Tanya+Johnson

__Week 10 (April 19th, 2012)__ you haven't found the final answer ! Good job with the explanations !

Please read the problem, we need to find the velocities after the collisions, not before and after

Note: you haven't used the fact that the collision is elastic

-- Professor**

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. **

---   No update found for week 9 - professor  --- __Week 08 (March, 12th, 2012)__


 * 01. A loaded sled weighing 85 N rests on a plane inclined at an angle of 20⁰ with the horizontal. The coefficents of static and kinetic friction between the inclined plane and the sled are 0.25 and 0.15 respectively. **
 * a). What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane **
 * b). What is the minimum force required to start the sled moving upward? **
 * c). What value of F (force) is required to move the sled up the plane at constant velocity? **



__Week 07 ( 05th of March, 2012)**__

01 . An object is hung from a spring balance attached to the celing of an elevator cab. The balance reads 65 N, when the cab is standing still. What is the reading when the cab is moving upward, (a) with a constant speed of 8 m/s and
 * so it is known because of the given information that the acceleration of the car is equal to 0 m/s^2 because of how it has a constant acceleration.
 * Next, because the Fnet is equal to zero the equation can be assumed to be
 * TsinB + mgsinC = ma
 * where B and C are the given angles for the respective forces
 * next you can simplify the problem down to using trig, where B is equal to 90 and C is equal to 270.
 * T - mg = m(0)
 * Then solve for T.
 * **T = mg**

(b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s²
 * for this equation we are given the acceleration, which because of the fact that it is decelerating gives us an acceleration of -2.4m/s^2
 * Next you set up the same equation as in the first problem
 * TsinB + mgsinC = ma
 * where B and C are the given angles for their respective forces
 * next you can simplify the problem down using trig, where B is equal to 90 and C is equal to 270.
 * T - mg = m(-a)
 * Then solve for T.
 * T = -ma + mg
 * **T = m(-a + g)**

__ **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? **


 * first create a table of the given information, which is Δx, ax, v0y, Δy, and ay.
 * while Δy is not explicitly stated, it can be assumed that the initial velocity in the y is 0 m/s because of how that would be the max height.
 * next, you need to solve for time which you can do using the formula - Δy = v0y*t - (1/2)(ay)*t^2 which can then be re-written to be
 * t = sqrt(Δy/(.5*ay))
 * then you solve for t, which will be a positive time.
 * upon finding the time it takes for the fish to hit the water, you can then use the formula Δx = v0x*t + (1/2)(ax)*t^2 which can then be rewritten as
 * Δx = v0x*t because the acceleration in the, unless otherwise stated, is always 0 m/s^2 which would cancel out the second part of the problem.
 * then you can rewrite the problem as v0x = Δx/t and then solve for the v0x which is equivalent to the general initial velocity because of how the v0y does not have a value other than zero.

**//Week 3//**
 * 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? **
 * for part (b) of this problem you have to do a similar set of steps as we did in part (a).
 * so start off by creating a new table of the given values, which vary from the original problem because now we have Δy, ay, ax, v0y, and v0x.
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;">as in the previous problem it can be assumed that v0y is 0 m/s because of how the fish is dropped from the maximum height possible.
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px; line-height: 22px;">next, we (once again) need to solve for time and we can use the same equation as we did before - <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;"><span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;">Δy = v0y*6 - .5(ay)*t^2 and then because of how v0y = 0, the first term cancels out and we're left with the equation - Δy = -.5(ay)*t^2
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;"><span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;">then you rearrange the equation so that it is t = sqrt( Δy/.5ay)
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px; line-height: 22px;">then solve for t.
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px; line-height: 22px;">upon finding the time it takes for the fish to hit the water, you can then use the formula <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;"><span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;">Δx = v0x*t + (1/2)(ax)*t^2, then because of how the acceleration in the x is 0 m/s the second term cancels out and we're left with Δx = v0x*t
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;"><span style="color: #222222; font-family: Arial,sans-serif; font-size: 15px;">then you solve for Δx.
 * <span style="display: block; font-family: Verdana,Arial,sans-serif;">1. A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 10.5 m/s at an angle 54º above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.3 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s² . What is the height of the building? **
 * <span style="font-family: Verdana,Arial,sans-serif;">First list all of the given variables in tables for the components of X and the components of Y. Also draw a diagram that utilizes the given information to help create a better understanding of the situation.
 * <span style="font-family: Verdana,Arial,sans-serif;">Next, you need to find the components of the initial velocity, which you do by setting v0x equal to |v0|cos(Θ) and v0y equal to |v0|sin(Θ) thus getting the X and Y components of the initial velocity.
 * <span style="font-family: Verdana,Arial,sans-serif;">Next use the equation <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = <span style="font-family: Verdana,Arial,sans-serif;">v0xt + .5axt to find the time the rock takes to hit the ground, due to the fact that it does not state otherwise it can be assumed that the acceleration of the X component is zero (0) thus creating the formula:
 * <span style="background-color: #ffffff; color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = v0xt
 * <span style="font-family: Verdana,Arial,sans-serif;">which is equivalent to t = <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x/v0x
 * Then solve for 't' (time).
 * Finally you can use the <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">y = v0y*t + .5g*t where g is the gravitational constant (9.8 m/s^2) and then you solve for <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">y.

