**Linear Momentum**

Momentum, what fun.

p = mv

F = p/t

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1.** Linear momentum is defined as the product of the mass of a body
and its velocity.

p = mv.. A 5kg duck waddles at a speed of 5m/s. p = mv, p = 5kg*5m/s. p = 25kg*m/s.

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**Problem 1: **A large crate is placed on a scale that measures
things in kilograms. A duck decides that he's had enough of his chicken slaves
and wants to get rid of them. Chickens, being rodents, multiplied quite quickly
and seemed to form an army of clones. It's too late. The duck can only hope
to throw out as many chickens as he can at a rate R. These chickens, being "clones,"
all have an identical mass of m. Luckily, the duck has the good fortune of knowing
that chickens can be sold for 500$ per chicken at the mart. If the chickens
hit the crate with a perfectly inelastic collision, what will the scale read
at time t, as chickens are being dropped? How much will this crate be worth?
The crate has a mass of 50kg, and the chickens are dropped from a height h.

Determine the momentum of a chicken falling. Knowing v^2=v0^2 + 2a(x-x0), we
can conclude from the that the chickens will be falling with a velocity of *root(2gh)*.
To find the momentum, (p = mv) it would just be mass*velocity, or *m(root(2gh)).*
Now, the scale exerts and equal and opposite force to the falling chicken. So
using F = p/t, we find that with time being 1/R and p being m(root(2gh)), F
= *m(root(2gh)R.*

Now, the expression for the force of the chickens that have already fallen onto the scale can be written as Rtmg.

Putting it altogether, keeping in mind that we want the answer as an expression
of mass,

*(m(sqrt 2gh)R + Rtmg)/g.*

And of course, we want to know how much the crate will be worth, and it will
simply be,

*500*(m(sqrt 2gh)R + Rtmg)/g dollars.*

**Problem 2: **An 9.00 g bullet is fired into a 2.10 kg block initially at rest at the edge of a frictionless table of height 1.00 m. The bullet remains in the block, and after impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

m1v1 + m2v2 = (m1+m2)v'.

v' = d/t.

the horizontal velocity of the block bullet object after collision given by horizontal displacement divided by..

y = 1/2at^2

t = sqrt(2y/g)

the time it takes for the block bullet object to fall a distance.

m1v1 = (m1+m2)*d/(sqrt(2y/g)

v1 = (m1+m2)*d/(m1*sqrt(2y/g))

the algebraic solution

v1 = (0.009+2.1)*2/(0.009*sqrt(2*1/9.8));

v1 =
a whopping 1037.45m/s. Rather nonsensical but get the method down!

**Problem 3: **After falling from rest at a height of 7.45 m, a 0.564 kg ball rebounds upward, reaching a height of 5.65 m. If the contact between ball and ground lasted 4.72 ms, what average force was exerted on the ball? (Thanks n99)

F = dp/dt.

We will want to focus on solving for **v** and **v0**, the velocity of the ball as it leaves the ground and the velocity of the ball as it hits the ground, respectively.

v^2 = v0^2 + 2ad.

Since it is dropped, v0 = 0.

v^2 = 2(-9.81)(0-7.45)

v ~= -12.09m/s. (negative because it's going down, this is really the v0 referred to above).

Now find v referred to above by solving for v0 of the ball rebounding.

At max height, v = 0.

0 = v0^2 + 2(-9.81)(5.65-0).

v0 ~= 10.53m/s. (this is really the v referred to above).

Now we can find dp.

dp = dmv => 0.564kg*(~10.53-~(-12.09))m/s => -12.7577kg*m/s.

dt was given, 4.72ms.

F is then ~2702.90N in magnitude.