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Question-1

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 gravitating objects, and free of electric and magnetic fields.

Solution:
(a) As drop of rain is failing with a constant speed, acceleration is zero, and in accordance with first law of motion, the net force on the drop of rain is zero.
(b) A cork is floating on water, its weight is balanced by the up thrust, the upward force namely buoyancy Therefore, net force on cork is zero.
(c) As the kite is held stationary no net force acts on it. The force exerted by the air on the kite is balanced by the tension produced in the string. 
(d) As the car is moving with a constant velocity, the net force on the car is zero.
(e) As force due to gravity, electric and magnetic fields are zero, no net force acts on the electron.

Question-2

A pebble of mass 0.05 kg 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 alter if the pebble were thrown at an angle of say 45° with the horizontal direction?

Solution:
As the acceleration is always in the vertically downward direction, the direction of force wil also be vertically down in all the three above cases.
(a) When the pebble moves vertically downwards, the magnitude and direction of the net force on the pebble is the same in all the three cases and is given by

Mass of the pebble 'm' = 0.05 kg
Therefore the net force 'F' = ma = 0.05 x 9.8 = 0.49 N

The net force is 0.49 N vertically downwards in all the cases i.e (a), (b) and (c). If the pebble were thrown at an angle of 45o with horizontal, the pebble will have components of velocity along horizontal and vertical. at the highest point the the pebble will not be at rest but will have a horizontal component given by u cos 450 where u is the speed with which the pebble was thrown.

Question-3

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 1ms-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.

Solution:
(a) The net force F = ma = 0.1 x 9.8 N = 0.98 N vertically downwards.
     The net force F acting on the stone is 0.98 N vertically downwards

(b) In this case, as the train is moving with constant speed, it will not affect the motion of the stone ,The magnitude and direction of the net force is same as (a)

(c) The magnitude and direction of the net force is same as (a)

(d) The net force F = ma = 0.1 x 1 N = 0.1 N
     The net force is 0.1 N along the direction of the motion of the train.

Question-4

One end of a string of length l is connected to a particle of mass m and other to a small peg on a smooth horizontal table. If the particles moves in a circle with speed v the net force on the particle (directed towards the centre) is: 
        (i) T       (ii)          (iii)         (iv) 0
        T is the tension in the string. [Choose the correct alternative).

Solution:
The tension acting towards the centre and along the radius provides the necessary centripetal force, i.e.
T =
The particle, if unrestrained by the string, will continue to move in a straight line without any change in velocity. The inertia force is internal to the particle. The weight of the particle = mg and the corresponding reaction of the table are balancing each other and do not figure in the net force. Any net external force acting on a particle should produce an acceleration. The only net force on the particle is T and the centripetal acceleration is the only acceleration on the particle. Hence the answer is (i).

Question-5

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?

Solution:
The magnitude of retarding force 'F' = 50 N
Mass of the body 'm'                      = 20 kg
Acceleration 'a' =  F / m
                      = 50 / 20
                      =- 2.5 m/s2 ( the negative sign meaning retardation)
Velocity of the body 'v' = 15 m/s
Therefore the time taken by the body to stop is given by
                     t = (v - u)/a = 6 s
The body takes 6 s to come to a stop.

Question-6

A constant force acting on a body of mass 3kg changes its speed from 2 ms-1 to 3.5 ms-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution:
Mass of the body 'm' = 3 kg
Initial velocity 'u'        = 2 ms-1
Final velocity 'v'          = 3.5 ms-1
Time taken 't'             = 25 s
Acceleration 'a'    = 0.06 m/s2
Force 'F' = m x a
            = 3 x 0.06
            = 0.18 N
Therefore the magnitude and direction of the force is 0.18 N along the direction of motion.

Question-7

A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body?

Solution:
Mass of the body 'm'               = 5 kg
Force acting on the body    'F1' = 8 N
                                      'F2' = 6 N
Resultant force 'F' = = 10 N
Using the relation F = ma
Acceleration 'a' = 2 m/s2

                                                                                
The acceleration will be produced along the direction of F,

or θ = cos ( 4/ 5 ) with 8N force.
tan θ = F2/ F1 = 6/8 = ¾ therefore
θ = 37o.

