1
FORCE AND NEWTON’S LAWS OF MOTION
EFFECTS OF FORCE
To define force first of all one has to see the effects of force. By ‘effects of force’ we mean what force can do
or what changes a force can bring about. Effects of Force :
A force can produce the following effects : (i) A force can move a stationary body. (ii) A force can stop a moving body.
(iii) A force can change the speed of a moving body. (iv) A force can change the direction of a moving body. (v) A force can change the shape (and size) of a body.
Based on the effects of force, it may be defined as : Force is a pull or push, which changes or tends to change the state of rest or of uniform motion of a body or changes its direction or shape.
(a ) Mathematical Representation of Force :
Mathematically, force F is equal to the product of mass, m of a body and acceleration a, produced in the body due to that force.
i.e. F = ma
Where a = final velocity – initial velocity/time
(b) Units of Force : (i) In C.G.S. system : F = ma gram × cm/s
2 = dyne
If m = 1 gram, a = 1 cm/s2, then F = 1 dyne
When a force is applied on a 1 gram body and the acceleration produced in the body is 1 cm/s2 then the
force acting on the body will be one dyne. (ii) In S.I. system :
F = ma kg × m/s2 = Newton
If m = 1 kg and a = 1 m/s2 then by F = ma,
F = 1 × 1 = 1 kg × m/s2 = 1 Newton.
If a force is applied on a body of mass 1 kg and acceleration produced in the body is 1 m/s2 then the
force acting on the body will be one Newton.
Relationship between the newton and dyne
1 N = 1 kg × 1 m s–2 = 1000 g × 100 cm s–2 = 100000 g cm s–2 = 105 dyne Thus 1 N = 105 dyne ILLUSTRATIONS
1. A force produces an acceleration of 5.0 cm/s2 in a body
of mass 20g. Then find out the force acting on the body in Newton.
Sol. Acceleration of the body, a = 5 cm/s = 0.05 m/s Mass of the body, m = 20 g = 0.02 kg
F = ma F = 0.05 × 0.02 = 10–3 N
2. A force of 15 N acts on a body of mass 5 kg for 2s. What is the change in velocity of body ?
Sol. Given : F = 15 N , t = 2s , m = 5 kg F = ma a = m F = 5 15 = 3 m/s2 a = t u v v – u = at = 3 × 2 = 6m/s RESULTANT FORCE
Many forces may be simultaneously applied on a body, for example- several persons may jointly make an effort to move a heavy body, each person pushes it i.e. each person applies a force on it. t is also possible that a
stronger man pushes that body hard enough and produces same acceleration in it. f a single force
acting on a body produces the same acceleration as produced by a number of forces, then that single force is called the resultant force of these individual forces .
FUNDAMENTAL FORCES
All forces observed in nature such as muscular force, tension, reaction, friction, weight, electric, magnetic, nuclear, etc., can be explained in terms of only following four basic interactions.
(a ) Gra vit ational Force :
The force of interaction which exists between two particles of masses m1 and m2, due to their masses is called gravitational force. The gravitational force acts over long distances and does not need, any intervening medium. Gravitational force is the weakest force of nature.
( b) Elec t roma gne tic Forc e :
Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force
2 (i) These can be attractive or repulsive.
(ii) These are long range forces.
(iii) These depend on the nature of medium between the charged particles.
(iv) All macroscopic forces (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and force experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions and repulsions between atoms/molecules.
( c ) Nuc lear Forc e :
It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, fission, and fusion, etc. results because of unbalancing of nuclear forces. It acts within the nucleus that too upto a very small distance. It does not depends on charge and acts equally between a proton and proton, a neutron and neutron, and proton and neutron, electrons does not experience this force. It acts for very short distance order of 10–15 m.
(d) Weak Force :
It acts between any two elementary particles. Under its action a neutron can change into a proton emitting an electron and a particle called antineutrino. The range of weak force is very small, in fact much smaller than size of a proton or a neutron.
It has been found that for two protons at a distance of 1 fermi :
FN:FEM:FW:FG::1:10–2:10–7:10–38
On the basis of contact forces are classified into two categories
(i) Contact forces
(ii) Non contact or field forces (a) Contact force :
Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other. e.g. Normal, Tension etc.