// 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 **// // ii. Find the magnitudes of ** A ** and ** B **// // iii. What is the angle between A and B//
 * //Week 2//**
 * to find A add (**A**+**B**)+(**A**-**B**) to which you will get 2**A** because +**B** and -**B** will cancel each other out
 * divide the solution for 2**A** by 2 because you solved for 2**A** rather than just a single **A** and you will get the vector for **A**
 * then substitute in the answers for **A** and solve for **B**
 * to find the magnitudes of **A** and **B** use Pythagorean theorem setting up two right triangles for **A** and **B** utilizing the vectors as the two sides
 * use the formula: sqrt(x^2 + y^2) = M (the hypotenuse)
 * then you will have the magnitude for both vectors
 * to find the angle between A and B you need to define an angle as Θ then use trig
 * I usually set up to solve for the angle utilizing tangent
 * so the formula would be tan(Θ) = j^/i^ but because we don't know what the angle Θ is you need to utilize basic trig skills to transform the equation into:
 * arctan(j^/i^) = Θ
 * then solve for Θ

//** Week 1 **//
 * 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? **
 * <span style="font-family: Arial,sans-serif;">First list all of the known variables and state the unknowns
 * <span style="font-family: Arial,sans-serif;">v = 305 m/min
 * <span style="background-color: #ffffff; font-family: Arial,sans-serif;">vo <span style="font-family: Arial,sans-serif;"><span style="font-family: Arial,sans-serif;">& v 1 = 0 m/min
 * <span style="font-family: Arial,sans-serif;">a = 1.22 m/ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">s <span style="background-color: white; font-family: Arial,sans-serif;">2
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = 190m
 * <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">t = ?
 * <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">Use the equation v^2=(v0)^2+2a( <span style="color: #222222; font-family: Arial,sans-serif;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x) because it does not contain time which is the only parameter that is not given.
 * <span style="font-family: Arial,sans-serif;">Then I converted all of the unites to SI unites, which meant the only variable that required conversion was v.
 * v = 305 m/min
 * v = 305 m/60s
 * v = 5.0833 m/s
 * <span style="font-family: Arial,sans-serif;">Then I substituted the numbers in for their respective variables
 * <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">(5.0833)^2=(0)^2+2(1.22)( <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x)
 * Then evaluate the equation, solving for <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = <span style="background-color: #ffffff; font-family: Arial,sans-serif;">[(5.0833)^2 - (0)^2]/[2(1.22)]
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = [25.8399 - 0]/2.44
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = [25.8339]/2.44
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">x = 10.5901 m
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **
 * First I decided to use the formula v = v0 + at
 * Then I re-ordered the equation to be (v - v0)/a = t because the problem requires us to solve for t.
 * Then I substituted the numbers in for their respective variables
 * (5.0833 - 0)/(1.22) = t
 * Then evaluate the equation, solving for t.
 * 5.0833/1.22 = t
 * t = 4.166 is the time that it takes to reach the top
 * next multiple 4.166 seconds by 2 to get the total time to end
 * total time = 4.166 x 2
 * total time = 8.332 seconds

No need to include numerical values in the final answer -- check my page
 * 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? **
 * First you need to define the variables.
 * <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">y = 50m
 * t1 = ?
 * t2 = t1 - 1.00s
 * v0 = 0 m/2
 * v = ?
 * a = 9.81 m/s^2
 * Next you need to identify the formula tat you are going to use and I decided to use the formula <span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">y = v0t + (1/2)a*t1^2
 * Next you solve for t:
 * 50 = 0(t) + (1/2)(9.81)*t1^2
 * 50 = (1/2)(9/81)*t1^2
 * t1^2 = 50/[(1/2)(9.81)]
 * t1^2 = 10.1937
 * t1 = sqrt(10.1937)
 * t1 = 3.19275 s
 * ii). What is the initial speed of the second stone? **
 * Next you define t2 as t2 = t1 - 1s
 * And then you solve for t2 in the equation of t2 = 3.19275s - 1s
 * t2 = 2.19275s
 * Then you utilize the same equation as in the previous question (<span style="color: #222222; font-family: Arial,sans-serif; font-size: 11pt;">Δ <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">y = v0t + (1/2)a*t1^2) except instead of using t1 you use t2 and solve for the initial velocity.
 * Then solve for v0
 * 50 = v0(2.19275) + (1/2)(9.81)(2.1975^2)
 * 50 - (1/2)(9.81)(2.1975^2) = 2.19275v0
 * 50 - 4.905(4.82901) = 2.19275v0
 * 50 - 23.6863 = 2.19275v0
 * 26.3137 = 2.19257v0
 * v0 = 26.3137/2.19257
 * v0 = 12.0003 m/s
 * Professor's Note : Excelent effort - Like the explanations very much !! keep up the good work **