Kalkidan-Z

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 1**

1) a=v/t t=v/a t=305/(1.22*60) =4.23611 i) s=1/2(1.22)(4.23611)^2 =10.94622 m ii) t=((2s)/a)^(1/2) =((2*190)/1.22)^(1/2) = 17.64866 sec i) To find A and B we can use substitution or adding the two equations. When we add the two equations we get A+A+B-B=(6-4)i+(1+7)j The B can cancel out and we can find A - We can find B by substituting the A in either of the two above formulas ii) We find the magnitude of A by squaring, adding, then by finding the square root of the i and j components. -We can find the magnitude of B by squaring, adding, then by finding the square root of the i and j components iii) cos^(-1)(-7*2^(1/2)/34)=1.86623
 * Solution ** Given: A+B=6i+j A-B=-4i+7j

__**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? x=15.3, Vo=10.5, ay=9.8 1)Vox=Vocos(54) 2)Voy=0 ---** Incorrect V0y= V0 sin 54 ** 3)find the time using X=Vox+1/2at^2....a=0 4) find y(height of building) substiting the time we found in step 3 to y=1/2gt^2 **Professor's Note: pay attention to the comment in red, also visit my page for more hints for tomorrow Quiz**
 * Solution**
 * we dont need** mass



__**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? 1) X-Xo = Vot - 1/2at^2 ......(a=0) Vo = (X-Xo)/t 2) Y-Yo= Voyt - 1/2 at^2....(Voy=0 at max) t= sqrt (2(Y-Yo)/a) 3) Vo = (X-Xo)/sqrt (2(Y-Yo)/a) 1) X-Xo = Vot - 1/2at^2 ......(a=0) 2) t= sqrt (2(Y-Yo)/a) 3) X- Xo = Vo*sqrt (2(Y-Yo)/a)
 * __Solution:__**
 * a)** a= 9.81m/s^2, Y-Yo = 5.4 m, X-Xo = 8m, Vo=?
 * b)** a= 9.81m/s^2, Y-Yo = 2.7 m, X-Xo = ?, Vo= (X-Xo)/sqrt (2(Y-Yo)/a**)...(Vo of a)**

__ **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



I can't comment on the document -- but Good job !! -- professor's Note Week 9 (April 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,

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