! 2!

SECTION A Attempt all questions. Bubble in answers on the scantron sheet

and circle your answer on these sheets. [1] The prefix “pico” means: A 10-15 B 10-12 C 10-9 D 10-6 E 10-3. ________________________________

[2] The frequency of vibration ( [f ]⇒ [T]−1 ) of a mass m attached to the end of a spring, with a spring

constant k ( [k]⇒ [M][T]−2 ), is related to m and k by a relationship of the form

f = (constant)makb, where the constant is dimensionless. What are the exponents a and b?

A a = −1 / 2;b = 1 / 2. B a = 1 / 2;b = −1 / 2. C a = 1;b = −1 / 2 . D a = 1 / 2;b = 1 / 2 . E a = −1 / 2;b = −1.

________________________________ [3] If you could count numbers at the rate of two per second, how long would it take you to count to one billion without stopping? About ...

A 10.3 years. B 15.9 years. C 18.7 years. D 31.7 years. E 54.6 years. ________________________________ [4] An object accelerates uniformly from a speed of 30 m/s to 40 m/s in 5 s and then moves at a constant speed of 40 m/s for an additional 3 s. What is the average speed of the object over the total time interval?

A 4.5 m/s. B 27.5 m/s. C 33.7 m/s. D 35.0 m/s. E 36.9 m/s. ________________________________ [5] Two cars, a Ford and a Chevy, start moving towards each other on a divided highway from two points 1.0 km apart. If the speed of the Ford is 18.0 m/s and the speed of the Chevy is 28.0 m/s, how far are they from the starting point of the Ford, when they pass each other?

A 247m. B 303m. C 391m. D 415m. E 609m. ________________________________ [6] A good baseball pitcher can throw a fastball at about 90 mph (40.3 m/s). He starts the ball’s motion from rest 45 cm behind his head and releases it at its final speed 76 cm in front of his head. Neglecting any motion of his body (apart from his arm) what is the acceleration of the ball during the throw, assuming it is constant?

A 62 m/s2. B 380 m/s2. C 670 m/s2. D 930 m/s2. E 2180 m/s2. ________________________________ [7] A rocket is fired vertically from rest with a net acceleration of 20 m/s2. After 4s the motor is switched off but the rocket continues to coast upwards. What is the maximum height of the rocket above ground level?

A 486 m. B 408 m. C 327 m. D 316 m. E 160 m. ________________________________ [8] The vector equation that represents the relationship among vectors

!A , !B and

!C is

A !A +!B −!C = 0. B

!A −!B −!C = 0. C

!A +!B +!C = 0.

D !C −!A −!B = 0. E

!A −!B +!C = 0.

! 3!

[9] An object starts from rest at the origin. If it experiences an acceleration a = 0.1(t2 − t) m/s2, what is its speed after 3 s?

A 0.22 m/s. B 0.33 m/s. C 0.45 m/s. D 0.60 m/s. E 0.90 m/s. ________________________________ [10] Given three vectors,

!A = (6i − 3j) ,

!B = (−3i + 4 j) and

!C = (2i + 5j) , what is the magnitude of the

vector !D = 2

!A +!B −!C ?

A 5.7 B 7.0 C 7.8 D 9.9 E 11.4 ________________________________ [11] Given two vectors

!X = 3i + 4 j and

!Y = 3i + 2 j , what is the angle between

!X and

!Y ?

A 19.4!. B 26.2!. C 33.7!. D 53.1!. E 70.4!. ________________________________ [12] An object starts from the origin,

!r" = (0,0) at t = 0 , with an initial velocity !v" = (2.0i − 3.0 j) m/s .

If the object moves in the x-y plane with an acceleration of !a = 1.2i m/s2 , what is the average velocity

of the object between t = 0 and t = 3 s? A (7.6i − 6.0 j) m/s. B (7.6i + 6.0 j) m/s. C (3.8i + 3.0 j) m/s.

D (3.8i − 3.0 j) m/s. E (−6.0i + 7.6 j) m/s. ________________________________ [13] A ship is traveling due east at 10 km/h. A second ship, starting from the same harbor at the same

time, is heading in a direction 30! east of north. What must be the speed of the second ship if it always due north of the first ship?

A 10 km/h. B 12.5 km/h. C 15 km/h. D 17.5 km/h. E 20 km/h. ________________________________ [14] An airplane takes off from the origin. The x-axis points east; the y-axis points straight up. If the instantaneous position and velocity vectors at some later time are

!r = (1.61×106 i + 9.14 ×106 j) m and !v = +224i m/s,

respectively, then the plane is most likely A just touching down. B in level flight in the air. C ascending.

D descending. E taking off. ________________________________ [15] A softball player throws the ball to first base, a distance of 22.5 m. If the ball is thrown at an angle of 36! to the horizontal and is caught at the same height it was thrown, what was the initial speed of the ball?

A 10.5 m/s. B 12.6 m/s. C 15.2 m/s. D 17.7 m/s. E 19.4 m/s. ________________________________ [16] An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her claws manages to wriggle loose and falls into a lake 5.00 m below. What is the speed of the fish when it hits the water?

A 9.90 m/s. B 10.3 m/s. C 11.8 m/s. D 12.7 m/s. E 14.2 m/s. ________________________________

! 4!

