SATHEE: Laws of Motion Exercise

Laws of Motion Exercise

Question:

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.(a) Just after it is dropped from the window of a stationary train.(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h.(c) Just after it is dropped from the window of a train accelerating with 1 ms^−2.(d) Lying on the floor of a train which is accelerating with 1 ms^−2, the stone being at rest relative to the train. Neglect air resistance throughout

Answer:

(a) Magnitude: 9.8 N (downwards) Direction: Downwards

(b) Magnitude: 0 N Direction: None

(d) Magnitude: 9.8 N (downwards) Direction: Downwards

Question:

A man of mass 70 kg stands on a weighing scale in a lift which is moving. (a) Upwards with a uniform speed of 10 ms^−1. (b) Downwards with a uniform acceleration of 5 ms^−2. (c) Upwards with a uniform acceleration of 5 ms^−2. What would be the reading on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:

(a) In this case, the reading on the scale would be 770 N, which is equal to the man’s weight plus the force of the lift moving upwards.

(b) In this case, the reading on the scale would be 665 N, which is equal to the man’s weight minus the force of the lift accelerating downwards.

(c) In this case, the reading on the scale would be 875 N, which is equal to the man’s weight plus the force of the lift accelerating upwards.

(d) In this case, the reading on the scale would be 980 N, which is equal to the man’s weight plus the force of gravity acting on the lift as it falls freely.

Question:

A block of mass 15kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5ms−2 for 20s and then moves with uniform velocity. Discuss the motion of the block as viewed by, a) a stationary observer on the ground. b) an observer moving with the trolley.

Answer:

A) A stationary observer on the ground would view the block as having an acceleration of 0.5ms-2 for 20s and then remaining at a constant velocity.

B) An observer moving with the trolley would view the block as remaining stationary for 20s and then having a constant velocity.

Question:

A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h.The mass of the train is 106 kg. What provides the centripetal force required for this purpose?

Answer:

Calculate the linear velocity of the train: Linear velocity = 54 km/h = 54 x (1000/3600) m/s = 15 m/s
Calculate the centripetal force required: Centripetal force = Mass x Velocity2/Radius = 106 x (15)2/30 = 735 N

Question:

A body of mass 0.40 kg moving initially with a constant speed of 10 ms^−1 to the north is subject to a constant force of 8.0 N directed towards the south for 30s. Take the instant the force is applied to be t=0, the position of the body at that time to be x=0, and predict its position at t=5s,25 s,100s.

Answer:

At t=5s: The body has a velocity of -8m/s due to the force. The displacement of the body is -40m.

At t=25s: The body has a velocity of -8m/s due to the force. The displacement of the body is -200m.

At t=100s: The body has a velocity of -8m/s due to the force. The displacement of the body is -800m.

Question:

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lower most point for ω≤√g/R. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω=√2g/R ? Neglect friction.

Answer:

The small bead on the wire loop is subject to the combined forces of gravity and centripetal force.
The centripetal force is equal to mω^2R, where m is the mass of the bead and R is the radius of the loop.
The gravitational force is equal to mg, where g is the acceleration due to gravity.
The two forces must be equal and opposite for the bead to remain at its lowermost point, i.e. mω^2R = mg.
Solving for ω, we get ω = √g/R.
Therefore, for ω ≤ √g/R, the bead will remain at its lowermost point.
For ω = √2g/R, the angle made by the radius vector joining the centre to the bead with the vertical downward direction is 45°.

Question:

A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. In which of the following cases will the rope break. When the monkey (a) Climbs up with an acceleration of 6 ms^−2. (b) Climbs down with an acceleration of 4 ms^−2. (c) Climbs up with a uniform speed of 5 ms^−1. (d) Falls down the rope nearly freely under gravity. (Ignore the mass of the rope)

Answer:

Answer: (a) Climbs up with an acceleration of 6 ms^−2.

Explanation: The tension in the rope is equal to the mass of the monkey multiplied by its acceleration. Therefore, the tension in the rope when the monkey climbs up with an acceleration of 6 ms^−2 is 40 kg x 6 ms^−2 = 240 N. Since this is greater than the maximum tension of 600 N that the rope can withstand, the rope will break.

