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G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
G12 Momentum P1
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G12 Momentum P1

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A set of slides created to teach G12 Momentum P1 to learners at Bishops Diocesan College in Cape Town.

A set of slides created to teach G12 Momentum P1 to learners at Bishops Diocesan College in Cape Town.

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  • 1. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Momentum & Impulse K Warne
  • 2. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net 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. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net 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 in the direction of the velocity 2. Calculate the momentum of a 500 tonne ship moving at 1 km.hr-1. p = mV  = (500 x 103)*(1/3.6)  = 138 888.89 kg.m.s-1 in the direction of the velocity  3. Calculate the momentum of a car which has a mass of 1500 kg and a kinetic energy of 200 kJ. Ek = ½ mv2  p = mV  200000= ½(1500)v2  = (1500)*(16.33)  v = √ (200000*2/1500) = 24 494.90 kg.m.s-1  in the direction of = 16.33 m.s-1 the velocity MOMENTUM Examples
  • 4. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net (Newtons 2nd Law) Newton 2: If an object experiences a resultant force it will ......................... in the ............... of that force. m Fres .. . . . . . . .a  Fres = m.a Fres = resultant force (...) M = mass (...) a = acceleration (...) F = M = 2000kg a: 0 16m.s-1 in 10s
  • 5. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net (Newtons 2nd 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 Fres .. . . . . . . . a  Fres = m.a Fres = resultant force (N) M = mass (kg) a = acceleration (m.s-2) F = m.a = (2000).(1.6) = 3200Nin direction of motion M = 2000kg a = (16-0)/10 = 1.6 m.s-2 v t  
  • 6. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Newtons 2nd 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? ANSWER >>
  • 7. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Newtons 2nd & Momentum: The applied …………….. Force is ……. to the ………………………………….., and that this change is in the ……………. of the ………… …………... Fnett = This is derived from Newton's 2nd Law the nett force is …………………… to the change in momentum. m Fres .. . . . . . . .a  Fres = ……. Fnet  p t
  • 8. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net 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 ………………. Fres Fres
  • 9. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net 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 nett force on the rocket if the rocket took 10s to reach the new speed. 3. The impulse experienced by the rocket. ANSWER >>
  • 10. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net 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 ma 2kg mb 3kg Final 5m/s -6m/s va = -2m/s vb = -1.33m/s A B
  • 11. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Elastic & Non-elastic Collisions 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.
  • 12. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Change in Momentum & Impulse • Pi = mvi = (1)(5) = 5kg·m·s-1 • pf = mvf = (1)(-3) = -3kg·m·s-1 • p = pf – pi = -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.
  • 13. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience

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