28) In the figure on the right, a 4.0 kg ball is attached to the end of a 1.6 m rope, which is fixed at O. The ball is held at A, with the rope horizontal, and is given an initial downward velocity. The ball moves through three quarters of a circle and arrives at B, with the rope barely under tension. The initial velocity of the ball, at A, is closest to: A) 6.9 m/s B) 4.0 m/s C) 7.9 m/s D) 5.6 m/s E) 6.3 m/s

Solution The motion of the ball is always perpendicular to the tension in the rope. Therefore, the tension does no work on the system and the total mechanical energy of the ball is conserved. EA = EB

→ mgR +12mvA

2 = mg(2R) + 12mvB

2 → vA2 = 2gR + vB

2

Newton’s 2nd law for the ball at point B:

mg + T = m vB2

R T ≈ 0 → vB = gR → vA = 3gR = 3(9.8m/s2 )(1.6m) = 6.9m/s

29) Two objects, one of mass m and the other of mass 2m, are dropped from the top of a

building. When they hit the ground:

A) the heavier one will have four times the kinetic energy of the lighter one B) both will have the same kinetic energy C) the heavier one will have one-fourth the kinetic energy of the lighter one D) the heavier one will have twice the kinetic energy of the lighter one E) the heavier one will have half the kinetic energy of the lighter one

Solution

Ki +Ui = K f +U f → K f =Ui = mgh Same h→ double the mass will double K f

30) In the figure above, a 1.82 kg block is held in place against the spring by a 83 N horizontal external force. The external force is removed, and the block has a velocity

v1 = 1.2 m/s after separation from the spring. The block descends a ramp and has a

velocity v2 = 1.8 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction is 0.32. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops. The force constant of the spring is closest to

A) 1170 N/m B) 2100 N/m C) 760 N/m D) 2630 N/m E) 1310 N/m Solution Initially, the system has no kinetic energy. When the external force is removed, elastic energy stored in the spring is converted to kinetic energy. Therefore, 12mvA

2 =12kx2

The initial compression of the spring can be obtained from the external force used to keep the block stationary:

Fext = kx→12mv1

2 =12F2 k→ k = F2 (mv1

2 ) = (83N)2 / (1.82kg × (1.2m/s)2 ) = 2630N/m

31) The potential energy function for a particle as a function of position x is shown in the

graph below.

Rank the x component of the force (including signs!) acting on the particle associated with this potential energy function at points x = –2 m, x = –1 m and x = 5 m A) Fx (–2 m) < Fx (–1 m) < Fx (5 m) B) Fx (–1 m) < Fx (–2 m) < Fx (5 m) C) Fx (–1 m) < Fx (5 m) < Fx (–2 m) D) Fx (5 m) < Fx (–2 m) < Fx (–1 m) E) Fx (5 m) < Fx (–1 m) < Fx (–2 m) Solution The force is the negative of the slope of the potential energy function F = −dU/dx. At x = −2m, Fx > 0; x = −1m, Fx = 0; x = +5m, Fx < 0.

32) After a completely inelastic collision between two objects of equal mass, each having initial speed the two move off together with speed What was the angle between their initial directions? A) B) C) D) 112 E) 141 Solution Call the final direction of the joined objects the positive x axis. A diagram of the collision is shown. Momentum will be conserved in both the x and y directions. Note that vA

= vB= v and ′v = v 3 .

py

: − mv sinθ1+ mv sinθ

2= 0 → sinθ

1= sinθ

2 → θ

1= θ

2

px

: mv cosθ1+ mv cosθ

2= 2m( ) v 3( ) → cosθ

1+ cosθ

2= 2

3

cosθ1+ cosθ

2= 2cosθ

1= 2

3 → θ

1= cos−1 1

3= 70.5o = θ

2

θ1+ θ

2= 141o

33) Determine the character of the

collision sketched in the figure. The masses of the blocks, and the initial and final velocities of the blocks are given in the figure. The collision is