Question-8

The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheelers is 400 kg and the mass of the driver is 65 kg?

Solution:
Initial velocity 'u' = 36 km/h = 10 m/s
Mass of the vehicle 'm1'       = 400 kg
Mass of the driver 'm2'         = 65 kg
...Total mass 'M'                 = m1 + m2
                                        = 400 + 65
                                        = 465 kg
                    Time taken 't' = 4s
Acceleration 'a'  = -2.5 m/s2 (negative sign indicates retardation)
Average retarding force 'F' = m x a
                                       = 465 x (-2.5)
                                       = 1162.5 N. (Acting opposite in direction to the motion of the vehicle)
Retarding force =1162.5N.

Question-9

A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5 ms-2. Calculate the initial thrust (force) of the blast?

Solution:
Mass of the rocket 'm' = 20,000 kg ( given)
Initial acceleration 'a1' = 5 ms-2
Acceleration due to gravity 'g' = 9.8 ms-2
... Total acceleration 'a' = a1 + g = 5 + 9.8 = 14.8 ms-2
... The initial thrust of the blast 'F' = m x a = 20,000 x 14.8 = 296000 = 2.96 x 105 N.

Question-10

A particle of mass 0.4 kg moving with a constant speed of 10 ms-1 to the north is subject to a constant force of 8 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the particle at that time to be x = 0, and predict its position at t = -5s, 25s, 100s?

Solution:
Mass of the particle 'm' = 0.4 kg
Initial velocity 'u' = 10 ms-1
Force acting on the particle 'F' = -8 N

(i) At time t = 0, x(0) = 0, and F = -8 N
    At t = -5 s, no force was acting, so a = 0
    xt - x0 = ut + 1 / 2 at2
    xt - 0 = 10
× (-5) = -50 m
    At time t = -5 s, the body was at -50 m

(ii) Position at t = 100 s, xt = ?
     a = = -20 ms-2
     Since the force acts towards south, F is taken as -ve. F acts for 30s so the body retards for 30 s. After 30 s it moves with constant speed.
    Position at 100s = position at 30s + distance traveled in 70s
    Position at 30s, xt = x0 + ut + at2
                              = 0 + 10 x 30 - (20 x 302)
                              = 300 - 9000
                              = -8700 m

    Velocity at the end of 30 s = u + at = 10 + (-20) x 30 = 10 - 600 = -590 ms-2

    Distance traveled in 70 s = Uniform velocity x time = (-590) x 70 = -41,300 m

    Position of the body at 100 s = - 8700 + (-41,300) = - 50, 000 m.

Question-11

A truck starts from rest and accelerates uniformly with 2 ms-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).

Solution:
Initial velocity 'u' = 0
Acceleration due to gravity 'g'       = 9.8 m/s2
Acceleration of the car 'a'             = 2 ms-2
Distance travelled by the stone 'h' = 6 m
Time at which stone is dropped 't' = 10s
Velocity of the car at time 10s 'v1' = u + at
                                                = 0 + 2 x 10
                                                = 20 m/s2
A stone is dropped from the car moving with a horizontal velocity 20 ms-1, and then the stone has the same
horizontal velocity at 11 s.
(a) It's vertical velocity after 1 s is
     v2 = u + at
     where ut is the initial vertical velocity
         = 0 + 9.8 x 1
         = 9.8 ms-1
     Resultant velocity 'v' =
                                   =
                                   = 22.27 ms-1

     Let a be the angle made by v with v1
     tan
α = = = 0.49   
         
α = 26° 
(b) Acceleration at 11 s = 9.8 ms-2
     There is no horizontal acceleration.

Question-12

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speeds of the bob at its mean position are 1 ms-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.

Solution:
(a) The velocity of the bob at its extreme position is zero. Therefore, when the string is cut, it will fall down vertically under the effect of acceleration due to gravity
(b) When the bob is in the mean position, its horizontal component of velocity = 1m/s. and it acts tangentially. Therefore, the bob will trace a parabolic path and will fall down along a parabolic path.

