Session 1 - Physics Mechanics and Energy PDF

Session 1 Physics MECHANICS AND ENERGY  MECHANICS Concepts of Motion, Forces, and Newton’s Laws of Motion Question No. 1: A stone is thrown horizontally and follows the path XYZ shown. The direction of the acceleration of the stone at point Y is:...

Session 1 Physics MECHANICS AND ENERGY  MECHANICS Concepts of Motion, Forces, and Newton’s Laws of Motion Question No. 1: A stone is thrown horizontally and follows the path XYZ shown. The direction of the acceleration of the stone at point Y is: a. ↓ b. → c. A d. t e. a SOLUTION: ANSWER: a The acceleration of the stone at point Y is directed vertically downward. In projectile motion, the only force acting on the object (ignoring air resistance) is gravity, which causes an acceleration that is always directed downward. Question No. 2: If an object moves with a constant speed in a circular path, what can be said about its acceleration? A) It is zero B) It is directed towards the center of the circle C) It is directed away from the center of the circle D) It depends on the radius of the circle SOLUTION: ANSWER: b When an object moves with a constant speed in a circular path, its velocity is continuously changing direction, even though its speed remains constant. This change in the direction of velocity indicates that the object is accelerating. This type of acceleration is called centripetal acceleration, and it always points towards the center of the circle. This inward acceleration is what keeps the object moving in a circular path.  Question No. 3: A young girl wishes to select one of the frictionless playground slides, illustrated below, to give her the greatest possible speed when she reaches the bottom of the slide. Which of the slides illustrated in the diagram below should she choose? A) A B) B C) C D) D E) It does not matter; her speed is the same for each slide SOLUTION: ANSWER: e Applying This to the Diagram All slides start at the same height (2.5 m) and end at the same lower height (0.5 m). Therefore, regardless of the shape or length of the slide, the change in height is the same, and thus the final speed will be the same for each slide. Question No. 4: A stone thrown from the top of a tall building follows a path that is: a. Circular b. made of two straight line segments c. Hyperbolic d. Parabolic e. a straight line https://www.doubtnut.com/qna/501548798 SOLUTION: ANSWER: d https://www.doubtnut.com/qna/501548798 When a stone is thrown from the top of a building, it follows a parabolic trajectory due to the influence of gravity, assuming air resistance is negligible. This is because the stone has both a horizontal and vertical component of motion, resulting in a curved path that forms a parabola. Question No. 5: Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth and one on the Moon. Which of the following three statements is/are true? I. The horizontal distance traveled by the bullet is greater for the Moon. II. The flight time is less for the bullet on the Earth. III. The velocity of the bullets at impact are the same. a. III only b. I and II only c. I and III only d. II and III only e. I, II, III SOLUTION: ANSWER: b I. The horizontal distance traveled by the bullet is greater for the Moon. This statement is true. Since the Moon has less gravity compared to Earth, the bullet will take longer to fall to the ground, allowing it to travel a greater horizontal distance before impact. II. The flight time is less for the bullet on the Earth. This statement is true. Because Earth's gravity is stronger, the bullet will reach the ground faster than it would on the Moon. III. The velocity of the bullets at impact are the same. This statement is false. The vertical component of velocity will be different due to the different gravitational accelerations. On Earth, the bullet will have a higher vertical speed upon impact compared to the Moon because of the stronger gravity.  Question No. 6: A feather and a lead ball are dropped from rest in vacuum on the Moon. The acceleration of the feather is: A) more than that of the lead ball B) the same as that of the lead ball C) less than that of the lead ball D) 9.8 m/s2 E) zero since it floats in a vacuum Images: https://learnmoredeeply.com/when-you-drop-a-heavy-metal-ball-and-a-feather-from-a-certain-height-whic h-one-does-fall-first https://www.slashgear.com/920618/whats-actually-under-the-moons-surface-according-to-nasa/  SOLUTION: ANSWER: b On the Moon, there is no atmosphere, which means no air resistance. In a vacuum, all objects experience the same acceleration due to gravity regardless of their mass, shape, or material. This principle was famously