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HallsofIvy

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" I am assuming that something small hitting something bigger would not conserve any momentum because of their differences."elissadi said:

Since that completely contradicts what any textbook would tell you

Yes, a "small object" moving at any speed has greater momentum (in magnitude at least) than a larger object motionless because the large object has none at all. However, remember that momentum is a

Suppose an object of mass 1kg is moving at 2 m/s. It collides with a motionless object of mass 2 kg.

Before the collision, the total momentum is (1 kg)(2 m/s)+ (2kg)(0 m/s)= 2 kg m/s. Let v

Notice that that is one equation in 2 unknowns. One way for it to be satisfied is to have v

Yet another would be for the two masses to "stick" together having the same velocity: v

The reason there are an infinite number of solutions is that we haven't considered the kinetic energy ((1/2)mv

v

Now solve the two equations v

v

From the first equation, v

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In this case it has transferred all its momentum to the larger mass.

The larger mass will now move. But with a speed slower than the smaller ball initially had.

Momentum => M_(initial) * Velocity_(initial) = M_(finall) * Velocity_(final).

Or a situation could arise where the smaller ball continues with a lesser speed in another direction and the larger ball also begins to move but with an even smaller speed.

That is... If you sum up the momentum before a collision and then afterwards they will be equal.

Momentum is

Though this is not the case for kinetic energy.

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Momentum is a vector quantity defined as the mass of an object multiplied by its velocity. Momentum is conserved when there is no net external force acting on the system. Conservation simply means constant in time.elissadi said:Can anyone explain to me in simple terms about momentum and conservation of momentum.

look at the momentum for the two objects? One has momentum the other does not. Which is which? Why does one have momentum while the other doesn't?elissadi said:I understand the terms, but don't understand how it relates to a smaller object sliding sideways into a non-moving larger object.

Yes!elissadi said:Before the smaller object hits the larger, doesnt it have a larger momentum?

Through the collision. When they touch, momentum is transferred. This is a good question.elissadi said:How can the smaller object transfer its momentum to a larger object which has mass but no velocity?

Well it all depends if there are any external forces present. Are there? It depends on the set-up. In a typical textbook set-up we usually assume that there are no external forces unless we are explicitly told so.elissadi said:I am assuming that something small hitting something bigger would not conserve any momentum because of their differences.

Let us say that we have a comet of mass m1 moving at a velocity v1i towards a stationary asteroid of mass m2. We are in a universe where they are the only 2 objects. Let us also assume both are spheres and the collision is head on. What is the velocity of asteroid after the collision if we measure the velocity of the comet to be v1f in the same direction as v1i after the collision?

Solution:

use conservation of momentum.

[tex] \vec{p}_{1i} + \vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f} [/tex]

Which means:

[tex] m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = m_1 \vec{v}_{1f} + m_f \vec{v}_{2f} [/tex]

but:

[tex] v_{2i} = 0 [/tex]

Which means that the initial momentum of the asteroid is zero also.

now just solve for the final velocity of the asteroid:

[tex] \vec{v}_{2f} = \frac{m_1}{m_2} ( \vec{v}_{1i} - \vec{v}_{1f} )[/tex]

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This is untrue.H_man said:Momentum isalwaysconserved!

Momentum is only conserved if there is no net external force acting on the system.

Proof:

[tex]

\Sigma_i \vec{F}_i = \frac{d\vec{p}}{dt}

[/tex]

if

[tex]

\Sigma_i \vec{F}_i = 0

[/tex]

then

[tex]

\frac{d\vec{p}}{dt} = 0

[/tex]

which means

[tex] \vec{p} = constant [/tex]

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