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__**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** i.

A+B = 6.0 iˆ + 1.0 jˆ ---> 1 A-B = - 4.0 iˆ + 7.0 jˆ ---> 2

By adding 1 and 2 we will get 2A = 2 iˆ + 8 jˆ By dividing on 2 A = 1 iˆ + 4 jˆ

by taking 3 >> into 1 we will get B= 5 iˆ - 3 jˆ

ii.
 * A| = sqrt((1)^2+(4)^2
 * B| = sqrt((5)^2+(-3)^2

iii. we will use the rule tanQ=(B/A) and then by taking tan inverse(B/A) and we will get the answer for the angle.

Professor's Note: Good job on i and ii. Please check on number iii in my page ! --- __**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?

we have the initial velocity, the angle, and the acceleration of the horizontal ( x ) is zero and for the vertical ( y ) is 9.8 m/s².also we have (x-x0). first we use the equation (x-x0) =( V.x )( cos theta )*t and solve for t we will find the time.

then we will use (y-y0) = ( V.y )( sin theta )*t - (1/2)(g)(t²) and will know (y-y0) which is the height of the building.

**Professor's Note : Good, check my page for more hints for tomorrow's Quiz**

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__**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- Using the equation (y-y0) = (v0y)t+1/2at^2, we have (y-y0) and v0y and a , the first t will cancels out because v0y = 0m/s , so we can solve for t and find the time. after that using the equation

(x-x0) = (v0x)t+1/2at^2 we can solve for v0x and find it. then by taking the square root of v0x^2+v0y^2 we will find the initial velocity.

b- using the equation (y-y0) = (v0y)t+1/2at^2, we have (y-y0) and v0y and a, the first t will cancels out because v0y = 0m/s so we can solve for t and find the new time.

then by using (x-x0) = (v0x)t+1/2at^2, we have v0x, t and a so we can do the calculations and find (x-x0).

__ **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.

The balance reads the weight of the object, W=mg=65N, g=9.8 m/s ² a- A=0 because it moves with a constant speed. T-mg= ma where a = 0 and mg is 65. So we can solve for T and find it b- A=-2.4 because it moves down (-y), T-mg=ma , we have mg and a , m we can find it by dividing the weight over the gravity. Professor's Note : very good, you have done a great job !

__** 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?

a. we have the equation fs = Ms.Fn, we have the M static = 0.25 and Fn = mgcos(theta). the sum of F = 0

then F + fs - mg.sin(theta) = 0 we can solve it for F and find the least magnitude of the force.

b. we have the equation fs = Ms.Fn, we have the M static = 0.25 and Fn = mgcos(theta)

then F - fs - mg.sin(theta) = 0 we can solve it for F and find the minimum force required to start the sled moving upward.

c. we have the equation fk = Mk.Fn, we have the M kinetic = 0.15 and Fn = mgcos(theta).

then F - fk - mg.sin(theta) = 0 ,, ( a = 0 because of the constant velocity )

we can solve it for F and find the value of F (force) is required to move the sled up the plane at constant velocity

__** Week 09 (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

i) Find the work done by the cord's force on the block

ii) Find the work done by the gravitational force on the block

iii) Find the kinetic energy of the block

iv) Find the speed of the block



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__** 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

Good job with explaining parts !

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**