A) elastic B) characterized by an increase in kinetic energy C) perfectly inelastic D) inelastic E) not possible because momentum is not conserved Solution Check that momentum is conserved: Check that collision is elastic by checking that the relative velocity of approach for the two blocks before the collision is equal to their velocity of separation after the collision: (1.8m/s) – (−0.2m/s) = (1.4m/s) – (−0.6m/s) Can double-check answer by kinetic energy calculation:

12 (4 kg)(1.8 m/s)2 + 1

2 (6 kg)(−0.2 m/s)2 = 12 (4 kg)(−0.6 m/s)2 + 1

2 (6 kg)(1.4 m/s)2

θ2  θ1  

(4 kg)(1.8 m/s) + (6 kg)(−0.2 m/s) = (4 kg)(−0.6 m/s) + (6 kg)(1.4 m/s)  

34) A 1.2 kg spring-activated toy bomb slides on smooth surface along the x-axis with a

speed of 0.50 m/s. At the origin O, the bomb explodes into two fragments. Fragment 1 has a mass of 0.4 kg and a speed of 0.9 m/s along the negative y-axis. The magnitude of the velocity of fragment 2 after the explosion is closest to:

A) 0.75 m/s B) 1.2 m/s C) 1.8 m/s D) 0.93 m/s E) 0.87 m/s

Solution: Conservation of momentum in the x- and y- directions:

x : (1.2 kg)(0.5 m/s) = (0.4 kg)(0) + (0.8 kg)v2 cosθ → v2 cosθ = 0.75 m/sy : 0 = (0.4 kg)(−0.9 m/s) + (0.8 kg)v2 sinθ → v2 sinθ = 0.45 m/s→ tanθ = 0.6→θ = 31° and v2 = 0.87 m/s

35) A 55-kg woman and an 80-kg man stand 10.0 m apart on frictionless ice. If each

holds one end of a rope, and the man pulls on the rope so that he moves 2.5 m, how far from the woman will he be now?

A) 2.5m B) 3.9m C) 4.3m D) 5.2m E) 6.1m Solution 1 Measure all distances from the original position of the woman

x

CM=

mW

xW+ m

Mx

M

mW+ m

M

=55 kg( ) 0( ) + 80 kg( ) 10.0 m( )

135 kg= 5.9 m from the woman

Since there is no force external to the man-woman system, the CM will not move, relative to the original position of the woman. The woman’s distance will no longer be 0, and the man’s distance has changed to 7.5 m.

Solution 2

xCM =mW xW + mM xMmW + mM

= const ⇒ mWΔxW + mMΔxM = 0⇒ ΔxW = −mM

mW

ΔxM

Note, the woman is moving in opposite direction to the man! New distance between them is

10.0m − ΔxW + ΔxM( ) = 10.0m − 1+ mM

mW

⎛⎝⎜

⎞⎠⎟ΔxM = 10.0m − 1+ 80kg

55kg⎛⎝⎜

⎞⎠⎟2.5m( ) = 3.9m

36) In the figure below, a 60 cm length of uniform wire, of 60 g mass, is bent into a right

triangle. The x- and y-coordinates of the center of mass, in cm, are closest to: A) (10, 3) B) (8, 5) C) (8, 3) D) (9, 4) E) (10, 5) Solution

(xcm , ycm ) = mi (xi , yi )

i∑ mi

i∑ = 10

60 (0,5) + 2460 (12,0) + 26

60 (12,5) = (10,3)

37) In the figure on the right, two blocks, of masses 2.00 kg and 3.00 kg, are connected

by a light string, which passes over a pulley of moment of inertia 4.00 × 10-3 kgm2 and radius 5.00 cm. The coefficient of kinetic friction for the table top is 0.300. The blocks are released from rest. After the upper block has moved 0.600 m, its speed is closest to:

A) 1.40 m/s B) 1.22 m/s C) 1.95 m/s D) 5.44 m/s E) 3.19 m/s Solution In this problem, gravitational PE is converted to KE of blocks and pulley. There is some loss in mechanical energy in the system due to kinetic friction.

