Matt-Stoutz

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. **


 * 1. First you need to find the velocity of the ball just before it hits the block by using Conservation of Energy.**
 * .5mv^2 + mgh = .5mv^2 + mgh (0 PE at bottom)**
 * Plugging in the values you can find the final velocity of the ball. The initial velocity cancels out as well as the final PE.**


 * 2. Next you need to find the speed of the ball just after the collision.**


 * Conservation of momentum is used next.**
 * m1vo1 +m2vo2 = m1vf1 + m2vf2**
 * Initial velocity of m2 cancels out since it starts at rest.**
 * You get vf1 = (m1vo1 - m2vf2) / m1 for equation #1.**


 * KE initial = KE final since the collision is elastic**
 * .5m1vo^2 + .5m2vo^2 = .5m1vf^2 + .5m2vf^2**
 * The initial velocity of m1 is 0 so that cancels, both m2 also cancel.**
 * You end up with Vf1= sqrt(Vo^2 + .5vf^2) which is equation #2.**


 * Next you plug #1 (vf1) into the conservation of momentum equation to get vf2, which is the final velocity of the block.**
 * Then you use either equation to find the final velocity of the ball.**


 * __ 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 **

W = [M*g + M*(g/4)]*(-d) J
 * i.) **
 * 1. Fy=M*ay is used to find the tension in the cord. **
 * Tcos(0) + Mgcos(180) = M*a **
 * T - M*g = M*a **
 * T = M*g + M*a **
 * 2. W = F*d*cos(theta) is used in the next step. T is substituted in for F **
 * W = T*d*cos(180) **
 * W = (M*g + M*a)*(-d) **
 * The acceleration is then substituted in to get.... **

W = M*g*d J
 * ii.) **
 * To find the work done by the gravitational force, W = F*d*cos(theta) is used **
 * You get W = Mg*d*cos(0) **


 * iii.) **
 * To find the KE, the equation used is KE = 0.5*M*v^2 **
 * You can plug in the velocity found in part iv. **
 * You then get KE = .5*M*[ (2Mdg) / 8 ] **
 * This simplifies to **KE = (M*g*d)/4 J

V = sqrt(2*(g/4)*d) m/s
 * iv.) **
 * To find the speed, V^2 = Vo^2 + 2*a*d is used **
 * Vo cancels out b/c the initial speed is 0 **

__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.) To find the force required to prevent the sled from slipping down the ramp, the sum of the forces in the x direction is found using F=ma **
 * Ax=0 m/s² because the block is staying in the same spot and not moving at all. **
 * Fs, max + Fcos0 + mgcos(180+70) + Fn*cos(90)=m(0)**
 * Fs, max is replaced by μsFn, so now you get...**
 * μsFn + Fcos0 + mgcos(180+70) + Fn*cos(90)=m(0)**
 * Fn is found by Fn=mgcos(20)**
 * And you can solve for F to find the applied force.**


 * b.) Since the sled is about Ax=0 m/s², since it is not moving. **
 * You then use F=ma in the x-direction to find the force to start to move the block. **
 * Fcos0 + mgcos(180+70) + Fs*cos(180)=0**
 * Fs, max is replaced by μsFn**
 * Fn is found by Fn=mgcos(20)**
 * F-μsFn + mgcos(180+70)=0**


 * c.) Since the sled will be moved up the inclined plane at constant velocity, we know that Ax=0 m/s² **


 * The kinetic force of friction is used since the sled will be in motion this time.**


 * Ff(180) + Fcos0 + mgcos(180+70) + Fn*cos(90)=m(0)**
 * Ff, max is then replaced by μkFn**
 * Fn is found by Fn=mgcos(20)**
 * You then get -μk*Fn + Fcos0 + mgcos(180+70))=m(0) to find the force required to move the sled 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 **


 * a.) **
 * 1. To find the weight, Newton's Second Law is used: F=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 the mass*acceleration side is equal to 0. **
 * Fn*sin(90)+mg*sin(270)=0 **
 * 5. The weight will remain the same even with the movement of the car. **