( b ) Field forc e :
Force which acts on an object at a distance by the interaction of the object with the field produced by other object is called field force. e.g. Gravitational force, Electro magnetic force etc.
DETAILED ANALYSIS OF CONTACT FORCE
( a ) Normal force (N) :
It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force.
e.g.1 A table is placed on Earth as shown in figure
Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure.
e.g.2 A boy pushes a block kept on a frictionless surface.
Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force.
A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block.
Normal force exerted by block on the surface of inclined plane is shown in figure. Here normal force is a component of weight of the body perpendicular to the inclined surface i.e. N = mgsin
3 3. Two blocks are kept in contact on a smooth surface as
shown in figure. Draw normal force exerted by A on B.
Sol. In above problem, block A does not push block B, so there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero. Note :
• Normal is a dependent force it comes in role when
one surface presses the other.
(b) Tension :
Tension is the magnitude of pulling force exerted by a string, cable, chain, rope etc. W hen a string is connected to a body and pulled out, the string said to be under tension. It pulls the body with a force T, whose direction is away from the body and along the length of the string. Usually strings are regarded to be massless and unstretchable, known as ideal string.
Note : (i) Tension in a string is an electromagnetic force and it arises only when string is pulled. If a massless string is not pulled, tension in it is zero. (ii) String can not push a body in direct contact. (c ) Forc e Exert ed by spring :
A spring is made of a coiled metallic wire having a definite length. When it is neither pushed nor pulled then its length is called natural length.
At natural length the spring does not exert any force on the objects attached to its ends.f the spring is pulled at the ends, its length becomes larger than its natural length, it is known as stretched or extended spring. Extended spring pulls objects attached to its ends.
A B A B A B Normal spring Stretched spring Compressed spring
Spring force on A Spring force on B
Spring force on A Spring force on B
If the spring is pushed at the ends, its length becomes less than natural length. It is known as compressed spring. A compressed spring pushes the objects attached to its ends.
F x F F x F F = 0 spring in natural length does not exerts any force on its ends
F = – kx x = compression in spring F = – kx ;k = spring constant or stiffness constant (unit = N/m) x = extension in spring Fext Fext
Note : Spring force is also electromagnetic in nature :
(d) Friction force :
When a body is moving on a rough surface resistance to the motion occurs because of the interaction between the body and its surroundings. We call such resistance as force of friction. Friction is also considered as component of contact force which acts parallel to the surfaces in contact.
(i) Origin of friction : The frictional force arises due to molecular interactions between the surfaces at the points of actual contact. When two bodies are placed one over other, the actual area of contact is much smaller then the total surface areas of bodies. The molecular forces starts operating at the actual points of contact of the surfaces. Molecular bonds are formed at these contact points. When one body is pulled over the other, these bonds are broken, and the material get deformed and new bonds are formed. The local deformation sends vibrations into the bodies. These Vibrations ultimately dumps out and energy of vibrations appears as heat. Hence to start or carry on the motion, there is a need of force.
Body 1
Body 2
Actual area of contact (ii) Statics and Kinetic Frictions :
• Experiment :
(A) Consider a block placed on a table, and a small force F1 is acted on it. The block does not move. It indicates that the frictional force fs starts acting in opposite direction of applied force and its magnitude is equal of F1(figure b). That is for the equilibrium of
4 the block, we have
F1– fs = 0 or F1 = fs
The force of friction when body is in state of rest over the surface is called static friction (fs).
(B) As the applied force increases the frictional force also increases. When the applied force is increased up to a certain limit (F2) such that the block is on the verge of motion. The value of frictional force at this stage is called limiting friction flim (figure c).
(C) Once the motion started, the smaller force is now necessary to continue the motion (F3) and thus frictional force decreases. The force of friction when body is in state of motion over the surface is called kinetic or dynamic friction fk (figure d).
(iii) More about frictional force : (A) About static friction
1. The limiting friction depends on the materials of the surfaces in contact and their state of polish.
2. The magnitude of static friction is independent of the apparent area of contact so long as the normal reaction remains the same.
3. The limiting friction is directly proportional to the magnitude of the normal reaction between the two surfaces i.e. flim= SN. Here
s is coefficient of static friction.
