Professor's+page+for+Problems+and+Help

__**Week 2 (Jan 30th 2012 )**__
 * IMPORTANT --- DO NOT EDIT MY PAGE ! **


 * 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?**

**Solutions posted by the professor** i.
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest?**
 * --> The cab starts from 0 m/s and reaches a maximum speed 305 m/min **
 * note: m/min is not the SI unit for speed **
 * --> **** we must convert "m/min" to "m/s" **
 * --> If we look at the first part of the problem, it says moving from rest to a maximum speed at a constant acceleration **
 * Thus, we have v0, v and a and we are looking for (x-x0) **
 * Find a kinematic equation with the above four variables. We don't have "t", its best to use an equation without "t" to find (x-x0) **


 * ii. Now that we have (x-x0) we can use another equation to find "t" **
 * -- > the problem is symmetric, we can double the time for the half way run (from the origin to the maximum speed) **


 * second method **
 * ii. --> from rest to maximum speed (0 m/s to 305 m/min) find the time "t" using a = 1.22 m/s² **
 * --> from maximum speed back to rest (305 m/min to 0 m/s) find the time "t" using a = - 1.22 m/s² **
 * --> add the two times together to find the total **

---

i. ** When something is dropped from above --> the initial velocity, v0 = 0 m/s **
 * 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?**
 * ii). What is the initial speed of the second stone?**
 * --> the first stone falls under the gravitational acceleration so, a = 9.8 m/s² (acceleration is a vector and we must consider the direction with the magnitude) **
 * --> consider downward as the positive direction **
 * --> a = 9.8 m/s², v0 = 0 m/s, y-y0 = 50 m [ you can replace (x-x0) by (y-y0) and use the same kinematic equations] **
 * --> use the proper equation to find the time "t" [you may have to solve a quadratic equation to get the time] **

ii. ** The time for the second stone = ( time taken by the first stone - 1 ) **
 * --> acceleration = 9.8 m/s² (still the stone is under gravity ) **
 * --> y-y0 = 50 m, time = time taken by the first stone - 1 , a = 9.8 m/s² **
 * --> find the proper equation to get v0 **

__**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** ii. Find the magnitudes of **A** and **B** iii. What is the angle between **A** and **B** **A** + **B** = 6.0 iˆ + 1.0 jˆ -(1) **A** - **B** = - 4.0 iˆ + 7.0 jˆ ---(2)
 * __Solution steps:__**
 * Note:** We discussed a similar problem in the class,

-- > add equation (1) and (2) to eliminate **B**, now solve for **A** --> subtract equation (2) from (1) to eliminate **A** and solve for **B** --> Use the definition for dot-product**, A.B =** /A//B/cos ø --(3) also, **A.B** = AxBx + AyBy + AzBz --(4) combine (3) and (4) to find the angle,ø

Added to this, please go through what we have gone through in the class (examples !)

__**Week four (13 th February, 2012)**__ 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?

**Solution in steps :** We are given that the initial velocity is 10.5 m/s with an angle of 54° above the horizontal. We are going to need to find the x and y components of this.

=
‍ ‍- Acceleration in the x-direction is 0 m/s ² and in the y-direction its 9.8m/s² =====

=
‍ ‍- When you fill out the chart of known data, you have few unknowns, final velocity, time and the displacements along x and y directions. Since time will be the same in both x and y directions, it is easier to pick an equation to use that does not have the final velocity term. =====

=
‍ ‍(x-x­o) = Voxt + ½ at² remember the ½(ax)t² cancels out since acceleration is 0 m/s² =====

- Once time is calculated, you can now plug it into the y-direction equation and solve for height (y-y­o) (y-yo­) = (Voy)t + ½ (ay)t²

**Hint** : Also learn to find the final velocity. Learn to reverse the problem and solve for the final velocity and time.

__**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?


 * Note: I HAVE NOTED YOUR WORK AND ASSIGNED THE POINTS. THIS WEEK I DID NOT ADD ANY COMMENTS ON YOUR PERSONAL PAGES.**
 * ALL PARTICIPANTS DID A GOOD JOB -- I AM HAPPY FOR YOU . BE PREPARED FOR THE QUIZ AS WELL AS FOR A REVIEW **
 * The first thing that should be found is the time the object spent in the air. Because the fish is dropped there is no initial velocity in the y direction. This means that to find the time in the air, the height of the bird and the gravitational acceleration are plugged into yf-yi=1/2at ² . t is then equal to sqrt(2(yf-yi)/a). The distance the fish traveled in the x direction can then be divided by this time to find the bird's initial speed.

