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__**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, we must make note that the mass of the rock is irrelevant here. Gravity has the same effect on any mass, and therefore does not affect projectile motion. We must also decide which vertical direction will be considered positive. This will affect the sign in front of the acceleration of gravity and will determine whether our vertical displacement is positive or negative. Either way, our final answer will be the same because we are finding the height of the building.

The first action we must take is finding the horizontal component of velocity. This is found by multiplying the magnitude of the given velocity vector (10.5 m/s) by the cosine of the angle theta (54 degrees) in relation to the positive x axis. (Remember that the initial horizontal velocity does not change because there is no horizontal acceleration).

Vx0 = 10.5 * cos(54)

Now we can use this horizontal component to find the total amount of time that the rock is in the air because we are given the total horizontal displacement (15.3 m). We will use the equation shown below...

horiz. disp. = Vx0 * t (Ax=0)

Next, we must find the initial vertical component of velocity. This is found by multiplying the magnitude of the given velocity vector (10.5 m/s) by the sine of the angle theta (54 degrees) in relation to the positive x axis.

Vy0 = 10.5 * sin(54)

Lastly, we can use this initial Vy component, the acceleration of gravity, and the total time the rock is in the air to determine the total vertical displacement.

vert. disp. = Vy0 * t + (1/2) * (Ay) * t^2 (Ay=acceleration of gravity)

The vertical displacement tells us the vertical change in position of the rock from the point that it was launched from the edge of the building to the point where it hits the ground. Therefore, the the vertical displacement of the rock is equivalent to the height of the building.

__**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) The first step is to use the equation **vert. disp. = Vy0 * t + (1/2) * (Ay) * t^2** to solve for t. Vy0 is zero because the fish is being dropped.

We can now use the equation **horiz. disp. = Vx0 * t** to solve for Vx0, which is also the pelican's initial speed.

b) With a new height, we will use the equation **vert. disp. = Vy0 * t + (1/2) * (Ay) * t^2** to solve for a new t. Vy0 is zero.

We now use the equation **horiz. disp. = Vx0 * t** again using the new time and same initial horizontal velocity to solve for the distance the fish would travel horizontally before hitting the water.

__ **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

There are 2 forces acting on the hanging object. One is the tension of the material attaching it to the spring balance on the elevator ceiling acting upward, the other is essentially the weight of the object acting downward, which is equivalent to mg (mass*acceleration from gravity). While the object is hanging still, the 2 forces are equal. The balance is measuring only one of the forces. We can then set 65 N equal to mg, and find the mass of the object.

a) While the elevator is moving upward at a constant velocity, all of the elements regarding the forces acting on the object remain the same. The acceleration does not change (it is zero), so the force and the reading of the balance is still 65 N.

b) Multiply the mass of the object by the deceleration (-2.4 m/s^2). Subtract this force from the weight of the object (65 N) to get the new reading of the balance.

__ 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

1) i) The cord's force on the block can be called the force of tension and named Ft. The angle between Ft and the direction of displacement is 180 degrees. The work done by the cord's force on the block is found by Wt=Ft*d*cos(180) or Wt=-Ft*d.

ii) The gravitational force on the block can be given by Fg or M*g. The angle between Fg and the direction of displacement is 0 degrees. The work done by the gravitational force on the block can found by Wg=M*g*d*cos(0) or Wg=M*g*d

iii) The Ke of the block is equivalent to the net work done on the block. So the Ke of the block can be found by adding Wt and Wg.

iv) The speed of the block can be found using the equation Ke=Kef-Kei. The initial velocity is zero so Ke=Kef and Ke=(1/2)*M*v^2, taking Ke from part iii.

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.

My Work: As the ball falls from its original position, the energy is conserved. So we can use the equation: (1/2)(m)(vi)^2 + (m)(g)(hi) = (1/2)(m)(vf)^2 + (m)(g)(hf) vi=0; hi=.7 meters; hf=0; m=.5 kg Solve for vf which is the velocity of the ball immediately before it hits the block

When the ball strikes the block, momentum is conserved. So we can use the equation: (m1)(v1i) + (m2)(v2i) = (m1)(v1f) + (m2)(v2f) m1 & m2 are .5 kg & 2.5 kg; v2i=0; v1i = vf from previous equation There are two unknown variables here: v1f & v2f So we need another equation with the same unknown variables Because the collision is elastic, Ke is the same before and after the collision We can use: (1/2)(m1)(v1i)^2 + (1/2)(m2)(v2i)^2 = (1/2)(m1)(v1f)^2 + (1/2)(m2)(v2f)^2 All variables here are the same as in the conservation of momentum equation Unknowns are also v1f & v2f. Solve for one. Substitute it into the cons. of momentum equation and find value of the other. Substitute this value into either equation to find the the remaining unknown. v1f is the speed of the ball just after the collision. v2f is the speed of the ball just after the collision.