We can write, s = N
flim
(B) About kinetic friction :
1. The kinetic friction depends on the materials of the surface in contact.
2. It is also independent of apparent area of contact as long as the magnitude of normal reaction remains the same.
3. Kinetic friction is almost independent of the velocity, provided the velocity is not too large not too small.
4. The kinetic friction is directly proportional to the magnitude of the normal reaction between the surfaces.
fk = k N. Here k is coefficient of kinetic friction.
We can write, k =
N fk
• There are two types of kinetic frictions:
(i) Sliding friction : The force of friction when one body slides over the surface of the another body is called sliding friction.
(ii) Rolling friction : When a wheel rolls without slipping over a horizontal surface, there is no relative motion of the point of contact of the wheel with respect to the plane. Theoretically for a rolling wheel the frictional force is zero. This can only possible when bodies in contact are perfectly rigid and contact of wheel with the surface is made only at a point. But in practice no material body is perfectly rigid and therefore bodies get deformed when they pressed each other. The actual area of their contact no longer remains a point, and thus a small amount of friction starts acting between the body and the surface. Here frictional force is called rolling friction. It is clear from above discussion that rolling friction is very much smaller than sliding friction.
flim > fkinetic > frolling. Note :
s and k are dimensionless quantities and
independent of shape and area of contact . It is a property of two contact surfaces. s will always be
greater than k .Theoretical value of can be o to but
5 ( a ) C ons e rv a t i v e Forc e :
A force is said to be conservative if the amount of work done in moving an object against that force is independent on the path. One important example of conservative force is the gravitational force. It means that amount of work done in moving a body against gravity from location A to location B is the same whichever path we may follow in going from A to B. This is illustrated in figure.
A force is conservative if the total work done by the force on an object in one complete round is zero, i.e. when the object moves around any closed path (returning to its initial position).
A force is conservative if there is no change in kinetic energy in one complete round. KE = 0
This definition illuminates an important aspect of a conservative force viz. Work done by a conservative force is recoverable. Thus in figure, we shall have to do mgh amount of work in taking the body from A to B. However, when body is released from B, we recover mgh of work.
Other examples of conservative forces are spring force, electrostatic force etc.
( b) Non-C ons e rv a t iv e Forc e :
A force is non-conservative if the work done by that force on a particle moving between two points depends on the path taken between the points.
The force of friction is an example of non-conservative force. Let us illustrate this with an instructive example. Suppose we were to displace a book between two points on a rough horizontal surface (such as a table). If the book is displaced in a straight line between the two points, the work done by friction is simply FS where : F = force of friction ;
S = distance between the points.
However, if the book is moved along any other path between the two points (such as a semicircular path), the work done by friction would be greater than FS. Finally, if the book is moved through any closed path, the work done by friction is never zero, it is always negative. Thus the work done by a non-conservative force is not recoverable, as it is for a conservative force.
GALILEO’S EXPERIMENTS Experiment 1 :
It was observed by Galileo that when a ball is rolled down on an inclined frictionless plane its speed increases, whereas if it is rolled up an inclined frictionless plane its speed decreases .If it is rolled on a horizontal frictionless plane the result must be between the cases describe above i.e. the speed should remain constant. It can be explain as :
v’
v
v’ = v
moving down : speed increases moving up : speed decreases moving horizontal : speed remains constant Experiments 2 :
When a ball is released on the inner surface of a smooth hemisphere, it will move to the other side and reach the same height before coming to rest momentarily. f the hemisphere is replaced by a surface
shown in figure(b) in order to reach the same height the ball will have to move a larger distance.
(a) h (b) h (c) v v
If the other side is made horizontal, the ball will never stop because it will never be able to reach the same height, it means its speed will not decrease. It will have uniform velocity on the horizontal surface. Thus, if unbalanced forces do not act on a body, the body will either remain at rest or will move with a uniform velocity. It will remain unaccelerated.
Newton concluded the idea suggested by Galileo and
6 NEWTON’S FIRST LAW OF MOTION
Every body remain in its state of rest or uniform motion in a straight line unless it is compelled by some external force.
It means a body remain unaccelerated if and only if, the resultant force on it is zero.
In such a case the body is said to be in equilibrium.
