Linear Momentum

Momentum, what fun.
p = mv
F = p/t

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.

2. Newtons second law rewritten in terms of momentum is Net force = change in momentum divided by change in time. Fnet = /\p//\t.

3. The law of conservation of momentum states that the before momentum is equal to the momentum after. m1v1+m2v2 = m1v1`+m2v2`. A 5kg duck waddles 5m/s into a 4 kg chicken which is initially at rest. The duck ends up moving at a speed of 1m/s in it’s original direction. m1v1+m2v2 = m1v1`+m2v2`. 5*5 + 0 = 5*1 + 4v2`. 20/4 = v2` = 5m/s

4. Impulse equals the change in momentum, or force times the change in time. Change in momentum equals change in time * force. The product of the force F times the time t over which the force acts is called the impulse.

5. The law of conservation of momentum is m1v1+m2v2 = m1v1`+m2v2`. The law of conservation of energy states the energy before = energy after. 1/2mv^2 + 1/2mv^2 = 1/2mv^2 + 1/2mv^2. ( 1, 2, 1`, 2`). v1 + v2 = v2` +v1 holds true for elastic collision. To solve the ballistic pendulum problems, m1v1+m2v2 = (m1+m2)sqrt(2gh), derived from GPE = KE and conservation of momentum.

6. Elastic collision is where total energy is conserved. This means that /\KE = /\KE. From thism we can derive that v1+v2=v2`+v1, regardless of mass. Inelastic collision would be when the total energy is not conserved. Thus, KE1+KE2 = KE1` + KE2` + other energy.

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.