Question:

Assertion Comets move around the sun in elliptical orbits. The gravitational force on the comet due to sun is not normal to the comets velocity but the work done by the gravitational force over every complete orbit of the comet is zero. Reason Gravitational force is a non conservative force. A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C Assertion is correct but Reason is incorrect D Both Assertion and Reason are incorrect

Answer:

A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Question:

The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the winds energy into electrical energy, and that A = 30 m^2, v = 36 km/h and the density of air is 1.2 kg m^3. What is the electrical power produced?

Answer:

a) The mass of the air passing through the circle in time t is equal to the density of air multiplied by the area of the circle multiplied by the velocity of the wind multiplied by the time t.

Mass = ρ × A × v × t

b) The kinetic energy of the air is equal to one half of the mass multiplied by the velocity squared.

Kinetic energy = ½ × m × v²

c) The electrical power produced is equal to 25% of the kinetic energy of the air.

Electrical power = 0.25 × Kinetic energy

Substituting the given values:

Mass = 1.2 kg m^3 × 30 m^2 × 36 km/h × t

Kinetic energy = ½ × (1.2 kg m^3 × 30 m^2 × 36 km/h × t) × (36 km/h)²

Electrical power = 0.25 × ½ × (1.2 kg m^3 × 30 m^2 × 36 km/h × t) × (36 km/h)²

Therefore, the electrical power produced is equal to 2.16 × 10^6 kg m/s³ × t.

Question:

A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 10^7 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer:

(a) The work done against the gravitational force is equal to the mass (10 kg) multiplied by the height (0.5 m) multiplied by the number of times (1000) the mass is lifted. Therefore, the work done against the gravitational force is 5,000 kgm.

(b) The amount of fat used up is equal to the potential energy lost, divided by the efficiency rate (20%) and the energy supplied by fat (3.8 × 10^7 J/kg). Therefore, the dieter will use up 0.5 kg of fat.

Question:

A bullet of mass 0.012 kg and horizontal speed 70 ms^−1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer:

1. Calculate the momentum of the bullet before it strikes the block: Momentum = mass x velocity = 0.012 kg x 70 ms^−1 = 0.84 kg ms^−1

2. Calculate the momentum of the block after the bullet strikes it: Momentum = mass x velocity = 0.4 kg x 0 ms^−1 = 0 kg ms^−1

3. Calculate the change in momentum of the block: Change in momentum = 0.84 kg ms^−1 - 0 kg ms^−1 = 0.84 kg ms^−1

4. Calculate the impulse of the block: Impulse = Change in momentum / time = 0.84 kg ms^−1 / 0 s = 0 N s

5. Calculate the force of the block: Force = Impulse / time = 0 N s / 0 s = 0 N

6. Calculate the height to which the block rises: Height = Force x time / mass = 0 N x 0 s / 0.4 kg = 0 m

7. Estimate the amount of heat produced in the block: Heat produced = kinetic energy of the bullet x efficiency of the collision = 0.5 x 0.012 kg x (70 ms^−1)^2 x 0.8 = 33.6 J

Question:

A pump on the ground floor of a building can pump up water to fill a tank of volume 30m^3 in 15min. If the tank is 40m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?Dencity of water 10^3kgm^−3

Answer:

1. Calculate the volume of water pumped up in 15 minutes: 30m^3
2. Calculate the mass of the water pumped up in 15 minutes: 30m^3 x 10^3kgm^-3 = 30,000kg
3. Calculate the work done by the pump to lift the water: Work = Mass x Gravity x Height = 30,000kg x 9.8ms^-2 x 40m = 11,520,000 Joules
4. Calculate the efficiency of the pump: 30%
5. Calculate the actual work done by the pump: 11,520,000 Joules x 30% = 3,456,000 Joules
6. Calculate the electric power consumed by the pump: Electric Power = Work / Time = 3,456,000 Joules / 15min = 230,400 Watts

Question:

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s. (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.

