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Momentum and Collisions

Momentum and Collisions. Momentum. The linear momentum of an object of mass m moving with a velocity v is the product of the mass and the velocity. Momentum = mass X velocity p=mv Units kg ●m/s. Momentum Practice.

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Momentum and Collisions

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  1. Momentum and Collisions

  2. Momentum • The linear momentum of an object of mass m moving with a velocity v is the product of the mass and the velocity. • Momentum = mass X velocity • p=mv • Units kg●m/s

  3. Momentum Practice • A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? • Given: m=2250 kg • v= 25 m/s • p=mv 2250kg x 25 m/s= • 5.6x104 kg●m/s

  4. Changing Momentum • A change in momentum takes force and time • F=m(Δv/Δt) • Newtons original formula • F=Δp/Δt • Δp=mvf-mvi • Force= change in momentum during time interval

  5. Impulse • Impulse is the force for the time interval. • Impulse-momentum theorem is the expression FΔt= Δp • If you extend the time of impact you reduce the amount of force.

  6. Force and Impulse • A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in .30 s. Find the force exerted on the car during the collision. • Given m=1400 kg, Δt= .30 s • vf=0 m/s vi=15 m/s • What formula can I use?

  7. Impulse-Momentum • FΔt= Δp • Δp=mvf-mvi • FΔt=mvf-mv I am looking for force • F=mvf-mv/ Δt Now I just plug in my numbers • (1400*0-1400*-15)/.30= 7.0*104 N to the east

  8. Stopping Distance • A 2240 kg car traveling west slows down uniformly from 20.0 m/s to 5.0 m/s How long does it take the car to decelerate if the force on the car is 8410 to the east? How far does the car travel during the deceleration?

  9. What am I given? • M= 2240 kg • vi= 20.0 m/s to the west = -20.0 m/s • vf= 5.0 m/s to the west = -5.0 m/s • F= 8410 N to the east so it stays positive • Unknown Δt and Δx

  10. Impulse Momentum Theorem • FΔt= Δp • Δp=mvf-mvi • Δt=mvf-mvi /F • ((2240*-5)-(2240*-20))/8410 • Δt=4.0s

  11. Displacement of the vehicle • To find the change in distance • Average velocity= change in d/change in t • Average velocity =(initial +final)/2 • Since these are equal I can combine them • Δx/Δt=(vf+vi)/2 • Solving for x=((vf+vi)/2) *Δt • x=((-20+-5)/2)*4 • x=-50 so it would be 50m to the west

  12. Conservation Of Momentum • When I have two objects A and B • p(Ai)+p(Bi)=p(Af)+p(Bf) • Total initial momentum=total final momentum

  13. Conservation of Momentum • A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right what is the final velocity of the boat? • Given: m of person= 76 kg vi=0 m/s,vf=2.5 to the right • m of boat= 45 kg vi=0 vf=?

  14. m1v1i+m2v2i=m1v1f+m2v2f Because the boater and the boat are initially at rest m1v1i + m2v2i=0 Rearrange to solve for final velocity of boat v2f = (-m1/m2)/v1f -76/45 * 2.5=4.2 to the right -4.2

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