Review for Physics I final (Answers to problems)

 

(The answers to these problems were worked quickly and not checked so may contain errors.  If you find any errors please email me.  The velocity on one problem was missing, it is included below.  We will discuss these problems Monday and Tuesday)

 

            Review for Physics I final                              Fall 2007

 

Final will be 8 AM, Wednesday, December 5.  Students who wish to do so may begin their final exam at 7:00.

 

You may bring to the final one 8” x 11" page which may be written on both sides.  There are no restrictions on what you put on the sheet.

 

Be able to state in complete and comprehensible English sentences the principles of physics as applied to specific problems.  Just writing “conservation of energy” or “vertical motion is independent of horizontal motion” will get you less than half credit on questions of this type.  Your statement should be of the type “The kinetic energy of the pendulum just after the ball strikes it will be equal the potential energy of the pendulum at the top of its swing.  The assumption is that the entire mass of the pendulum is at the position of the ball at the end of the pendulum.”

 

Know what the principles of the conservation of linear momentum, angular momentum and energy tell us about the laws of physics in space and time.

 

Be able to draw clearly labeled diagrams including free body and net force diagrams.

 

Be able to show clearly all the steps in the solution of all problems.

 

Topics for the final exam - You are certain to have at least one problem from the following areas:

Conservation of energy (including the work-energy theorem, rotational motion and heat energy problems with heats of evaporation and fusion)

Conservation of (linear and angular) momentum

Vectors: addition, subtraction

Newton’s laws of motion

Rectilinear motion - velocity, acceleration, f = ma, friction

Elastic and inelastic collisions

Projectile motion

 

Selected problems from the following areas will also be on the final.

Gravity - weightlessness, (orbital motion)

Rotational motion - centripetal force - moment of inertia – (torque - center of gravity)

Vector products (dot product, cross product)

Statics

 

The above list does not include all possible problems but the overwhelming majority of problems will be from these areas.  The tests we had are an indication of what is most important.  Be sure you are able to work and completely understand all the problems from our tests.

 

Practice problems:  To benefit from these problems, show your statement of the principle and your solution to Dr. Gault.  Answers are available upon request.

 

A 220 lb man is sitting beside his 140 lb girlfriend in her Ferrari holding a protractor that has a string with a rock tied to it.  When the car accelerates, he notices that the string with the rock makes an angle of 35 degrees with the vertical.  How many seconds does it take to go from 0 to 60 mph? Draw a figure, State the principle, Solve the problem. 60 mph = ? m/s  (show how to convert)

There are several ways to describe the principle on this problem.  The two forces acting on the rock (gravity + string) must produce a net force in the horizontal direction.

 

 

 

 

 


35º

 

mg

 
 

               

 

The acceleration is 6.86 m/s2

 

 

v = v0 + at  so t = v/a

 

The time to attain 26.8 m/s is 3.9 seconds.

 

What velocity must the space shuttle have to stay in orbit 100 km above the earth’s surface?  How much should the velocity change to move 100 m further out from the earth?

 Draw a figure, State the principle, Solve the problem

Principle: In order for the space shuttle (ms)  to stay in orbit the force of gravity must supply the centripetal force:

 

 

 

 

 

To remain at a radius of 6,480,000 m, the velocity must be 7845.597 m/s

To remain at a radius of 6,480,100 m, the velocity must be 7845.537 m/s

 The shuttle needs to slow down .06 m/s to move 100 m further from the Earth.

 

A 500 kg car is traveling west at 30 m/s when struck by a drunken driver in a 1000 kg pickup traveling 50 m/s, 25 degrees north of east.  The two cars stick together after the collision. What is the velocity of the vehicles after the collision?  Is this collision elastic or inelastic?  How can you tell for sure? Draw a figure, State the principle, Solve the problem

 

Final velocity of the truck and car stuck together is  24.6 m/s at an angle of 34.9 º.

 

Explain what would have happened (and sketch the results) if the above collision would have been elastic.

 

If the collision had been elastic, the vehicles would have bounced off each other with no damage to either and no loss of energy.  The sum of the momentum after the collision would have been the same as the above result.

 

 

 

 

 

 

 

 

 


A 200 g block of iron with a temperature of 25ºC slides down a 30 degree incline that is 5 m long and has a coefficient of friction of 0.2.  How fast is the block going when it reaches the bottom?  Assuming that all the heat energy goes into the iron block, what is the temperature of the block at the bottom of the incline?  Draw free body and net force diagrams, State the principle, solve the problem.

Principle: The potential energy of the block at the top of the incline is changed into K.E. and heat energy (due to the friction).

 


5m

 
The height of the block at the beginning is 2.5m.

 


30°

 

FN

 

 
P.E. = K.E. + friction work  (work = force·distance)

 


mgh = ½ mv2+5m· μFN = ½ mv2+ 5m·μmgcos30°

 

where FN = mg cos 30°

 

v2 = 49 – 17

 

v = 5.66 m/s

 

The velocity of the block at the bottom of the incline is 5.66 m/s.

 

The energy lost to friction is 5m·μmgcos30° = 1.7 joules = mcFeΔT

The temperature of the block at the bottom of the incline (CFe = 450 j/kg-ºC)

 

so ΔT is 25.00188 ºC

 

After the block reaches the level surface, the coefficient of friction remains the same.  How far will the block slide before coming to a stop?  State the principle, solve the problem.

 

Principle:  The K.E. of the block at the bottom of the incline will do work against friction.

 

½ mv2 = F·d where the force is μmg.

 

The block will travel 8.17 m before coming to a stop.

 

A bow has a 300 newton pull which is constant throughout the drawback of the arrow.  How far would one draw back the string to shoot a 50 gram arrow 200 m straight up into the air?  If the arrow were aluminum and were shot into a tree with the same drawback; assuming half the energy ended up in the arrow and that its original temperature was 20 Celsius, what would the final temperature of the arrow be? 

 

The bow must be drawn back 32.6 cm to shoot the arrow 200 m high.

 

If the arrow is shot into the tree and half the energy goes into the arrow, its temperature will be 21.09 ºC  (The specific heat of aluminum is 900 j/(g -ºC)

 

The arrow in the above problem is shot at an angle of 20 degrees with the vertical from the top a 50 m building.  How far will the arrow go?  Write the principle necessary to solve the problem. Solve the problem.

 

Initial velocity of the arrow is 62.6 m/s2

Time to top of path is 6 sec.  Max height is 226 m.  Total flight time is 12.79 s, the distance traveled is 273.8 m.

 

A comet is traveling v0 when it is at its closest to the sun (3 x 107 km).  How fast will it be going when it is at its furthest point from the sun (1.5 x 1010 km)?  Write the principle.  Solve the problem.

Principle: The angular momentum of the comet is the same at all times, including when it is close to the sun and far form the sun.

 

mr1v1 = mr2v2

 

v = 0.002 v0

 

A rifle fires a 100 gram bullet with a temperature of 100º with a velocity of 100 m/s into 1.5 kg of water that is also at 100º. How much of the water is changed to steam?  Write the principle and solve the problem.

 

0.2216 grams of water will be changed to steam.

 

A 100 g weight will stretch a spring 15 cm.  Write the equation of motion for a 20 g weight placed on the spring and displaced 2 cm.  Also find the frequency and the period of the oscillations of the 20 g weight.

 

x = 2 Cos(18.1 t) cm or x  = 0.02 Cos(18.1 t) m.  k = 6.5 N/m, f = 2.88/s, period = 0.348 seconds