INERTIA
(a ) De finition of Ine rtia :
The tendency of the body to oppose the change its states of rest or uniform motion in a straight line is called inertia. Newton’s first law of motion is also called
law of inertia.
( b) D e s c r ipt ion :
It follows from first law of motion that in absence of any external force, a body continues to be in its state of rest or in uniform motion along a straight line. In other words, the body cannot change by itself its position of rest or of uniform motion.
(c ) Ine rt ia De pe nds upon Mas s :
We know that it is difficult to move a heavier body than the lighter one. Similarly it is difficult to stop a moving heavier body than a lighter body moving with the same velocity. Thus, we conclude that mass of the body is the measure of inertia, more the mass, more the inertia.
TYPES OF INERTIA
There are three types of Inertia which are :
(a ) Inert ia of Rest :
The tendency of the body to oppose the change in its state of rest when some external unbalance force is applied on it, is called the inertia of rest.
Example based on Inertia of rest :
A person sitting in a bus falls backwards when the bus suddenly starts. The reason is that lower part of his body begins to move along with the bus but the upper part of his body tends to remain at rest due to inertia of rest.
(b) Inert ia of Mot ion :
The tendency of the body to oppose its state of motion when some unbalance forces are applied on it, is called the inertia of motion.
Example based on Inertia of motion :
A man carelessly getting down a moving bus falls forward, the reason being that his feet come to rest suddenly, whereas the upper part of his body retains the forward motion.
(c) Inertia of Direction :
The tendency of a body to oppose any change in its direction of motion is known as inertia of direction. Example based on Inertia of direction :
Tie a stone to one end of a string and holding other end of the string in hand, rotate the stone in a horizontal circle. If during rotation, the string breaks at certain stage, the stone is found to fly off tangentially at that point of the circle.
String Breaks
String breaks, stone goes away tangentially
Definition of force from first law of motion :
According to first law of motion, if there is no force, there is no change in state of rest or of uniform motion. In other words, if a force is applied, it may change the state of rest or of uniform motion. If the force is not sufficient, it may not produce a change but only try to do so. Hence force is that which changes or tries to change the state of rest or of uniform motion of a body in straight line.
7 MOMENTUM
D e finit ion :
Momentum of a particle may be defined as the quantity of motion possessed by it and it is measured by the product of mass of the particle and its velocity. Momentum is a vector quantity and it is represented by p mv p Unit of mome nt um :
(In C.G.S. system) p = mv gram × cm/s = dyne × s
(In M.K.S. system)p = mv kg × m/s = Newton × s
4. A ball of mass 100 gm. is moving with a velocity of 15 m/s. Calculate the momentum associated with the ball.
Sol .
Mass of the ball = 100 gm. = 1000
100 kg. = 0.1 kg.
Velocity of the ball = 15 m/s
So, momentum = mass of the ball × velocity of
the ball
= 0.1 kg. × 15 m/s
= 1.5 kg. m/s
NEWTON’S SECOND LAW OF MOTION
The rate of change of momentum of a body is directly proportional to the applied unbalanced forces i.e. Rate of change of momentum Force applied
Let a body is moving with initial velocity u and after applying a force F on it, its velocity becomes v in time t.
Initial momentum of the body p1 = mu Final momentum of the body p2 = mv Change in momentum in time t is mv – mu
So rate of change of momentum = t
mu – mv
But according to Newton’s second law,
t mu – mv F or F t ) u – v ( m Here, t u – v = a (acceleration) So Fma
or F = kma (Here k is proportionality constant. If 1N force is applied on a body of mass 1 kg and the acceleration produced in the body is 1 m/s2, then
1 = k × 1 × 1 or k = 1
Hence, F = ma
So the magnitude of the resultant force acting on a body is equal to the product of mass of the body and the acceleration produced. Direction of the force is same as that of the acceleration.
UNITS OF FORCE
( a) In C.G.S. Sys t e m :
F = ma gm × cm/s2 = Dyne Definition of one dyne :
If m = 1 gm, a = 1 cm/s2, then F = 1 dyne.
When a force is applied on a body of mass 1 gram and the acceleration produced in the body is 1 cm/s2, then
the force acting on the body will be one dyne.
