Newton used the concept of momentum to Newton used the concept of momentum to explain the results of collisionsexplain the results of collisions

Momentum Momentum = mass x velocity= mass x velocity

p p = m v= m v

Units :Units :

p (kg m/s) p (kg m/s) = m (kg) x (m/s)= m (kg) x (m/s)

Note since velocity is a Note since velocity is a vectorvector quantity, (both quantity, (both magnitude and direction) then magnitude and direction) then momentummomentum is also is also a a vectorvector quantity quantity

When objects collide, assuming that there are no When objects collide, assuming that there are no external forces, then momentum is always external forces, then momentum is always conserved.... Definition :conserved.... Definition :

When two or more objects interact, the total When two or more objects interact, the total momentum remains constant provided that momentum remains constant provided that there is no external resultant forcethere is no external resultant force

Mass 75 kgVelocity 4m/s

Mass 50 kgVelocity 0m/s

Mass 125 kgVelocity ??? m/s

When objects hit each other the resulting When objects hit each other the resulting collision can be considered to be either elastic or collision can be considered to be either elastic or inelastic. Momentum and inelastic. Momentum and totaltotal energy are always energy are always conserved in both cases. conserved in both cases.

Elastic :Elastic : momentum conserved, kinetic momentum conserved, kinetic energy conserved, total energy energy conserved, total energy conservedconserved

Inelastic :Inelastic : momentum conserved, kinetic momentum conserved, kinetic energy energy NOTNOT conserved, total energy conserved, total energy conservedconserved

In an Inelastic collision some of the kinetic energy is In an Inelastic collision some of the kinetic energy is converted to other forms of energy (often heat & Sound)converted to other forms of energy (often heat & Sound)

Two trolleys on an air track are fitted with Two trolleys on an air track are fitted with repelling magnets. The masses are 0.1kg and repelling magnets. The masses are 0.1kg and 0.15kg respectively. When they are released the 0.15kg respectively. When they are released the lighter trolley moves to the left at 0.24m/s. What lighter trolley moves to the left at 0.24m/s. What is the velocity of the heavier trolleyis the velocity of the heavier trolley

A ball of 0.6kg moving at 5m/s collides with a A ball of 0.6kg moving at 5m/s collides with a larger stationary ball of mass 2kg. The smaller ball larger stationary ball of mass 2kg. The smaller ball rebounds in the opposite direction at 2.4m/srebounds in the opposite direction at 2.4m/s

Calculate the velocity of the larger ballCalculate the velocity of the larger ball

Is the Collision elastic or inelastic. Explain your Is the Collision elastic or inelastic. Explain your answeranswer

Definition :Definition :

The impulse of a force is defined as the product of The impulse of a force is defined as the product of the force and the time which the force acts forthe force and the time which the force acts for

The impulse = FThe impulse = Ft = t = mvmv

The impulse of the force acting upon an object is The impulse of the force acting upon an object is equal to the change of momentum for the forceequal to the change of momentum for the force

An object of constant mass m is acted upon by a An object of constant mass m is acted upon by a constant force F which results in a change of constant force F which results in a change of velocity from u to vvelocity from u to v

From the 2From the 2ndnd law law F = (mv – mu )/tF = (mv – mu )/t

Rearranging : Ft = Rearranging : Ft = mv – mu mv – mu

Graphically.....Graphically.....

time

forc

e

F

t

Area under graph “Ft” = change of momentum

A train of mass 24,000kg moving at a velocity of A train of mass 24,000kg moving at a velocity of 15m/s is stopped by a braking force of 6000N. 15m/s is stopped by a braking force of 6000N. Calculate :Calculate :

1.1. The initial momentum of the trainThe initial momentum of the train

2.2. The time taken for the train to stopThe time taken for the train to stop

An aircraft with total mass 45,000kg accelerates An aircraft with total mass 45,000kg accelerates on the runway from rest to 120m/s at which on the runway from rest to 120m/s at which point it takes off. The engines provide a constant point it takes off. The engines provide a constant driving force of 120kN. Calculate the gain in driving force of 120kN. Calculate the gain in momentum and the take of timemomentum and the take of time

The velocity of a car of mass 600kg was reduced The velocity of a car of mass 600kg was reduced from a speed of 15m/s by a constant force of from a speed of 15m/s by a constant force of 400N which acted for 20s and then by a constant 400N which acted for 20s and then by a constant force of 20N for a further 20s.force of 20N for a further 20s.