Question-13

A man weighs 70 kg 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-1
 (c) upwards with a uniform acceleration of 5 ms-2.
 What would be the reading of the scale in each case? 
 (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Solution:
The forces acting on the man are,
1. Weight of the man acting vertically downward
2. The reaction R of the weighing machine acting vertically up
(a) When the lift moves upwards with a uniform velocity
                 Ma = R - Mg, 
     Acceleration, a = 0 for uniform velocity
    
R = Mg = 70 × 9.8 = 686 N
    
The reading shown by the machine = = 70 kg f
(b) When the lift is moving downwards with a uniform acceleration of 5 ms-2, we have,
      Ma = Mg - R
      R   = M (g - a) = 70(9.8 - 5)
           = 70
× 4.8 = 336 N
     The reading shown by the machine is = 34.29 kg f
(c) When the lift moves upwards with a uniform acceleration of 5 ms-2
     Ma = R - Mg
     R   = M (g + a) = 70 (9.8 + 5) 
          = 70
× 14.8 = 1036 N
     The reading shown by the machine is = 105.71 kg f
(d) When the lift falls freely under gravity, a = g
     Ma   = Mg - R
     or R = M (g - a) = M (g - g)
         R = 0 since the normal reaction is zero, the machine will not register any reading.

Question-14

Figure below shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4s, 0 < t > 4s? (b) impulse at t = 0 and t = 4 s ? ( Consider one-dimensional motion only).

                   

Solution:
Mass of the particle, m =4 kg,
(a) Force acting on the particle for the interval t < 0 , the particle is not moving, hence no force acts on it.
For t > 4s, x has a constant value 3m, as given by AB, the body is again at rest, ... no force is acting on the body.  
During the interval, 0 < t < 4s, shown by OA in the figure the speed being constant, acceleration is zero. hence . no force acts on the body.
(b) Impulse = Final momentum - initial momentum
     Initial momentum just before t = 0 is 0
     Final momentum just after t = 0 is 4
× 3/4 = 3 kgms-1
    
... (Speed = slope of OA) 
     (Impulse)t = 0 = 3 - 0 = 3kgms-1
   
Similarly, (Impulse)t=0 = 4s = Final momentum - Initial momentum
                                    = 0 - 3 = -3Kgms-1 check font

        

Impulse at t = 4s
Before t = 4s, momentum = 3 kgms-1
After t = 4s, momentum = 0
Therefore change of momentum = 0 - 3 kgms-1
and Impulse = -3 kgms-1.

Question-15

A horizontal force of 500 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a light string. What is the tension in the string? Does the answer depend on which mass end the pull is applied?

              

Solution:
Let us assume that the force is applied, to 20kg mass as shown in the figure.
Let T1 be the tension in the string between the two masses.
The horizontal force F = 500N
Mass m1                   = 10 kg
Mass m2                   = 20 kg
Therefore the acceleration produced in the body of total mass m1 + m2 = 30 kg
We know that acceleration produced in the body of F = ma = 500 = 50 ms-2
                                                                                                  
30       3
Therefore acceleration a = = 500/30 = 50/3 ms-2 = 16.67 ms-2




The tension T1 is acting on the mass m2 = 10 kg which is also under the acceleration a
(a) When 500N pull is applied on 20 kg, tension T1 of the string due to the mass m1 = T1 = m1 x a
= 10 x 16.67
= 167 N

(b) When 500N pull is applied on 10kg tension T2 of the string due to the mass m2 = T2 = m2 x a
= 20 x 16.67
= 333 N.

Question-16

Two masses 8 kg and 12 kg are connected at the 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.

Solution:
Consider two masses m1 and m2 connected to the ends of an inextensible string passing over a smooth frictionless pulley as shown in the fig. Let T be the tension in the string which is uniform throughout. The heavier mass (say m1) moves downward with an acceleration 'a'.
            


a =     
                   = 4 / 20 = 1.96 m/s2                       

The resultant downward force while considering the heavier mass m1 is given by
m1 g - T = m1 a ---------(i)
The resultant upward force while considering mass m2 is given by
T - m2g = m2a ------------- (ii)
Adding (i) and (ii) we have,
(m1 - m2) g = (m1 + m2)a
or a = g ------------ (iii)
Putting the value of 'a' in (i) and simplifying we get,

T = ------------ (iv)
Substituting the given values, we get                                     
T =      
    = 84.08
    = 84 N.