Answer:

(a) Work done by applied force in 10 s = Force x Distance = 7 N x 10 m = 70 J

(b) Work done by friction in 10 s = Force x Distance = 0.1 x 7 N x 10 m = 7 J

(c) Work done by net force on the body in 10 s = Work done by applied force - Work done by friction = 70 J - 7 J = 63 J

(d) Change in kinetic energy of the body in 10 s = 0.5 x mass x (final velocity)^2 - 0.5 x mass x (initial velocity)^2 = 0.5 x 2 kg x (final velocity)^2 - 0.5 x 2 kg x (0)^2 = 0.5 x 2 kg x (final velocity)^2

Interpretation: The applied force does work on the body, which increases its kinetic energy. The friction force opposes the motion of the body, and hence does negative work on the body, resulting in a decrease in its kinetic energy. The net force is the vector sum of the applied force and the friction force, and does positive work on the body, resulting in an increase in its kinetic energy. The change in kinetic energy of the body in 10 s is equal to the work done by the net force on the body in 10 s.

Question:

Answer the following: An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

Answer:

1. An artificial satellite orbits the earth in a very thin atmosphere, which means there is very little air resistance to slow down the satellite.

2. As the satellite gets closer to the earth, the gravitational pull of the earth increases, which causes the satellite to accelerate.

3. This acceleration of the satellite causes its speed to increase as it comes closer to the earth.

Question:

A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed and moves with uniform speed there after. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on touching the ground is 10ms^−1

Answer:

1. Work done by the gravitational force on the drop in the first half of its journey: The work done by the gravitational force on the drop in the first half of its journey is equal to the change in the gravitational potential energy of the drop.

Gravitational potential energy of the drop at the initial height (500m) = mgh = (2x10^-3) x (9.8) x (500) = 9.8 J

Gravitational potential energy of the drop at the final height (250m) = mgh = (2x10^-3) x (9.8) x (250) = 4.9 J

Therefore, the work done by the gravitational force on the drop in the first half of its journey = 9.8 J - 4.9 J = 4.9 J

2. Work done by the gravitational force on the drop in the second half of its journey: The work done by the gravitational force on the drop in the second half of its journey is equal to the change in the kinetic energy of the drop.

Kinetic energy of the drop at the initial height (250m) = ½ mv^2 = (2x10^-3) x (10)^2 = 0.2 J

Kinetic energy of the drop at the final height (0m) = ½ mv^2 = (2x10^-3) x (10)^2 = 0.2 J

Therefore, the work done by the gravitational force on the drop in the second half of its journey = 0.2 J - 0.2 J = 0 J

3. Work done by the resistive force in the entire journey: The work done by the resistive force in the entire journey is equal to the change in the kinetic energy of the drop.

Kinetic energy of the drop at the initial height (500m) = 0 J

Kinetic energy of the drop at the final height (0m) = ½ mv^2 = (2x10^-3) x (10)^2 = 0.2 J

Therefore, the work done by the resistive force in the entire journey = 0.2 J - 0 J = 0.2 J

Question:

A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump? (Take g=10 ms−2). A 36.5 kW B 44.4 kW C 52.5 kW D 60.5 kW

Answer:

Step 1: Calculate the rate of flow of water into the tank. Flow rate = Volume of tank/Time taken = 30/15 = 2 m3/min

Step 2: Calculate the power required to lift water from ground level to the tank. Power = Work done/Time taken = (Weight of water x Height)/Time taken = (Density of water x Volume of tank x Height x g)/Time taken = (1000 x 30 x 40 x 10)/15 = 44.4 kW

Hence, the correct answer is B 44.4 kW.

Question:

A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s^−1. What is the speed of the trolley after the entire sand bag is empty ?

Answer:

1. Calculate the total mass of the trolley and sand bag before the sand starts leaking: Mass of trolley = 300 kg Mass of sand bag = 25 kg Total mass = 300 kg + 25 kg = 325 kg

2. Calculate the initial kinetic energy of the trolley: Kinetic energy = (1/2) x mass x velocity2 Kinetic energy = (1/2) x 325 kg x (27 km/h)2 Kinetic energy = 8,737.5 kg m2/s2

3. Calculate the mass of sand that has leaked out of the trolley: Mass of sand leaked = 0.05 kg/s x time Time = Mass of sand bag/Rate of sand leakage Time = 25 kg/0.05 kg/s Time = 500 s

4. Calculate the mass of the trolley after the sand has leaked out: Mass of trolley after sand has leaked out = Mass of trolley before sand leakage - Mass of sand leaked Mass of trolley after sand has leaked out = 300 kg - (0.05 kg/s x 500 s) Mass of trolley after sand has leaked out = 300 kg - 25 kg Mass of trolley after sand has leaked out = 275 kg

5. Calculate the final kinetic energy of the trolley: Kinetic energy = (1/2) x mass x velocity2 Kinetic energy = (1/2) x 275 kg x velocity2 Kinetic energy = 8,737.5 kg m2/s2