[17]

________________________________ [18]

A 0.11 m/s2. B 0.29 m/s2. C 0.38 m/s2. D 0.44 m/s2. E 0.67 m/s2.

_______________________________ SECTION B

Choose TWO from the following three free response questions. Write your solutions on this question test and write the answers on the lines provided.

[19] A stone is dropped from a balloon that is descending at a uniform rate of 12.0 m/s when it is 1000 m above the ground. You can ignore air resistance. (a) How long does it take the stone to reach the ground? t = __________13.1____________ s. (b) What is the speed of the stone as it strikes the ground? v = ____________140.5____________ m/s. (c) What is the height of the balloon when the stone strikes the ground? h = ________842.8_______ m. SOLUTION: (a) Take the upward direction as the positive direction. For the stone:

v! = −12.0 m/s, v = ?, a = −9.81 m/s2, y! = 1000 m, y = 0, t = ?

We have: (y − y!) = v!t +

12

at2, i.e., −1000 m = (−12.0 m/s)t +12

(−9.81 m/s2)t2.

After rearrangement: 4.91t2 +12.0t −1000 = 0.

Solutions are: t =

−12.0 ± (12.0)2 − 4(4.91)(−1000)2 × 4.91

= 13.1 s or −15.6 s.

Take the positive root, i.e., t = 13.1 s.

(b) v = v! + at = (−12.0 m/s) + (−9.81 m/s2)(13.1 s) = −140.5 m/s. Therefore, the speed is 140.5 m/s. OR

m

F

30

f

A force F = 5.0 N is applied at an angle of 30! to the horizontal on an object with mass m = 3.0 kg , as shown. The surface produces a retarding force f = 3.0 N. What is the resulting acceleration of the object?

An object of mass 2.0 kg on a horizontal, frictionless table in the x-y plane, is acted on by two forces,

!F1 = 4.0 N and

!F2 = 2.5 N . What is the magnitude of the resultant

acceleration? A 2.5 m/s2. B 3.5 m/s2. C 4.4 m/s2. D 5.0 m/s2. E 6.9 m/s2. !

! 5!

v2 = v!2 + 2a(y − y!), i.e., v2 = (−12.0 m/s)2 + 2(−9.81 m/s2)(−1000m).

∴v = ±140.5 m/s.

(c) In 13.1 s the balloon descends (12.0 m/s)(13.1 s) = 157.2 m. Therefore, it is

1000 m −157.2 m = 842.8 m, above the ground when the stone lands. _______________________________ [20] After graduating from FAU with a physics degree, your parents decide to take you to one of Palm Beach’s finest restaurants to celebrate and they order a bottle of champagne. Tilting the bottle at an angle of 35! the cork comes flying out with an initial velocity of 8 m/s. (a) How much time does it take the cork to reach the highest point in its trajectory?

t = __________0.468__________ s.

(b) How high did the cork go as measured from the point where it left the bottle?

h = _________1.07___________ m.

(c) A person at a nearby table catches the cork at the same height it was released. How far away from the bottle were they sitting?

d = ____________ 6.13___________ m. SOLUTION:

(a) v!x = v! cos35! = 6.55 m/s and v!y = v! sin35! = 4.59 m/s.

In the vertical direction vy = v!y − gt,

i.e., t =

4.59 m/s − 0

9.81 m/s2= 0.468 s.

(b) Also in the vertical direction vy

2 = v!y2 +2a(y − y!) = v!y

2 − 2g(y − y!),

i.e., h = (y − y!) =

v!y2

2g =

(4.59 m/s)2

2(9.81 m/s2)= 1.07 m.

(c) In the horizontal direction, the range is d = R =

v!2 sin 2θ!

g=

(8.0 m/s)2 sin70!

9.81 m/s2= 6.13 m.

OR

d = v!x (2t) = 2(6.55 m/s)(0.468 s) = 6.13 m.

! 6!

[21]

(a)What is the acceleration vector of the object?

!a = ___+ 0.66______ i, ____+1.18_____ j m/s2.

(b)What is its velocity vector after 3.0 s?

!v = ____+1.98_____ i, ___+ 3.54______ j m/s.

(c) What is the position vector after 3.0 s?

!r = ____+ 2.97 _____ i, ___+ 5.31______ j m.

SOLUTION: Resolve the forces into components.

FRx =

!F1 cosθ −

!F3 = 6.0 N × 0.94 − 4.0 N = 1.64 N.

FRy =

!F2 −

!F1 sinθ = 5.0 N − 6.0 N × 0.34 = 2.95 N.

∴!FR = (1.64i + 2.95j) N.

(a) Acceleration vector is

!a =

!FRm

=1

2.5 kg(1.64i + 2.95j) m/s2 = (0.66i +1.18 j) m/s2.

(b) The velocity vector is

!v = !v" +!at = (0.66i +1.18 j)(3.0) = (1.98i + 3.54 j) m/s.

(c) The position vector is !r = !r" +

!v"t +12!at2 =

12

(0.66i +1.18 j)(3.0)2 = (2.97i + 5.31j) m. !!

A 2.5 kg object sits, at rest, on a horizontal, frictionless table. It is acted on by three forces,

!as shown, which sets it in motion.!