Question-17

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

Solution:
If M is the mass of the nucleus at rest. Initial momentum of the nucleus = 0. Let and be the velocities of two disintegrated parts (masses M1 and M2 respectively) therefore 0 = M1 + M2 . Since and have opposite signs therefore, M1 and M2 move in opposite directions.

Question-18

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

Solution:
Initial momentum of the ball before collision = mv= 0.5x (-6) = -0.30kg-m/s
The ball rebounded with same speed after collision hence its direction changes,
Mass of the ball 'm' = 0.05 kg
Velocity of each ball before collision 'u' = 6 ms-1
Velocity after collision = -u = -6 ms-1
Change in velocity = u - (-u)
= 2u = 12 ms-1
Impulse = Change in momentum
= m x 2u
= 0.05 x 12
= 0.6 kg ms-1.

Question-19

A shell of mass 0.02 kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80 ms-1, what is the recoil speed of the gun?

Solution:
Mass of the shell m = 0.02 kg
Mass of the gun M= 100 kg
Speed of the shell v= 80 ms-1
Recoil speed of the gun V is given by

= -0.016 m/s.

Question-20

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

Solution:
Speed of the ball = 54 km/h chk alignment next line
= = 15 ms-1
The change in momentum will be obtained by finding the resultant of two momenta inclined at an angle of 45° to each other.
Initial momentum of the ball A = m x u = 0.15 x 15 kgms-1
= 2.25 kgms-1
Final momentum B of ball will also be 2.25 kgms-1 because there is no change in the speed of the ball.
The resultant of two momentum


= 4.16 kgms-1
The impulse imparted is the change in momentum and is equal to 4.16 kgms-1 and will act along the bisector of the initial and final directions.

Question-21

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

Solution:
Mass of the stone, M = 0.25 kg
Centripetal acceleration, a = v2 / R [v = tangential velocity]
Angular speed, w = 40 rev/min
= 40
× 2 p / 60 rad/s
= 1.33 p rad/s
Tension in the string, T = M a = M v2 / R
= M R w 2 [v =Rw ]
= 0.25
× 1.5 × 1.332 p 2 N
= 6.54 N
If the tension is to be a maximum of 200 N
200 = 0.25
×1.5 ×w max2
w max2 = 200/0.375
Maximum angular speed, w = 23.09 m/s.

Question-22

If in the previous 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 jerks 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?

Solution:
When the stone is being whirled around, the tension of the string is acting radially. The tension provides the necessary centripetal force for the particle to move in the circle. At the moment the string is cut, the particle is moving with a tangential velocity. Then there is no other force acting on the particle. It therefore continues to maintain the tangential velocity, which it had at the time of separation. Alternative (b) is correct.

Question-23

For ordinary terrestrial experiments, which of the observers below are inertial and which non-inertial:
(a) a child revolving in a giant wheel,
(b) A driver in a sports car moving with a constant high speed of 200 km/h on a straight road,
(c) The pilot of an aero plane which is taking off,
(d) A cyclist negotiating a sharp turn,
(e) The guard of a train which is slowing down t step at a station.

Solution:
(a) A child revolving in a giant wheel has centripetal acceleration. Hence is a non-inertial frame
(b) The driver in a sports car moving with a constant high speed (200 km h-1) is not accelerating, hence is an inertial frame.
(c) When the aeroplane takes off, it has an accelerated motion. Therefore, the pilot of an aeroplane, which is taking off is a non-inertial frame
(d) While negotiating a sharp turn, there is change in direction of motion of the cyclist and hence the motion is an accelerated one. Therefore, a cyclist negotiating a sharp turn is non-inertial observer.
(e) When a train is slowing down to stop at a station, its motion is retarding. Therefore, the guard of a train, who is slowing down to stop at a station is non-inertial observer.

Question-24

One often comes across the following type of statement concerning circular motion: ‘A particle moving uniformly along a circle experiences a force directed towards the centre (centripetal force) and an equal and opposite force directed away from the centre (centrifugal force). The two forces together keep the particle in equilibrium’ Explain what is wrong with this statement.

