MidtermI 11012010 Solution

Physic (I) Midterm I (2010/11/01) Total:110 points

Part. I : Multiple Choice (3 points per question; subtotal 45 points) Identify the choice that best completes the statement or answers the question.

____ 1. A particle moving along the x axis has a position given by x = 54t  2.0t3 m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?

a. The particle remains at rest after t = 3.0 s. b. The particle no longer accelerates after t = 3.0 s.

c. The particle can be found at positions x < 0 m only when t < 0 s. d. All of the above are correct.

e. None of the above is correct.

____ 2. A stunt pilot performs a circular dive of radius 800 m. At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s2. What acceleration does the pilot have at point B?

a. (50i + 20j) m/s2 b. (20i  50j) m/s2 c. (20i + 50j) m/s2 d. (20i + 50j) m/s2 e. (50i + 20j) m/s2

____ 3. When the vector sum of three co-planar forces, , and , is parallel to , we can conclude that and

a. must sum to zero.

b. must be equal and opposite.

c. must have equal and opposite components perpendicular to . d. must have equal and opposite components parallel to .

e. must have equal and opposite components parallel and perpendicular to .

____ 4. A 0.30-kg mass attached to the end of a string swings in a vertical circle (R = 1.6 m), as shown. At an instant when  = 50, the tension in the string is 8.0 N. What is the magnitude of the resultant force on the mass at this instant?

a. 5.6 N b. 6.0 N c. 6.5 N d. 5.1 N e. 2.2 N

____ 5. An object moving along the x axis is acted upon by a force Fx that varies with position as shown. How much work is done by this force as the object moves from x = 2 m to x = 8 m?

a. 10 J b. +10 J c. +30 J d. 30 J e. +40 J

____ 6. A 10-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 80 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block when it is 13 cm from its equilibrium position?

a. 0.85 m/s b. 0.89 m/s c. 0.77 m/s d. 0.64 m/s e. 0.52 m/s

____ 7. A 0.04-kg ball is thrown from the top of a 30-m tall building (point A) at an unknown angle above the horizontal. As shown in the figure, the ball attains a maximum height of 10 m above the top of the building before striking the ground at point B. If air resistance is negligible, what is the value of the kinetic energy of the ball at B minus the kinetic energy of the ball at A (KB KA)?

a. 12 J b. 12 J c. 20 J d. 20 J e. 32 J

____ 8. A small lead sphere of mass m is hung from a spring of spring constant k. The gravitational potential energy of the system equals zero at the equilibrium position of the spring before the weight is attached. The total mechanical energy of the system when the mass is hanging at rest is

a. kx2. b.  kx2. c. 0. d. + kx2. e. +kx2.

____ 9. Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value k. At the highest point,

a. KA > KB. b. KA = KB. c. KA < KB.

d. The work done by F on A is greater than the work done by F on B. e. The work done by F on A is less than the work done by F on B.

____ 10. The law of conservation of momentum applies to a collision between two bodies since a. they exert equal and opposite forces on each other.

b. they exert forces on each other respectively proportional to their masses. c. they exert forces on each other respectively proportional to their velocities. d. they exert forces on each other respectively inversely proportional to their masses. e. their accelerations are proportional to their masses.

____ 11. A ball of mass mB is released from rest and acquires velocity of magnitude vB before hitting the ground. The ratio of the impulse delivered to the Earth to the impulse delivered to the ball is

b. . c. . d. 1 e. .

____ 12. Two bodies of equal mass m collide and stick together. The quantities that always have equal magnitude for both masses during the collision are

a. their changes in momentum. b. the force each exerts on the other. c. their changes in kinetic energy. d. all of the above.

e. only (a) and (b) above.

____ 13. A constant force is applied to a body that is already moving. The force is directed at an angle of 60 degrees to the direction of the body's velocity. What is most likely to happen is that

a. the body will stop moving.

b. the body will move in the direction of the force.

c. the body's velocity will increase in magnitude but not change direction.

d. the body will gradually change direction more and more toward that of the force while speeding up.

e. the body will first stop moving and then move in the direction of the force.

