1. MOMENTUM Definition: Momentum (Symbol : ….) is defined as the product of the ………….. and ……………. of a moving body. Momentum p = units: ……………. N.B. Since …………… is a vector quantity, momentum is also a ……………… quantity. V = 16 m.s -1 Mass = 2000kg

2. MOMENTUM Definition: Momentum (Symbol : p) of an object is the product of the mass and velocity of a moving body. Momentum p = m.v = (2000)(16)` = 32 000 kgms -1 forwards units: kg. m.s -1 N.B. Since velocity is a vector quantity, momentum is also a vector quantity. V = 16 m.s -1 forwards Mass = 2000kg

3. Calculate the momentum of a cyclist who has a mass of 75 kg and rides a bike of mass 15 kg at a speed of 20 km.hr -1. Calculate the momentum of a 500 tonne ship moving at 1 km.hr -1. Calculate the momentum of a car which has a mass of 1500 kg and a kinetic energy of 200 kJ. MOMENTUM Examples

4. 1.Calculate the momentum of a cyclist who has a mass of 75 kg and rides a bike of mass 15 kg at a speed of 20 km.hr -1. p = mv  = (15k+75)*(20/3.6)  = 500 kg.m.s -1  2.Calculate the momentum of a 500 tonne ship moving at 1 km.hr -1. p = mV  = (500 x 10 3 )*(1/3.6)  = 138 888.89 kg.m.s -1  3.Calculate the momentum of a car which has a mass of 1500 kg and a kinetic energy of 200 kJ. E k = ½ mv 2  p = mV  200000= ½(1500)v 2  = (1500)*(16.33)  v = √ (200000*2/1500)= 24 494.90 kg.m.s -1  = 16.33 m.s -1 MOMENTUM Examples

5. (Newtons 2 nd Law) Newton 2: If an object experiences a resultant force it will......................... in the............... of that force. m F res......... a  F res = m.a F res = resultant force (...) M = mass (...) a = acceleration (...) F = M = 2000kg a: 0  16m.s -1 in 10s

6. (Newtons 2 nd Law) Newton 2: If an object experiences a resultant force it will accelerate in the direction of that force. Eg a 2 tonne car accelerates from rest to 16m.s -1 in 10 s. m F res......... a  F res = m.a F res = resultant force (N) M = mass (kg) a = acceleration (m.s -2 ) F = m.a = (2000).(1.6) = 3200N in direction of motion M = 2000kg a = (16-0)/10 = 1.6 m.s -2

7. Newtons 2 nd Law & Momentum Examples Calculate the acceleration of a 5 kg box which is pushed by a 20 N force with a 2 N frictional force acting against it. What applied force would be required to accelerate a 75 kg person vertically upward at 5 m.s -2 ? What would be the mass of an object that is accelerated by a 100 N applied force from rest to 10 m.s -1 in 5s against a 3 N frictional force?

8. Newtons 2 nd Law & Momentum Examples Calculate the acceleration of a 5 kg box which is pushed by a 20 N force with a 2 N frictional force acting against it. What applied force would be required to accelerate a 75 kg person vertically upward at 5 m.s -2 ? What would be the mass of an object that is accelerated by a 100 N applied force from rest to 10 m.s -1 in 5s against a 3 N frictional force?

9. Newtons 2 nd & Momentum: The applied …………….. Force is ……. to the ………………………………….., and that this change is in the ……………. of the ………… …………... F nett = This is derived from Newton's 2nd Law the nett force is …………………… to the change in momentum. m F res......... a  F res = …….

10. Newtons 2 nd & Momentum: The applied resultant Force is equal to the rate of change of momentum, and that this change is in the direction of the applied Force. Fres = mv f - mv i  t This is derived from Newton's 2nd Law the nett force is directly proportional to the change in momentum. m F res......... a  F res = m.a

11. Rate of Change of momentum examples What is the rate of change of momentum of a 2 tonne mass that experiences a nett force of 1 N over a 10 s time interval? What is the rate of change of momentum of a 5 kg object accelerating at 2m.s -2. Calculate the resultant force on an object which starts from rest and gains a momentum of 200 kg.m.s -1 in 5s?

12. Rate of Change of momentum examples What is the rate of change of momentum of a 2 tonne mass that experiences a nett force of 1 N over a 10 s time interval? What is the rate of change of momentum of a 5 kg object accelerating at 2m.s -2. Calculate the resultant force on an object which starts from rest and gains a momentum of 200 kg.m.s -1 in 5s?