(b) In S.I. System :
F = ma kg × m/s2 = Newton
Definition of one Newton :
If m = 1 kg and a = 1 m/s2 then by, F = ma
F = 1 × 1 = 1 kg × m/s2 = 1 N.
If a force is applied on a body of mass 1 kg and acceleration produced in the body is 1 m/s2, then the
force acting on the body will be one Newton.
(c) Kilogram Force (kgf) :
Kilogram force (kgf) or Kilogram weight (kg. wt.) is force with which a mass of 1 kg is attracted by the
earth towards its centre.
1kgwt = 1kgf = 9.8 N (d) Gram Forc e (gf) :
Gram force or gram weight is the force with which a mass of 1 gram is attracted by the earth towards its centre.
1gwt = 1gf = 981 dyne
Abou both the units are called gravitational unit of force. Relation between Newton and dyne.
We know : 1 N = 1kg × 1ms-2
or 1 N = 1000 g × 100 cms-2
or 1 N = 105 g cms-2 = 105 dyne
1 N = 105 dyne
5. A force of 20N acting on a mass m1 produces an acceleration of 4 ms–2. The same force is applied on
mass m2 then the acceleration produced is 0.5 ms–2.
What acceleration would the same force produce, when both masses are tied together ?
Sol. For mass m1: F = 20N, a = 4 ms–2
then m1 = a F = 4 20 = 5 kg For mass m2 : F = 20N, a = 0.5 ms–2
8 then m2 = a F = 5 . 0 20 = 40 kg When m1 and m2 are tied together : Total mass = m1 + m2 = 45 kg, F = 20N then a = ) m m ( F 2 1 = 45 20 = 0.44 ms–2 IMPULSE OF FORCE
A large force acting for a short time to produce a finite change in momentum is called impulsive force. The product of force and time is called impulse of force. i.e., Impulse = Force × Time
or I = Ft
The S.I. unit of impulse is Newton-second (N-s) and the C.G.S unit is dyne- second (dyne-s)
Im pu ls e a nd M ome nt um :
From Newton’s second law of motion
Force, F = t p p2 1 or Ft = p2– p1
i.e., Impulse = Change in momentum
This relation is called impulse equation or momentum-impulse theorem. It has an important application in our everyday life.
IMPULSE DURING AN IMPACT OR COLLISION The impulsive force acting on the body produces a change in momentum of the body on which it acts. We know, Ft = mv – mu, therefore the maximum force
needed to produce a given impulse depends upon time. If time is short, the force required in a given impulse or the change in momentum is large and vice-versa.
NEWTON’S THIRD LAW
(a) Statement :
The law states that “ To every action there is an equal
and opposite reaction“. Moreover, action and reaction
act on different bodies.
( b) D emonst rat ion :
Two similar spring balances A and B joined by hook as shown in the figure. The other end of the spring balance B is attached to a hook rigidly fixed in a rigid wall.
Demonstration- Newton’s third law of motion
The other end of the spring balance A is pulled out to the left. Both balances show the same reading (20 N) for the force.
The pulled balance A exerts a force of 20N on the balance B. It acts as action, B pulls the balance A in opposite direction with a force of 20 N. This force is known as reaction.
We conclude that action-reaction forces are equal and opposite and act on two different bodies.
NO ACTION IS POSSIBLE WITHOUT REACTION Examples :
(i) A nail cannot be fixed on a suspended wooden ball. (ii) A paper cannot be cut by scissors of single blade. (iii) A hanging piece of paper cannot be cut by blade. (iv) Writing on a hanging page is impossible. (v) Hitting on a piece of sponge does not produce reaction. You do not enjoy hitting.
ACTION AND REACTION ARE NOT BALANCED Action and reaction, though equal and opposite are not balanced because they act on two different bodies. In case when they act on two different bodies forming a single system, they become balanced.
ANY PAIR OF EQUAL AND OPPOSITE FORCES IS NOT AN ACTION–REACTION PAIR
Consider a book kept on a table. We have seen that the table pushes the book in the upward direction. Then why does not the book fly up? It does not fly up because there is another force on the book pulling it down. This is the force exerted by the earth on the book, which we call the weight of the book. So, there are two forces on the book– the normal force, N acting upwards, applied
by the table and the force, W acting downwards, applied by the earth. As the book does not accelerate, we conclude that these two forces are balanced. In other words, they have equal magnitudes but opposite directions. V V V
N
N=W
Can we call N the action and W the reaction ? We cannot. This is because, although they are equal and
9 opposite, they are not forces applied by two bodies on
each other. The force N is applied by the table on the book, its reaction will be the force applied by the book on the table. Weight W is the force applied by the earth on the book, its reaction will be the force applied by the book on the earth.
