1) You and your friend are hiking when you stumble upon a ghost town. You find a

Question

1) You and your friend are hiking when you stumble upon a ghost town. You find a well that has no pulley or rope. You take a stone and drop it into the well and you and Callie wait for the splash. You time it and it took 4 seconds. How deep is the well?

2) A horizontal disk of mass 0.12 kg and radius 0.13m rotates freely about a vertical axis through its center with an angular speed of 5 rad/s. A mass of 0.022kg of sticky matieral drops vertically onto the disk from above and sticks to the edge of the disk. (a) What is the angular speed of the disk immediately after the material system and teh disk slows to a stop. (b) Find the time it takes the disk to stop if the magnitude of the angular acceleration is 2rad/s^2. (c) Find the number of turns the disk completes from the time the material sticks to the time the disk stops.

3) Two masses M=2kg and m=1kg slide without friction down the sides of a hemisphereical bowl with height h=0.563m. Each mass starts with zero velocty at the lip, which is at the height h above the bottom. They start at the same time, move in opposite directions, and collide elastically at the bottom of the bowl. (a) by how much will the lighter mass overshoot the lip after this classic collision? (b) How high from teh bottom of the bowl will the masses move if they stick together on contact?

4) a uniform thin rod of weight W is supported horizontally by two bricks at its ends. At t=0 one of these bricks is kicked out quickly. Calculate the force on the other support immediately thereafter - it should be in terms of W.

5)A very massive object almost infinitely massive moving with velocity v1=100m/s hits a very light object -massless- at rest. The collision is classic. What is the final velocity of the very light object?

6)By how much will the earth slow down if its two ice caps melt. Find out how much water the two ice caps contain and teh conservation of angular momentum. Give your answer in seconds.

Explanation / Answer

Number 1)

Apply d = vot + .5at^2

d = (0) + (.5)(9.8)(4^2)

d = 78.4 m

Number 2)

a) Immediately after the object sticks, the velocity will go down and can be calculated from conservation of angular momentum

L = Iw, so Iw = Iw

(.5mr^2)(w) = (.5mr^2 + mr^2)(w)

.5(.12)(.13^2)(5) = [(.5)(.12)(.13)^2 + (.022)(.13^2)](w)

w = 3.66 rad/s

When the two finally stop, the angular speed will be zero

b) Apply wf = wo + (alpha)t

0 = 3.66 + (-2)(t)

t = 1.83 sec

c) wf^2 = wo^2 + 2(alpha)(theta)

0^2 = 3.66^2 + (2)(-2)(theta)

theta = 3.35 rad

Divide by 2pi = .533 revolutions

Number 3)

a)

They will hit at the bottom at the same time, and same speed

v = sqrt(2gh) = sqrt(2)(9.8)(.563) = 3.32 m/s

The initial KE = .5(2)(3.32)^2 + (.5)(1)(3.32)^2 = 16.53 J

The final KE must be the same

The momentum conservation

2(3.32) - 1(3.32) = (2v) + v'

3.32 = 2v + v'

v' = 3.32 - 2v

For final KE 16.53 = .5(2)(v)^2 + .5(v')^2

16.53 = v^2 + .5v'^2

16.53 = v^2 + .5(3.32 - 2v)^2

16.53 = v^2 + 5.51 - 6.64v + 2v^2

3v^2 -6.64v - 11.02 = 0

Use quadratic equation for roots

roots are 3.31 and -1.11

Since the velocity of the large block moves backwards, we will use -1.11 m/s

Thus the other block can be found - v' = 3.32 -(2)(-1.11) = 5.54 m/s

Thus we will use -1.11 and 5.54

Now mgh = .5mv^2

(9.8)(h) = .5(5.54)^2

h = 1.566 m from the bottom

1.566 - .563 = 1.0 m above the lip

b)

If they stick together

Conservation of momentum

2(3.32) + 1(-3.32) = 3v

v = 1.106 m/s

mgh = .5mv^2

(9.8)(h) = (.5)(3)(1.106)^2

h = .187 m

Above are the answers to the first three questions in your multi post question.

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