13. IMPULSE Impulse is the ……… of the ………………….. applied and the time over which it is applied. Impulse = ………. t Impulse changes …………………. Fres = ma = m( ) = / Fres.  t = …………. = ……… Impulse = change in ………………. F res

14. IMPULSE Impulse is the product of the resultant force applied and the time over which it is applied. Impulse = F res.  t t Impulse changes momentum Fres = ma = m(  v /  t ) = mv-mu /  t Fres.  t = mv - mu =  P Impulse = change in momentum F res

15. Increase in Momentum Example: A rocket with a mass of 2000 tones traveling at 100m.s -1 is accelerated to a velocity of 300m.s -1 when the second stage of the rocket engine fires, on reaching this speed the rocket has burned up another 1500kg of fuel. Calculate: 1. the change in momentum of the rocket and 2.the magnitude of the thrust of the rocket if the rocket took 10s to reach the new speed. 3.The impulse experienced by the rocket.

16. Increase in Momentum A rocket with a mass of 2000 tones traveling at 100m.s -1 is accelerated to a velocity of 300m.s -1 when the second stage of the rocket engine fires, on reaching this speed the rocket has burned up another 1500kg of fuel. Calculate: 1. the change in momentum of the rocket 2.the magnitude of the thrust of the rocket if the rocket took 10s to reach the new speed. 3.The impulse experienced by the rocket.

17. MOMENTUM CONSERVA TION Principle: It states that the ……………………… of an …………. system remains …………………. in both ………………………………….. TOTAL momentum before crash = TOTAL momentum after …..= …… CHANGE IN MOMENTUM change in momentum = ……………………………………………….  p = ……………………….. A B Crash! Initial m a 2kg m b 3kg Final 5m/s-6m/s v a = -2m/sv b = -1.33m/s A B

18. MOMENTUM CONSERVATION Law of Conservation of Momentum: It states that the total linear momentum of an isolated system remains constant in both magnitude and direction. A B Crash! Initial m a 2kg m b 3kg TOTAL momentum before crash = TOTAL momentum after Final P i = P f (m a V a + m b V b ) i =(m a V a + m b v b ) f 5m/s-6m/s v a = -2m/sv b = -1.33m/s (2)(5) + (3)(-6) = (2)(-2) + (3)(-1.33) CHANGE IN MOMENTUM = 0 change in momentum = final momentum - initial momentum p = mv f - mv i = 0!! A B -8 = -7.99

19. Elastic & Non-elastic Collisions 1.………… on impact. 2.…………… on impact. Collisions can either be: a.Elastic (……. is …………….. and no energy lost) b.Inelastic (…………… is …………… - some is converted to heat, sound etcJ N.B. Momentum is conserved in any collision between two objects, but Kinetic energy is conserved only in a perfectly elastic collision.

20. Elastic & Non-elastic Collisions 1.Combine on impact. 2.Separate on impact. Collisions can either be: a.Elastic (Kinetic Energy Ek = 1 / 2 mv 2 is conserved and no energy lost) b.Inelastic (Kinetic Energy is not conserved - some is converted to heat, sound etc N.B. Momentum is conserved in any collision between two objects, but Kinetic energy is conserved only in a perfectly elastic collision.

21. Change in Momentum & Impulse 1kg  p …………… FORCE  t 0.05s A 1 kg ball moving with a velocity of 5m.s -1 to the right collides with a wall and bounces back with a velocity of 3m.s -1. If the collision takes 0.05s calculate the force exerted by the wall on the ball.

22. Change in Momentum & Impulse P i = mv i = (1)(5) = 5kg·m·s -1 p f = mv f = (1)(-3) = -3kg·m·s -1  p = p f – p i = -3 – 5 = -8 kg·m·s -1  p = 8kg ·m·s -1 in opposite direction 1kg 5m.s -1 1kg 3m.s -1  p 8kg.m.s -1 FORCE  t 0.05s Impulse = F x  t =  p 8 = F x (0.05) F = 8 / (0.05) = A 1 kg ball moving with a velocity of 5m.s -1 to the right collides with a wall and bounces back with a velocity of 3m.s -1. If the collision takes 0.05s calculate the force exerted by the wall on the ball.

23. Momentum Change in Momentum  State Newton’s Second Law (N2) in terms of momentum.  Express Newton 2 in symbols.  Explain the relationship between net force and change in momentum.  Calculate the change in momentum when a resultant force acts on an object and its velocity: (Rocket 10 000kg) o Increases in the direction of motion (e.g. 2 nd stage rocket engine fires). Velocity goes from 150m.s -1 to 400m.s -1 o Decreases (e.g. brakes are applied). 900kg car at 16m.s -1 stops. o Reverses its direction of motion, e.g. a soccer ball kicked back in the direction it came from. 450g. 10m.s -1 right, kicket 20m.s -1 to LEFT.  Draw vector diagrams to illustrate the relationship between the initial momentum, the final momentum and the change in momentum in each of the above cases.  State the condition necessary for the conservation of momentum in a system.  Apply the conservation of momentum to collisions of two objects moving in one dimension (along a straight line). Calculate the momentum of a 60kg man standing on a frictionless scateboard when he throws a 5kg bag of sugar forward at 12m.s -1