So, although N and W are equal and opposite, they do not form an action–reaction pair.
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
By Newton’s second law, the rate of change of
momentum is equal to the applied force.
time momentum in Change = Force Change in momentum = F × t If F = 0 then, Change in momentum = 0
If the force applied on the body is zero then its momentum will be conserved, this law is also applicable on the system. If in a system the momentum of the objects present in the system are P1, P2, P3... and external force on the system is zero, then–
P1 + P2 + P3 +... = Constant
NOTE : If only internal forces are acting on the system
then its linear momentum will be conserved.
( a ) The Law of Conservation of Linear Momentum by Third Law of Motion :
Suppose A and B are two objects of masses m1 and m2 are moving in the same direction with velocity u1 and u2 respectively (u1 > u2). Object A collides with object B and after time t both move in their original direction with velocity v1 and v2 respectively.
The change in momentum of object A = m1v1– m1u1
u1 u2 m1 m2 before collision (u > u )1 2 The force on B by A is F1 = time momentum in Change F1 = t u m – v m1 1 1 1 ...(1)
The change in momentum of object B = m2v2– m2u2
The force on A by B is F2 = time momentum in Change = t u m – v m2 2 2 2 ...(2) v1 v2 m1 m2 after collision
By Newtons third law, F1 = –F2
t u m – v m1 1 1 1 = t u m – v m – 2 2 2 2
m1v1– m1u1 = –m2v2 + m2u2 or m1u1 + m2u2 = m1v1 + m2v2or Initial momentum = Final momentum
6. Two ice hockey players, suitably padded collide directly with each other and immediately become entangled. One has a mass of 120 kg and is travelling at 2 m/s, while the other has a mass of 80 kg and is travelling at 4 m/s towards the first player at what speed do they travel after they become entangled?
Sol. m1u1 + m2u2 = m1v1 + m2v2 120 × 2 – 80 × 4 = (120 + 80) 240 – 320 = 200 V – 80 = 200 V V = – 5 2 m/s
SOME ILLUSTRATION ON CONSERVATION OF MOMENTUM
(a) Recoil of Gun :
A loaded gun (rifle) having bullet inside it forming one system is initially at rest. The system has zero initial momentum.
v V
When the trigger (T) is pressed, the bullet is fired due to internal force of explosion of powder in cartidge inside. The bullet moves forward with a high velocity and the gun move behind (recoils) with a lesser velocity.
Let the bullet and the gun have masses m and M respectively. Let the bullet move forward with velocity v and the gun recoils with velocity V.
Then final momentum of the gun and bullet is MV + mv By the law of conservation of momentum–
Initial momentum of the system = Final momentum of the system.
0 = MV + mv
or V = –
M mv
Hence the recoil velocity of gun = M mv
and the velocity of the gun is = –
M mv
10 (b) The Working of a Rocke t :
the momentum of a rocket before it is fired is zero. When the rocket is fired, gases are produced. These gases come out of the rear of the rocket with high speed. The direction of the momentum of the gases coming out of the rocket is in the downward direction. Thus, to conserve the momentum of the system i.e., (rocket + gases), the rocket moves upward with a momentum equal to the momentum of the gases. So, the rocket continues to move upward as long as the gases are ejected out of the rocket. Thus a rocket works on the basis of the law of conservation of momentum. 7. A bullet of mass 0.01 kg is fired from a gun weighing
5.0 kg. If the speed of the bullet is 250 m/s, calculate the speed with which the gun recoils.
Sol. V = 5 250 1 = – 50 m/s SYSTEM
Two or more than two objects which interact with each other form a system.
Classification of forces on the basis of boundary of system :
(a) Internal Forces : Forces acting with in a system among its constituents.
(b) External Forces : Forces exerted on the constituents of a system by the outside surroundings are called as external forces.
