# MidtermI 11012010 Solution

July 15, 2017 | Author: Rolf Wang | Category: Force, Friction, Collision, Mass, Velocity

#### Description

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. b. c. d. e.

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

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. b. c. d. e. ____

5.6 N 6.0 N 6.5 N 5.1 N 2.2 N

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

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

12 J 12 J 20 J 20 J 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 a. 0.

Physic (I) Midterm I 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

(2010/11/01)

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 vi. Beginning at t = 0, when the object passes position x = 0, it experiences a net resistive force proportional to the square of its speed: Fnet  mkv 2ˆi , 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. (c) Find the object’s velocity as a function of position. Solution: (a)

Fnet   mkv 2  m

dv dt

1 dv   kdt v2 v (t ) 1 t  dv   k  dt 2 vi 0 v 1 1    kt vi v(t ) 

 v(t ) 

vi vi  kt

(b) v

i dx  dt 1  i kt

dt 1  kdt 0 dx  0 i 1  i kt  k 0 1 i i kt x

t

t

x0  x

x0 

t 1 ln 1  i kt  0 k

1 ln 1  i kt   ln 1 k

x

1 ln 1  i kt  k

(c) We have ln(1 + υ0kt) = kx

(5 points) (5points)

1  0 kt  ekx so  

i   kxi  i e kx   1  i 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  U ( x )  4        ,  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., dU ( x) 0 dx 12 6  12 12 6 6  dU ( x) d        =>  4         4   13  7  =0 dx dx  x   x   x   x

 12 12 6 6  4   7   0  xeq  (2)1/6  13 x xeq   eq

(b) d 2U ( xeq ) dx 2 

 12(13) 12 6(7) 6   4    x14 x8  x  x  eq

4  42 12 156 6  [  ]  270 8  0 2 12 6  4 2 

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, F1 

dp d (m ) dm d   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,

dm d  0 and m 0 dt dt

Since the mass per unit length is uniform, we can express each link of length dx as having a mass dm: dm 

M 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.

F1  

dm  M  dx  M      dt  L  dt  L

 2  

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

 2  2gx (from kinematics), hence F1 

2Mgx L

The links already on the table have a total length x, and their weight is supported by a force F2: F2 

Mgx L

Hence, the total force on the chain is

4. In polar coordinate system the position vector is described by r  rrˆ , where rˆ is the unit vector along the radial direction which can be expressed in terms of Cartesian coordinates as rˆ  cos  iˆ  sin  ˆj (Note:    (t ) is a function of time

t ). d d ˆ (a) Express ˆ in terms of Cartesian coordinates and calculate rˆ and  ; dt dt express your results in terms of polar coordinate rˆ and ˆ .

(b) Apply the results of (b) and calculate v  polar coordinates.

dr dv and a  in terms of dt dt

(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  iˆ  cos ˆj

d ˆ    cos   iˆ  sin   ˆj   (cos  iˆ  sin  ˆj )   rˆ dt

r  cos  iˆ  sin  ˆj d rˆ   sin   iˆ  cos   ˆj   ( sin  iˆ  cos  ˆj )  ˆ dt

(c ) v

d d d r rˆ  r   ˆrr ˆ r r r ˆ   rˆ   dt dt dt

 

d d  ˆr r  rˆ  ˆ r r ˆ r   v  dt dt  2  r ˆr r ˆ  rˆ   rˆ rˆ  r  r  r 2 ˆ r r 2 r ˆ  a

 

r

ˆr ˆ r d ˆr  d t

5. An object of mass m1 hangs from a string that passes over a very light fixed pulley P1 as shown in Figure below. The string connects to a second very light pulley P2. A second string passes around this pulley with one end attached to a wall and the other to an object of mass m2 on a frictionless, horizontal table. (a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations? (5points) (b) Find expressions for the tensions in the strings. (5 points) (c) the accelerations a1 and a2 in terms of the masses m1 and m2, and g. (5points)

Solution: (a)

Pulley P2 has acceleration a1. Since m2 moves twice the distance P2 moves in the same time, m2 has twice the acceleration of P2, i.e., a2 = 2a1. (b) From the figure, and using

 F  ma :

m1 g  T1  m1a1

(1)

T2  m2 a2  2m2 a1 T1  2T2  0

(2) (3)

Equation (1) becomes m1g  2T2 = m1a1. This equation combined with Equation (2) yields

T2  m2   2m2    m1 g m2  2  T2

m1m2 2m2  m1 1 2

g

and T2

m1m2

g

m2  m1 1 4

(c) From the values of T2 and T1 we find that a2 

T2 m1 g  m2 2m  1 m 2 1 4

and a1 

m1 g 1 a2  2 4m2  m1

ANS FIG. P5.34