Beth+W

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. **

you haven't found the final answer ! I wish if you have explained the steps, reasoning why you are using each equations instead of giving the final answer

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**

__Week 08 (March, 12th, 2012)__

kinetic friction is when the object is moving
 * 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. **

would use the static coefficent becuase you want to find the force it takes before the sled starts to move. Fs,max=Coefficent of static * Fn Fs, max=Coefficent of static* Wcos(20)
 * a). What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane **

use kinetic friction because the sled is now moving max=Fkcos(0)+Fncos(90)+mgcos(250) Fx=Fk+mgcos(250) Fx=coefficentof kinetic*85+mgcos(250)
 * b). What is the minimum force required to start the sled moving upward? **

may=Fksin(0)+Fnsin(90)+mgcsin(250) m(0)=Fn+Wsin(250)=Fy Fnet=sqrt(Fx^2+Fy^2) if there is a constant velocity then acceleration equals zero. so the sum of the forces equals zero. so when an object is in equilibruim , there is no acceleration because there is a constant velocity or the object is at rest. Fnet=ma Fnet=0
 * 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 (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s² > ** T-mg=m(-a) ** > ** T=m(-a)+mg ** > ** T=m(-a+g) **
 * **we know that when the cart is stationary the acceleration is zero. so if something is at a constant speed there is also no acceleration.**
 * **Tsin( θ)+ mgsin(θ)=ma **
 * ** once you put in the θ and know that a=0, then T=mg **
 * **so a=-2.4m/s^2 because it is decelerating**
 * **Tsin( θ)+ mgsin(θ)=m(-a) **

Professor's Note : This is exactly what I was looking for -- Good job !

__ **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)Since we know** Δy ** and the acceleration, we use the equation ** Δ ** y=Voy*t+.5*ay*t^2. since the pelican is flying at the max height we know that Voy=0 so that part of the equation cancels out. then solve for t. **
 * ** then use Δx=Vox*t+.5*a*t^2. the horizontal acceleration is equal 0 so the second part of the equation cancels out and solve for Vox. **
 * ** b) the ** Δ ** y changes and use the Vox that was found in part a. **
 * **find t the exact same way we did in part a.**
 * **this time when we use the equation** ** Δx=Vox*t+.5*a*t^2, we just cancel the second part because the acceleration still equals 0 and just solve for the **
 * ** Δx. **

__ **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?
 * **First draw a projectile graph that states the data given.**
 * first thing to do is find the components of the initial velocity by using Vox=Vo*Cos θ and Voy=Vo*Sin ** θ, this gives you the x and y components of the initial velocity. **
 * to find the time we use the equation Δx= Voxt+1/2at^2
 * **the acceleration of the x componenent is zero so 1/2at^2 is equal to 1/2(0)t^2. the zero causes this part of the equation cancels out.**
 * with the rest of the equation which is Δx= Voxt, rearrange the equation to get t= Δx/Vox
 * **the the height of the building is also known as Δy. the equation used is Δy=Voyt+1/2at^2 **
 * ** the acceleration for ay is the acceleration of gravity. **
 * ** plug in the information. **
 * ** and done :) **

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

__ **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 **
 * **add (A+B)+(A-B) so then the B's cancel out and you are left with 2A**
 * **divide both sides by 2 and then you get what A is equal too**
 * **plug in A in (A+B)**
 * **solve for B**
 * **make triangle, have A and B as the sides.**
 * **use the Pythagorean theorem to solve for the hypotenuse**
 * **theorem: C=sqrt(A^2+B^2)**
 * **arctan(J^/i^)**

Professor's Note: Good work ! First two parts are correctly done ! Explain part iii bit more or visit my page for a different method


 * __Week 2 (Jan 30th 2012 ) __**

** 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. **

Δx=190 Vo=0m/s V=305m/min a=1.22m/s^2 t=?

** i). How far does the cab move while accelerating to the maximum speed starting from rest? ** converting 305m/min into SI units: 305m/min*1min/60s= 5.08m/s V^2=Vo^2+2a( Δx) (5.08)^2=0^2+2(1.22)(Δx)  25.8064=2.44Δx  Δx=10.5764m

** ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? ** V=Vo+at 5.08=0+1.22t time equals 4.16393 seconds but that only accounts for the trip to the max speed to get the total time from start to end i would multiply the time by two 4.16393*2=Total time Total time=8.32787 ** -- **

** 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. ** Δx=50m a=9.8m/s^2 Vo for first stone=0m/s t for both=? Vo for second stone=?

** i). How long does it take for the first stone to reach the water surface? ** Δx=Vot+1/2at^2 50=0t+1/2(9.8)t^2 50=1/2(9.80t^2 50=4.9t^2 t^2=10.2041 t=3.19438s

** ii). What is the initial speed of the second stone? ** Δx=Vot+1/2at^2j The speed of the second stone is one second less than the first stone  First stone - 1 second=second stone  3.19438-1=2.19438s  50=Vo(2.19438)+1/2(9.8)(2.19438)^2

50=Vo(2.19438)+1/2(9.8)(4.81532) 50=2.19438Vo+23.595 26.405=2.19438Vo Vo=12.033m/s


 * Professor's Note : Good effort !! Do not include numerical values in the final answer -- check my page-- **