two balls X and Y move along a horizontal frictionless surface as Illustrated in figure below the two balls Collide at point B the balls stick together and then travel along the horizontal surface in a direction at right angles to line AB as shown in figure above a by considering the components of momentum in the direction from A to B show that the ball X has a mass of 4 kg we break the velocity 99.6 m/s into vertical and horizontal components as follows 9.6 sin 60° and 9.6 60° as shown horizontally the total momentum is conserved horizontal momentum of the ball X before the Collision equals MX * 3.0 horizontal momentum of the ball y before the Collision equals -2.5 * 9.6 cos 60° the Nega sign indicates the left Direction horizontal momentum of two balls after the Collision equals zero because they move only in the vertical we get the mass MX equals for 0 kg this is shown be calculate the common speed V of the two balls after the Collision vertically the total momentum is conserved vertical momentum of the ball X before the Collision equals zero because it moves in only horizontal vertical momentum of the ball y before the Collision equals 2.5 * 9.6 in 60° vertical momentum of two balls after the Collision equals 6.5 * V we get the speed V equal 3.2 m/ second for two significant figures C determine the difference between the initial kinetic energy of ball X and the initial kinetic energy of ball y kinetic energy of ball X before the Collision equals half of 4 * 3 SED resulting in 18 Jewel kinetic energy of ball y before the Collision equals half of 2.5 * 9.6 SAR resulting in 115.2 Jew so the difference in kinetic energy equals 97 jeel for two significant figures types of collisions when the ball mass m a with the speed UA collides with the ball Mass MB with the speed u b after the Collision the ball mass m a moves with the speed V A and the ball mass m b moves with the speed v b total momentum before the Collision like this total momentum after the Collision like this total kinetic energy before the Collision like this total kinetic energy after the Collision like this there are two types of the Collision as elastic collisions and inelastic collisions elastic collisions the momentum is conserved during elastic collisions so total momentum before the Collision equals total momentum after the Collision the kinetic energy is conserved during elastic collisions so total kinetic energy before the Collision equals total kinetic energy after the Collision the relative speed of approach is equal to the relative speed of Separation during the elastic collisions in elastic collisions the momentum is conserved during inelastic collisions so total momentum before the Collision equals total momentum after the Collision the kinetic energy is not conserved during inelastic collisions so total kinetic energy before the Collision is more than total kinetic energy after the Collision the relative speed of approach is not equal to the relative speed of Separation during the inelastic collisions when the ball a with a speed 5 m/s to the right and the ball B with a speed 8 m/s to the right so the relative speed equals 8 - 5 is equal to 3 m/s when the ball a with a speed for m/s to the left and the ball B with a speed 6 m/s to the left so the relative speed equals 6 minus for is equal to 2 m/s when the ball a with a speed for me/ second to the right and the ball B with a speed 5 m/s to the left so the relative speed equals for plus 5 is equal to 9 m/s when the ball a with a speed 2 m/s to the left and the ball B with a Speed 3 m/s to the right so the relative speed equals 2 + 3 is equal to 5 m/s therefore the relative speed between two objects is calculated as follows subtracting their speeds when they move in same directions adding their speeds when they move in opposite directions exam style question four two spheres approach each other along the same straight line their speeds are U1 and U2 before collision and V1 and V2 after Collision in the direction shown below which equation is correct if the Collision is perfect elastic the relative speed of approach is equal to the relative speed of Separation during elastic Collision the relative speed of approach is U1 + U2 the relative speed of separation is V1 minus V2 so D is correct exam style question five the diagram shows two spherical masses approaching each other headon at an equal speed U one has mass 2 m and the other has mass m Which diagram showing the situation after the Collision shows the result of an elastic Collision the relative speed of approach is equal to the relative speed of Separation during elastic Collision the relative speed of approach is U + u = 2 U A the relative speed of separation is 2 U this is correct B the relative speed of separation is 5 u/ 6 this is incorrect C the relative speed of separation is U over 2 this is incorrect D the relative speed of separation is zero this is incorrect exam style question six a state Newton's third law of motion if body a exerts a force on body B then body B will exert a force on body a of equal magnitude but in the opposite direction b a block X of mass MX slides in a straight line along a horizontal frictionless surface as shown in figure below the block X moving with speed 5V collides headon with a stationary block y of mass my y the two blocks stick together and then move with