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NOTES 14 - Topic 2 - Mechanics *

NOTES 14 - Topic 2 - Mechanics * - -------------------------------------------------------------------------------- 2 .2.10 Define linear momentum and impulse

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NOTES 14 - Topic 2 - Mechanics *

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  1. NOTES 14 - Topic 2 - Mechanics* --------------------------------------------------------------------------------- 2.2.10 Define linear momentum and impulse Momentum (p) - the vector quantity resulting from the inertia of a moving mass; momentum = “inertia-in-motion”; momentum equals mass x velocity; p = mv = Ns For a 1.0 kg object moving at 1.0 ms-1, p = mv = (1.0 kg) (1.0 ms-1) = 1.0 kgms-1 = 1.0 Ns; Impulse (J) - an unbalanced force exerted for a specific time causes an object to accelerate; F = ma F = m (∆v / ∆t) F ∆t = m ∆v F ∆t = impulse = J = N•s m ∆v = ∆p = change in momentum = Ns “Exerting an unbalanced force for a period of time causes a change in the velocity of an object.”

  2. 2.2.11 Find impulse from a Force-Time graph On a Force-Time graph, the area under the curve represents force x time which is measured in Ns and is momentum. Calculate the total impluse shown by the graph in the space below: Area of Triangle #1 = (1/2) base x height = Area of Rectangle = length x width = Area of Triangle #2 = (1/2) base x height = Total Area = Total Impulse =

  3. 2.2.12 State the Law of Conservation of Linear Momentum Law of Conservation of Momentum The total momentum before a collision or explosion equals the total momentum after a collision or explosion. mvbefore = mvafter m1v1 + m2v2 = m1v1’ + m2v2’

  4. A. Collisions 1.Inelastic Collisions - colliders are permanently “deformed” and are not restored (do not snap back) to original shape...they stick together: m1v1 + m2v2 = (m1+ m2) vf

  5. 2. Elastic Collisions - colliders are deformed but are restored (do snap back) to original shape... m1v1 + m2v2 = m1v1’ + m2v2’

  6. B. Explosions - a central force is exerted simultaneously on at least two non-moving objects and causes them to move off in opposite directions at speeds inversely proportional to their masses... 0 = m1v1’ + m2v2’ m1v1’ = - m2v2’

  7. 2.2.13 Solve problems involving momentum and impulse Sample Problem 1 - impulse (show solution in NB) A 0.50 kg rubber ball moving at 10. ms-1 strikes a brick wall and rebounds with a speed of 8.0 ms-1in the opposite direction. What impulse was applied to the ball by the wall? Given: Unknown: Equation:

  8. Sample Problem 2 - inelastic collision (show solution in NB) A 20,000. kg railroad boxcar moving at 10.0 km h-1 couples with another boxcar at rest with a mass of 10,000. kg. At what velocity will the 2 cars move after the coupling? Given: Unknown: Equation:

  9. Sample Problem 3 - elastic collision (show solution in NB) A 0.50 kg billiard ball moving at 5.0 ms-1 strikes a motionless 5.0 kg bowling ball and bounces off at speed of 2.0 ms-1 in the opposite direction. At what speed will the bowling ball be moving? Given: Unknown: Equation:

  10. Sample Problem 4 - “explosion” (show solution in NB) A 100. kg astronaut is floating alongside the International Space Station at a distance of 10. m when he discovers his safety tether is not attached. Rather than wait the 60 minutes it will take a colleague to “suit up” and come out to rescue him, he decides to throw his 1.0 kg hammer away and drift back to the station. If he throws the hammer with a speed of 5.0. ms-1, how long will it take him to reach the ISS? Given: Unknown: Equation:

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