demonstrated by Galileo and is explained by Newton's laws of motion. On the Moon, the acceleration due to gravity is about 1.6 m/s2. Since both the feather and the lead ball are in free fall in a vacuum, they will fall with the same acceleration. Therefore, their accelerations are equal. This outcome is different from what happens on Earth, where air resistance affects objects differently based on their shape and mass, causing a feather to fall more slowly than a lead ball in the presence of air. https://www.youtube.com/watch?v=frZ9dN_ATew https://www.youtube.com/watch?v=E43-CfukEgs Question No. 7: A block weighing 200 N lies on a plane inclined 20° with the horizontal. What force tends to pull the block down the plane? A) F=200/cos 20° B) F=200/sin 20° C) F = (200)(sin 20°) D) F = (200)(cos 20°) SOLUTION: ANSWER: c N m N Fx = Wx Θ F = Wy Θ = 200 W = 200 N Θ Θ W = 200 N sin Θ = opposite / hypotenuse sin Θ = Fx 200 N Fx = 200 N (sin 200) Question No. 8: A steel ball is attached to a string and is swung in a circular path in a horizontal plane as illustrated in the accompanying figure. At the point P indicated in the figure, the string suddenly breaks near the ball. If these events are observed from directly above as in the figure, which path would the ball most likely follow after the string breaks? a. A b. C c. B d. E SOLUTION: ANSWER: c When the string breaks, the ball will no longer be constrained to move in a circular path. Instead, it will move in a straight line tangent to the point where the string broke (point P) due to inertia. This is because, when the string breaks, the ball follows the tangent to the circular path at the point where the string broke, which is represented by path B. Question No. 9: A plane traveling north at 200 m/s turns and then travels south at 200 m/s. The change in its velocity is: a. Zero b. 200 m/s north N 200 m/s c. 200 m/s south d. 400 m/s north e. 400 m/s south 200 m/s S  SOLUTION: ANSWER: e To determine the change in velocity, we consider both the magnitude and direction of the initial and final velocities. 1. Initial velocity: The plane is traveling north at 200 m/s. 2. Final velocity: The plane then turns and travels south at 200 m/s. Since velocity is a vector quantity, we must take the direction into account: The initial velocity is +200 m/s (north). The final velocity is −200 m/s (south). The change in velocity (Δv) is: Δv = vfinal − vinitial = −200 m/s − 200 m/s = −400 m/s The magnitude of the change is - 400 m/s. Question No. 10: The airplane shown is in level flight at an altitude of 0.50 km and a speed of 150 km/h. At what distance d should it release a heavy bomb to hit the target X? Take g = 10 m/s2 a. 150 m b. 295 m c. 420 m d. 2550 m e. 15, 000 m SOLUTION: ANSWER: c The time it takes for the bomb to hit the ground is the same time the airplane (and thus the bomb, before it is released) travels horizontally. This is because the bomb maintains the same horizontal velocity as the airplane at the moment it is released, and there is no horizontal acceleration acting on the bomb (assuming air resistance is negligible). The horizontal distance travelled by airplane at the same time the bomb hit the target is d = vt d = (150 km/h) (1 h / 3600 s)(1000 m/1 km) (10.10 s) d = 420 m Question No. 11: A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45◦. With what speed was it thrown? a. 14 m/s b. 20 m/s c. 28 m/s d. 32 m/s e. 40 m/s SOLUTION: ANSWER: b Vinitial = Vx at the moment of impact, Voy = 0 The vertical velocity vy just before impact Vy = Voy + gt Vx 45 0 Vy = 0 + (9.8 m/s2) (2.02 s) Vy 450 Vy = 19.79 m/s Since the ball strikes the ground at a 450 angle, this means that the horizontal and vertical components of the velocity are equal at the moment of impact. Vx = Vy = 19.79 m/s = Vinitial Question No. 12: A dart is thrown horizontally toward X at 20 m/s as shown. It hits Y 0.1 s later. The distance XY is: a. 2 m b. 1 m c. 0.5 m d. 0.1 m e. 0.05 m SOLUTION: ANSWER: e Since the dart is thrown horizontally, its initial vertical velocity is 0. The vertical distance (Yfinal = distance XY ) it falls at time t = 0.1 s is given by: Question No. 13: A girl wishes to swim across a river to a point directly opposite as shown. She can swim at 2 m/s in still water and the river is flowing at 1 m/s. At what angle θ with respect to the line joining the starting and finishing points should she swim? a. 30◦ b. 45◦ c. 60◦ d. 63◦ e. 90◦ SOLUTION: ANSWER: a To determine the angle θ at which the girl should swim, we must account for both her swimming speed and the current of the river. She needs to swim at an angle upstream to counteract the downstream flow of the