MOMENTUM AND COLLISIONS

MOMENTUM AND COLLISIONS

LINEAR MOMENTUM Momentum = Mass x Velocity p=mv The SI unit for momentum is kg·m/s Momentum and velocity are in the same

direction Is a vector

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Using the equation p=mv At the same velocity, as mass increases

– momentum increases At the same mass, as velocity increases

– momentum increases

Example You are driving north, a deer with mass

of 146 kg is running head-on toward you with a speed of 17 m/s. Find the momentum of the deer.

p=mv p=146 kg * 17 m/s p= 2500 kg·m/s to the south

CHANGING MOMENTUM A change in momentum takes time and force For example soccer – when receiving a pass

it takes force to stop the ball It will take more force to stop a fast moving

ball than to stop a slow moving ball A toy truck and a real truck moving at the

same velocity, it will take more force to stop the real truck than to stop the toy truck

IMPULSE Impulse is the applied force times the

time interval FΔt Impulse = FΔt Force is reduced when the time interval

increasesExample – giving a little when you catch a

ballTrampoline

IMPULSE-MOMENTUM THEOREM From Newton’s second law (it will never go away)

And the equation for acceleration: a=v/t We can find the equation for force in terms of

momentum F=Δp/Δt Force = change in momentum / time interval We can also rearrange this equation to find the

change in momentum in terms of external net force and time

Δp=FΔt, Δp=FΔt=mvf-mvi

Example A 0.50 kg football is thrown with a

velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020 s. What is the force exerted on the ball by the receiver?

p=mv Δp=FΔt=mvf-mvi

vi= 15 m/s vf=0.0 m/s t=0.020 s F*0.020 s = 0 – 0.50 kg*15 m/s F= 380 N

IMPULSE-MOMENTUM THEOREM AND STOPPING DISTANCE The impulse-momentum theorem can be

used to determine the stopping distance of a car or any moving object

Use Δp=FΔt So Δt=Δp/F=mvf-mvi/F Then when you find time you can use distance = average velocity * time Δx=1/2(vf+vi) Δt

Example A 2240 kg car traveling to the west slows down

uniformly to rest from 20.0 m/s. The decelerating force on the car is 8410 N to the east. How far would the car move before stopping?

Δt=Δp/F=mvf-mvi/F Δt=0-2240kg*20.0m/s / 8410 N Δt = 5.33 s Δx=1/2(vf+vi) Δt Δx=1/2(0+20.0m/s)*5.33 x=53.3 m to the west

Conservation of Momentum

What happens to momentum when two or more objects interact?

First you have to consider total momentum of all objects involved. This is the sum of all momentums

Like energy momentum is conserved Conservation of Momentum: Total initial momentum = total final momentum m1v1i + m2v2i = m1v1f + m2v2f

Example A boy on a 2.0 kg skateboard initially at

rest tosses an 8.0 kg jug of water in the forward direction. If the jug has a speed of 3.0 m/s relative to the ground and the boy and the skateboard move in the opposite direction at 0.60 m/s, find the mass of the boy.

m1v1i + m2v2i = m1v1f + m2v2f

m1=mass of jug = 8.0 kg m2= mass of boy and skateboard= 2.0 kg + x V1i = 0, v2i = 0 V1f = 3.0 m/s, v2f=0.60 m/s 0=8.0kg*3.0m/s forward + (2.0kg +x)*.60m/s back 24 kg·m/s = (2.0 kg + x)*.60 m/s 40 kg = 2.0 kg + x x= 38 kg

Newton’s Third law and Collisions If the forces exerted in a collision are equal and

opposite And the time each force is exerted would be the same Than the impulse on each object in a collision would

be equal and opposite Since impulse is equal to the change in momentum

than the change in momentum would be equal and opposite

So if one object gained momentum after a collision than the other object must lose the same amount of momentum

Collisions When two objects collide and then move

together as one mass, the collision is called a Perfectly inelastic collision These are easy situations because the two

objects become pretty much one object after the collision

So the conservation of momentum equation becomes

m1v1i + m2v2i = (m1 + m2)vf

The two objects will have the same final velocity

Example A grocery shopper tosses a 9.0 kg bag

of rice into a stationary 18.0 kg grocery cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and the bag?

m1v1i + m2v2i = (m1 + m2)vf

0 + 9.0kg * 5.5 m/s = (18kg + 9kg)vf

50. kg m/s = 27*vf

1.9 m/s

Kinetic Energy and Inelastic Collisions

Total kinetic energy is not conserved in inelastic collisions, it does not remain constant

Some of the energy is converted into sound energy and internal energy as the objects deforms during the collision

This is why it is called inelastic, elastic usually means something that can keep its shape or return to its original shape

In physics elastic means that the work done to deform an object is equal to the work done to return to its original shape

In inelastic collisions some of the work done on the inelastic material is converted to other forms of energy such as heat or sound

Example A 0.25 kg arrow with a velocity of 12 m/s

to the west strikes and pierces the center of a 6.8 kg target.What is the final velocity of the combined

mass?What is the decrease in kinetic energy

during the collision?

m1v1i + m2v2i = (m1 + m2)vf

0 + 0.25 kg *12 m/s = (0.25 kg+ 6.8 kg)v vf=0.43 m/s to the west

KEi = 0+ ½ 0.25*(12m/s)2

= 18 J KEf = ½ (7.1 kg)*(0.43 m/s)2

= .66 J Change in kinetic = 17 J

Elastic Collisions In an elastic collision, two objects collide

and return to their original shapes with no loss of total kinetic energy.

The two objects move separately Both total momentum and total

kinetic energy are conserved

Since both are conserved than these equations apply

m1v1i + m2v2i =m1v1f +m2v2f

½ m1v1i2 + ½ m2v2i

2 = ½ m1v1f2 + ½ m2v2f

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Example A 16.0 kg canoe moving to the left at

12.5 m/s makes an elastic head-on collision with a 14.0 kg raft moving to the right at 16.0 m/s. After the collision, the raft moves to the left at 14.4 m/s. Disregard any effects of the water.Find the velocity of the canoe after the

collision.Verify your answer by calculating the total

kinetic energy before and after the collision.

Collisions in Two or more Dimensions Conservation of momentum still applies Use your vectors. Momentum is conserved in all directions pxi=pxf

pyi=pyf

mAvAxi +mBvBxi = mAvAxf +mBvBxf

mAvAyi +mBvByi = mAvAyf +mBvByf

Center of Mass (CM) The point at which all mass of an

object can be considered to be located

An object can be considered a point or small particle no matter what the size or shape of the object

Center of mass moves just as a particle moves no matter what the object does

Finding the center of mass of an object Define a coordinate system

(preferably one that would make the math easy) Center of mass can be found by finding

the sum of all the masses times their respective distance from the defined origin and dividing by the sum of all the masses

BA

BBAAcmx mm

xmxm

Center of gravity (CG) The point at which the force of gravity

can be considered to act Gravity acts on all parts of the object But for determining translational motion

we can assume that gravity acts in one particular spot

Same spot as the center of mass