Physics 214

Review Test 2

 

Here is a sampling of the type of problems that will appear on the test.  I will

post answers later.  Remember, the test is next Tuesday, the 27th.

 

1)  A baseball is struck by Sammy Sosa at 140.0 ft/s at 38 degrees above the horizontal

from a point 3 feet above home plate.

Will the ball clear the left field wall that is 8 feet high, and located 395 feet from

home plate?

 

A: Two dimensional motion problem that boils down to finding y, when x = 395 feet.

 

Use:

 

First, find t when x = 395, with initial velocity 140, and angle 38 degrees, you should

get t = 3.6 sec.  Plug this into the second equation with g = 32 ft/s/s to get y = 106 feet.

This is an upper deck home run.

 

 

2) A particle travels in a circular path of circumference .2 m, and makes 100 revolutions

per second.  What is its centripetal acceleration.

 

A:  Use    with r = .2 m.

 

To find v you need to convert from rev/sec to m/sec:

 

So, a = 79000 m/s/s

 

 

3) When you twirl a weight tied to the end of a rope you experience a force pulling

away from you.  What is the name of this force and what creates it?

 

A: This is what is commonly referred to as “centrifugal” force.  It is the force that

opposes the centripetal force that the weight experiences from the rope changing

its direction (and, hence, its velocity resulting in an acceleration).  This opposite

force is predicted by Newton’s third law.

 

 

4) A raft is floating west at 4 km/hour relative to land.  To a person on the raft, a hang

glider appears to move away at 8 km/hour at 40 degrees S of W.  What is the glider’s

velocity relative to land?

A: The easiest way to solve this reference frame problem is to change all velocity

vectors to unit vector notation.

 

VRG = velocity of the raft with respect to the ground = 4i km/hour (west is pos. x-dir)

 

VHR  = velocity of the hang glider with respect to the raft = 8cos(40)i – 8sin(40)j

= 6.1i – 5.1j.

 

We want VHG  = velocity of the hang glider with respect to the ground.

 

 

                        G            R

 

                                          VHR

 


                               VRG           VHG

                                               

 

 

 


Looking at the reference frames, we want VHG = VRG  + VHR  = 10.1i – 5.1j

 

Or, as components: 11.3 km/hour 26.8 degrees S of W.

 

 

5)   A 10 kg picture is suspended with two wires, as shown.  Find the tension in each

wire.

 


                                                                                    T1                    T2

                  480             550

 

 

 


                                                                                                                                                                                                                                                            W

 

 

A: From the free body diagram and using the standard x-y coordinates:

 

 

 

Two equations, two unknowns gives: T1 = 57.5 N, T2 = 67.6 N

 

 

6)  A particle of mass .5 kg has an initial velocity of 2ij m/s.  It is acted on by a force

3i + 3j for 4 sec.  What is its final velocity?

 

 

A:  From kinematics

 

We know the initial velocity and the time; we need the acceleration.  Use F = ma.

 

a = (3i + 3j)/.5 = 6i + 6j

 

v = 2i j + (6i + 6j)(4) = 26i + 23j

 

 

 

 

7)   Two blocks with masses m1 = 3 kg, and m2 = 1 kg are connected

by a massless rope, and slide on a frictionless surface as in the figure.  Find

the acceleration of the system and the tension in the rope.

 


                                                                                                x

                                   

                        m1

     m2                                                                                                600

 

                                                                                                            W

                                          300                                                      

 

 

 A: From m2:    I made the pos-y direction downward.

 

From m1:   pos-x to the left. 

 

Two equations with two unknowns.  T = 11 N      a = -1.2 m/s/s  it is moving down the

ramp

 

 

 

 

 

 

 

 

 

 

8)   A 2 kg hammer is set on a roof for which = .34.  The roof has a pitch of

40 degrees.  Will the hammer stay on the roof or slide off?  What would happen

if a 10 kg tool box was placed on the roof?

 

 


                                                            N         f

 

 

                                    x                      50   W

 

 

 

A:

 

 

 

We wish to know if the static frictional force is less than the component

of the weight pulling on the hammer. If it is, the hammer will stay in place.

 

That is, is

 

Well,   so examine: 

 

The force from gravity overcomes the friction so it will slide.  So will

a tool box if it has the same coefficient of static friction.  Notice that

mass, or weight, really has nothing to do with the result. It only depends

upon  the coefficient of static friction and the angle of the roof.

 

 

 

 

 

 

 

 

9)  A 3 kg block is acted on by a 25 N force that acts 30 degrees below the horizontal

as shown.  Take the coefficient of static friction to be .47, and kinetic friction to be .2.

Does the block move if it is initially at rest?  If it moves to the right, what is its

acceleration?

 


                                    30

 

 

A:

 

At rest, is the static friction greater than the force?

 

 

So, it will move.

 

Since it moves, we now use the kinetic friction to find:

 

 and solve to get: a = 5.3 m/s/s

 

 

10) Describe the concept of apparent weight.

 

A: An upward force acting on a body, such as an elevator going up, will add to the

normal force and from Newton’s third law appear to increase the body’s weight.

The reverse occurs when the force is decreased, for instance, when an elevator

accelerates downward.