We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy

See Pricing

Hire a Professional Writer Now

The input space is limited by 250 symbols

Choose 3 Hours or More.
Back
2/4 steps

How Many Pages?

Back
3/4 steps

Back
Get Offer

# Physics Essay

Hire a Professional Writer Now

The input space is limited by 250 symbols

Write my paper

UIC PHYSICS 105 Spring 2013 Practice Exam 1

MULTIPLE CHOICE QUESTIONS (2 points each) Clearly circle the letter of the best answer MCQ 1: The figure to the right represents the position of a particle as it travels along the x-axis. At what value of t is the velocity of the particle equal to zero? (A) 1 s Answer: velocity = slope of x vs t line (B) 2 s slope = 0 at t = 3 s (C) 3 s (D) 4 s

Don't use plagiarized sources. Get Your Custom Essay on
Physics
Just from \$13,9/Page

MCQ 2: A runner runs around a track consisting of two parallel lines 96 m long connected at the ends by two semicircles with a radius of 49 m.

She completes one lap in 100 seconds. What is her average velocity? (A) 2.5 m/s ∆ (B) 5.0 m/s Answer: 0 m/s ∆ ∆ (C) 10 m/s (D) 0 m/s MCQ 3: You drop a very bouncy rubber ball. It falls, and then it hits the floor and bounces right back up to you. Which of the following represents the v vs. t graph for this motion?

Answer: Initially, the ball is falling down, so its velocity must be negative (if UP is positive).

Its velocity is also increasing in magnitude as it falls. Once it bounces, it changes direction and then has a positive velocity, which is also decreasing as the ball moves upward. MCQ 4: Two objects are dropped from a bridge, an interval of 1.0 s apart. During the
time that both objects continue to fall, their separation Object 1: , . (A) increases. (B) decreases. Object 1: , . (C) stays constant. If t1 > t2 , , , , , , (D) increases at first, but then stays constant. then and the separation between two objects x1 – x2 increases

Page 2 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

MCQ 5: Given that and that each other? (A) they are perpendicular to each other (B) they are parallel and in the same direction (C) they are parallel but in the opposite direction (D) they are at 45° to each other

, how are vectors and oriented with respect to Answer: The only time vector magnitudes will simply add together is when the direction does not have to be taken into account (i.e., the direction is the same for both vectors). In that case, there is no angle between them to worry about, so vectors A and B must be pointing in the same direction.

MCQ 6: You are adding vectors of length 20 and 40 units. What is the only possible resultant magnitude that you can obtain out of the following choices? Answer: (A) 18 The minimum resultant occurs when the vectors are (B) 37 opposite, giving 20 units. The maximum resultant occurs (C) 64 when the vectors are aligned, giving 60 units. Anything in (D) 100 between is also possible for angles between 0° and 180°. MCQ 7: A train is moving in the west direction with a velocity 15 m/s. A monkey runs on the roof of the train against its motion with a velocity 5 m/s with respect to train. Take the motion along west as positive. The velocity of train with respect to monkey is Answer: (A) 5m/s velocity of train=15m/s (B) 5 m/s velocity of monkey relative to the train=5 m/s (C) 15 m/s Now velocity of monkey relative to the train =  velocity of (D) 15 m/s train with respect to monkey So velocity of train with respect to monkey=5 m/s MCQ 8: Two particles start from the origin of the horizontal x-y plane. Particle A moves along +x axis with speed v. Particle B moves along y axis with the same speed v. The relative velocity of particle B with respect to A is
Answer: (A) m/s Velocity of A is m/s (B) m/s m/s Velocity of B is (C) 0 m/s  velocity of B with respect to A m/s (D) m/s MCQ 9: A rock of mass m is thrown horizontally off a building from a height h, as shown in the figure to the right. The speed of the rock as it leaves the thrower’s hand at the edge of the building is v0. How much time does it take the rock to travel from the edge of the building to the ground? (A) / Answer: / (B)  2 / Velocity of A is (C) 2 / (D) 2 /