Sketch a force v time graphSketch a force v time graph

Calculate the initial momentum of the carCalculate the initial momentum of the car

Use your Force v time graph to establish the Use your Force v time graph to establish the change in momentumchange in momentum

Show the final velocity of the car is 1m/sShow the final velocity of the car is 1m/s

During the Y11 course of study, it was discussed how During the Y11 course of study, it was discussed how many car safety features such as seatbelts, crumple many car safety features such as seatbelts, crumple zones and air bags increase safety by making the crash zones and air bags increase safety by making the crash “last longer”“last longer”

During our Y12 presentations, change in momentum was During our Y12 presentations, change in momentum was connected to car safety. Now taking it further and connected to car safety. Now taking it further and considering the impulse of a force :considering the impulse of a force :

The impulse = FThe impulse = Ft = t = mvmv

For a given crash the mass & velocity of the For a given crash the mass & velocity of the vehicle are defined. vehicle are defined. By By increasingincreasing t we t we decreasedecrease the force acting on the occupantsthe force acting on the occupants

We have seen that momentum is a vector We have seen that momentum is a vector quantity since it’s related to velocity which is a quantity since it’s related to velocity which is a vector quantity. vector quantity. direction is important and direction is important and therefore we need a “sign” convention to take therefore we need a “sign” convention to take this into account.this into account.

If we consider a ball with mass If we consider a ball with mass mm hitting a wall hitting a wall and rebounding normally, (i.e. at 90°): and rebounding normally, (i.e. at 90°):

Initial velocity = +uInitial momentum = +mu

Towards the wall we take as positive

Away from the wall we take as negative

If we assume there is no loss of speed after the If we assume there is no loss of speed after the impact then considering the change in impact then considering the change in momentum...momentum...

Ft = final momentum – initial momentumFt = final momentum – initial momentum

Ft = -mu – (+mu)Ft = -mu – (+mu)

F = -2mu /tF = -2mu /t

Final velocity = -uFinal momentum = -mu

When the impact is oblique, (i.e. At an angle, not When the impact is oblique, (i.e. At an angle, not normally at 90°): normally at 90°):

In this case we use the normal components of the In this case we use the normal components of the velocity. Initially, this is +(u cos velocity. Initially, this is +(u cos ). Similarly this ). Similarly this will give an overall change in momentum of :will give an overall change in momentum of :

Ft = -2mu cos Ft = -2mu cos

Initial velocity = +uInitial momentum = +mu

A squash ball is released from rest above a flat A squash ball is released from rest above a flat surface. Describe how the energy changes is i) it surface. Describe how the energy changes is i) it rebounds to the same height, ii) It rebounds to a rebounds to the same height, ii) It rebounds to a lesser heightlesser height

If the ball is released from a height of 1.20m and If the ball is released from a height of 1.20m and rebounds to a height of 0.9m show that 25% of rebounds to a height of 0.9m show that 25% of the kinetic energy is lost upon impactthe kinetic energy is lost upon impact

A shell of mass 2kg is fired at a speed of 140m/s A shell of mass 2kg is fired at a speed of 140m/s from a gun with mass 800kg. Calculate the recoil from a gun with mass 800kg. Calculate the recoil velocity of the gunvelocity of the gun

A molecule of mass 5.0 x 10A molecule of mass 5.0 x 10-26-26 kg moving at a kg moving at a speed of 420m/s hits a surface at right angles and speed of 420m/s hits a surface at right angles and rebounds at the opposite direction at the same rebounds at the opposite direction at the same speed. The impact lasted 0.22ns. Calculate:speed. The impact lasted 0.22ns. Calculate:

i)i) The change in momentumThe change in momentum

ii)ii) The force on the moleculeThe force on the molecule

Repeat the last molecule question. This time the Repeat the last molecule question. This time the molecule strikes the surface at 60° to the normal molecule strikes the surface at 60° to the normal and rebounds at 60° to the normal.and rebounds at 60° to the normal.

Angles can be measured in both degrees & radians :Angles can be measured in both degrees & radians :

The angle in radians is defined as the arc length / the radius

For a whole circle, (360°) the arc length is the circumference, (2r)

360° is 2 radians

Arclength

r

Common values :

45° = /4 radians90° = /2 radians180° = radians

Note. In S.I. Units we use “rad”

How many degrees is 1 radian?

Angular velocity, for circular motion, has Angular velocity, for circular motion, has counterparts which can be compared with linear counterparts which can be compared with linear speed speed s=d/ts=d/t..

Time (t) remains unchanged, but linear distance (d) Time (t) remains unchanged, but linear distance (d) is replaced with angular displacement is replaced with angular displacement measured measured in radians.in radians.

Angular displacement

r

r Angular displacement is the number of radians moved

Consider an object moving along the arc of a circle Consider an object moving along the arc of a circle from A to P at a constant from A to P at a constant speedspeed for time t: for time t:

Definition : The rate of change of angular displacement with time

“The angle, (in radians) an object rotates through per second”

= / t

Arc length

r

r

P

A

This is all very comparable with normal linear speed, (or velocity) where we talk about distance/time

Where is the angle turned through in radians, (rad), yields units for of rad/s

The period T of the rotational motion is the time The period T of the rotational motion is the time taken for one complete revolution (2taken for one complete revolution (2 radians). radians).