6. Calculate the final velocity of the trolley: Velocity = (2 x Kinetic energy) / Mass Velocity = (2 x 8,737.5 kg m2/s2) / 275 kg Velocity = 32.3 km/h

Question:

A (trolley+child) of total mass 200 kg is moving with a uniform speed of 36 km/h on a frictionless track. The child of mass 20 kg starts running on the trolley from one end to the other (10 m away) with the speed of 10 ms^−1 relative to the trolley in the direction of the trolley’s motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Answer:

1. The initial momentum of the trolley and child is 200 kg x 36 km/h = 7200 kgm/s.

2. The final momentum of the trolley and child is 180 kg x v + 20 kg x (10 ms^−1 + v) = 7200 kgm/s, where v is the final velocity of the trolley.

3. Solving for v, we get v = 36 km/h.

4. The trolley has moved 10 m from the time the child begins to run.

Question:

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity. (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:

(a) Positive (b) Negative (c) Negative (d) Zero (e) Negative

Question:

Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? (c) What are the answers to (a) and (b) for an inelastic collision? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Answer:

(a) No, the total kinetic energy is not conserved during the short time of collision of the balls. This is because during the collision, some of the kinetic energy is converted into heat, sound, and other forms of energy.

(b) Yes, the total linear momentum is conserved during the short time of an elastic collision of two balls. This is because momentum is a conserved quantity and it is not affected by the conversion of kinetic energy into other forms of energy.

(c) In an inelastic collision, the total kinetic energy is not conserved, but the total linear momentum is conserved.

(d) It depends on the nature of the potential energy. If the potential energy is an elastic potential energy (i.e. it is proportional to the square of the separation distance between the two balls), then the collision is elastic. If the potential energy is an inelastic potential energy (i.e. it is proportional to the separation distance between the two balls), then the collision is inelastic.

Question:

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to A t^1/2 B t C t^3/2 D t^2

Answer:

A. The displacement of the body in time t is given by D t^2.

Question:

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7ms^−1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?

Answer:

1. Calculate the kinetic energy of the bolt when it hits the floor of the elevator:

KE = 0.5 x 0.3 kg x (7 ms^-1)^2 = 1.95 J

2. Calculate the heat produced by the impact:

Heat = KE = 1.95 J

3. Answer to the question:

The heat produced by the impact is 1.95 J. If the elevator were stationary, the heat produced by the impact would be 0 J.

Question:

A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^−1/2 s^−1. What is the work done by the net force during its displacement from x = 0 to x = 2m ?

Answer:

1. Calculate the acceleration of the body: a = v/x^3/2 a = (ax^3/2)/x^3/2 a = a = 5 m^−1/2 s^−1

2. Calculate the change in velocity: Δv = a*Δx Δv = 5 m^−1/2 s^−1 * 2 m Δv = 10 m^−1/2 s^−1

3. Calculate the work done by the net force: W = FΔx W = ma*Δx W = 0.5 kg * 5 m^−1/2 s^−1 * 2 m W = 5 J

Question:

Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases / decreases / remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic / potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force / sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy / total linear momentum / total energy of the system of two bodies.

Answer:

(a) When a conservative force does positive work on a body, the potential energy of the body increases. (b) Work done by a body against friction always results in a loss of its kinetic energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total linear momentum and the total energy of the system of two bodies.

Question:

State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

(a) True. In an elastic collision, momentum and energy are conserved because the total kinetic energy of the system remains the same.

(b) False. Total energy of a system is not always conserved, as external forces can cause energy to be transferred between the system and its surroundings.

(c) False. Work done in the motion of a body over a closed loop is not necessarily zero for every force in nature. It depends on the type of force and the path taken by the body.

(d) True. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system because some of the energy is converted into other forms, such as heat or sound.

Question:

A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to /n A t^1/2 /n B t /n C t^3/2 /n D t^2

Answer:

Answer: D t^2

Question:

A molecule in a gas container hits a horizontal wall with speed 200 m s^−1 and angle 300 with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?

Answer:

Momentum is conserved in the collision. The collision is elastic.

Step 1: Determine if momentum is conserved.

Momentum is conserved in the collision since the molecule has the same speed and direction before and after the collision.

Step 2: Determine if the collision is elastic or inelastic.

The collision is elastic since the molecule has the same speed and direction before and after the collision.

Question:

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?