Page 3 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

MCQ 10: You’re on the street, trying to hit a friend with a water balloon. He sits in his dorm room window above your position and is aiming at you with his water balloon. You aim straight at him and shoot and he does the same in the same instant. Do the water balloons hit each other? (A) yes, they hit Answer: (B) maybe – it depends on the speeds of the shots Both water balloons are aimed straight at each other (C) the shots are impossible and both still fall with the same acceleration – g!! (D) no, they miss MCQ 11: A spring-loaded gun can fire a projectile to a height h if it is fired straight up. If the same gun is pointed at an angle of 45° from the vertical, how far from the gun will the projectile hit the ground? (A) /4 Answer: Maximum height of a projectile is found from 0 at max. height and (B) /√2 , 2 and gives sin /2 . Fired straight up,  = 90° (C) , (D) 2 2 . So, ∆ / sin 2 ∙ 45° = 2 and we have MCQ 12: From rest, we step on the gas of our Ferrari, providing a constant force F for 40 m, speeding it up to a final speed 50 km/hr. If the same force would be applied for 80 m, what final speed would the car reach? A) 100 km/hr Answer: In the first case, the acceleration acts over a distance x = 40 m, to give a final B) 90 km/hr speed of v2 = 2ax (why is there no term?). In the 2nd case, the distance is doubled, so C) 70 km/hr the speed increases by a factor of √2. D) 50 km/hr MCQ 13: What can you say about the force of gravity Fg acting on a heavy stone and a light feather? A) Fg is greater on the feather Answer: The force of gravity (weight) depends on the mass B) Fg is greater on the stone of the object!! The stone has more mass, therefore more C) Fg is zero on both due
to vacuum weight. D) Fg is equal on both always MCQ 14: A force F acts on mass m1 giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2 = 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? A) 3/2 a1 Answer: Mass m2 must be (1/2)·m1 because its acceleration was 2a1 with the same B) 1/2 a1 force. Adding the two masses together gives (3/2)·m1, leading to an acceleration of C) 4/3 a1 (2/3)·a1 for the same applied force. D) 2/3 a1 MCQ 15: A flatbed truck is carrying a 20.0-kg crate along a level road. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place? Answer:  0.4 ∙ 9.81 3.92 m/s2 (A) 3.92 m/s2 (B) 8.00 m/s2 (C) 78.5 m/s2 (D) 196 m/s2 Page 4 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

MCQ 16: A mass, M, is at rest on a frictionless surface, connected to an ideal horizontal spring that is unstretched. A person extends the spring 30 cm from equilibrium and holds it by applying a 10 N force. The spring is brought back to equilibrium and the mass connected to it is now doubled to 2M. If the spring is extended back 30 cm from equilibrium, what is the necessary force applied by the person to hold the mass stationary there? (A) 20.0 N Answer: Based on F = k·Δx. The mass attached to the spring does not change the spring constant so the (B) 14.1 N (C) 10.0 N (D) 7.07 N

MCQ 17: Two equal-mass rocks tied to strings are whirled in horizontal circles as shown in the figure to the right. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? A) T2 = 1/4 T1 Answer: The centripetal force in this case is given by the tension, so T = mv2/r. For the B) T2 = 1/2 T1 same period, we find that v2 = 2v1 (and this term is squared). However, for the C) T2 = T1 denominator, we see that r2 = 2r1 which gives us the relation T2 = 2T1. D) T2 = 2 T1 MCQ 18: box is being pulled up a rough incline by a rope connected to a pulley and a second box as shown in the figure to the right. How many forces are doing work on the first box? A) one force Answer: Answer: B) two forces Any force not
perpendicular to the motion will do work: C) three forces N does no work, T does positive work, f does negative D) four forces work, and mg does negative work. MCQ 19: Car #1 has twice the mass of car #2, but they both have the same kinetic energy. How do their speeds compare? Answer: Since the kinetic energy is 1/2 mv2, and the mass of car #1 is greater, then car (A) 2v1 = v2 #2 must be moving faster. If the ratio of m1/m2 is 2, then the ratio of v2 values must also (B) √2 v1 = v2 be 2. This means that the ratio of v2/v1 must be the square root of 2. (C) 4v1 = v2 (D) v1 = v2 MCQ 20: The work W0 from 50 to 150 km/hr? (A) 3 W0 (B) 4 W0 (C) 8 W0 (D) 9 W0 accelerates a car from 0 to 50 km/hr. How much work is needed to accelerate the car Answer: Let’s call the two speeds v and 3v, for simplicity. We know that the work is given by: ∆ Case #1: Case #2: 3 0 8 8