Substituting into : = /t

= 2 / T

T = 2 /

From our earlier work on waves we know that the period (T) & frequency (f) are related T = 1/f

f = / 2

Considering the diagram below, we can see that Considering the diagram below, we can see that the linear distance travelled is the arc lengththe linear distance travelled is the arc length

Linear speed (v) = arc length (AP) / t

v = r /t

Substituting... ( = / t)

v = r

Arc length

r

r

P

A

A cyclist travels at a speed of 12m/s on a bike A cyclist travels at a speed of 12m/s on a bike with wheels which have a radius of 40cm. with wheels which have a radius of 40cm. Calculate:Calculate:

a.a. The frequency of rotation for the wheelsThe frequency of rotation for the wheels

b.b. The angular velocity for the wheelsThe angular velocity for the wheels

c.c. The angle the wheel turns through in 0.1s inThe angle the wheel turns through in 0.1s in

i radians ii degrees i radians ii degrees

The frequency of rotation for the wheelsThe frequency of rotation for the wheels

Circumference of the wheel is 2Circumference of the wheel is 2r r

= 2= 2 x 0.4m = 2.5m x 0.4m = 2.5m

Time for one rotation, (the period) is found usingTime for one rotation, (the period) is found using

s =d/t rearranged for ts =d/t rearranged for t

t = d/s = T = circumference / linear speedt = d/s = T = circumference / linear speed

T = 2.5/12 = 0.21sT = 2.5/12 = 0.21s

f = 1/T = 1/0.21 = f = 1/T = 1/0.21 = 4.8Hz 4.8Hz

The angular velocity for the wheelsThe angular velocity for the wheels

Using T = 2Using T = 2 / / , rearranged for , rearranged for

= 2= 2 /T /T

== 2 2 /0.21 /0.21

== 30 rad/s 30 rad/s

The angle the wheel turns through in 0.1s inThe angle the wheel turns through in 0.1s in

i radians ii degreesi radians ii degrees

Using Using = = / t re-arranged for / t re-arranged for

= = tt

= 30 x 0.1= 30 x 0.1

= 3 rad = 3 rad

= 3 x (360°/2= 3 x (360°/2) ) = 172°= 172°

If an object is moving in a circle If an object is moving in a circle with a constant speed, it’s with a constant speed, it’s velocity is constantly changing....velocity is constantly changing....

Because the direction is Because the direction is constantly changing....constantly changing....

If the velocity is constantly If the velocity is constantly changing then by definition the changing then by definition the object is acceleratingobject is accelerating

If the object is accelerating, then If the object is accelerating, then an unbalanced force must existan unbalanced force must exist

Velocity v

acceleration

We can substitute for angular velocity....We can substitute for angular velocity....

a = va = v22/r/r

From the last lesson we saw that:From the last lesson we saw that:

v = rv = r (substituting for v into above) (substituting for v into above)

a = (ra = (r))22/r/r

a = ra = r22

In exactly the same way as we can connect force In exactly the same way as we can connect force f f and acceleration and acceleration aa using Newton’s 2 using Newton’s 2ndnd law of law of motion, we can arrive at the centripetal force motion, we can arrive at the centripetal force which is keeping the object moving in a circlewhich is keeping the object moving in a circle

f = mvf = mv22/r/r

oror

f = mrf = mr22

Any object moving in a circle is acted upon by a Any object moving in a circle is acted upon by a single resultant force towards the centre of the single resultant force towards the centre of the circle. We call this the centripetal forcecircle. We call this the centripetal force

The wheel of the London Eye has a diameter of The wheel of the London Eye has a diameter of 130m and takes 30mins for 1 revolution. 130m and takes 30mins for 1 revolution. Calculate:Calculate:

a.a. The speed of the capsuleThe speed of the capsule

b.b. The centripetal accelerationThe centripetal acceleration

c.c. The centripetal force on a person with a The centripetal force on a person with a mass of 65kgmass of 65kg

The speed of the capsule :The speed of the capsule :

Using v = rUsing v = r

we know that we do a full revolution (2we know that we do a full revolution (2 rad) rad) in 30mins (1800s)in 30mins (1800s)

v = (130/2) x (2v = (130/2) x (2 / 1800) / 1800)

v = 0.23 m/sv = 0.23 m/s

The centripetal acceleration:The centripetal acceleration:

Using a = vUsing a = v22/r/r

a = (0.23)a = (0.23)22 / (130/2) / (130/2)

a = 0.792 x 10a = 0.792 x 10-4-4 m/s m/s22

The centripetal force:The centripetal force:

Using f = maUsing f = ma

F = 65 x 0.792 x 10F = 65 x 0.792 x 10-4-4

F = 0.051 NF = 0.051 N

An object of mass 0.15kg moves around a circular An object of mass 0.15kg moves around a circular path which has a radius of 0.42m once every 5s at path which has a radius of 0.42m once every 5s at a steady rate. Calculate:a steady rate. Calculate:

a.a. The speed and acceleration of the objectThe speed and acceleration of the object

b.b. The centripetal force on the objectThe centripetal force on the object

During the last lesson we saw that an object During the last lesson we saw that an object moving in a circle has a constantly changing moving in a circle has a constantly changing velocity, it is therefore experiencing acceleration velocity, it is therefore experiencing acceleration and hence a force towards the centre of and hence a force towards the centre of rotation.rotation.

We called this the centripetal force: The force We called this the centripetal force: The force required to keep the object moving in a circle. In required to keep the object moving in a circle. In reality this force is provided by another force, reality this force is provided by another force, e.g. The tension in a string, friction or the force e.g. The tension in a string, friction or the force of gravity.of gravity.

Consider a car with mass Consider a car with mass mm and speed and speed vv moving moving over the top of a hill...over the top of a hill...

mg

r

S

At the top of the hill, the support force S, is in the At the top of the hill, the support force S, is in the opposite direction to the weight (mg). It is the opposite direction to the weight (mg). It is the resultant between these two forces which keep resultant between these two forces which keep the car moving in a circlethe car moving in a circle

mg – S = mvmg – S = mv22/r/r

If the speed of the car increases, there will If the speed of the car increases, there will eventually be a speed eventually be a speed vv00 where the car will leave where the car will leave the ground (the support force S is 0)the ground (the support force S is 0)

mg = mvmg = mv0022/r/r vv00 = (gr) = (gr)½½

Any faster and the car will leave the groundAny faster and the car will leave the ground

A car with mass 1200kg passes over a bridge with a radius A car with mass 1200kg passes over a bridge with a radius of curvature of 15m at a speed of 10 m/s. Calculate:of curvature of 15m at a speed of 10 m/s. Calculate:

a.a. The centripetal acceleration of the car on the bridgeThe centripetal acceleration of the car on the bridge

b.b. The support force on the car when it is at the topThe support force on the car when it is at the top

The maximum speed without skidding for a car with The maximum speed without skidding for a car with mass 750kg on a roundabout of radius 20m is 9m/s. mass 750kg on a roundabout of radius 20m is 9m/s. Calculate:Calculate:

a.a. The centripetal acceleration of the car on the The centripetal acceleration of the car on the roundaboutroundabout

b.b. The centripetal force at this speedThe centripetal force at this speed

A car is racing on a track banked at 25° to the A car is racing on a track banked at 25° to the horizontal on a bend with radius of curvature of horizontal on a bend with radius of curvature of 350m350m

a.a. Show that the maximum speed at which the Show that the maximum speed at which the car can take the bend without sideways car can take the bend without sideways friction is 40m/sfriction is 40m/s

b.b. Explain what will happen if the car takes the Explain what will happen if the car takes the bend at ever increasing speedsbend at ever increasing speeds

A car on a big dipper starts from rest and descends A car on a big dipper starts from rest and descends though 45m into a dip which has a radius of though 45m into a dip which has a radius of curvature of 78m. Assuming that air resistance & curvature of 78m. Assuming that air resistance & friction are negligible. Calculate:friction are negligible. Calculate:

a.a. The speed of the car at the bottom of the dipThe speed of the car at the bottom of the dip

b.b. The centripetal acceleration at the bottom of The centripetal acceleration at the bottom of the dipthe dip

c.c. The extra force on a person with a weight of The extra force on a person with a weight of 600N in the train600N in the train

A swing at a fair has a length of 32m. A passenger of A swing at a fair has a length of 32m. A passenger of mass 69kg falls from the position where the swing is mass 69kg falls from the position where the swing is horizontal. Calculate:horizontal. Calculate:

a.a. The speed of the person at the lowest pointThe speed of the person at the lowest point

b.b. The centripetal acceleration at the lowest The centripetal acceleration at the lowest pointpoint

c.c. The support force on the person at the lowest The support force on the person at the lowest pointpoint

A wall of death ride at the fairground has a radius A wall of death ride at the fairground has a radius of 12m and rotates once every 6s. Calculate:of 12m and rotates once every 6s. Calculate:

a.a. The speed of rotation at the perimeter of the The speed of rotation at the perimeter of the wheelwheel

b.b. The centripetal acceleration of a person on The centripetal acceleration of a person on the perimeterthe perimeter

c.c. The support force on a person of mass 72kg at The support force on a person of mass 72kg at the highest pointthe highest point