FREE BODY DIAGRAM
A free body diagram consists of a diagrammatic representations of single body or a subsystem of bodies isolated from surroundings showing all the forces acting on it.
Steps for F.B.D.
Step 1 : Identify the object or system and isolate it from other objects, clearly specify its boundary.
Step 2 : First draw non-contact external force in the diagram, generally it is weight.
Step 3 : Draw contact forces which acts at the boundary of the object of system. Contact forces are normal , friction, tension and applied force. In F.B.D, internal forces are not drawn only external are drawn. 8. A block of mass ‘m’ is kept on the ground as shown in
figure.
(i) Draw F.B.D. of block.
(ii) Are forces acting on block forms action- reaction pair.
(iii) If answer is no, draw action reaction pair. Sol.(i) F.B.D. of block
(ii) ‘N’ and mg are not action -reaction pair. Since pair
act on different bodies, and they are of same nature.
(iii) Pair of ‘mg’ of block acts on earth in opposite
direction.
mg earth
and pair of ‘N’ acts on surface as shown in figure.
N
9. Two sphere A and B are placed between two vertical walls as shown in figure. Draw the free body diagrams of both the spheres.
A B
Sol.F.B.D. of sphere ‘A’ :
F.B.D. of sphere ‘B’ :
(exerted by A)
Note : Here N
AB and NBA are the action - reaction pair
11 10. Draw F.B.D. for systems shown in figure below.
Sol.
TRANSLATORY EQUILIBRIUM
When several forces acts on a body simultaneously in such a way that resultant force on the body is zero, i.e.,
F= 0 with F
=
i
Fthe body is said to be in translatory equilibrium. Here it is worthy to note that :
(i) As if a vector is zero all its components must vanish i.e. in equilibrium as -F= 0 with F=
i F= 0 xF
= 0 ;
Fy= 0 ; zF
= 0So in equilibrium forces along x axes must balance each other and the same is true for other directions. If a body is in translatory equilibrium it will be either at rest or in uniform motion. If it is at rest, equilibrium is called static, otherwise dynamic.
Static equilibrium can be divided into following three types :
( a ) St a ble e quilibrium :
If on slight displacement from equilibrium position a body has a tendency to regain its original position it is said to be in stable equilibrium. In case of stable equilibrium potential energy is minimum and so center of gravity is lowest.
O
(b) Unstable equilibrium : If on slight displacement from equilibrium position a body moves in the direction of displacement, the equilibrium is said to be unstable. In this situation potential energy of body is maximum and so center of gravity is highest.
O
(c) Neutral equilibrium : If on slight displacement from equilibrium position a body has no tendency to come back to its original position or to move in the direction of displacement, it is said to be in neutral equilibrium. In this situation potential energy of body is constant and so center of gravity remains at constant height.
.
( a ) Ne wtons 2n d
law of motion :
The rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction of the applied force. In relation
F
= ma the force F
stands for the net external force. Any internal force in the system is not to be included in F
.In S.I. the absolute unit of force is newton (N) and gravitational unit of force is kilogram weight or kilogram force (kgf.)
Note : The absolute unit of force remains the same everywhere, but the gravitational unit of force varies from place to place because it depends on the value of g.
( b ) App lic a t ions o f Ne w t on’s 2n d L a w (i) When objects are in equilibrium :
Steps to solve problem involving objects in equilibrium :
Step 1 : Make a sketch of the problem.
Step 2 : Isolate a single object and then draw the free-body diagram for the object. Label all external forces acting on it.
Step 3 : Choose a convenient coordinate system and resolve all forces into rectangular components along x and Y direction.
Step 4 : Apply the equations
Fx 0and
Fy 0. Step 5 : Step 4 will give you two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations12 for those unknown quantities.
Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities.
11. A ‘block’ of mass 10 kg is
suspended with string as
shown in figure. Find tension in the string.
(g = 10 m/s2).
Sol.F.B.D. of block
For equilibrium of block along Y axis
Fy 0T – 10 g = 0
T = 100 N
12. The system shown in figure is in equilibrium. Find the magnitude of tension in each string ; T1 , T2, T3 and T4. (g = 10 m/s2).