Question:

Explain why (a) A horse cannot pull a cart and run in empty space. (b) Passengers are thrown forward from their seats when a speeding bus stops suddenly. (c) It is easier to pull a lawn mower than to push it. (d) A cricketer moves his hands backwards while holding a catch.

Answer:

(a) A horse cannot pull a cart and run in empty space because the force of friction between the ground and the cart’s wheels would slow the horse down.

(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly because the momentum of the bus is greater than the passengers’ inertia, so the passengers are forced forward.

(c) It is easier to pull a lawn mower than to push it because the force of friction between the ground and the mower’s wheels is less when the mower is pulled than when it is pushed.

(d) A cricketer moves his hands backwards while holding a catch because this motion helps absorb the shock of the ball hitting the cricketer’s hands, thus reducing the risk of injury.

Question:

Answer:

The equation of motion for the bead on the wire loop is given by mR2θ¨=T-mg, where m is the mass of the bead, R is the radius of the wire loop, θ is the angular displacement of the bead from the lower most point and T is the tension in the wire.
Solving the equation of motion for the angular frequency ω, we get ω2=g/R-T/mR2.
For ω≤√g/R, the tension in the wire T is greater than mg, and the bead remains at its lower most point.
For ω=√2g/R, the tension in the wire is equal to mg, and the angle made by the radius vector joining the centre to the bead with the vertical downward direction is 45°.

Question:

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Answer:

A nucleus is composed of positively charged protons and neutrons, which are held together by the strong nuclear force.
When a nucleus decays, it releases energy in the form of particles or radiation.
The energy released by the decay is equal to the mass of the nucleus before the decay, minus the mass of the two daughter nuclei.
This energy is conserved and must be transferred to the daughter nuclei in the form of kinetic energy.
Since the total kinetic energy of the system must remain constant, the two daughter nuclei must move in opposite directions in order to conserve momentum.

Question:

Two billiard balls each of mass 0.05kg moving in opposite directions with speed 6ms^−1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Answer:

Calculate the momentum of each ball before the collision: Momentum of Ball 1 = 0.05kg x 6ms^-1 = 0.3kgms^-1 Momentum of Ball 2 = 0.05kg x 6ms^-1 = 0.3kgms^-1
Calculate the impulse imparted to each ball due to the other: Impulse of Ball 1 = Momentum of Ball 2 - Momentum of Ball 1 = 0.3kgms^-1 - 0.3kgms^-1 = 0kgms^-1

Impulse of Ball 2 = Momentum of Ball 1 - Momentum of Ball 2 = 0.3kgms^-1 - 0.3kgms^-1 = 0kgms^-1

Question:

Answer:

(a) A horse cannot pull a cart and run in empty space because the horse does not have enough energy to pull the cart and still have enough energy to run in the empty space.

(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly because of inertia. When the bus is speeding, the passengers are moving forward at the same speed as the bus. When the bus stops suddenly, the passengers’ inertia causes them to continue moving forward, resulting in them being thrown forward from their seats.

(c) It is easier to pull a lawn mower than to push it because when you pull a lawn mower, you are using your body’s own momentum to move the lawn mower forward. When you push a lawn mower, you are using your body’s own energy to move the lawn mower forward, which requires more energy.

(d) A cricketer moves his hands backwards while holding a catch because this helps to reduce the impact of the ball on his hands. By moving his hands backwards, the cricketer is able to absorb some of the force of the ball and reduce the impact on his hands.

Question:

Answer:

Answer: (a) Climbs up with an acceleration of 6 ms^−2.

The tension in the rope will be equal to the mass of the monkey multiplied by the acceleration, which is 40 kg x 6 ms^−2 = 240 N. This is less than the maximum tension of 600 N, so the rope will not break.

(b) Climbs down with an acceleration of 4 ms^−2.

The tension in the rope will be equal to the mass of the monkey multiplied by the acceleration, which is 40 kg x 4 ms^−2 = 160 N. This is less than the maximum tension of 600 N, so the rope will not break.