m2gd − µk m1gd = 12 m2v

2 + 12 m1v

2 + 12 I(v R)2

v =2 m2 − µkm1( )gdm2 + m1 + I / R

2 = 1.40m/s

38) The figure on the right shows scale drawings of four objects, each of the same mass

and uniform thickness. Which has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing? In each case the axis passes through point P at the center of the object. (Cases A and C – uniform disk; case B – uniform thin rim with thin, massless spokes; case D - uniform rod)

A) A B) B C) C D) D E) The moment of inertia is the same for all of these objects Solution

Cases A and C: uniform disks I = 12 MR2 RA < RC → I A < IC

Case D: rod about CM I = 13 MR2 → ID < IC

Case B: most of the mass concentrated on the rim I ≈ MR2 → IB > IC

39) Consider the system of point masses shown below. What is the moment of inertia

with respect to the “Axis” shown in the figure? A) 5 kg m2 B) 10 kg m2 C) 12 kg m2 D) 20 kg m2 E) 24 kg m2 Solution The masses along the axis of rotation do not contribute to the moment of inertia. Calculating the moment of inertia of the remaining masses: I = miri

2 = 2(1kg)(2 2)2 m2 + 2(1kg)( 2)2 m2∑ = 20 kg m2

40) A sphere of radius 20.0 cm and mass 1.80 kg starts from rest and rolls without

slipping down a 30.0º incline that is 10.0 m long. What is the ratio of translational to rotational KE (KETrans / KERot ) at the bottom?

A) 1.0 B) 1.5 C) 2.0 D) 2.5 E) 3.0 Solution: The sphere is rolling without slipping, so vCM

= ωR .

KEtrans

KErot

=12

MvCM2

12

ICMω 2

=12

MvCM2

12

25

MR2( ) vCM2

R2

= 2.5

41) A light triangular plate OAB is in a horizontal plane. Three forces, F1 = 7 N, F2 = 3 N, and F3 = 8 N, act on the plate, which is pivoted about a vertical axis through point O. Considering the counterclockwise rotation as positive, the sum of the torques about the vertical axis through point O, acting on the plate due to forces F1, F2, F3, is closest to: A) zero B) −0.30 N m C) 0.30 N m D) 1.2 N m E) −1.2 N m

Solution:

τ = rF⊥ τ∑ = −(0.6 m)(F1 cos30°) + (0.8 m)F2 + (0)F3 = −1.2 N m

42) A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought

uniformly to rest by a frictional torque of 1.20 Nm. If the mass of the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest?

A) 933 B) 745 C) 1866 D) 3732 E) 467 Solution The torque on the rotor will cause an angular acceleration given by α = τ I . The torque and angular acceleration will have the opposite sign of the initial angular velocity because the rotor is being brought to rest. The rotational inertia is that of a solid cylinder. Substitute the expressions for angular acceleration and rotational inertia into the equation ω

2 = ωo2 + 2αθ , and solve for the angular displacement.

ω 2 = ωo2 + 2αθ ⇒ θ =

ω 2 −ωο2

2α=

0 −ωο2

2 τ I( ) =−ω

ο2

2 τ 12

MR2( ) =−MR2ω

ο2

4τ

=− 4.80 kg( ) 0.0710 m( )2

10,300revmin

⎛⎝⎜

⎞⎠⎟

2π rad1 rev

⎛⎝⎜

⎞⎠⎟

1 min60 s

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

2

4 −1.20 Nim( ) = 5865 rad1 rev

2π rad⎛⎝⎜

⎞⎠⎟

= 933 rev

43) In the figure below, a given force F is applied to a rod in several different ways. In

which case is the torque due to F about the pivot P greatest?

A) 5 B) 2 C) 1 D) 4 E) 3

Solution

τ = rF⊥ Case 1 has the largest r and the largest F⊥

44) A small 1 kg ball is attached to the end of a rod 1 m long. If the system balances at a

point on the rod 0.25 m from the end holding the mass, what is the mass of the rod? A) 1/4 kg B) 1/2 kg C) 1 kg D) 2 kg E) 4 kg Solution The total torque about the pivot must be zero! The CM of the rod is at its center, 0.25 m to the right of the pivot. This must balance the ball, which is the same distance to the left of the pivot. Thus, the masses must be the same! Answer is 1 kg. 45) A 2.1-m-radius merry-go-round is rotating freely with an angular velocity of 0.80

rad/s. Its total moment of inertia is 1760 kg ⋅m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. What is the angular velocity of the merry-go-round now?

