Explosion and Collision

A space probe explodes in flight into three equal portions. One portion continues along the original line of flight. The other two go off in directions each inclined at 60º to the original path. The energy released in the explosion is twice as great as the kinetic energy possessed by the probe at the time of the explosion. Determine the kinetic energy of each fragment immediately after the explosion.

explosion1

Analyze the problem:

With the word explosion we should know that it could involve conservation of momentum and kinetic energy.

What are the given information?

  1. Three fragments of exploded probe that have equal masses
  2. Direction of probe is φ=0º with the horizontal.
  3. Directions of motion of each fragment after explosion are: fragment 1 (θ1=60º with the horizontal), fragment 2 (θ2=0º with the horizontal), fragment 3 (θ3=-60º with the horizontal).
  4. Total energy released during explosion is twice the total kinetic energy of the probe.

What are the needed information?

  1. The final kinetic energy for each fragment of the probe immediately after explosion.

SOLUTION:

explosion1-2explosion1-3

Two Blocks on Inclined plane

“Physics Principles with Applications”, Douglas C. Giancoli, 7th Edition, Global Edition

44. Two blocks are connected by a light string passing over a pulley of radius 0.15 m and moment of inertia I. The blocks move (toward the right) with an acceleration of 1.00 m/s2 along their frictionless inclines (See figure below). (a) Draw free-body diagrams for each of the two blocks and the pulley. (b) Determine FTA and FTB, the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, I.

Giancoli 8-44

Analyze the problem:

The problem has two blocks on inclined surfaces and they are attached by an inelastic string (so they should have the same acceleration) the pulley’s mass is not ignored (we know that because the problem mentioned that there is a moment of inertia).

What are the given information?

  1. Block A mass, mA=8 kg
  2. Block B mass, mB=10 kg
  3. Acceleration, a = 1 m/s2
  4. Inclination Angle 1, θ1=32º
  5. Inclination Angle 2, θ2=61º
  6. Radius of pulley, R= 0.15 m

What are the needed information?

  1. Body-diagram draw for the two blocks and the pulley.
  2. FTA and FTB tension forces
  3. The net torque on the pulley
  4. The moment of inertia of the pulley

SOLUTION:

Giancoli 8-44-1

Giancoli 8-44-2

Newton’s laws and weightlessness

Source: “Physics for Scientist and Engineers” Serway and Jewett, 6th edition

6.25 A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person’s mass, and (c) the acceleration of the elevator.

Analyze the problem:

The problems mentions two stages of motion: (1) starting of motion and (2) stopping of motion for an elevator. The given forces are the normal forces or the readings of the scale.

What are given information:

  1. Normal force (N1) when elevator is starting = 591 N
  2. Normal force (N2) when elevator is stopping = 391 N

What the problem need:

  1. Weight of person
  2. Person’s mass
  3. Acceleration of the elevator

SOLUTION:

Now, we have to recall Newton’s laws to be able to analyze the motion of man in the elevator.

By the starting of motion an acceleration of the elevator will be directed toward a specific direction and when it stops it will change to the other opposite direction.

6-25c.jpg

Newton’s Laws and Circular Motion

From Serway and Jewett Physics for Scientists and Engineers with modern physics, 9th edition

6.11 A 4.00-kg object is attached to a vertical rod by two strings, as in Figure bellow. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.

Serway 6.11 problem.jpg

Analyze the problem:

The problems states that it wants tension force in string for the above setting. Newton’s laws should be used beside the dynamic of circular motion concepts.

What are given information:

  1. Strings lengths (L) = 2 m
  2. Height (h) = 3 m
  3. Mass of object (m) = 4 kg
  4. Speed of object (v) = 6 m/s

What the problem need:

  1. Tension of upper string
  2. Tension of lower string

 

SOLUTION:

serway-6-11-problem-sol-aFrom above figure we can see that we have isosceles triangle hence we can have right triangle by drawing a perpendicular line on the base of triangle to get sides lengths 2 m, and 3/2=1.5 m, so:

Serway 6.11 problem main sol.jpg

Then:

The tension in the upper string and  the lower string

Serway 6.11 problem main sol2.jpg

 

 

 

 

Simple Dynamics of Circular Motion Problem

From Serway and Jewett Physics for Scientists and Engineers with modern physics, 9th edition

Serway 6.8 problem.jpg

Let’s analyze the wording of the problem:

  1. A conical pendulum is a swinging pendulum that can rotate around an axis. The bob (the mass) will have circular path at all angles θ, but here we need to investigate when the angle is 5°.
  2. When a body moves is circular motion it must have at least a radial acceleration (centripetal acceleration).
  3. The mass is connected by a wire, the wire will have a tension (because it is used to suspend the bob). The tension should have x- and y-components.

What the information give in problem:

  1. The bob’s mass (m) = 80 kg
  2. Length of wire (L) = 10 m
  3. Angle of pendulum (θ) = 5°

What the problem need:

  1. The x- and y-components of tension force.
  2. The radial acceleration

 

SOLUTION:

  • The x- and y-components of tension force.
    • The x- and y-components of the tension force are:

serway-6-8-problem-sol-aa

  • The radial acceleration

Serway 6.8 problem -sol b.jpg

Newton’s Laws of Motion

From Young and Freedman’s University Physics with Modern Physics textbook, 14th edition

Problem 5.11.jpg

Let’s analyze the wording of the problem:

  1. An astronaut inside a rocket that moves vertically upward “An astronaut is inside…rocket…blasting off vertically”. The motion is in 1 dimension (y-axis).
  2. The speed the rocket needs to reach is 331 m/s and as quick as possible but reaching that speed in an acceleration above 4g could make the astronauts blackout.
  3. by reading the word thrust it will definitely mean using Newton’s laws of motion.

What are the given information:

  1. Rocket mass (M) = 2.25×10^6
  2. Initial velocity of rocket (v0)=0 m/s
  3. Final velocity of rocket (vf)=331 m/s
  4. Maximum possible acceleration of rocket = 4g
  5. Maximum possible acceleration of the rocket will give us the quickest blast to 331 m/s without astronauts blackout, then it is constant through out the motion of rocket until it reaches 331 m/s.

The problem needs:

  1. Maximum initial thrust (the force that takes the rocket up).
  2. The force the rocket exert on astronaut (in terms of astronaut’s weight).
  3. The shortest time the rocket can reach the speed of sound.

 

THE SOLUTION:

  1. Maximum initial thrust (the force that takes the rocket up).

Problem 5.11-sol(a).jpg

2.The force the rocket exert on astronaut (in terms of astronaut’s weight).

problem-5-11-solb

3. The shortest time the rocket can reach the speed of sound.

problem-5-11-solc

Newton’s Laws Problem

From Young and Freedman’s University Physics with Modern Physics Textbook, 14th edition

problem-4-28

Let’s analyze the wording of the problem:

  1. “…bullet…strikes a large tree…” and “…assume a constant retarding force…”, gives hints that Newton’s laws would be involved.
  2. “What force… the tree exert on the bullet?” means Newton’s 2nd and 3rd laws will be used.

What information given in this problem:

  1. Speed of bullet = 350 m/s
  2. Penetration depth = 0.13 m = distance traveled
  3. Bullet mass = 1.8 g
  4. 0.22-caliber rifle not an important information for solving the problem.

What the problem need:

  1. The time required for the bullet to stop inside the tree.
  2. The force the tree exert on the bullet.

SOLUTION:

problem-4-28-sol1