Momentum and Collisions Momentum n n The linear

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Momentum and Collisions

Momentum and Collisions

Momentum n n The linear momentum of an object of mass m moving with

Momentum n n 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 n n n A 2250 kg pickup truck has a velocity of

Momentum Practice n n n 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 2250 kg x 25 m/s= 5. 6 x 104 kg●m/s

Changing Momentum n n n A change in momentum takes force and time F=m(Δv/Δt)

Changing Momentum n n n 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

Impulse n n n Impulse is the force for the time interval. Impulse-momentum theorem

Impulse n n n 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.

Force and Impulse n n A 1400 kg car moving westward with a velocity

Force and Impulse n n 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?

Impulse-Momentum n n n FΔt= Δp Δp=mvf-mvi FΔt=mvf-mv I am looking force F=mvf-mv/ Δt

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

Stopping Distance n A 2240 kg car traveling west slows down uniformly from 20.

Stopping Distance n 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?

What am I given? n n n M= 2240 kg vi= 20. 0 m/s

What am I given? n n n 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

Impulse Momentum Theorem n n n FΔt= Δp Δp=mvf-mvi Δt=mvf-mvi /F ((2240*-5)-(2240*-20))/8410 Δt=4. 0

Impulse Momentum Theorem n n n FΔt= Δp Δp=mvf-mvi Δt=mvf-mvi /F ((2240*-5)-(2240*-20))/8410 Δt=4. 0 s

Displacement of the vehicle n n n n To find the change in distance

Displacement of the vehicle n n n n 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 50 m to the west

Conservation Of Momentum n n n When I have two objects A and B

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

Conservation of Momentum n n n A 76 kg boater, initially at rest in

Conservation of Momentum n n n 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=?

m 1 v 1 i+m 2 v 2 i=m 1 v 1 f+m 2

m 1 v 1 i+m 2 v 2 i=m 1 v 1 f+m 2 v 2 f Because the boater and the boat are initially at rest m 1 v 1 i + m 2 v 2 i=0 Rearrange to solve for final velocity of boat v 2 f = (-m 1/m 2)/v 1 f -76/45 * 2. 5=4. 2 to the right -4. 2

n Two train cars of mass 8000 kg are traveling toward each other. One

n Two train cars of mass 8000 kg are traveling toward each other. One at a speed of 30 m/s to the west and the other at a speed of 25 m/s to the east. They crash and become entangled, what is their final velocity and direction?

n n n m 1 =8000 kg m 2= 8000 kg v 1=-30 m/s

n n n m 1 =8000 kg m 2= 8000 kg v 1=-30 m/s v 2=25 m/s m 1 v 1+m 2 v 2=mtotvtot