Newton’s Laws of Motion

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

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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

 

 

Simple Newton’s laws problem

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

problem-4-2

Find the hints of type of problem:

  1.  “… producing the force vectors…” the two words “force vector” hints using Newton’s law.
  2. “…find the magnitude and direction of the resultant…” then Newton’s 2nd law would be used.

The given information:

  1. Force vector 1 magnitude = 985 N and direction 31° with the positive x-axis.
  2. Force vector 2 magnitude =788 N and direction 32° with the positive y-axis (it means 90°+32° = 122° from positive x-axis counter clockwise).
  3. Force vector 3 magnitude = 411 N and direction 53° with the negative x-axis (it means 180°+53° = 233° from positive x-axis counter clockwise).

What the problem need:

  1. The x- and y-axis components of each force vector.
  2. The resultant force vector (magnitude and direction).

SOLUTION:

  1. The x- and y-axis components of each force vector are:

problem-4-2-sol1

2. The resultant force vector (magnitude and direction)

problem-4-2-sol2

From looking at the vector of the resultant force equation, the vector is at the first quarter and that makes the angle in the first quarter from the positive x-axis counter clockwise.

Newton’s Laws Problem

From past exam (Kuwait University), Summer 2016 2nd midterm.

summer-2016-2nd-midterm-sp10-1

The wording of the problem:

  1. Objects with masses that are in physical contact (…Two small blocks…connected on a large block…”, a hint that it might be a problem involves Newton’s laws.
  2. “…Horizontal force F…”, means Newton’s laws would be used.
  3. “…small blocks remain stationary relative to M?”, Would mean that Newton’s 3rd law would be used.

The given information:

  1. Small blocks masses (m1 and m2) = 1 kg
  2. The large block mass (M) = 10 kg
  3. m1 and m2 connected by massless robe through a massless and frictionless pulley.
  4. Block M on wheels and its surface is frictionless (m1 and m2 not affected by frictional force).

The needed information:

  1. What could be the force F (shown above) if we do not want m1 and m2 to move at all?

SOLUTION:

The forces of interest shown in the figure below:

summer-2016-2nd-midterm-sp10-sol1

 

summer-2016-2nd-midterm-sp10-sol2

 

Dynamics of Circular Motion Problem

Source: Past exam (Kuwait University) 2nd Midterm, Fall Semester 2015

A pilot of mass m=60 kg in a jet aircraft executes a loop-the-loop, as shown in the figure. The aircraft moves in a vertical circle of radius 800 m at a constant speed of 200 m/s. Determine the force (in N) exerted by the seat on the pilot at the top of the loop.

fall-2015-2nd-mitterm-sp7-1

Let’s analyze the paragraph above:

  1. Aircraft doing a vertical circular loop “…aircraft executes a loop-the loop,…. moves in a vertical circle…”. Then the problem is a circular motion problem. Knowledge about radial acceleration should be present.
  2. “…The force exerted by the seat on the pilot…” means it is a normal force. Normal force commonly does not present alone.. we should think about other forces that are present as well.

What are the given information:

  1. Pilot mass (m) = 60 kg
  2. The circular loop radius (R) = 800 m
  3. The speed of the plane (v) = 200 m/s
  4. Tangential acceleration (at)=0 m/s2 (clue: the two words “constant speed”).

What the problem need:

  1. The normal force on the pilot at the top (shown in figure above)

 

SOLUTION:

The pilot diagram at top of the loop:

fall-2015-2nd-mitterm-sp7-sol1

The normal force (n) directed from seat to pilot, hence when the aircraft at top of loop and pilot is up side down the normal force is directed downward.

The pilot weight (Fw) directed from pilot toward the ground, hence it is directed downward.

From the below image, with a constant speed, the total acceleration rises from the circular motion is the radial acceleration (ar).

fall-2015-2nd-mitterm-sp7-sol2

fall-2015-2nd-mitterm-sp7-sol3

 

 

 

Projectile Problem

From Young and Freedman’s University Physics with modern physics, 14th edition

problem-3-19

problem-3-19-2

 

Lets analyze the wording of the problem:

  1. A coin will be thrown into the air to fall into a dish “…you toss a quarter into a dish…”.. then it is a projectile problem.
  2. The coin should land into a dish that is in a higher height than were the coin tossed is. We can take the height of the tossed coin as y1=0 m (because the height of dish needed is from where the coin left the hand) and the height of the dish as y2=h.

What are the given information:

  1. initial velocity (Vo) = 6.4 m/s.
  2. Angle of projected coin = 60°
  3. Δx=2.1 m
  4. The height where the coin (y1) assumed to be 0 m

What the problem need:

  1. The height where the dish is (y2=h).
  2. The vertical component of the final velocity.

SOLUTION:

  1. The projected coin path could be seen in the below diagram:

problem 3.19 sol1.jpg

problem 3.19 sol2.jpg

 

 

Relative velocity problem

A problem from past exam of physics 1 (Kuwait University)
First midterm, First semester (Fall semester) 2001

Two ships A and B, leave port at the same time. Ship A travels due east with a speed of 20 km/h with respect to the earth and ship B travels 48 degrees north of east with a speed of 30 km/h with respect to the earth. Determine the velocity of A relative to B.

 

lets analyze the wording of the problem:

  1. By looking at  ” velocity of A relative to B” and “Ship A travels …. with respect to the earth” … then it is a relative velocity problem

The given information in problem:

  1. Ship A travels to east with speed 20 km/h relative to ground (earth).
  2. Ship B travels to the direction of north of east by a 48° with a velocity of 30 km/h relative to the ground (earth).

What the problem need:

  1. The velocity of A relative to B

SOLUTION:

The diagram below shows the directions of ship A and ship B.

relative-velocity-1st-2001

Ship A velocity is 20 km/h in the direction of positive x-axis with an angle zero degree with the x-axis that is why it is shown as a horizontal velocity. The only component of the velocity is in the direction of positive x-axis.

Ship B has a direction that is 48° from the positive x-axis, hence this ship’s velocity has a component in the x-axis and a component in the y-axis.

relative-velocity-1st-2001-sol