A) 0.24 rad/s B) 0.48 rad/s C) 0.72 rad/s D) 0.80 rad/s E) 0.90 rad/s Solution The angular momentum of the merry-go-round and people combination will be conserved because there are no external torques on the combination. This situation is a totally inelastic collision, in which the final angular velocity is the same for both the merry-go-round and the people. Subscript 1 represents before the collision, and subscript 2 represents after the collision. The people have no initial angular momentum.

L1= L

2 → I

1ω

1= I

2ω

2 →

ω2= ω

1

I1

I2

= ω1

Im-g-r

Im-g-r

+ Ipeople

= ω1

Im-g-r

Im-g-r

+ 4 Mperson

R2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 0.80rad s( ) 1760kgim2

1760kgim2 + 4 65 kg( ) 2.1 m( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0.48rad s

46) In the figure on the right, a carousel has a radius of 3.0 m and a moment of inertia of 8000 kg m2. The carousel is rotating unpowered and with negligible friction with an angular velocity of 1.2 rad/s. An 80-kg man runs with a velocity of 5.0 m/s, on a line tangent to the rim of the carousel, overtaking it. The man runs onto the carousel and grabs hold of a pole on the rim. The change in angular velocity of the carousel is closest to: A) +0.04 rad/s B) +0.06 rad/s C) −0.07 rad/s D) +0.08 rad/s E) −0.05 rad/s

Solution No external torque. Conservation of angular momentum gives

Iω i + mvr = (I + mr 2 )ω f →ω f = 1.24 rad/s

47) In the figure on the right, a very rapidly spinning bicycle

wheel is suspended from a string attached to the axle. In the subsequent motion, the point P on the axle will

A) remain where it is B) move in a circle in the B direction C) move in a vertical circle in the upward direction D) move in a circle in the D direction E) move in a vertical circle in the downward direction

Solution For the case of a rapidly spinning bicycle wheel, the torque acting on the system controls the motion of the system. The weight of the bicycle wheel is opposed by the upward tension in the string leading to a torque with axis pointing in the B direction. The angular momentum of the bicycle wheel is pointing radially outward by the right hand rule. The external torque acting on the body will rotate the system in the B direction leading to a precession of the bicycle wheel in the B direction.

(Torque in B direction leads to rotation of in B direction)

48) A 100-kg non-uniform beam, 6.0 m long, is

loosely pinned at the pivot at P. A 400 kg block is suspended from the end of the beam at A. The beam forms a 30 o angle with the horizontal, and is supported by a cable, 4.0 m long, between points D and B. Point B is 4.0 m from P, and point D is 4.0 m above P. The center of mass of the beam is at point C, which is 2.0 m from P. The tension in the cable is closest to:

A) 5300 N B) 5600 N C) 5800 N D) 6400 N E) 4900 N Solution The total torque on the beam around the pivot P must be zero. Note that PBD is an equilateral triangle, so the angle PBD is 60 degrees.

∴ (2)(mg cos30°) + (6)( Mg cos30°) = (4)(T cos30°) → T = 6370 N 49) A 15-cm-long tendon was found to stretch 3.7 mm by a force of 13.4 N. The tendon

was approximately round with an average diameter of 8.5 mm. Calculate the Young’s modulus of this tendon.

A) 5.4 × 106 N/m2 B) 5.4 × 107 N/m2 C) 9.6 × 106 N/m2 D) 9.6 × 107 N/m2 E) 9.6 × 107 N/m Solution The Young’s Modulus is the stress divided by the strain.