 * 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**
 * Fn*sin(90)+mg*sin(270)=m*(-a)**
 * 4. The mass is found by dividing the weight by g-9.81 m/s² (W=m*g) **
 * 5. You then plug in the all the values you have to find the weight as the elevator decelerates as it moves upward. **


 * Matt : By far the best job !! --- very good, keep up the good work !! -- professor **

__Week 5__
 * 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? **


 * You can assume that Voy=0m/s since the pelican is flying in a horizontal direction.**
 * a.)**
 * First you need to find the time the fish is in the air by using the equation** Y-Yo=Voy*t-.5*ay*t^2 **and Voy*t cancels out since the initial y component of the velocity is 0m/s.**
 * Using the time found, you can plug that back into** X-Xo=Vox*t-.5ax*t^2 **to find the initial velocity**. **(-.5ax*t cancels out since the acceleration in the x direction is 0m/s^2)**


 * b.)**
 * First you need to find the time the fish is in the air for the new height using the equation** Y-Yo=Voy*t-.5*ay*t^2 **(Voy*t cancels out since the initial y component of the velocity is 0m/s)**
 * Using the time found and using the initial velocity already found in part a, you can plug those into** X-Xo=Vox*t-.5ax*t^2 **to find the horizontal distance the fish travels (also -.5ax*t cancels out since the acceleration in the x direction is 0m/s^2).**

__Week 4__


 * 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? **


 * Steps:**


 * First you need to find the horizontal component of the velocity using:** Vx=Vo*cosA


 * You know that ax=0 and ay=9.81**


 * Using the horizontal component of the velocity to find the time the rock is in the air using** X-Xo=Vox*t+(1/2)*ax*t^2


 * Using the time (the ball is in the air) found, you plug that into the equation:** Vy=Vo*sinA-g*t **to find the vertical component of the velocity**


 * With the vertical component of velocity, you can finally use that in the equation** Y-Yo=Vo*sinA*t-(1/2)*g*t^2 **to find the height of the building.**

Professor's Note **: Good effort ! Please be specific on finding time part, mention ax = 0 and ay = -9.8 m/s² -- visit my page for more hints !**

__Week 3__


 * 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.)**
 * First, you need to combine (B+A) and (A-B).**
 * It is set up like this:(B+A) + (A-B) = (6.0i^ + 1.0j^) + (-4.0i^ + 7.0j^)**
 * Solve for B, then plug the answer back into one of the first 2 equations to get A.**


 * ii.)**
 * To find the magnitude, the Pythagorean Theorem is used, taking the square root of A^2 + B^2**


 * iii.)**
 * To find the angle between A and B, combine the equation: A.B= |A| |B|cosx and A.B= AxBx+AyBy+AzBz**
 * You then get |A| |B| cosx=AxBx+AyBy**
 * Solve for angle x (angle between A and B)**


 * Professor's Note : Good work ! please check the answer for part iii in my page ! **


 * Week 2 **
 * 1.) **
 * i.) First you need to convert the max speed (305m/min) into m/s. You get 5.083m/s. **
 * Next, you use the equation V^2=Vo^2+2a(X-Xo) to get the distance covered to reach the max speed. **
 * V=5.083m/s **
 * Vo=0m/s **
 * a=1.22m/s^2 **
 * X-Xo=? **
 * Plugging those in, you find the distance it takes to reach the max speed to be **10.59m


 * ii.) To find the total time it takes to travel the 190m, you use **V=Vo+at
 * You get t=4.17sbut you need to double that number since that finds only the time for half the trip.you end up with **t=8.34s


 * 2.) **
 * i.) To find how long it takes the first stone to hit the ground, you use the equation Y-Yo=Vo*t+.5*a*t^2Given:Y-Yo=50ma=-9.81m/s^2Vo=0m/st=?Plugging those in, you find that it takes the first stone 3.19s to hit the water surface. **


 * ii.) First, subtract 1s from 3.19 to get 2.19s in order to get the time the 2nd stone is in the air. **
 * Then plug in that time into the equation Y-Yo=Vo*t+.5*a*t^2. **
 * Y-Yo=50m **
 * a=-9.81m/s^2 **
 * t=2.19 **
 * Plugging those in, you find the initial velocity of the stone to be **12.1m/s.**