Page 5 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

SHORT PROBLEMS Clearly circle the letter of the best answer SP1: A cat walks in a straight line, which we shall call the x-axis with the positive direction to the right. As an observant physicist, you make measurements of this cat’s motion and construct a graph of the feline’s velocity as a function of time as shown in the graph below. (a) [4 points] What is the cat’s velocity at t = 7.0 s? Answer: From graph: , , ,

= – 1.33 cm/s2

At t = 7 s

8

1.33 ∙ 7

1.3 cm/s

(A) – 0.67 cm/s

(B) – 0.98 cm/s

(C) – 1.3 cm/s

(D) – 1.6 cm/s

(b) [6 points] What distance does the cat move from t = 0 to t = 7.0 s? Answer: The area under curve from t = 0 to t = 6 is equal to the distance the cat moved in positive x direction. ∙

24 m

The area under curve from t = 6 to t = 7 is equal to the distance the cat moved in negative x direction. | . |∙

0.65 m 24 0.65 24.7 m

Thus, the total distance

(A) 22.3 m

(B) 24.7 m

(C) 28.4 m
Page 6 of 11

(D) 32.1 m

UIC PHYSICS 105 Spring 2013 Practice Exam 1

SP2: A bird flies in the xy-plane with a velocity vector given by 2.4 2.0 4.0

The positive y-direction is vertically upward. At t = 0 the bird is at the
origin. (a) [5 points] How far from the origin will be the bird after flying for the first 10 s? Answer: x – axis: y – axis:  , , , ,

2.4 m/s, 0 m/s,

2.0 m/s2 4.0 m/s2 2.4 ∙ 10 0.5 ∙ 2.0 ∙ 10 76 m

0 ∙ 10 76

0.5 ∙ 4.0 ∙ 10

200 m

200 = 214 m

(A) 58 m

(B) 124 m

(C) 176 m

(D) 214 m

(b) [5 points] What is the bird’s altitude (y-coordinate) as it flies over x = 0 for the first time after t = 0? Answer: ,

0 or 0  2
,

,

0

,

/

or

2 ∙ 2.4/
,

2.0 = 2.4 s 0 ∙ 2.4 0.5 ∙ 4.0 ∙ 2.4 11.5 m

The bird’s altitude after it flies for 2.4 s is

(A) 5.7 m

(B) 8.7 m

(C) 11.5 m

(D) 14.1 m

Page 7 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

SP3: A projectile is fired at a wall a horizontal distance d = 50.0 m away. The projectile is launched at angle θ0 = 60o above the horizontal and strikes the wall at a height h = 10.0 m above the ground as shown in the Figure to the right. (a) [5 points] Determine the initial speed, v0, of the projectile. Answer: We can find the v0 by using the equation for projectile trajectory: ∆ ∆  ∙ tan , , = . ∙ ° ∙ °

2

cos

= 25.3 m/s

(A) 6.71 m/s

(B) 8.24 m/s

(C) 18.7 m/s

(D) 25.3 m/s

(b) [5 points] How long is the projectile in the air? Answer: ,

cos cos ∙ , x = d 
. ∙ °

3.95 s

(A) 2.57 s

(B) 3.95 s

(C) 8.25 s

(D) 10.1 s

Page 8 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

SP4: A massless spring is attached to the ceiling of an elevator, initially at rest, as shown in the figure. By attaching an object to the free end of the spring, the spring stretches a distance ∆ = 3.0 cm. Take the positive direction upward as shown. (a) [6 points] The elevator now travels with acceleration of magnitude a = g/2 pointing upwards. What is the new
extension ∆ of the spring? Answer: While elevator is at rest, the spring extends under the action of the object’s weight. When object is in equilibrium, the elastic force equilibrates its weight: ∆ 0 The elevator with its contents is not in equilibrium anymore and elevator together with its contents travels with acceleration a pointing upwards. In particular, the object itself travels with a pointing upwards. ∆ ∆ 1 1 ∆ = 1 3.0 4.5 cm