river and end up directly opposite her starting point. We need to find the angle θ at which she should swim relative to the straight line across the river (perpendicular to the river flow) so that her horizontal velocity (the component counteracting the river's flow) exactly cancels out the river's current. To swim directly across, her horizontal component must exactly Vx cancel the river’s flow: Vx = Vriver Vy 2 m/s sin Θ = 1 m/s V = 2 m/s sin Θ = opposite / hypotenuse Sin Θ = ½ Θ = sin -1 0.5 sin Θ = __Vx_ 2 m/s Θ = 300 V = 2 m/s (sin Θ) Question No. 14: Two blocks are connected by a string and pulley as shown. Assuming that the string and pulley are massless, the magnitude of the acceleration of each block is: a. 0.049 m/s2 b. 0.020 m/s2 c. 0.0098 m/s2 d. 0.54 m/s2 e. 0.98 m/s2  SOLUTION: ANSWER: e FBD W = m 1g W = m 2g The net force is due to the difference in the weights of the two blocks: Fnet = m2g – m1g Using Newton's second law: Fnet = (m2 – m1) g Fnet = mtotal (a) Fnet = (0.11 kg –.09 kg) 9.8 m/s2 Fnet = 0.196 N a = 0.196 N / 0.20 kg a = 0.98 m/s2 Question No. 15 : At what angle should the roadway on a curve with a 50 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is icy (and the frictional force is zero)? a. 0 b. 16◦ c. 18◦ d. 35◦ e. 73◦ Source: https://dynref.engr.illinois.edu/avb.html SOLUTION: ANSWER: b The angle at which the roadway should be banked to allow cars to negotiate the curve at 12 m/s without friction is approximately 16°.  ENERGY Understanding Different Forms of Energy, Conservation of Energy, and Energy Transformations Question No. 1: Which of the following groups does NOT contain a scalar quantity? a. velocity, force, power b. displacement, acceleration, force c. acceleration, speed, work d. energy, work, distance e. pressure, weight, time  SOLUTION: ANSWER: b A scalar quantity is defined by its magnitude alone, whereas a vector quantity is defined by both magnitude and direction. Let's examine each group to identify which one contains only vector quantities or at least one vector quantity: 1. velocity, force, power 1. Velocity and force are vectors (they have direction), while power is a scalar. 2. displacement, acceleration, force 1. Displacement, acceleration, and force are all vectors. 3. acceleration, speed, work 1. Acceleration is a vector, but speed and work are scalars. 4. energy, work, distance 1. All of these are scalars. 5. pressure, weight, time 1. Weight is a vector (it has direction), while pressure and time are scalars. The correct answer is: displacement, acceleration, force, as it is the only group where all the quantities are vectors. Question No. 2: The amount of work required to stop a moving object is equal to: a. the velocity of the object b. the kinetic energy of the object c. the mass of the object times its acceleration d. the mass of the object times its velocity e. the square of the velocity of the object SOLUTION: ANSWER: b When you apply a force to stop the object, the work done is equal to the change in kinetic energy (from its initial value to zero). Question No. 3: Which one of the following five quantities CANNOT be used as a unit of potential energy? a. watt·second b. gram·cm/s2 c. Joule d. kg·m2/s2 e. ft·lb  SOLUTION: ANSWER: b To determine which quantity cannot be used as a unit of potential energy, let's evaluate each option and check if it is equivalent to the unit of energy (joules). A. watt·second 1. A watt is a unit of power (joules/second\text{joules/second}joules/second). 2. Multiplying it by seconds gives: watt×second=joule\text{watt} \times \text{second} = \text{joule}watt×second=joule. 3. Therefore, this can be used as a unit of energy. B. gram·cm/s² 4. This is a unit of force in the CGS system, known as a dyne. 5. Force (dyne) is not a unit of energy. Energy requires multiplying force by a distance (work done). 6. Therefore, this cannot be used as a unit of potential energy. C. joule 7. A joule is the SI unit of energy and can definitely be used as a unit of potential energy. D. kg·m²/s² 8. This is equivalent to a joule (the SI unit of energy) because energy in SI units is expressed as kg\cdotpm²/s²\text{kg·m²/s²}kg\cdotpm²/s². 9. Therefore, this can be used as a unit of potential energy. E. ft·lb 10. This is a unit of energy in the Imperial system, called a foot-pound. 