____ 14. The first of two identical boxes of mass m is sitting on level ground. The second box is sitting on a ramp that makes an angle with the ground. When a force of magnitude F is applied to each box in a direction parallel to the surface it is on, upwards on the box on the ramp, neither box moves. Which statement comparing the friction force on the box on the level, fL, to the friction force on the box on the ramp, fR, is correct?

a. fR = fL. b. fR > fL. c. fR < fL.

d. The coefficient of static friction is needed to determine the correct answer.

e. Depending on the values of the coefficient of static friction, the angle of elevation of the ramp, the mass of the boxes, and the applied force, answers (a), (b), and (c) are each a possible correct answer.

____ 15. When a car goes around a circular curve on a level road without slipping, a. no frictional force is needed because the car simply follows the road.

b. the frictional force of the road on the car increases when the car's speed decreases. c. the frictional force of the road on the car increases when the car's speed increases. d. the frictional force of the road on the car increases when the car moves to the outside of

the curve.

e. there is no net frictional force because the road and the car exert equal and opposite forces on each other.

Physic (I) Midterm I (2010/11/01)

Answer Section

MULTIPLE CHOICE 1. ANS: E 2. ANS: C 3. ANS: C 4. ANS: C 5. ANS: C 6. ANS: A 7. ANS: A 8. ANS: B 9. ANS: A 10. ANS: A 11. ANS: D 12. ANS: E 13. ANS: D 14. ANS: E 15. ANS: C

Part II. (Subtotal 65 points)

1. For t < 0, an object of mass m experiences no force and moves in the positive x

direction with a constant speed v

i

. Beginning at t = 0, when the object passes

position x = 0, it experiences a net resistive force proportional to the square of its

speed:

F

_net

 

mkv

2

ˆ

i , where k is a constant.

(a) Calculate that the speed of the object after t = 0 as a function of timt. (5

points)

(b) Find the position x of the object as a function of time. (5 points)

(c) Find the object’s velocity as a function of position. (5points)

Solution:

(a)

2 2 ( ) 2 ₀

1

1

1

1

( )

( )

i net v t t v i i i

dv

F

mkv

m

dt

dv

kdt

v

dv

k

dt

v

kt

v

v t

v

v t

v

kt

 





 



 

















(b)

1 i i dx v dt kt





   0 0 0 1 1 1 x t t i i i i kdt dt dx kt k kt









   











0 0 1 ln 1 t x i x kt k



 





1 0 ln 1 _i ln 1 x kt k



  _   _





1 ln 1 _i x kt k



 

(c)

We have ln(1 + υ

0

kt) = kx

0 1



ktekx

so

1 kx i i i kx i e kt e













     

2. The potential energy associated with the force between two neutral atoms in

a molecule can be modeled by the Lennard–Jones potential energy function:

12 6

( )

4

U x

x

x







_{ }

_{ }



 



_{ }



_{ }



 

 









,

where x is the separation of the atoms,



>0 and



>0. The function U(x) contains

two parameters



and



that are determined from experiments.

(a) Find the equilibrium position in terms of



and



. (5 point)

(b) Determine if the equilibrium position is stable or unstable? (5 point)

Solution:

(a)

Equilibrium position exists for a separation distance at which the potential

energy of the system of two atoms (the molecule) is a minimum, i.e.,

( ) 0 dU x dx 

=>

12 6 _{12 6} 13 7

( )

12

6

4

4

dU x

d

dx

dx

x

x

x

x











_{ }

_{ }



_

_

_

 



_{ }



_{ }



 

_



_

 

 













=0

12 6 1/6 eq 13 7 eq eq 12 6 4 0 x (2) x x





_

_  _  _      

(b)

2 12 6 2 14 8 12 6 2 12 6 8 ( ) _{12( 13) 6( 7)} 4 4 42 156 [ ] 270 0 4 2 eq eq x x d U x dx x x

















      _  _         

3. A chain of length L and total mass M is released from rest with its lower end just

touching the top of a table as shown in Figure below. Find the force exerted by the

table on the chain after the chain has fallen through a distance x as shown in

Figure below. (Assume each link comes to rest the instant it reaches the table.)