(A) 1.5 cm

(B) 3.0 cm

(C) 4.5 cm

(D) 6.0 cm

(b) [4 points] The elevator now travels with acceleration of the same magnitude , but pointing downwards. What is the new extension ∆ of the spring? Answer: Now, the acceleration is pointing downwards and ∆ Thus, ∆ 1 ∆ 1 3.0 1.5 cm

(A) 1.5 cm

(B) 3.0 cm

(C) 4.5 cm

(D) 6.0 cm

Page 9 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

SP5: Block A with weight 3W, slides down an inclined plane S of slope angle  = 36.9° at a constant speed while plank B, with weight W, rests on top of A. The plank is attached by cord to the wall as shown in the figure below. (b)
[4 points] Find the normal force exerted by the plane S on the block A in terms of  and W (weight). Answer: Block A y – axis: ∑  3 3 cos cos cos 4W cos 0 y NA A x 3wSin 3w fBA fAS

(A) 2W·cos

(B) 2W·sin

(C) 4W·cos

(D) 4W·sin

(b) [6 points] If the coefficient of kinetic friction is the same between A and B and between S and A, determine its value Answer: Block A x – axis: ∑ 3 cos 3  sin tan 3 cos tan 36.9° 0.45 cos 0 sin cos 0

(A) 0.15

(B) 0.30

(C) 0.45

(D) 0.60

Page 10 of 11

UIC PHYSICS 105 Spring 2013 Practice Exam 1

SP6: An object of mass m = 1 kg is attached to the free end of a massless spring, initially unstretched as shown in the figure to the right. The dependence of the restoring force of the spring on stretching (x > 0) and compression ( x < 0 ) is shown by the thick line in the graph at the right below. The hatched regions are quarter circles of radius 2, such that 2 2 0 0 2 2 Hints: (i) Use the graphical interpretation of work in a force versus position plot. (ii) This is a special spring that does NOT obey Hooke’s Law . (a) [5 points] Calculate the work done by the restoring force when object moves from x = 0 m to x = 2 m. Answer: Work is area under the curve from x = 0 m to x = 2 m The only force producing work in the system is the elastic force, because weight and displacement are perpendicular at all times. The work done by the elastic force is transferred to the object in the form of kinetic energy. Negative work done by the elastic force signifies a decrease in kinetic energy. Object has maximum velocity when it passes through the equilibrium position x = 0 m. 2 ∙2 ∙2 ∙2 Area under the curve from x = 0 m to x = 2 m: Work done by the elastic force on this portion is negative: 0.86 J (A) – 0.43 J (B) 0.43 J (C) – 0.86 J

0.86 J

(D) 086 J

(b) [5 points] If the spring is initially compressed to x = – 2 m and then released, what is the velocity of the object when it reaches the position x = 0 m ? Answer: The spring compressed to x = −2 m acquires a potential elastic energy. Upon release, the elastic force does positive work and releases this energy in the form of kinetic energy. At x = 0 m elongation reaches zero, the entire elastic energy is converted into kinetic energy and the velocity of the object is maximum. Apply the work-energy theorem: ∆ . At x = 0, 0.86 J  ∙ .

1.31 m/s

(A) – 1.31 m/s

(B) 1.31 m/s

(C) – 2.62 m/s
Page 11 of 11

(D) 2.62 m/s

### Cite this Physics Essay

Show less • Use multiple resourses when assembling your essay
• Get help form professional writers when not sure you can do it yourself
• Use Plagiarism Checker to double check your essay