Solution:
A particle in uniform motion in a circular path has to have a force to move it in a path other than in a straight line. (By Newton’s first law, it will continue to move in a straight line unless acted upon by an external force). This force is the centripetal force = mv2/R directed radially towards the centre. The net force on the particle is not zero and the particle is not in equilibrium. This statement is not true for an inertial observer, that is, an observer from outside the frame of the system, who can perceive only the forces external to the body.
The statement makes some sense to an observer rotating with the particle. He perceives the particle to be at rest and imagines a centrifugal force relative to the particle radially outward balancing the centrifugal force directed inwards. The centrifugal force should not be conceived in an inertial frame.

Question-25

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) a cricketer moves his hands backwards when holding a catch?

Solution:
(a) The reaction force of the ground on the feet of the horse, makes the horse move along with the cart. As the space is empty, there is no reaction. So the horse cannot move and therefore it cannot pull a cart.
(b) When the speeding bus stops suddenly, the e lower part of the body of the passengers come to rest along with the bus but the upper part of the body remains in a state of motion on account of 'inertia of motion'.
(c) The fact that while catching a cricket ball, a cricketer moves his hand backwards, can be explained on the basis of impulse. If he catches the ball abruptly, he will have to apply a large retarding force for a small time, but while moving the hands backwards, he applies a small retarding force, but for a longer time. The reaction on his hand is also less, thereby decreasing the hurt on the palms.

Question-26

Figure below shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle? What is the magnitude of each impulse?

Time in seconds

Solution:
From the figure it is clear that the
1. the speed of the particle changes direction every 2 seconds.

2. In both the directions, the magnitude of the speed is the same as is given by, the slope of the line OA and AB

3.The slope shows the speed = 2/2 =1cm/s

This speed of the particle is constant but velocity is not constant because the direction reverses every 2 seconds. However the magnitude of the velocity remains the same.

The impulse is change in momentum. At points like O and A there is a change in momentum. The velocity of the particle changes from 1 cms-1 to -1 cms-1, i.e., by 2 cm s-1 or 0.02 ms-1

The time between two consecutive impulses received by the aorticle is 2 s
Therefore impulse = change in momentum
= m x change in velocity
= 0.04 kg x 0.02 ms-1
= 8 x 10-4 ms-1

The graph may represent rebounding of a particle between two walls situated at x = 0 and x = 2 cm. Thus the particle receives an impulse of magnitude 8 x 10 -4 kg ms-1 after every two second.

Question-27

Figure below shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? ( Mass of the man = 65 kg).


Solution:

When the man is stationary, w.r.t the belt, the net force acting on him is given by
Acceleration of belt, a = 1m/s-2
Net force on the man = mass (m) of man x a = 65 x 1 = 65 N
Coefficient of static friction between man's shoes and belt, ms = 0.2
The various forces acting are,

(i) weight f the man acting vertically downward = mg = 65 x 9.8 = 637 N.

(ii) Normal reaction of the belt (equal and opposite to the weight of the man), R = 637N.

(iii) The force due to maximum acceleration of rthe belt, F = ma = 65a

(iv) Maximum frictional force,
F' = m mg = 0.2 x 65 x 9.8
The amn would remain stationary, for the value of acceleration, a such that, At equilibrium F = F'
65a = 0.2 x 65 x 9.8
a = ( 0.2 x 65 x 9.8 ) / (65)
a = 1.96ms-2.

Question-28

Here T1, T2 (and v1, v2) denote the tension in the string (and the speed of the stone) at the lowest
and the highest point respectively.

 


Solution:
(a) Lowest point = mg – T1 and Highest point = mg + T1
The three forces acting on the mass are mg (always downward), T1 and T2 (along the string towards the centre), and mv2/R, the centripetal force. The force on the lowest point = T2 - mg (radially inward) = mv2/R. The force experienced by the stone at the lowermost point is T2 - mg. Similarly, the force experienced at the highest point = T1 - mg. Alternative (i) is therefore correct.

Question-29

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms-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.