The tension in the rope will be equal to the mass of the monkey multiplied by the acceleration, which is 40 kg x 0 ms^−2 = 0 N. This is less than the maximum tension of 600 N, so the rope will not break.

(d) Falls down the rope nearly freely under gravity.

The tension in the rope will be equal to the mass of the monkey multiplied by the acceleration due to gravity, which is 40 kg x 9.8 ms^−2 = 392 N. This is less than the maximum tension of 600 N, so the rope will not break.

Question:

Give the magnitude and direction of the net force acting on (a) A drop of rain falling down with a constant speed. (b) A cork of mass 10 g floating on water. (c) A kite skillfully held stationary in the sky. (d) A car moving with a constant velocity of 30 km/h on a rough road. (e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Answer:

(a) Magnitude: 0 N, Direction: Downward (b) Magnitude: 0 N, Direction: Upward (c) Magnitude: 0 N, Direction: Stationary (d) Magnitude: Dependent on the force of friction, Direction: Opposite to the direction of motion (e) Magnitude: 0 N, Direction: Stationary

Question:

A disc revolves with a speed of 33^1/3 rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Answer:

A disc is revolving with a speed of 33^1/3 rev/min, and has a radius of 15 cm.
Two coins are placed at 4 cm and 14 cm away from the centre of the record.
The coefficient of friction between the coins and the record is 0.15.
The coin which is placed at 14 cm away from the centre of the record will revolve with the record, as it is closer to the centre and has a higher coefficient of friction.

Question:

A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Answer:

Calculate the centripetal force acting on the man: Fc = mv2/r = (70 kg)(200 rev/min)(2π radians/rev)(3 m) = 84,640 N
Calculate the static friction force needed to keep the man stuck to the wall: Fs = μsFc = (0.15)(84,640 N) = 12,696 N
Calculate the minimum rotational speed of the cylinder needed to keep the man stuck to the wall: vmin = √(Fs/mr) = √(12,696 N/[(70 kg)(3 m)]) = 40.3 rev/min

Question:

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer:

Determine the total mass of the system. Total mass = 8 kg + 12 kg = 20 kg
Calculate the acceleration of the system. Acceleration = F/m = Tension/Total mass = Tension/20 kg
Calculate the tension in the string. Tension = m x a = 20 kg x Acceleration = 20 kg x (F/m) = 20 kg x (Tension/20 kg) = Tension

Question:

A pebble of mass 0.05kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45o with the horizontal direction? Ignore air resistance.

Answer:

a) During its upward motion, the net force on the pebble is in the upward direction with a magnitude of mg, where m is the mass of the pebble and g is the acceleration due to gravity. Therefore, the net force on the pebble during its upward motion is in the upward direction with a magnitude of 0.05kg * 9.8m/s^2 = 0.49N.

b) During its downward motion, the net force on the pebble is in the downward direction with a magnitude of mg. Therefore, the net force on the pebble during its downward motion is in the downward direction with a magnitude of 0.05kg * 9.8m/s^2 = 0.49N.

c) At the highest point where the pebble is momentarily at rest, the net force on the pebble is zero.

Do your answers change if the pebble was thrown at an angle of 45o with the horizontal direction?

No, the answers do not change. The net force on the pebble during its upward and downward motion will still be in the upward and downward direction, respectively, with a magnitude of 0.05kg * 9.8m/s^2 = 0.49N.

Question:

A constant force acting on a body of mass 3.0kg changes its speed from 2.0m s^−1 to 3.5m s^−1 in 25s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Answer:

Calculate the change in velocity (Δv): Δv = 3.5 m/s - 2.0 m/s = 1.5 m/s
Calculate the acceleration (a): a = Δv/t = 1.5 m/s / 25 s = 0.06 m/s^2
Calculate the force (F): F = ma = 3.0 kg × 0.06 m/s^2 = 0.18 N
Determine the direction of the force: The direction of the force is the same as the direction of the motion of the body.