Step 1: Draw a Diagram Step 2: Make a table of known values Step 3: Start by looking for the final velocity in the Y direction. Voy = 0 m/s, g = 9.81 m/s2 Use V² = Voy² + 2g( D Y), D Y = 5.4 m Step 4: Use the two known vertical velocities in the equation V = V0y + gt to find time t.  Step 5: Time t is the same for the X component of velocity. Use D X = V0xt to find the initial velocity of the pelican. The bird is traveling horizontally, so the speed is the same as the initial horizontal velocity Step 2: Make a table of new values Step 3: Vy² = Voy² + 2g D Y, Find Vy. Step 4: Use Vy to find t from Vy //=// Voy + gt Step 5: Use D X = (Vox)t to find D X, which is the same as R, the horizontal distance the fish will fall.
 * Again the first thing to be found is the time the fish was in the air. This can be found by using sqrt(2(yf-yi)/a)=t. Then the initial speed of the pelican from the part a can be multiplied by the time the fish was in the air to find the horizontal displacementraw a modified diagram

**Week 06 -- EXAM WEEK --- No wikis**

__ **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 (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s²

Solutions:

**visit Matt- S, and S_Alhawal's pages too**

a.) 1. To find the weight, Newton's Second Law is used: Fnet = ma  2. It is used in the y direction since the elevator is going up or down only.  3. The sum of the y components is equal to mass*acceleration  4. The acceleration (ay) is 0m/s² because of the constant speed, thus the right hand side of the equation (m * a) equal to 0.   -- > T*sin(90) + mg*sin(270) =0, where T is the tension.   T - mg = 0  T = mg  5. The tension will remain the same even with the movement of the car. The scale reading (Tension) will be the same as before.

b.) a= -2.4m/s² (acceleration is negative because it is decelerating. W=65N 1. F=ma is used again in the y-direction. 2. This time the acceleration matters. 3. Once again, t he sum of the y components is equal to mass*acceleration T*sin(90)+mg*sin(270)=m*(-a) T - mg = - ma T = mg - ma  T = m (g - a ) **Note:** in this equation a = 2.4 m/s² negative sign is already taken care of ! 4 . The mass is found by dividing the weight by 9.81 m/s² (W=m*g) 5. Tension of the rope is the reading of the scale.

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__** 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?


 * Steps to solve the problem: **

1. First sketch a force diagram

showing the steps to solve part (b)



Pay attention to the force diagram.

F cos (0) + fs cos (180) + mg cos (250) + Fn cos (90) = 0 F - µsFn + mg cos(250) = 0 -- (1) [ Note: fs,max = µsFn]
 * Since the sled is attempting to move up along the incline, and since we are considering the minimum required force, take maximum static friction acting in the downward direction.
 * Sled is not moving yet, so "a" along the plane = 0 m/s^2
 * Thus, the sum of forces along the plane = m (0)
 * The sum of forces along the direction perpendicular to the plane = m (0), "a" along y direction = 0

F sin (0) + fs sin (180) + mg sin (250) + Fn sin (90) = 0 mg sin (250) + Fn = 0 (2)
 * Find Fn using equation (2), and substitute in equation (1)
 * Finally find the minimum force required to start the sled moving


 * ** I scanned through your pages. Most of you are doing a wonderful job !! **
 * ** Note: coefficient of friction, µ, and the friction force, f, are two different things (related but not the same thing) !! **
 * ** Hala did an impressive job and I recommend you to visit her page. **

Week 9 (April 6th, 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,

Class, please note that I went through the pages and noted down the work you submitted. I don't have time to comment on each page this week, but will do it over the weekend.


 * Steps to solve the problem: **
 * Draw a picture -- Important ! **

A. Find the work done by the cord's force on the block W= F·dcosƟ Need to find F F=ma Tcos(0) + mgcos(180)=ma T= mg+ma W= **T·d** **answer in joules**


 * When taking the dot product between the **T **and the displacement** d **the angle between them is 180**


 * W = mg (1.25)d cos (180) = - 1.25 mgd **


 * B. Find the work done by the gravitational force on the block **
 * W=F·dcosƟ **
 * F=mg **
 * W=mgdcos(0) **
 * W=mgd **


 * D. Find the speed of the block **
 * V²=Vo² +2a(Δd) **
 * Vo is zero **
 * V= Sqrt(2(g/4)d) **


 * C. Find the kinetic energy of the block **
 * KE =1/2 mv² **
 * Use the v from part d **
 * KE=1/2 m (2(g/4)d) **


 * Now that all of you have taken the midterm two, here goes the Quiz-help problem **
 * Now that all of you have taken the midterm two, here goes the Quiz-help problem **

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. **


 * Steps : **

I went through some answers, and did not find the correct final answer or the steps anywhere.

(Just found out Steve G has the right approach - please visit his page as well !) **

Here are my steps:

1. Draw a diagram

2. Use the conservation of energy equations, accounting the lowest poison (where the block is) as the zero potential level

--> initial velocity of the ball = 0 m/s thus KE (initial) = 0 --> Initial potential energy = ugh where h is the length of the cord --> final KE = 1/2 mv² --> final potential energy =

using the above information and placing them in, (KE + PE )i = (KE + PE)f

you can find the final velocity of the ball, v, before it hits the block

3. Now we are considering the conservation of momentum

m1v1i + m2v2i = m1v1f + m2v2f (1) we know, m1, m2, v1i and v2i we don't know, v1f or v2f --> to find two un-knowns we need two equations

4. We consider elastic collision

(Total KE )before = (Total KE)after

(1/2m1v1i² +1/2m2v2i²) = (1/2m1v1f² +1/2m2v2f²) (2)

again we know m1, m2, v1i and v2i

Now solve the two equations to find the v2f and v1f


 * Enjoy the video**

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