11. It can be used as a unit of potential energy in the Imperial system. Question No. 4: No kinetic energy is possessed by: a. a shooting star b. a rotating propeller on a moving airplane c. a pendulum at the bottom of its swing d. an elevator standing at the fifth floor e. a cyclone SOLUTION: ANSWER: d Question No. 5: An elevator is rising at constant speed. Consider the following statements: I. the upward cable force is constant II. the kinetic energy of the elevator is constant III. the gravitational potential energy of the Earth-elevator system is constant IV. the acceleration of the elevator is zero V. the mechanical energy of the Earth-elevator system is constant. a. all five are true b. only II and V are true c. only IV and V are true d. only I, II, and III are true e. only I, II, and IV are true SOLUTION: ANSWER: e Let's analyze each statement in the context of an elevator rising at constant speed: I. The upward cable force is constant: This is true because the elevator is moving at a constant speed, implying that the upward force from the cable balances the downward force due to gravity. There is no net force, so the cable force remains constant. II. The kinetic energy of the elevator is constant: This is true because the elevator moves at constant speed. Kinetic energy depends on the speed (and mass) of the elevator, and since speed is constant, the kinetic energy remains constant. III. The gravitational potential energy of the Earth-elevator system is constant: This is false. As the elevator rises, its height increases, which increases the gravitational potential energy (which depends on height). IV. The acceleration of the elevator is zero: This is true because the elevator is moving at a constant speed. Constant speed means no change in velocity, so the acceleration is zero. V. The mechanical energy of the Earth-elevator system is constant: This is true if we consider that the mechanical energy (sum of kinetic and potential energy) remains constant when no non-conservative forces (like friction or air resistance) are doing work. In this case, the increase in gravitational potential energy is balanced by the work done by the cable. Based on the analysis, statements I, II, IV, and V are true. Question No. 6: A woman lifts a barbell 2.0 m in 5.0 s. If she lifts it the same distance in 10 s, the work done by her is: a. four times as great b. two times as great c. the same d. half as great e. one-fourth as great SOLUTION: ANSWER: c The work done by the woman is determined by the formula: Work = Force × Displacement or Work = ΔGPE = mghf - mghi Since the distance lifted and the weight (force) of the barbell remain the same in both cases, the work done depends only on these two factors. The time taken (whether it's 5 seconds or 10 seconds) does not affect the amount of work done, as work is independent of time. Therefore, the work done in lifting the barbell at the same distance in 10 s is: the same. Question No. 7: A 2-kg block is thrown upward from a point 20 m above Earth’s surface. At what height above Earth’s surface will the gravitational potential energy of the Earth-block system have increased by 500 J? a. 5 m b. 25 m c. 46 m d. 70 m e. 270 m SOLUTION: ANSWER: c To determine the height at which the gravitational potential energy (GPE) increases by 500 J, we use the formula for gravitational potential energy: ΔGPE = mghf – mghi = mg (hf –hi) Given: ΔGPE = 500 J hf = ? m = 2 kg hi = 20 m 500 J = (2 kg) (9.8 m/s2) (hf – 20 m) 500 J = 19.6 kg m/s2 hf – 392 kg m2/s2 500 J + 392 J = 19.6 kg m/s2 hf 892 J = 19.6 kg m/s2 hf 45.51 = h Question No. 8: A man pulls a 100-N crate up a frictionless 30◦ slope 5 m high, as shown. Assuming that the crate moves at constant speed, the work done by the man is: a. −500 J b. −250 J c. 0 J d. 250 J e. 500 J SOLUTION: ANSWER: e FBD N Fx = Wx m N Θ F = Wy Θ = 300 Wx F= W = 100 N 0 Θ=30 sin Θ = opposite / hypotenuse W = 100 N sin Θ = Fx 100 N The work done by the man, W = F. d cos (Θ) Fx = 100 N (sin 300) = 50 N sin Θ = opposite / hypotenuse W = F. d cos (Θ) ? W = (50 N) (10 m) cos 00 d = h=5m sin Θ = h W = 500 J d Θ = 300 d = 5 m/(sin 300) = 10 m Question No. 9: An escalator is used to move 20 people (60 kg each) per minute from the first floor of a department store to the second floor, 5 m above. Neglecting friction, the power required is approximately: a. 100 W b. 200 W c. 1 000 W d. 2 000 W e. 60 000 W SOLUTION: ANSWER: c To find the power required, Power = Work / Time First, let's calculate the total work done in moving the 20 people. The work done to lift an object is given by: W = Δ GPE = mg(h2-h1) Each person has a mass of 60 kg, and there are 20 people: Mtotal = 60 kg/person × 20 people = 1200 kg So, the total work done is W = 1200 kg × 9.8 m/s2 × 5 m = 58 800 J Since this work is done every minute, we convert the time to seconds (1 minute = 60 seconds): Power = 58 800 J / 60 s = 980 W or approximately 1000 W Question No. 10: An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100-N weight