Question:

A stream of water flowing horizontally with a speed of 15ms^−1 gushes out of a tube of cross-sectional area 10^−2m^2, and hits a vertical wall normally. Assuming that it does not rebound from the wall, the force exerted on the wall by the impact of water is: A 1.25×103N B 2.25×103N C 3.25×103N D 4.25×103N

Answer:

Step 1: Calculate the mass flow rate of the water.

Mass flow rate = (density of water) x (speed) x (cross-sectional area)

Mass flow rate = (1000 kg/m3) x (15 m/s) x (10-2 m2)

Mass flow rate = 1.5 kg/s

Step 2: Calculate the force exerted on the wall.

Force = (mass flow rate) x (acceleration)

Force = (1.5 kg/s) x (9.8 m/s2)

Force = 14.7 N

Answer: A 1.25×103N

Question:

You may have seen in a circus a motorcyclist driving in vertical loops inside a death well (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

Answer:

At the uppermost point, the motorcyclist is experiencing centripetal force which is provided by the friction between the motorcycle tires and the wall of the death well. This force is directed towards the center of the death well, causing the motorcyclist to travel in a circular motion.
The minimum speed required to perform a vertical loop in a death well with a radius of 25 m is equal to the centripetal acceleration of the motorcycle, which is equal to the square of the radius of the death well divided by the time it takes for the motorcyclist to complete one revolution. Therefore, the minimum speed required is equal to 25 m/s.

Question:

One end of string of length l is connected to a particle of mass ’m’ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed ‘v’, the net force on the particle (directed towards center) will be (T represents the tension in the string) A Zero B T C T+mv^2/l D T−mv^2/l

Answer:

Answer: C) T+mv^2/l

Question:

A. A constant retarding force of 50 N applied to a body of mass 20 Kg moving initially with a speed of 15 m s−1. How long does the body take to stop? B. A constant force acting on a body of mass 3.0 Kg change its speed from 2.0 m s^−1 to 3.5 m s^−1 in 25 s. The direction of the motion of the body remain same, the magnitude of the force will be

Answer:

A. The body will take to stop when its velocity reaches 0 m/s.

We can use the equation F = ma to calculate the acceleration of the body, which is given by:

a = F/m = 50 N/20 kg = 2.5 m/s^2

Now, we can use the equation v = u + at to calculate the time taken for the body to stop, where u is the initial velocity of 15 m/s and a is the acceleration of 2.5 m/s^2.

t = (v - u)/a = (0 - 15) / 2.5 = -6 s

Therefore, the body will take 6 seconds to stop.

B. We can use the equation F = ma to calculate the force acting on the body, which is given by:

F = ma = (3.0 kg)((3.5 m/s - 2.0 m/s) / 25 s) = 0.24 N

Therefore, the magnitude of the force acting on the body is 0.24 N.

Question:

A rocket with a lift-off mass 20,000kg is blasted upwards with an initial acceleration of 5.0ms^−2. Calculate the initial thrust (force) of the blast.

Answer:

Use the equation F = ma, where F is the force (thrust), m is the mass, and a is the acceleration.
Substitute the given values: F = (20,000kg)(5.0ms^−2)
Solve: F = 100,000 N

Question:

A truck starts from rest and accelerates uniformly at 2.0ms^−2. At t=10s, a stone is dropped by a person standing on the top of the truck (6m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t=11s ? (Neglect air resistance.)

Answer:

a) Velocity of the stone at t=11s = 0 m/s

b) Acceleration of the stone at t=11s = -9.8 ms^-2 (due to gravity)

Question:

A bob of mass 0.1kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speed of the bob at its mean position is 1ms^−1.What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

Answer:

(a) If the string is cut when the bob is at one of its extreme positions, then the bob will move in a straight line along the same direction in which it was oscillating.

(b) If the string is cut when the bob is at its mean position, then the bob will move in a parabolic trajectory with an initial velocity of 1ms^−1.

Question:

A shell of mass 0.020kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80ms^−1. what is the recoil speed of the gun?

