Momentum and Momentum and CollisionsCollisions

MomentumMomentum

The linear momentum of an object of The linear momentum of an object of mass m moving with a velocity v is mass m moving with a velocity v is the product of the mass and the the product of the mass and the velocity.velocity.

Momentum = mass X velocityMomentum = mass X velocity p=mv p=mv Units kgUnits kg●m/s●m/s

Momentum PracticeMomentum Practice

A 2250 kg pickup truck has a velocity A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the of 25 m/s to the east. What is the momentum of the truck?momentum of the truck?

Given: m=2250 kgGiven: m=2250 kg v= 25 m/sv= 25 m/s p=mv 2250kg x 25 m/s=p=mv 2250kg x 25 m/s= 5.6x105.6x104 4 kgkg●m/s●m/s

Changing MomentumChanging Momentum

A change in momentum takes force A change in momentum takes force and timeand time

F=m(F=m(ΔΔv/v/ΔΔt)t) Newtons original formula Newtons original formula F=F=ΔΔp/p/ΔΔtt ΔΔp=mvp=mvff-mv-mvii

Force= change in momentum during Force= change in momentum during time intervaltime interval

ImpulseImpulse

Impulse is the force for the time Impulse is the force for the time interval.interval.

Impulse-momentum theorem is the Impulse-momentum theorem is the expression Fexpression FΔΔt= t= ΔΔpp

If you extend the time of impact you If you extend the time of impact you reduce the amount of force.reduce the amount of force.

Force and ImpulseForce and Impulse

A 1400 kg car moving westward with A 1400 kg car moving westward with a velocity of 15 m/s collides with a a velocity of 15 m/s collides with a utility pole and is brought to rest utility pole and is brought to rest in .30 s. Find the force exerted on in .30 s. Find the force exerted on the car during the collision.the car during the collision.

Given m=1400 kg, Given m=1400 kg, ΔΔt= .30 st= .30 s vvff=0 m/s v=0 m/s vii=15 m/s=15 m/s What formula can I use?What formula can I use?

Impulse-MomentumImpulse-Momentum

FFΔΔt= t= ΔΔpp ΔΔp=mvp=mvff-mv-mvii

FFΔΔt=mvt=mvff-mv I am looking for force-mv I am looking for force F=mvF=mvff-mv/ -mv/ ΔΔt Now I just plug in my t Now I just plug in my

numbersnumbers (1400*0-1400*-15)/.30= 7.0*10(1400*0-1400*-15)/.30= 7.0*1044 N to N to

the eastthe east

Stopping DistanceStopping Distance

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

What am I given?What am I given?

M= 2240 kg M= 2240 kg vvii= 20.0 m/s to the west = -20.0 m/s= 20.0 m/s to the west = -20.0 m/s vvff= 5.0 m/s to the west = -5.0 m/s= 5.0 m/s to the west = -5.0 m/s F= 8410 N to the east so it stays F= 8410 N to the east so it stays

positivepositive Unknown Unknown ΔΔt and t and ΔΔxx

Impulse Momentum TheoremImpulse Momentum Theorem

FFΔΔt= t= ΔΔpp ΔΔp=mvp=mvff-mv-mvii

ΔΔt=mvt=mvff-mv-mvi i /F/F ((2240*-5)-(2240*-20))/8410((2240*-5)-(2240*-20))/8410 ΔΔt=4.0st=4.0s

Displacement of the vehicleDisplacement of the vehicle

To find the change in distance To find the change in distance Average velocity= change in d/change in Average velocity= change in d/change in

tt Average velocity =(initial +final)/2Average velocity =(initial +final)/2 Since these are equal I can combine them Since these are equal I can combine them ΔΔx/x/ΔΔt=(vt=(vff++ vvii)/2)/2 Solving for x=((vSolving for x=((vff++ vvii)/2) *)/2) *ΔΔtt x=((-20+-5)/2)*4x=((-20+-5)/2)*4 x=-50 so it would be 50m to the westx=-50 so it would be 50m to the west

Conservation Of MomentumConservation Of Momentum

When I have two objects A and BWhen I have two objects A and B p(Ap(Aii)+p(B)+p(Bii)=p(A)=p(Aff)+p(B)+p(Bff)) Total initial momentum=total final Total initial momentum=total final

momentummomentum

Conservation of MomentumConservation of Momentum

A 76 kg boater, initially at rest in a A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of stationary 45 kg boat, steps out of the boat and onto the dock. If the the boat and onto the dock. If the boater moves out of the boat with a boater moves out of the boat with a velocity of 2.5 m/s to the right what velocity of 2.5 m/s to the right what is the final velocity of the boat?is the final velocity of the boat?

Given: m of person= 76 kg vGiven: m of person= 76 kg vii=0 =0 m/s,vm/s,vff=2.5 to the right=2.5 to the right

m of boat= 45 kg vm of boat= 45 kg vii=0 v=0 vff=?=?

mm11vv1i1i+m+m22vv2i2i=m=m11vv1f1f+m+m22vv2f2f

Because the boater and the boat are Because the boater and the boat are initially at rest minitially at rest m11vv1i 1i + m+ m22vv2i2i=0=0

Rearrange to solve for final velocity of Rearrange to solve for final velocity of boatboat

vv2f 2f = (-m= (-m11/m/m22)/v)/v1f1f

-76/45 * 2.5=4.2 to the right -4.2-76/45 * 2.5=4.2 to the right -4.2

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

mm1 =8000 kg m2= 8000 kg v1=-30m/s v2=25 m/s m1v1+m2v2=mtotvtot