Solution:
Mass of the helicopter m1 = 1000 kg
Acceleration a = 15 m/s2
The mass of passenger and crew m2 = 300 kg
Acceleration due to gravity = 10 m/s2

(a) Force on the floor by the crew and passengers = m2 x g + m2 x a
= (300 x 10) + (300 x 15)
= 3000 + 4500
= 7500 N
The force on the floor by the crew and passengers is 7500N direct downwards.

(b) Action of the rotor of the helicopter on the surrounding air is given by
= (m1 + m2)g + (m1 + m2)a
= ( m1 + m2 ) ( g + a )
= ( 1000 + 300 ) (10 + 15)
= 1300 + 25
= 32500 N
This force acts vertically downwards.

(c) Applying Newton's 3rd law of motion, we find that the force on the helicopter due to the surrounding air is 32500 N
acts vertically upwards.

Question-30

A stream of water flowing horizontally with a speed of 15 ms-1 gushes out of a tube of cross-sectional area 10-2 m2, and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Solution:
Let us calculate the mass of water flowing out per second and hitting the wall,
Volume of water coming out of A in one second, =
Volume of of water contained in a cylinder of length 15m and area of cross section0.01m2.
Initial velocity 'u' = 15 ms-1
Density of water r = 1000 kgm-3
Suppose the liquid flows from P to Q in t sec, the velocity V = l/t
Volume of cylinder PQ = A x l
Mass of liquid transported in one sec
                                                                            = A V
ρ
Here A = 10-2 m2
Velocity with which it strikes the wall, V = 15 ms-1
= A V
ρ
= 10-2 x 15 x 1000 check alignment previous line
= 150 kgs
-1
The momentum given to the wall in one second by water hitting it = momentum = mass x velocity
=150 x15 = 2250 n
The change of momentum in one second =force
F = = 150 (0 - 15) = - 2250 N
Force exerted by water on wall = 2250 N.

Question-31

Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m kg. 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 8th coin.
(c) The reaction of the 6th coin on the 7th coin.

Solution:
(a) Mass of each coin = m kg. there are three coins each of mass m on the 7th coin. The force on the 7th coin (counted from the bottom) due to the three coins on its top is 3mg. This force is directed vertically downward.

(b) The force on the 7th coin by the 8th coin is 3mg. i.e., each coin is placed one over the other. Hence the force exerted by the 8th coin on the 7th is also 3mg, and this force is directed downward.

(c) Four coins are placed on the 6th coin. Therefore force of 4 mg acts on the 6th coin downwards, the reaction of the 6th coin on the 7th coin is 4mg directed upwards opposite in direction to the force exerted on the 6th coin.

Question-32

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15o. What is the radius of the loop?

Solution:
Speed of the craft, v = 720 km/h = 720,000 / 3600 m/s = 200 m/s
Radial force on the craft to provide centripetal acceleration = m v2 / R [R = radius of the loop]

As per the above force diagram, tan 15o = (mv2/R) / (mg)
R = v2 /g tan 15o = 2002 / (9.8 x 0.268) m = 15,230 m
Radius of the loop = 15.23 km.

Question-33

A train rounds an unbanked circular bend 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 ? The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Solution:
Mass of the train, m = 106 kg
Radius of bend R = 30 m
Speed = 54 km/h = 54,000/3600 m/s = 15 m/s
The lateral thrust required for the centripetal force is provided by the outer rail. (The wheels have flanges inside). The outer wheels rub against the outer rails. Hence the outer rails wear faster.
Forces acting on the train is given in the sketch below.

tan
θ = (mv2/R) / mg = v2 / Rg = 152 / (30 × 9.8) = 0.765
Required angle of banking =
θ = tan-1 0.765 = 37.4o.

Question-34

"It is reasonable to consider earth as an inertial frame for many laboratory scale terrestrial experiments, but the Earth is a non-inertial frame of reference for astronomical observations". Explain this statement. Do you see how the same frame of reference is approximately inertial for one purpose, and non-inertial frame for other?

Solution:
Consider a ball thrown vertically upwards by a person standing on the ground. This person will observe that the ball begins to fall down vertically after sometime. But, to an observer sitting in a moving car, the same ball will appear to be following a parabolic path. So, the same event is described differently by two observers in two different frames of reference. In the first case the frame of reference is the earth which is not accelerating. In this frame of reference Newton's first law works. A reference frame that is not accelerating is called inertial frame of reference, because law of inertial works there. For all practical purposes, earth or any frame of reference set up on earth is regarded as an inertial frame of reference. On the other hand, to the observer sitting in a moving car, he is viewing the motion from an accelerating reference frame which is called the non-inertial reference frame. In this reference frame Newton's first law does not work.

