Momentum : Real World Examples

1. Learning Objectives Book Reference : Pages 4-17 Momentum : Real World Examples • To Summarise what has been learnt about momentum by looking at real world examples. • Rebound impacts including oblique impacts • Dropped balls • Explosions & Guns • Rockets & Water jets

2. We have seen that momentum is a vector quantity since it’s related to velocity which is a vector quantity.  direction is important and therefore we need a “sign” convention to take this into account. If we consider a ball with mass m hitting a wall and rebounding normally, (i.e. at 90°): Rebound Impacts 1 Towards the wall we can take as positive Away from the wall we can take as negative Initial velocity = +u Initial momentum = +mu

3. Rebound Impacts 2 Final velocity = -u Final momentum = -mu • If we assume there is no loss of speed after the impact then considering the change in momentum... • Ft = final momentum – initial momentum • Ft = -mu – (+mu) • F = -2mu /t

4. When the impact is oblique, (i.e. At an angle, not normally at 90°): Rebound Impacts 3   Initial velocity = +u Initial momentum = +mu • In this case we use the normal components of the velocity. Initially, this is +(u cos). Similarly this will give an overall change in momentum of : • Ft = -2mu cos

5. If we drop a perfectly bouncy ball onto a hard surface it should rebound to almost the original height it was dropped from. Kinetic energy just prior to impact will equal kinetic energy just after impact if it is a perfect elastic collision In this way we can make a connection between “elastic” and a perfectly bouncy ball Dropped Bouncing balls 1

6. When approaching “dropped ball” calculations one applies our SUVAT equations in a manner similar to the way we approached projectiles Dropped Bouncing balls 2 v = u + at (1) s = (u + v)t (2) 2 s = ut + ½at2 (3) v2 = u2 + 2as (4)

7. Explosion problems are categorised by all the components initially being at rest. Then after some event, two or more objects move apart. Since initially all objects are at rest, the total initial momentum is zero We use the “signed” nature of direction to again equate the total final momentum to zero Common examples include, trolleys or air track vehicles pushed part by springs or by repelling magnets The recoil in gun barrels is also a good example Explosions 1

8. Explosion problems can be tested with either sprung trolleys or air track vehicles: Explosions 2 Spring loaded bolt Block Block B A Trolley A Trolley B When the sprung bolt is released the two trolleys move apart in opposite directions. The blocks A and B are positioned such that the trolleys strike them at the 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 speeds of the trolleys which is turn is the inverse of the mass ratios.

9. When a rocket fires its engines the rocket gains momentum equal and opposite to the momentum of the hot exhaust gases. For example we could be told that a 20,000kg rocket at rest fires its rocket for 10s. The exhaust gases leave at 100kgs-1 at a speed of 1kms-1. Show that the final velocity of the rocket is 53ms-1 Rockets

10. Water of density 1000 kg m-3 flows out of a hose pipe with cross sectional area 7.2x10-4 m2 at a rate of 2.0 x 10-4 m3 per second. How much momentum is carried by the water each second? • First we need to consider the mass each second. Knowing the volume and density we can find the mass • = m/v m = v • Mass each second = 1000 x 2x10-4 = 0.2kg per second • Next we need to consider the velocity. We know the volume leaving each second and also the area. We effectively have a cylinder of length 2x10-4 / 7.2 x10-4 = 0.278 m. Hence the velocity is 0.278 ms-1 • Momentum each second = 0.2 x 0.278 = 0.056 kgms-1 Water jets

11. A squash ball is released from rest above a flat surface. Describe how the energy changes if i) it rebounds to the same height, ii) It rebounds to a lesser height If the ball is released from a height of 1.20m and rebounds to a height of 0.9m show that 25% of the kinetic energy is lost upon impact Problems 1...

12. A shell of mass 2kg is fired at a speed of 140ms-1 from a gun with mass 800kg. Calculate the recoil velocity of the gun • A molecule of mass 5.0 x 10-26 kg moving at a speed of 420ms-1 hits a surface at right angles and rebounds at the opposite direction at the same speed. The impact lasted 0.22ns. Calculate: • The change in momentum • The force on the molecule • Repeat the last question. This time the molecule strikes the surface at 60° to the normal and rebounds at 60° to the normal. Problems 2...