common speed v as shown one use conservation of momentum to show that the ratio m y or to MX is equal to 4.0 the momentum of block X before the Collision equals MX * 5 V the momentum of block y before the Collision equals zero because it is stationary the momentum of two blocks after the Collision equals mx + my y * V we simplify and rearrange the equation like this we get the ratio equals for 0 two calculate the ratio total kinetic energy of X and Y after Collision to Total kinetic energy of X and Y before Collision total kinetic energy of X and Y before the Collision is half of the sum MX and m y * V ^2 total kinetic energy of X and Y after the Collision is half of MX * 5 V ^ 2 we simplify the equation like this substituting the ratio of m y to MX is four we get the ratio of kinetic energy is 0.20 three state the value of the ratio in two for a perfect elastic Collision the ratio is one this is because the kinetic energy is conserved for a perfect elastic Collision meaning that total kinetic energy before the Collision equals total kinetic energy after the Collision C the variation with time T of the momentum of block X in B is shown in figure below block X makes contact with block Y at time t equal 20 milliseconds 1 describe qualitatively the magnitude and direction of the resultant Force if any acting on block X in the time interval t equal 0 to 20 milliseconds the resultant Force equals zero this is because the speed and momentum is constant t equal 20 milliseconds to tal 40 milliseconds the resultant force is constant and opposite direction to the movement or momentum this is because the gradient of the graph is the resultant force and its gradient is constant and negative two on figure above sketch the variation of the momentum of block y with time T from tal 0 to tal 60 milliseconds the momentum of block Y is 0 between t = 0 to 20 milliseconds because it is stationary so we draw the graph like this the momentum of block y increas es from 0 between tal 20 to 40 milliseconds to this this is because total momentum before the Collision is equal to total momentum after Collision total momentum before the Collision equals 5 total momentum after the Collision must be five momentum in explosion the conservation of momentum principle can be applied to explosions the total momentum of the system before the explosion is equal to the total momentum of the system after the explosion although there will be a significant increase in the total kinetic energy of the fragments an explosion involves a rapid release of energy causing an object to break apart into pieces that fly in different directions let's consider an object with mass m that is stationary before the explosion so its momentum is zero the object explodes into two fragments with masses M1 and M2 which move apart at velocities V1 and V2 2 respectively the total momentum after the explosion is the sum of the momenta of the two fragments M1 V1 plus M2 V2 momentum is conserved when during explosion so total momentum before explosion equals total momentum after explosion then 0 equal M1 V1 + M2 V2 and M1 V1 equal minus M2 V2 this equation tells us that the momentum of fragment M1 is equal in magnitude but opposite in direction to the momentum of fragment M2 as the rocket burns the fuel generating hot gases that are ejected at high speeds from the rocket this produces the momentum large amount of fast moving gases out of the back of the rocket so the rocket gains an equal amount of momentum in the opposite direction to that of the moving exhaust gases exam style question 7 a stationary explodes into two components of mass m and 2 m the compol lenters gain kinetic energies X and Y respectively what is the value of the ratio X to Y we give a velocity of mass m is VX and velocity of Mass 2 m is v y we find the velocity VX first using the principle of conservation of momentum total momentum before the explosion equals 0 due to it is stationary the momentum of mass m after the explosion equals m * VX the negative sign indicates the left Direction the momentum of Mass 2 m after the explosion equals 2 m * v y we rearrange and simplify the equation like this we get the velocity v x = 2 v y the kinetic energy xal half of m vx^ 2 the kinetic energy y equals half of 2 m v y or 2 we simplify the equation like this substituting VX = 2 v y like this we simplify the equation like this we get the ratio of x to y equal 2 exam style question 8 a stationary nucleus has nuclear number K the nucleus decays by emitting a proton with speed V to form a new nucleus speed U the new nucleus and the proton move away from one another in opposite directions what is the speed V in term of a and u the nucleon number is mass number and it can be represented Mass so mass of a proton equals 1 and Mass of a new nucleus equals a n minus1 we find the velocity V of a proton using the principle of conservation of momentum total momentum before decay equals 0 due to a nucleus is stationary the momentum of a proton after Decay equals -1 * V the negative sign indicates the left Direction the momentum of a new nucleus after Decay equals a -1 * U we get the velocity V equal a - 1 * U I hope you found this video helpful if you did I would be grateful if you would subscribe share like and leave a positive comment your support will encourage me to create more content thank you