1) A 0.14-kg baseball is dropped from rest from a height of 2.0 m above the grou

Question

1) A 0.14-kg baseball is dropped from rest from a height of 2.0 m above the ground. What is the magnitude of its momentum just before it hits the ground if we neglect air resistance?

2) A 40-kg uniform ladder that is 5.0 m long is placed against a smooth wall at a height of h = 4.0 m, as shown in the figure. The base of the ladder rests on a rough horizontal surface whose coefficient of static friction with the ladder is 0.40. An 80-kg bucket is suspended from the top rung of the ladder, just at the wall. What is the magnitude of the force that the ladder exerts on the wall?

3) A rolling wheel of diameter of 68 cm slows down uniformly from 8.4 m/s to rest over a distance of 115 m. What is the magnitude of its angular acceleration if there was no slipping?

4) A fire hose is turned on the door of a burning building in order to knock the door down. This requires a force of 1000 N. If the hose delivers 40 kg of water per second, what is the minimum velocity of the stream needed to knock down the door, assuming the water doesn't bounce back?

5) A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane?

6) A child is riding a merry-go-round that is turning at 7.18 rpm. If the child is standing 4.65 m from the center of the merry-go-round, how fast is the child moving?

7) To drive a typical car at 40 mph on a level road for one hour requires about 3.2

Explanation / Answer

1)first you need to calculate the velocity of the ball right before it hits the ground with the velocity and position equations

velocity = -9.8 (t) + Vo

Vo= 0 so v= 9.8(t)

to solve for t use the position equation

position = -4.9 t^2 +Vo(t) + So

solve for t when the position is zero

0 = -4.9 t^2 +0+2

t= 0.639 seconds

now solve for velocity

velocity = -9.8 *(0.639)

velocity = 6.26 m/s

multiply velocity and mass to get momentum

6.26m/s * 0.14 kg = .877 kg m/s

2)Using the sum of the squares of two sides equals the square of the hypotenuse, to find the lenght 'X' of the ladder's base from the wall..........Thus; X^2 + (4)^2=(5)^2 or X^2=25-16 X=3
The sides are a 3 4 5 triangle.
Using these ratios, to get the force exerted perpendicular to the wall by the hanging 80Kg, to be
80 x 3/4= 60Kg (force) or 60 x9.8 newtons=588 N.

the effective force at the top of the ladder, from its mass, is one half of the midpoint force or 15Kg or 15 x 9.8 =147 N
The total force exerted by the wall is the sum of the two, 588 +147= 735 N

4)force = change in momentum = change in mass of water * velocity of water stream

1000 = 40 * velocity
velocity = 1000/ 40 = 25 m/s

5)Total kinetic energy at the bottom = 0.5(1+0.4) mv^2 = mgh
V^2 = 7*9.8/0.7
V = 9.9m/s
? = V/r = 9.9/1.7 = 5.8rad/s

6) 7.18 rotation means 7.18 * 2 ? radians = 14.36? rad. in one minute = 0.23933? rad/s.

now use v = r*? (where ? is angular speed; r is radius and v is velocity)
thus v = 4.65 * 0.23933? m/s = 3.496 m/s ? 3.5 m/s.

7)the job is to calculate the I for a solid cylindrical disk of mass = 400 kg and radius = 1.2/2 = 0.6m.
KE of solid disk = 1/2MR

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