If your solution to stated problem does not match the given solution, you should self-critique per instructions at

 

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 

020.  `query 20 

Question:  `qExplain how we get the components of the resultant of two vectors from the components of the original vectors.

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Given Solution: 

`a** If we add the x components of the original two vectors we get the x component of the resultant. 

 

If we add the y components of the original two vectors we get the y component of the resultant. **

 

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Question:  `qExplain how we get the components of a vector from its angle and magnitude.

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Given Solution: 

`a** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis).

 

To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **

NOTE REGARDING THE CIRCULAR DEFINITION OF TRIGONOMETRIC FUNCTIONS VS. THE RIGHT-TRIANGLE DEFINITIONS: 

Students with a background in trigonometry often use the right-triangle definitions of sine and cosine (sine and cosine defined in terms of opposite and adjacent sides and hypotenuse), as opposed to the circular definition (using a coordinate system, with angles measured counterclockwise from the positive x axis--the definition used in this course).

The two definitions are pretty much equivalent and completely consistent.  The circular definition is a bit more general for two reasons:

In most applications it is your choice which definition you use. Some applications are easier if you use the right-triangle definition, others are easier of you use the circular definition, and some simply require the circular definition.

In developing this course I chose to express all trigonometric solutions in terms of one of the definitions, in order to avoid confusion for students with a weak background in trigonometry. If only one of the definitions is to be used, it must be the more general circular definition with its four simple rules

The circular definition is sufficient for Principles of Physics or General College Physics..

However General College Physics students are to have completed a year of precalculus or equivalent, which includes trigonometry, and it is expected that these students can reconcile the circular and right-triangle definitions and approaches, and understand the right-angle trigonometry in their text.

University Physics students are of course expected to already be familiar with trigonometry and the use of vectors (though in reality some refreshing is usually required, and is provided in the first chapter of the University Physics text).  However students at the level of University Physics should encounter no serious obstacle with the trigonometry.

In a nutshell, here is a summary of how the right-triangle definitions are related to the circular definitions:

On a circle of radius r centered at the origin, any first- or second-quadrant angle gives us a triangle in the upper half-plane having that base angle.

Thus

The definitions of the cosecant, secant and cotangent functions are then made in the usual manner.

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Question `prin `gen (Optional Openstax):  A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg.  Do not calculate the acceleration; use the techniques of the present chapter.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

It would be possible to calculate the average acceleration of the car and multiply this by the mass of the driver.  It is even recommended that you do so in order to verify the solution you obtain using present techniques.

The solution appropriate to the present chapter applies the impulse-momentum theorem, which states that the impulse of the net force acting on an object is equal to the change in the momentum of that object.

The impulse is F_ave `dt, the change in momentum is the change in the quantity m * v.

The object will be the 70 kg driver, whose initial momentum is 70 kg * 10 m/s = 700 kg m/s and whose final momentum is zero.

`dt is the .26 s time interval.

Solving

F_net * `dt = `d( m v )

for F_net we get

F_net = `d( m v ) / `dt = ( 0 kg m/s - 700 kg m/s) / (.26 s) = 2700 kg m/s^2 = 2700 Newtons.

This force is almost 4 times the weight of the person. 

A force of this magnitude could cause some bruising, but would be unlikely to cause any injury to a healthy person.  However note that 10 m/s is only about 23 miles/hour, so this is a relatively low-speed collision.  Forces in such a collision go up in proportion to the square of the speed; if the speed is doubled the forces would be about 4 times as great, and some degree injury from the seat belt becomes more likely.  However, a minor injury from the seat belt is much preferable to the injury that would result from being stopped by the steering wheel or from exiting the car through the windshield.

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Question:  `qprin phy, gen phy 7.02.  Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Given Solution: 

`aIf the direction of the velocity is taken to be positive, then the directio of the frictional force is negative.  A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. 

 

By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec.

 

The change in momentum is m * `dv, so we have

 

m `dv = impulse and

 

`dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s.

 

 

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Question:  `qgen phy  #7.12  23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Given Solution: 

`a**STUDENT SOLUTION:  Momentum conservation gives us

 

m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have

 

(.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v).  Solving for v:

 

(5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) 

 

.78kg m/s = 2kg * v

 

v = 1.38 kg m/s / (2 kg) = .69 m/s.

 

INSTRUCTOR COMMENT: 

 

It's probably easier to solve for the variable v2 ': 

 

Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get

 

m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get

 

v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2.  

 

Substituting for m1, v1, m2, v2 we will get the result you obtained.**

 

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Question `gen, `prin (Optional Openstax): 

A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board.

(a) Calculate the duration of the impact.

(b) What was the average force exerted on the nail?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

We can calculate the hammer's initial and final momentum, knowing its mass and its initial and final velocities.  If we can calculate the duration of the impact, then, we will be able to find the average force from the impulse-momentum theorem

F_net_ave * `dt = `d( m v ).

The force resisting the nail's penetration of the wood could be expected to increase as the nail penetrates further and further into the wood.  So we wouldn't expect the acceleration to be uniform.

However, to deal with a nonuniform acceleration would require calculus, so we will make the assumption that the acceleration is uniform.

The hammer's velocity decreases from 7 m/s to 0 over a displacement of 1 cm.  Its average velocity is 3.5 m/s, so to travel 1 cm the time required is

`dt = 1 cm / (3.5 m/s^2) = .01 m / (3.5 m/s^2) = .003 seconds, very approximately.

The hammer's initial momentum is .45 kg * 7 m/s = 3.2 kg m/s, approx., and (assuming that the hammer comes to rest while in contact with the nail) its final momentum is zero.  So its change in momentum is -about 3.2 kg m/s.

Solving F_net_ave * `dt = `d ( m v ) for F_net_ave we get

F_net_ave = `d( m v ) / `dt = -3.2 kg m/s / (.003 s) = -1000 kg m/s^2 = -1000 Newtons, approximately.

This is the average force acting on the hammer.  This force is exerted by the nail, and by Newton's Third Law is equal and opposite to the force exerted by the nail on the hammer.

The force exerted on the hammer is thus -1000 Newtons, a force of 1000 Newtons in the direction opposite its motion.  The force exerted by the hammer on the nail is equal and opposite to this, or 1000 Newtons.

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Question:  `q****    Univ. 8.75 (11th edition  8.70) (8.68 10th edition).  8 g bullet into .992 kg block, compresses spring 15 cm.  .75 N force compresses .25 cm.  Vel of block just after impact, vel of bullet?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a** The spring ideally obeys Hook's Law F = k x.  It follows that k = .75 N / .25 cm = 3 N / cm.

 

At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J.

 

The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. 

 

The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet.  Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s.  So we have

 

mBullet *  vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. **

 

 

 

 

 

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