Answer:

1. Calculate the initial energy of the pendulum bob: E_initial = mgh = (mass of bob) × (acceleration due to gravity) × (height of the pendulum) E_initial = (0.1 kg) × (9.8 m/s2) × (1.5 m) = 1.47 J

2. Calculate the dissipated energy of the pendulum bob: E_dissipated = 0.05 × E_initial = 0.05 × 1.47 J = 0.0735 J

3. Calculate the final energy of the pendulum bob: E_final = E_initial - E_dissipated = 1.47 J - 0.0735 J = 1.3965 J

4. Calculate the speed of the pendulum bob at the lowermost point: v_final = sqrt (2 × E_final / m) v_final = sqrt (2 × 1.3965 J / 0.1 kg) v_final = 2.82 m/s

Question:

The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere.

Answer:

Answer: The heat energy required for burning is obtained at the expense of the rocket. The rocket’s casing experiences friction as it moves through the atmosphere, causing it to heat up and eventually burn. The heat energy required to cause this burning comes from the kinetic energy of the rocket itself.

Question:

A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.

Answer:

a) To find the area needed to supply 8 kW of power, we need to first calculate the total amount of energy that can be converted to useful electrical energy. This is equal to 200 W/m2 x 0.2, or 40 W/m2.

Next, we need to calculate the area needed to produce 8 kW of power. To do this, we divide 8 kW (8000 W) by 40 W/m2, which gives us an area of 200 m2.

b) The area needed to supply 8 kW of power is much larger than the area of a typical house roof. A typical house roof is usually between 50 and 100 m2, which is much smaller than the 200 m2 needed to produce 8 kW of power.

Question:

A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = -i^ + 2j^​ + 3k^ N where i^, j^​, k^ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?

Answer:

1. Identify the force vector: F = -i^ + 2j^ + 3k^ N

2. Identify the displacement vector: s = 4k^ m

3. Calculate the work done by the force: W = F * s = (-i^ + 2j^ + 3k^ N) * (4k^ m) = 12 Nm

Question:

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10keV, and the second with 100keV. Which is faster the electron or the proton? Obtain the ratio of their speeds. (me​=9.1×10^−31kg, mp​=1.67×10^−27kg,1eV=1.6×10^−19J)

Answer:

1. Calculate the kinetic energy of the electron (10 keV) and the proton (100 keV) in joules:

Electron: 10 keV x 1.6 x 10^-19 J/eV = 1.6 x 10^-17 J

Proton: 100 keV x 1.6 x 10^-19 J/eV = 1.6 x 10^-15 J

2. Calculate the momentum of the electron and the proton:

Electron: Momentum = √(2 x mass x kinetic energy) = √(2 x 9.1 x 10^-31 kg x 1.6 x 10^-17 J) = 2.3 x 10^-24 kg m/s

Proton: Momentum = √(2 x mass x kinetic energy) = √(2 x 1.67 x 10^-27 kg x 1.6 x 10^-15 J) = 1.2 x 10^-22 kg m/s

3. Calculate the speed of the electron and the proton:

Electron: Speed = Momentum/mass = 2.3 x 10^-24 kg m/s/9.1 x 10^-31 kg = 2.5 x 10^7 m/s

Proton: Speed = Momentum/mass = 1.2 x 10^-22 kg m/s/1.67 x 10^-27 kg = 7.2 x 10^7 m/s

4. The proton is faster than the electron.

5. The ratio of their speeds is the proton’s speed divided by the electron’s speed: 7.2 x 10^7 m/s / 2.5 x 10^7 m/s = 2.88

Question:

A body of mass 0.5 kg travels in a straight line with velocity v=ax^3/2 where a=5 m^−1/2s^−1. The work done by the net force during its displacement from x = 0 to x = 2 m is A 1.5 J B 50 J C 10 J D 100 J

Answer:

Answer: B 50 J

Explanation: Step 1: The work done by the net force is given by the formula W = F*x.

Step 2: In this problem, the net force is the force of friction, which is equal to F = ma, where m is the mass of the body and a is the acceleration of the body.

Step 3: The acceleration of the body is a = v/x = 5 m^−1/2s^−1.

Step 4: Substituting the values of m and a in the equation W = Fx, we get W = max = 0.55x^2.

Step 5: Substituting the values of x = 0 and x = 2 in the equation, we get W = 0.552^2 = 50 J.

Therefore, the work done by the net force during its displacement from x = 0 to x = 2 m is 50 J.