Kyle+Wald


 * Week 10 (April 16th, 2012) **

01**. A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of it's path the ball strikes, a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find, a) the speed of the ball b) the speed of the block, both just after the collision

Hint: first draw a diagram. find the velocity of the ball, just before it hits the block.


 * Week 9 (March 26th, 2012)**


 * 1) A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. When the block has fallen a distance d,

After drawing/labeling a picture and gathering the "known values"...

> NOTE: You need to find F in order to find Work
 * Find work of the cords force on the block using the equation W = F · d cos(theta)

F = m a Tcos(0) + mg cos(180) = ma T = mg + ma W = T · d (J)

(Is this how you have been wanting me to show my answers? You always say no numerical values so I took them out.)

B. Find the work done by the gravitational force on the block.

Use the equation " W = F · d cos(theta) "

NOTE: You need to find F in order to find Work

F = m g W = mg · d cos(theta) W = mgd

C. Find kinetic energy of the block

K.E. = 1/2m v^2 K.E. = 1/2m (2(g/4)d)
 * Use K.E. = "1/2 mv ^2" (This is easier after you complete D)

D.Find speed (VELOCITY) of the block.

Use equation V^2 = V0^2 + 2a (d) NOTE: Don't bother taking the square root of V because you need V^2 for C NOTE: V0 is zero

V^2 = V0^2 + 2a (d) V = (2(g/4)d)


 * sorry for lack of color. I have an am pressed for time, tons of work.*

__**Week 05 ( 20th February, 2012)**__

A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. The acceleration of gravity is 9.81 m/s2. a) What was the pelican’s initial speed? b) If the pelican was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?

A)

After drawing/labeling a picture and setting up a table of values for the given problem...


 * Find time it takes for the fish to hit the water with " y - yo = Voy t + 1/2 ay t^2 "

5.4 = 0 + 1/2 (9.81) t^2 5.4 = 4.9 t^2 5.4 / 4.9 = t^2 t = sqrt (5.4 / 4.9) t = 1.05s

NOTE: Problem states horizontal path, this means solely Vox component for Vo
 * Find Vox by using " x-xo = Vox t + 1/2 ax t^2 "

8 = Vox (1.05) + 0 Vox = 8 / 1.05 Vox = 7.62 m/s

B)

After drawing/labeling a picture and setting up a table of values for the given problem...


 * Find time it takes for the fish to hit the water with " y - yo = Voy t + 1/2 ay t^2 "

2.7 = 0 + 1 / 2 (9.8) t^2 2.7 = 4.9 t^2 2.7 / 4.9 = t^2 t = sqrt (2.7 / 4.9) t = .74s


 * Find x-xo by using " x-xo = Vox t + 1/2 ax t^2 "

x-xo = 7.62 (.74) + 0 x-xo = 5.64 m

__**Week three (06th of February, 2012)**__

1. If **B** is added to **A** the result is 6.0 iˆ + 1.0 jˆ. If **B** is subtracted from **A**, the result is - 4.0 iˆ + 7.0 jˆ.

i. Find **A** and **B**


 * Add "(A + B) + (A - B)" equation to single out A

A + B = 6.0 i^ + 1.0 j^ A - B = -4.0 i^ + 7.0 j^  2A = 2.0 i^ + 8.0 j^  A = 1.0 i^ + 4.0 j^


 * Plug A into either equation and solve for B

A = 1.0 i^ + 4.0 j^ B = X + Y  C = 6.0 i^ + 1.0 j^

B = 5 i^ - 3 j^

ii. Find the magnitudes of **A** and **B**

After drawing/labeling a picture for the given problem...


 * NOTE: To find magnitude take the sqrt ((x^2) + (y^2))

IAI = sqrt ((1)^2) + (4)^2) = rad(17) = 4.12

IBI = sqrt ((5)^2) + (-3)^2) = rad(34) = 5.83

iii. What is the angle between A and B
 * Use "AxBx + AyBy + AzBz" to find A * B

A * B = (1 * 5) + (4 * -3) A * B = -7


 * Use "A * B = IAI * IBI cos (theta)" in order to find theta

-7 = 4.12 * 5.83 cos (theta) -7 = 24.02 cos (theta) -7 / 24.02 = cos (theta) arccos (-7 / 24.02) = theta theta = 106.94 degrees

**Professor's Note: Vey well done !! way to go Kyle. try not to provide the final answer in the numerical form**

__**Week 2 (Jan 30th 2012 )**__
 * 01. A certain cab has a total run of 190 m and a maximum speed of 305 m/min. It accelerates from rest and back to rest at 1.22 m/s².**


 * i). How far does the cab move while accelerating to the maximum speed starting from rest?**

After drawing/labeling a picture and setting up a table of values for the given problem...


 * Convert 305m/min to m/s

(305m/min)(1/60) = 5.08m/s


 * Use the equation "vx^2 = vox^2+a(x-xo)" in order to find the delta X.

5.08^2 = 0 + 2(1.22)(x-xo) 25.81 = 2.44(x-xo) 25.81 / 2.44 = (x-xo) (x-xo) = ** 10.58 m traveled during acceleration **


 * NOTE: the distance traveled during acceleration is the same distance during deceleration. (total = 21.16m)


 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest?**

After drawing/labeling a picture and setting up a table of values for the given problem...


 * NOTE: There are three different points of interests in the cabs travels. The acceleration from rest to max speed, the distance traveled at max speed, and the deceleration to rest once more. Each will have their times calculated and summed up for total time.


 * Let t1 = time to achieve max speed from rest and use the equation "vx = vox + at"

5.08 = 0 + 1.22t1 ---> **watch out with the final/initial velocities and the sign of "a" -- PROFESSOR** 5.08 / 1.22 = t1 t1 = 4.16s


 * NOTE: t1 and t3 are the same values due to acceleration and deceleration having the same speed of 1.22 m/s^2


 * Find t2 by simply subtracting the total distance traveled during t1 and t3 and dividng the answer by the max speed

190 - 21.16 = 168.84m t2 = 168.84 / 5.08 = 33.24s


 * Find  total time by using "t1 + t2 + t3"

4.16 + 33.24 + 4.16 = 41.56s total


 * 02. A stone is dropped into a river from a bridge 50 m above the water. Another stone is thrown vertically down 1.00 s after the first stone is dropped. The two stones st rike the water surface at the same time.**


 * i). How long does it take for the first stone to reach the water surface?**

After drawing/labeling a picture and setting up a table of values for the given problem...


 * Use the equation "y - yo = voyt + 1/2at^2" to find time of free fall1

50 = 0 + 1/2(9.8)(t^2)

50 / 4.9 = t^2

t = sqrt(50/4.9)

t = 3.19s


 * ii). What is the initial speed of the second stone?**

After drawing/labeling a picture and setting up a table of values for the given problem...


 * NOTE: The time for the second rock is 3.19 - 1 = 2.19s

> > 50 = voy(2.19) + 1/2(9.8)(2.19^2) > 50 = voy(2.19) + 23.52 > 26.48 = voy(2.19) > 26.48 / 2.19 = voy > voy = 12.09 m/s is the initial speed of the second stone
 * Use the equation "y - yo = voyt + 1/2at^2" to find time of free fall2


 * Professor's Note : Good job Kyle- Like the explanations a lot !! Try not include numerical values in the final answer. Keep up the good work !! -- also check my page **