Young's Modulus =

StressStrain

=F AΔL L

o

=13.4 N( ) π 1

2× 8.5 × 10−3 m( )2⎡

⎣⎢⎤⎦⎥

3.7 × 10−3 m( ) 15 × 10−2 m( ) = 9.6 × 106 N m2

50) The reason an astronaut in an Earth-orbiting satellite feels weightless is that

A) the astronaut’s acceleration is zero B) the astronaut is falling C) the astronaut is beyond the range of the earth’s gravity D) the astronaut is at a point in space where the effects of the moon’s gravity and the

earth’s gravity cancel E) this is a psychological effect associated with rapid motion

Solution The astronaut and the satellite experience Earth’s gravity, which provides the centripetal force for orbiting earth. Moon’s gravity also acts on them, but is much smaller. As the astronaut experiences the same centripetal acceleration as the satellite both move together. In the frame of reference of the satellite there is no force on the astronaut. Therefore, the astronaut is weightless. 51) The Moon does not crash into Earth because: A) the Moon is in Earth’s gravitational field B) the net force on the Moon is zero C) the Moon is beyond the main pull of Earth’s gravity D) the Moon is being pulled by the Sun as well as by Earth E) none of the above Solution The Moon is in Earth’s gravitational field, and the net force on it is not zero. The net force is radial (centripetal) force.

52) Consider an object that drops a distance h in a time of 66 s on the surface of the earth

(neglecting air effects). How long would it take the same object to drop the same distance on the surface of Pluto? The mass of Pluto is 1.1×1022 kg and its radius is 4.0×105 m. The mass of Earth is 5.97×1024 kg, and the radius of Earth is 6.38×106 m.

A) 120 s B) 66 s C) 34 s D) 96 s E) 54 s

Solution

h = 12 gt2 → t = 2h g ∴ tP tE = gE gP

g = GM R2 → tP tE = ( M E M P )(RP RE ) = 1.46 tP = 1.46tE = 1.46 ⋅66s = 96s 53) A newspaper article discussing the space program noted that it is easier to launch a

satellite into an eastward orbit than into a westward one. Is this true? A) Yes, this is true because launching to the east reduces wind resistance. B) No, this is not true because it is the upward component of velocity that is important in

reaching an orbit, not the horizontal component east and west. C) No, this is not true because the kinetic energy of the launch vehicle is independent of

the direction in which it is launched. D) No, this is not true because launching toward the east means an even greater launch

speed is needed to attain orbiting speed due to the earth’s rotation. E) Yes, this is true because the rotation of the earth toward the east gives the satellite

added speed, thereby reducing the speed required with respect to earth if orbital velocity is to be attained.

Solution A satellite needs to be launched with high speed such that its kinetic energy at launch is equal to the difference in potential energy of the gravitation at the altitude of the orbit and on the surface of the Earth plus the kinetic energy on the orbit. The actual launch velocity is the vector sum of the vertical speed and horizontal speed. The horizontal speed is the sum of Earth’s surface speed and the horizontal component of the launch velocity (with respect to Earth). Lunching in the direction of the surface velocity (i.e. eastward) gives the largest horizontal speed. The surface speed on Earth depends on latitude. It can be up to 1,040 mph at the equator.

54) An 18 gram bullet is shot vertically upwards into a 10 kg

block and quickly becomes imbedded in the block. If the block lifts upwards 9.0 mm, what was the initial kinetic energy of the bullet?

A) 0.0016 J B) 0.88 J C) 49 J D) 330 J E) 490 J Solution Let us denote the mass of the bullet = m and the mass of the block = M. Taking the initial position of the block as the reference point for gravitational potential, the final mechanical energy of the block is: Ef = (m+M)gh where h is the height lifted (= 9 mm). This will also be the kinetic energy of the block-bullet system in the instant after the bullet is embedded. Neglecting the impulse due to gravity over the short period of time the bullet is burrowing into the block, momentum is conserved during the inelastic collision between the bullet and the block. On the one hand, this momentum is given in terms of the post-collision KE of the block (i.e. Ef): pafter = 2(m + M )Ef and, on the other hand, in terms of the initial kinetic energy (Ki) of the bullet: pbefore = 2mKi

Equating these two: 2mKi = 2(m + M )Ef = 2(m + M )2gh

∴Ki =(m + M )2

mgh = (10kg)2

0.018kg(9.81 m/s2 )(9 ×10−3m) = 490J

Note, that it is reasonable to take the approximation M + m ≈ M throughout. Doing so does not change the answer significantly.