1.1. To Summarise what has been learnt about To Summarise what has been learnt about momentum by looking at real world examples.momentum by looking at real world examples.

• Rebound impacts including oblique impactsRebound impacts including oblique impacts

• Dropped ballsDropped balls

• Explosions & GunsExplosions & Guns

• Rockets & Water jetsRockets & Water jets

Book Reference : Pages 4-17Book Reference : Pages 4-17

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 can take as positive

Away from the wall we can 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

If we drop a perfectly bouncy ball If we drop a perfectly bouncy ball onto a hard surface it should onto a hard surface it should rebound to almost the original rebound to almost the original height it was dropped from.height it was dropped from.

Kinetic energy just prior to impact Kinetic energy just prior to impact will equal kinetic energy just after will equal kinetic energy just after impact if it is a perfect elastic impact if it is a perfect elastic collisioncollision

In this way we can make a In this way we can make a connection between “elastic” and a connection between “elastic” and a perfectly bouncy ballperfectly bouncy ball

When approaching “dropped ball” When approaching “dropped ball” calculations one applies our SUVAT calculations one applies our SUVAT equations in a manner similar to the equations in a manner similar to the way we approached projectilesway we approached projectiles

v = u + atv = u + at (1)(1)

s = s = (u + v)t(u + v)t (2)(2)22

s = ut + ½ats = ut + ½at22 (3)(3)

vv22 = u = u22 + 2as + 2as (4)(4)

Explosion problems are categorised by all the Explosion problems are categorised by all the components initially being at rest. Then after components initially being at rest. Then after some event, two or more objects move apart.some event, two or more objects move apart.

Since initially all objects are at rest, the total initial Since initially all objects are at rest, the total initial momentum is zeromomentum is zero

We use the “signed” nature of direction to again equate We use the “signed” nature of direction to again equate the total final momentum to zerothe total final momentum to zero

Common examples include, trolleys or air track vehicles Common examples include, trolleys or air track vehicles pushed part by springs or by repelling magnetspushed part by springs or by repelling magnets

The recoil in gun barrels is also a good example The recoil in gun barrels is also a good example

Explosion problems can be tested with either Explosion problems can be tested with either sprung trolleys or air track vehicles: sprung trolleys or air track vehicles:

Spring loaded bolt

Trolley A Trolley B

Block Block

BA

When the sprung bolt is released the two trolleys move When the sprung bolt is released the two trolleys move apart in opposite directions. The blocks A and B are apart in opposite directions. The blocks A and B are positioned such that the trolleys strike them at the positioned such that the trolleys strike them at the same moment. From s=d/t, since the time is identical, same moment. From s=d/t, since the time is identical, the ratio of the distances to the blocks is the same as the ratio of the distances to the blocks is the same as the ratio of the speeds of the trolleys which is turn is the ratio of the speeds of the trolleys which is turn is the inverse of the mass ratios.the inverse of the mass ratios.

When a rocket fires its engines the rocket gains When a rocket fires its engines the rocket gains momentum equal and opposite to the momentum equal and opposite to the momentum of the hot exhaust gases.momentum of the hot exhaust gases.

For example we could be told that a 20,000kg For example we could be told that a 20,000kg rocket at rest fires its rocket for 10s. The exhaust rocket at rest fires its rocket for 10s. The exhaust gases leave at 100kgsgases leave at 100kgs-1-1 at a speed of 1kms at a speed of 1kms-1-1..

Show that the final velocity of the rocket is 53msShow that the final velocity of the rocket is 53ms-1 -1

Water of density 1000 kg mWater of density 1000 kg m-3-3 flows out of a hose pipe with cross flows out of a hose pipe with cross sectional area 7.2x10sectional area 7.2x10-4-4 m m22 at a rate of 2.0 x 10 at a rate of 2.0 x 10-4 -4 mm33 per second. How per second. How much momentum is carried by the water each second?much momentum is carried by the water each second?

First we need to consider the mass each second. Knowing the First we need to consider the mass each second. Knowing the volume and density we can find the mass volume and density we can find the mass

= m/v = m/v m = m = v v

Mass each second = 1000 x 2x10Mass each second = 1000 x 2x10-4-4 = 0.2kg per second = 0.2kg per second

Next we need to consider the velocity. We know the volume Next we need to consider the velocity. We know the volume leaving each second and also the area. We effectively have a leaving each second and also the area. We effectively have a cylinder of length 2x10cylinder of length 2x10-4-4 / 7.2 x10 / 7.2 x10-4-4 = 0.278 m. Hence the velocity is = 0.278 m. Hence the velocity is 0.278 ms0.278 ms-1-1

Momentum each second = 0.2 x 0.278 = Momentum each second = 0.2 x 0.278 = 0.056 kgms0.056 kgms-1-1

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 if i) it surface. Describe how the energy changes if 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 140msA shell of mass 2kg is fired at a speed of 140ms-1-1 from a from a gun with mass 800kg. Calculate the recoil velocity of the gun with mass 800kg. Calculate the recoil velocity of the gungun

A molecule of mass 5.0 x 10A molecule of mass 5.0 x 10-26-26 kg moving at a speed of kg moving at a speed of 420ms420ms-1-1 hits a surface at right angles and rebounds at the hits a surface at right angles and rebounds at the opposite direction at the same speed. The impact lasted opposite direction at the same speed. The impact lasted 0.22ns. Calculate:0.22ns. Calculate:

i)i) The change in momentumThe change in momentumii)ii) The force on the moleculeThe force on the molecule

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