Answer:

Calculate the momentum of the shell: Momentum of the shell = mass of the shell x muzzle speed of the shell Momentum of the shell = 0.020kg x 80ms^−1 Momentum of the shell = 1.6kgms^−1
Calculate the momentum of the gun: Momentum of the gun = mass of the gun x recoil speed of the gun Momentum of the gun = 100kg x recoil speed of the gun
Since momentum must be conserved, the momentum of the shell must be equal to the momentum of the gun. Therefore, 1.6kgms^−1 = 100kg x recoil speed of the gun
Solve for the recoil speed of the gun: Recoil speed of the gun = 1.6kgms^−1 / 100kg Recoil speed of the gun = 0.016ms^−1

Question:

A batsman deflects a ball by an angle of 45o without changing its initial speed which is equal to 54km/h. what is the impulse imparted to the ball? (mass of the ball is 0.15kg)

Answer:

Step 1: Calculate the initial velocity of the ball. Initial velocity = 54 km/h = 15 m/s

Step 2: Calculate the final velocity of the ball. Final velocity = 15 m/s (same as initial velocity since the ball is deflected by an angle of 45o without changing its initial speed)

Step 3: Calculate the change in velocity (Δv). Δv = Final velocity - Initial velocity = 15 m/s - 15 m/s = 0 m/s

Step 4: Calculate the impulse imparted to the ball. Impulse = Mass of the ball x Change in velocity = 0.15 kg x 0 m/s = 0 Ns

Question:

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Answer:

To calculate the tension in the string, use the formula T = mv2/r, where T is the tension, m is the mass of the stone, v is the speed, and r is the radius of the circle.
Substituting the given values, we get T = (0.25 kg)(40 rev/min)(2π radians/rev)(1.5 m)2 = 75.398 N.
To calculate the maximum speed with which the stone can be whirled around, use the formula vmax = √(Tmax/m)r, where Tmax is the maximum tension the string can withstand, m is the mass of the stone, and r is the radius of the circle.
Substituting the given values, we get vmax = √(200 N/0.25 kg)(1.5 m) = 80 rev/min.

Question:

The speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

Answer:

Answer: (b) the stone flies off tangentially from the instant the string breaks.

Question:

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m/s^2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the(a) Force on the floor by the crew and passengers.(b) Action of the rotor of the helicopter on the surrounding air.(c) Force on the helicopter due to the surrounding air.

Answer:

(a) The magnitude of the force on the floor by the crew and passengers is equal to the sum of their masses multiplied by the vertical acceleration of the helicopter. This can be calculated as follows:

Force on Floor = (1000 kg + 300 kg) x 15 m/s^2 = 4500 N (downward)

(b) The action of the rotor of the helicopter on the surrounding air is a force in the opposite direction of the helicopter’s motion. This can be calculated as follows:

Force of Rotor on Air = (1000 kg + 300 kg) x 15 m/s^2 = 4500 N (upward)

(c) The force on the helicopter due to the surrounding air is equal to the sum of the masses of the helicopter and its passengers multiplied by the acceleration due to gravity. This can be calculated as follows:

Force of Air on Helicopter = (1000 kg + 300 kg) x 9.8 m/s^2 = 4900 N (downward)

Question:

Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of (a) The force on the 7th coin (counted from the bottom) due to all the coins on its top. (b) The force on the 7th coin by the eighth coin. (c) The reaction of the 6th coin on the 7th coin.

Answer:

(a) The force on the 7th coin (counted from the bottom) due to all the coins on its top is 10mg in the downward direction.

(b) The force on the 7th coin by the eighth coin is 1mg in the downward direction.

Question:

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 150. What is the radius of the loop?

Answer:

Convert the speed from kilometers per hour to meters per second: 720 km/h = 200 m/s
Calculate the centripetal acceleration required for the loop: a = (v^2)/r
Calculate the centripetal force required for the loop: F = m*a
Calculate the bank angle of the aircraft: θ = tan^-1 (v^2/rg)
Substitute the given values into the equation for the bank angle: θ = tan^-1 (200^2/rg)
Solve for the radius of the loop: r = (200^2)/(g*tan 150)
The radius of the loop is approximately 656 meters.

JEE NCERT Solutions (Physics)

Laws of Motion Exercise

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