(10 points)

Solution:

The force exerted by the table is equal to the change in

momentum of each of the links in the

chain.

By the calculus chain rule of derivatives,

1 ( ) dp d m dm d F m dt dt dt dt



_



   

We choose to account for the change in momentum of each link by

having it pass from out area of interest just before it hits the table, so

that,

0 dm dt





and

md 0 dt



_

Since the mass per unit length is uniform, we can express each link of

length dx as having a mass dm:

M

dm dx

L



The magnitude of the force on the falling chain is the force that will be

necessary to stop each of the elements dm.

2 1 dm M dx M F dt L dt L





   



  _{ } _{ }    

After falling a distance x, the square of the velocity of each link

2

2gx





(from kinematics), hence

1

2Mgx

F L



The links already on the table have a total length x, and their weight is

supported by a force F

2

:

2 Mgx F L 

Hence, the total force on the chain is

4. In polar coordinate system the position vector is described by

r rrˆ

, where

ˆr

is the unit vector along the radial direction which can be expressed in terms of

Cartesian coordinates as

rˆ cos



iˆsin



jˆ

(Note:

 

 ( )t

is a function of time

t

).

(a) Express ˆ



in terms of Cartesian coordinates and calculate

d rˆ

dt

and

ˆ

d dt



;

express your results in terms of polar coordinate

ˆr

and ˆ



.

(b) Apply the results of (b) and calculate

v dr

dt



and

a dv

dt



in terms of

polar coordinates. (5 points)

(5 points)

(c) When the system is under circular (but not uniform) motion, what are v

and a ? Explain the physics of your results. (5 points)

ANS.

(a)

ˆ _{sin i}ˆ _cos ˆ_j

ˆ _cos ˆ _sin ˆ _(cos ˆ _sin ˆ_{) ˆ} d i j i j r dt



 

 

 

 

  









 



ˆ

ˆ

cos

sin

r





i





j

ˆ ˆ ˆ ˆ ˆ

ˆ sin cos ( sin cos )

d r i j i j dt  

 

 

 

  













(c )

v d

 

r d

 

ˆr r ˆr r d

 

rˆ rˆ r rˆ r d t d t d t



     

 



 





 



2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 d d d a v r r r r r r r r r d t d t d t r r r r r r r r r r r r

 

 

 

 





 

 

 







 

       _   _           

5. An object of mass m

1

hangs from a string that passes over a very light fixed pulley

P

1

as shown in Figure below. The string connects to a second very light pulley P

2

.

A second string passes around this pulley with one end attached to a wall and the

other to an object of mass m

2

on a frictionless, horizontal table.

(a) If a

1

and a

2

are the accelerations of m

1

and m

2

, respectively, what is the

relation between these accelerations? (5points)

(b) Find expressions for the tensions in the strings. (5 points)

Solution:

(a)

Pulley P

2

has acceleration a

1

.

Since m

2

moves twice the distance P

2

moves in the same time,

m

2

has twice the acceleration of P

2

, i.e., a

2

= 2a

1

.

(b) From the figure, and using

1 1 1 1 2 2 2 2 1 1 2

:

(1)

2

(2)

2

0

(3)

F

ma

m g T

m a

T

m a

m a

T

T



 











Equation (1) becomes m

1

g



2T

2

= m

1

a

1

. This equation

combined with Equation (2) yields

2 2 2 1 2

2

2

T

m

m

m g

m



_



_









1 2 2 2 1 1 2 2  m m T g m m

and

1 2 2 2 1 1 4  m m T g m m

(c) From the values of T

2

and T

1

we find that

2 1 2 2 2 1 1 4 2    T m g a m m m

and

1 1 2 2 1 1 2 4 m g a a m m   

ANS FIG. P5.34