Chris_J

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 **




 * Start by Drawing a picture: **
 * Next: The velocity of the ball is found by using the Energy equations: PE = KE **
 * m*g*h = 1/2*m*v^2 **
 * After finding v at the bottom of the arc, when the ball is at its lowest point, you can now find the velocity of the block after the collision. **

**and solve for V**1f and **V**2f


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) Fmax = u Nf F of friction

b). F = u Nf + sin(20 mg c). F = u Nf + F - sin(20)mg

__ **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

a) Use F = ma to find the mass in the system by using a equal to 9.8m/s^2 Speed is constant so a = 0, plug that into F = ma ,

b) mass was found in part a, a = -2.4m/s, negative because it is slowing down. so now us F = ma, where it is now F = ma+mg and the negative sign is take care of.

I don't see the new work on the page !! -- professor. Will visit again Last week you did a wonderful job and loved the diagram and the explanation.

__**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) First: Use Y-Yo=Vy*t+1/2*a*t^2 where the Vy = 0 and solve for t, where Y-Yo = -5.4m

Plug in t into X-Xo=Vx*t+1/2*a*t^2 where 1/2*a*t^2=0 because there is no ax (ax = 0) So X-Xo=Vx*t or X/t = Vx

b) Begin by Y-Yo=Vy*t+1/2*a*t^2 where the Vy = 0 and solve for t, where Y-Yo = -2.7m

Now use Vy^2 = Voy^2 + 2gDy and find Vy by using the new t

Now us Dx = (Vox)t and solve for Dx

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?
 * Week 04**

First draw a picture. Find Vx and Vy by Vx = Vicos(54) Vy = Vsin(54) We know ax = 0 ay = g = -9.81m/s^2 Find t by equation x= Vx*t + 1/2*ax*t^2 but we use x= Vx*t because ax = 0, and x = 15.3

//*X = (y-yo)// Then plug in t to y=Vy+1/2*ay*t^2 and solve for y to get the height of the building

**Professor's Note : WOW !! great job Chris. check my page for more hints for tomorrow's Quiz**

i. A is given as **A =**- 4.0 iˆ + 7.0 jˆ because "**B** is subtracted from **A"** therefore only A is left so subtract**- 4.0 iˆ + 7.0 jˆ** from **6.0 iˆ + 1.0 jˆ** to get B's Vector.
 * 1. A + B =** 6.0 iˆ + 1.0 jˆ
 * A - B** = - 4.0 iˆ + 7.0 jˆ

ii. to get the magnitudes of **A and B, do the** Pythagorean theorem (R^2 = A^2 + B^2)

iii. to get the angles take the inverse tangent of the "j" component over the "i" component

Professor's Note : part one and two are correct. Part three, not clear ! See my page for details. -- I will arrange the Quiz on Monday for you --


 * Week 2**

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 * 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.**======