Question-35

A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in the figure. What is the action on the floor by the man in the two cases? If the floor yields to normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding.

Solution:
Let the action by the man on the floor be A and let f (= 25 x g ) be the force the man is exerting to lift the block.
In fig (a) it is clear that the man applies an upward force of 25 kg wt. According to Newton's third law of motion, there will be a downward reaction on the floor.
The action on the floor by the man = 50 kg wt. + 25 kg wt.
                                                        = 75 kg wt.
                                                        = 75 x 9.8 N = 735 N
In the fig. (b), it is clear that the man applies a downward force of 25 kg wt. According to Newton's third law, the reaction is in the upward direction.
In this case, action on the floor by the man = 50 kg wt. - 25 kg wt.
                                                                    = 25 kg wt
                                                                    = 25 x 9.8 N = 245 N
Since the action of the amn on the floor in case b is less than the force at which the floor yields( to a downward force of 700 N )therefore the man should adopt mode (b).

Question-36

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: the monkey?

(a) climbs up with an acceleration of 6 m s-2
(b) climbs down with an acceleration of 4 ms-2
(c) climbs up with a uniform speed of 5 ms-2
(d) falls down the rope nearly freely under gravity?

Solution:
The acceleration due to gravity acts vertically downward. This direction is taken as positive.
(a) When the monkey climbs up with an acceleration, then (T - ma) = mg
(where T represents the tension)
Therefore T = mg + ma = m (g + a)
              T = 40 (10 + 6) ms-1 = 640 N
The rope can withstand a maximum tension of 600N. So the rope will break.

(b) When the monkey is climbing down with an acceleration, then (T + ma) = mg
Therefore T = ( mg - ma )= m (g - a)
              T = 40 (10 - 6) ms-2 = 160 N
The rope will not break.

(c) When the monkey climbs up with uniform speed, then T = mg = 40 x 10= 400 N
The rope will not break.

(d) When the monkey is falling freely, a=g, T = m ( g-g) = 0
it would be a state of weightlessness. So, the tension will be zero and the rope will not break.

Question-37

Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid partition. The co-efficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally at A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B? What happens when the partition is removed? Does the answer to (b) change, when the bodies are in motion? Ignore difference between m s and m k?


Solution:
Mass of the body A, MA = 5 kg
Mass of the body B, MB = 10 kg
Co-efficient of friction = 0.15
Applied force, P = 200 N

(a) Force of friction, F = mR where R is the normal reaction.
R = (5 + 10) x 9.8 kgms-2
  = 147 kgms-2 or 147 N
F = 0.15 x 147 = 22.05 N
Reaction of partition = (P - F) = 200 N - 22.05 N = 177.95 N

(b) Force of friction on body A, FA = m x RA
Where RA is the normal reaction of body A.
But RA = MAg
Therefore FA = mMAg = 0.15 x 5 x 9.8 = 7.35 N
Force exerted by A and B = ( P - FA ) = 200 N - 7.35 N = 192.65 N
The force exerted by B on A is also 192.65 N
According to the third law of motion action reaction pair are equal and opposite,, hence action reaction forces between A and b are 193 N
When the partition is removed, the masses move together with an acceleration 'a' given by
a = net force/total force = 177.95 / 15 = 11.86 ms-2
Yes, when the bodies are in motion the answer to (b) changes.

Question-38

A block of mass 15 kg is placed on a long trolley. The co-efficient of friction between the block and trolley is 0.18. The trolley accelerates from rest with 0.5 ms-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?

Solution:
(a) Frictional force equal to 7.5 N in the direction of motion of the trolley keeps the block at rest relative to the trolley during its accelerated motion.

(b) Law of inertia not valid for the accelerated observer.

Question-39

The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end as shown in the figure. The co-efficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2ms-2. At what distance from the starting point does the box fall off the truck?