attached, the pointer indicates “40” on the scale as shown. Using a 200-N weight instead results in “60” on the scale. Using an unknown weight X instead results in “30” on the scale. The weight of X is: a. 10 N b. 20 N c. 30 N d. 40 N e. 50 N SOLUTION: ANSWER: e 60 30 200 N X A 100-N weight results in a scale reading of 40. A 200-N weight results in a scale reading of 60. Since the scale reading for the unknown weight X is 30, we can The difference in weight: calculate its weight based on the scale reading of 40 (which 200 N − 100 N = 100 N corresponds to 100 N): The difference in scale readings: The difference in scale readings: 40 − 30 = 10 60 − 40 = 20 Since each unit on the scale represents 5 N: Therefore, for every 20 units on the scale, the 10 × 5 N = 50 N weight changes by 100 N. This gives us a ratio: Therefore, the weight corresponding to a scale reading of 30 is: 100 N / 20 unit = 5 N per unit 100 N − 50 N = 50 N Question No. 11: A small object of mass m, on the end of a light cord, is held horizontally at a distance r from a fixed support as shown. The object is then released. What is the tension force of the cord when the object is at the lowest point of its swing? a. mg/2 b. mg c. 2mg d. 3mg e. mgr SOLUTION: ANSWER: d To find the tension in the cord when the object is at the lowest point of its swing, we need to consider two forces: 1. The gravitational force acting on the object 2. The centripetal force required to keep it moving in a circular path. Tension Fcentripetal Conservation of Mechanical Energy Fgravitational = mg Tension = Fgravitational + Fcentripetal Tension = mg + 2mg = 3mg Question No. 12: Suppose the string in the figure is 50 cm long. When the ball is released from rest, it swings along the dotted arc. How fast is it going at the lowest point in its swing? a. 2.0 m/s b. 2.2 m/s c. 3.1 m/s d. 4.4 m/s e. 6.0 m/s SOLUTION: ANSWER: c Question No. 13: A block is released from rest at point P and slides along the frictionless track shown. At point Q, its speed is: SOLUTION: ANSWER: d Principle of conservation of mechanical energy Question No. 14: Three identical blocks move either on a horizontal surface, up a plane, or down a plane, as shown below. They start with different speeds and continue to move until brought to rest by friction. They all move the same distance. Rank the three situations according to the initial speeds, least to greatest. a. The same for all cases b. 1, 2, 3 c. 1, then 2 and 3 tie d. 3, 1, 2 e. 2, 1, 3  SOLUTION: ANSWER: d We have three identical blocks moving on surfaces with friction, and they travel the same distance before coming to rest. However, they are moving on different inclines: 1.Block 1 moves on a horizontal surface. 2.Block 2 moves up an incline. 3.Block 3 moves down an incline. We'll rank them based on their initial speeds needed to cover the same distance against friction before stopping. 1.Horizontal Surface (Block 1): 1. The only force opposing the motion is the friction force. 2. The initial kinetic energy must be equal to the work done by friction over the distance. 3. It requires more speed than Block 3 but less than Block 2, as it only needs to overcome friction. 2.Up the Incline (Block 2): 1. Friction opposes the motion, just like in the horizontal case, but gravity also acts against the motion as the block moves up. 2. Therefore, the block needs more initial speed compared to the horizontal case to counter both friction and gravity over the same distance. 3. It needs the greatest initial speed, as it has to overcome both friction and gravity. 3.Down the Incline (Block 3): 1. Friction opposes the motion, but gravity assists it as the block moves down. 2. As a result, the block needs less initial speed compared to the horizontal case because gravity helps it move. 3. It requires the least initial speed (due to the assistance from gravity), it should be ranked first. Therefore: The correct ranking, from least to greatest initial speed, is: 3 (down the incline), 1 (horizontal), 2 (up the incline). Question No. 15: A 0.75-kg block slides on a rough horizontal table top. Just before it hits a horizontal ideal spring its speed is 3.5 m/s. It compresses the spring 5.7 cm before coming to rest. If the spring constant is 1200 N/m, the internal energy of the block and the table top must have: a. not changed b. decreased by 1.9 J c. decreased by 2.6 J d. increased by 1.9 J e. increased by 2.6 J SOLUTION: ANSWER: e Given: m = 0.75 kg (mass of the block) v = 3.5 m/s (initial speed of the block) k = 1200 N/m (spring constant) x = 0.057 m (compression distance converted from 5.7 cm to meters) V = 3.5 m/s x = 5.7 m m = 0.75 m = 0.75 kg kg