Aprojectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. at the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. the 1.0-kg part falls straight down after breakup with an initial speed of 10.0 m/s, the 0.7-kg part moves in the original forward direction, and the 0.3-kg part goes straight up.

(a) find the speeds of the 0.3-kg and 0.7-kg pieces immediately after the break-up.

(b) how high from the break-up point does the 0.3-kg piece go before coming to rest

a)

b)h=56.68 m

Explanation:

1) Notation and basic concepts

Projectile motion

When the motion is in two dimensions is known as projectile motion. The path that follows the object is known as trajectory.

Other important concepts that we will use to solve this problem are Conservation of momentum, this concept is a fundamental law of physics which states "that the momentum of a system is constant if there are no external forces acting on the system"

And the other important concept is conservation of energy a principle of physics that states "the energy of interacting bodies or particles in a closed system remains constant".

is the initial angle for the motion

is the initial velocity

is the velocity for the 1kg object

is the velocity for the 0.7kg object (variable of interest)

is the velocity for the 0.3kg object (variable of interest)

maximum height reached by the 0.3kg object

The figure attached describe the situation for the problem.

2) Part a

The momentum is defined as using this concept we can find the velocities required for part a

If we use this principle from the starting point and at the highest point for the fragments and we focus on the 0.7 kg, since this fragment follows the same x trajectory we have this

And we can solve for , on this case

Now for the 0.3 kg part we can use again conservation of momentum this part moves forward, the momentum for this particle upward need to be the same for the particle of 1kg moving downward, so we have the following formula

And solving for we got:

3) Part b

In order to find "How high from the break-up point does the 0.3-kg piece go before coming to rest" we can use conservation of energy when the fragment of 0.3kg begins to move upward we can set up this level as our reference height, at this starting point we have kinetic energy given by , where is the velocity founded on part a), and 0 potential energy since we are at the reference level. And we want to find the maximum height h that this part achieves, so at this final point the velocity for the particle would be 0 so we don't have kinetic energy but the potential energy would be maximum and given by , applying the balance of energy we got:

Cancelling the mass and solving for h we got

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

freezing

deposition

condensation

explanation;

melting also called fusion is the process where solid becomes liquid, sublimation is the process where solid becomes gaseous without first changing to liquid, deposition involves change from gaseous to solid state, freezing involves change from liquid to solid, boiling or evaporation involves change of liquid to gaseous state, while condensation is the change from gaseous state to liquid state.

endothermic reactions involves absorption of heat from the surroundings or the environment while exothermic reactions involves loss of heat energy to the surroundings.

in this case, sublimation, evaporation and melting are endothermic reactions (gain energy from the environment) while, freezing, deposition and condensation are exothermic (loss of energy to the surroundings) reactions.