Solution:
The maximum force of friction between the box and the surface of truck = F = m x R =m x Mg
Net force on the box = ma - force of friction = (ma - mR)
= (ma - mmg)
= m (a - mg)
= 40 (2 - 0.15 x 9.8)
= 40 (2 - 1.47)
= 21.2 N
The truck is accelerating with 2 ms-2 hence the relative acceleration between the box and the truck = 0.53 ms
-2
The box will start moving with a net acceleration of 0.53 ms-2 towards the open end of the ruck. To fall down the box will have to travel a distance of 5m.
Acceleration of the box relative to truck,
A = = 0.53ms-2
Let t be the time taken by the box to cover a distance of 5m
Then, 5 = 0 x t +(0.53)t2
        t2 = s2
If xt be the distance covered by the truck in time t, then
xt = 0 x
   = 18.87 m.

Question-40

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 two coins will revolve with the record?

Solution:
Speed of revolution of the disc, w = 100/3 rev/min = (100/3) x (2p /60) radians/s = 3.49 rad/s
For the coin to revolve with the record, the static friction force should be enough to resist the centripetal force.
If R = the radius where the coin will just slip, m = mass of the coin, and coefficient of friction = m = 0.15,
m w 2 R = m . m g …(1)
R = m g / w 2 = (0.15 x 9.8) / 3.492 m = 0.12 m
In the equation (1) above, LHS represents the centripetal force and the RHS the frictional force. If R (radius) is more than the value of balance (0.12 m) the centripetal force will be more than the frictional resistance and the coin will slip. This feature is independent of the mass of the coin.
A coin will slip when placed at 14 cm radius and will not slip at 4 cm radius.

Question-41

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?

Solution:
When the motorcyclist is at the highest point of the death-well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards.
These two forces are balanced by the outward centrifugal force acting on him.
R + mg = ---(i)
Here v is the speed of the motorcyclist and m is the mass of the motorcyclist (including the mass of the motor cycle).
Because of the balancing of the forces, the motorcyclist does not fall down.
The minimum speed required to perform a vertical loop is given by equation (i) when R=0.
mg = or = gr = 9.8 × 25 = 15.65 m/s.
So, the minimum speed required to perform a vertical loop is 15.65 m/s.

Question-42

A 70 kg man stands in contact against the wall of a 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?

Solution:
Forces acting on the man when he just begins to slip are marked out in the figure below.
Radius of the drum R = 3m
Coefficient of static friction = m = 0.15
Minimum angular speed = w rad/s
Reaction of the wall = N. This provides the centripetal force required.

N = m w 2 / R = m
× w 2 × 3 = 210 w 2
F = m N = 0.15
× m × w 2 × 3 = 0.45 m w 2. This should equal mg.
mg = 0.45 m w 2
w =
(9.8 / 0.45) = 4.67 rad/s = minimum angular speed.

Question-43

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

Solution:

 

The various forces acting on the bead are given below:

(i) Normal reaction (N) acting at right angles to the direction of the motion of the bead

(ii) weight (mg) of the head acting vertically downward

(iii) The centripetal force acting towards the centre of the circular path.
We can write, mg = N cos
θ -------(i)
and sin
θ = N sin θ -------(ii)
or mRw 2 sin
θ = N sin θ or N = mRw 2
Substituting the above equation in (i), we get
or cos
θ = or cos θ = --------(iii)
1 (cos θ 1)
Thus, the bead will remain at its lowest point for w

From equation (ii), cos
θ = or θ = 60°.

Question-44

A circular race track of radius 300 m is banked at an angle of 15° . If the coefficient of friction between the wheels of a race and the road is 0.2, what is the (a) optimum speed of the race car to avoid wear and tear on its tyres and the (b) maximum permissible speed to avoid slipping?

Solution:
On a banked road, the horizontal component of the normal reaction and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the component of the normal reaction is enough to provide the required centripetal force. In this case, the frictional force is not required. The optimum speed is given by
v0 = (rg tan θ )1/2

v0 = (300 × 9.8 tan 15° )1/2 m/s = 28.1 m/s
The maximum permissible speed is given by
vmax =
Substituting values and simplifying, we get vmax = 38.1 m/s.




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