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Note: When you are subtracting a vector, SWAP the vector then ADD it.

A 0.4 kg basketball bounces off the ground. The ball’s speed the moment it hits the ground is 20 ms -1 . The ball’s speed the moment it leaves the ground is 20 ms -1 . Calculate the ball’s initial momentum Calculate the ball’s final momentum

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Note: When you are subtracting a vector, SWAP the vector then ADD it.

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  1. A 0.4 kg basketball bounces off the ground. The ball’s speed the moment it hits the ground is 20 ms-1. The ball’s speed the moment it leaves the ground is 20 ms-1. Calculate the ball’s initial momentum Calculate the ball’s final momentum Calculate the change in momentum, Δp = pf – pi (*remember - vector subtraction)

  2. Note: When you are subtracting a vector, SWAP the vector then ADD it.

  3. Momentum Momentum is the amount of ‘oomph’ a moving object has. “The more ‘oomph’ the object has, the harder it is to stop” p = mv(vector quantity) p = momentum (measured in ________) m = mass (in kg) v = velocity (in ms-1)

  4. Example ONE A cricket ball of mass 500 g is bowled straight into the batter’s face at a speed of 35 ms-1. The batter blocks the ball with his bat, then the ball flies directly back towards the bowler’s face at 25 ms-1. Calculate the ball’s initial momentum Calculate the ball’s final momentum Calculate the ball’s change in momentum

  5. Example TWO A 120 kg man accidently falls off the roof of a building. The man’s velocity, the moment before he hits the ground, is 40 ms-1. He comes to a complete stop after he has hit the ground. Calculate his change in momentum.

  6. Impulse = “change in momentum” • Δp = pf – pi • = mvf – mvi • = m (vf – vi) • = m Δv • = m a Δt • = F Δt F = ma • Δv • a = • Δt

  7. Impulse = “change in momentum” • Δp = F Δt • “Change in momentum • is a result of force” • F = force • Δt = the duration of the force applied

  8. Example THREE • A driver tries to slow down a runaway car by pushing against its motion. The car’s mass is 2100 kg and its initial velocity is 3.0 ms-1. • Will the driver manage to stop the car if: • the resultant force on the car is 600 N (against the car’s motion) and, • the force is applied for 9.0 seconds?

  9. Example FOUR A car travelling at 20 ms-1crashes into a wall. The mass of the car is 1500 kg. It takes 0.4 seconds, between the moment the car comes into contact with the wall, and the moment the car comes to a complete stop. Calculate the change in momentum Calculate the force exerted on the car Calculate the force if; the wall was very soft, and it took 1.6 seconds to stop, instead of 0.4 seconds.

  10. The battle between “F” and “t” • Δp = F Δt • You can achieve the exact same change in momentum by: • Applying a great force over a short period of time or; • Applying a small force over a long period of time

  11. Question How do air bags work?

  12. With or without the airbag, the person would experience the same change in momentum • However, with the presence of the airbag, the person will stop over a longer period of time, because of the cushioning. Since Δp = F Δt, a longer stopping time means a smaller force is applied on the person, to give the same change in momentum. • Without the airbag, the person will stop over a very short period of time (almost instantaneously). Since Δp = F Δt, short stopping time means a much greater force is applied, to give the same change in momentum. • A greater force means the person is more likely to suffer greater injuries.

  13. Example FIVE – “Total momentum” Ball A has a mass of 400 g and is travelling at 4 ms-1. Ball B has a mass of 250 g and is travelling at 8 ms-1. Calculate the momentum of ball A Calculate the momentum of ball B Calculate the total momentum (*remember – vector addition)

  14. Example FIVE – continued A few moments later, ball A and ball B collide then each ball moves in the opposite direction afterwards. Ball A moves at 4.5 ms-1 and ball B moves at 5.6 ms-1. Calculate the total momentum Which is greater – the total momentum before the collision OR the total momentum after the collision?

  15. Conservation of Momentum Total momentum before a collision or explosion equals the total momentum after the collision or explosion. However, this is true only when no resultant external force is present, such as gravity or friction. m1u1 + m2u2 = m1v1 + m2v2 u = initial velocity, v = final velocity

  16. Collisions – elastic vs. inelastic One may ask – “Couldn’t we just use the conservation of energy principle, by calculating the total kinetic energy?” The answer is NO – because often kinetic energy is lost due to friction, meaning that the “total kinetic energy” is not conserved. However, there are collisions where “total kinetic energy” IS conserved. These collisions are called “elastic collisions”.

  17. Activity 10A (green book, pg. 123) Questions 1, 2, 3 and 6 only • Homework Booklet Worksheet TEN & ELEVEN • NCEA 2010 up to page 7 • By Tuesday 26th June

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