Sol.F.B.D. of 10 kg block
For equilibrium of block along Y axis. 0 Fy
T0 10g T0 = 10 g T0 = 100 N F.B.D. of point ‘A’ 0 Fy
30º T2 T1 x T0 A y T2 cos 30º = T0 = 100 N T2 = 3 200 N
Fx 0 T1 = T2 . sin 30º = 3 200 . 2 1 = 3 100 N. F.B.D. of point of ‘B’ 60º T4 T3 B x y 30º T2
Fy= 0 T4 cos 60º = T2 cos 30º T4 = 200 Nand
Fx= 0 T3 + T2 sin30º = T4 sin 60ºT3 = 3
200 N
13. Two blocks are kept in contact as shown in figure. Find :-(a) forces exerted by surfaces (floor and wall) on
blocks.
(b) contact force between two blocks.
Sol. A : F.B.D. of 10 kg block N1 = 10 g = 100 N ...(1) N2 = 100 N ...(2) F.B.D. of 20 kg block N2 = 50 sin 30º + N3 N3 = 100 – 25 = 75 N & N4 = 50 cos 30º + 20 g N4 = 243.30 N
14. Find magnitude of force exerted by string on pulley.
Sol B. F.B.D. of 10 kg block :
T = 10 g = 100 N F.B.D. of pulley :
13 Since string is massless, so tension in both sides
of string is same.
So magnitude of force exerted by string on pulley =
100
2
100
2 = 100 2N Note : Since pulley is in equilibrium position, so net forces on it is zero.
( i i ) Ac c e le r a t i ng Ob je c t s :
Steps to solve problems involving objects that are in accelerated motion :
Step 1 : Make a sketch of the problem.
Step 2 : Isolate a single object and then draw the free - body diagram for that object. Label all external forces acting on it. Be sure to include all the forces acting on the chosen body, but be equally careful not to include any force exerted by the body on some other body. Some of the forces may be unknown , label them with algebraic symbols.
Step 3 : Choose a convenient coordinate system, show location of coordinate axis explicitly in the free - body diagram, and then determine components of forces with reference to these axis and resolve all forces into x and y components.
Step 4 : Apply the equations
Fx= max &
Fy= may.Step 5 : Step 4 will give two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities.
Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities.
15. A force F is applied horizontally on mass m1 as shown in figure. Find the contact force between m1 and m2.
Sol.Considering both blocks as a system to find the common acceleration. Common acceleration a =
m1 m2
F ...(1) m1 m 2 F aTo find the contact force between ‘A’ and ‘B’ we draw
F.B.D. of mass m2. F.B.D. of mass m2
Fx= max N = m2 . a N =
1 2
2 m m F m 2 1 m m F a ce sin16. A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s2 by an external force F
0.
(a) What is F0 ?
(b) What is the net force on rope ?
(c) What is the tension at middle point of the rope ? (g = 10 m/s2)
Sol.For calculating the value of F0. F.B.D of whole system (a) 2m/s2 F0 10 g = 100 N F0 –100 = 10 × 2 F0 = 120 N ...(1)
(b) According to Newton’s second law, net force on
rope.
F = ma = 2 × 2
= 4N ...(2)
(c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown.
T – 4 g = 4 . 2
T = 48 N
17. A block of mass 50 kg is kept on another block of mass 1 kg as shown in figure. A horizontal force of 10 N is applied on the 1Kg block. (All surface are smooth). Find : (g = 10 m/s2)
14 (b) Force exerted by B on A. 50 kg 1 kg A B Sol.(a) F.B.D. of 50 kg N2 = 50 g = 500 N
along horizontal direction, there is no force aB = 0 (b) F.B.D. of 1 kg block :
N1 N2
10 N 1g
along horizontal direction 10 = 1 aA.
aA = 10 m/s2
along vertical direction N1 = N2 + 1g
= 500 + 10 = 510 N
18. One end of string which passes through pulley and connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g =9.8 m/s2)
Sol.Since string is pulled by 100 N force. So tension in the string is 100 N
F.B.D. of 10 kg block
100 – 10 g = 10 a
100 – 10 × 9.8 = 10 a
a = 0.2 m/s2.
19. A man of mass m stands on a platform of equal mass m and pulls himself by two ropes passing over pulleys as shown. If he pulls each rope with a force equal to half his weight, find his upward acceleration ?
Sol. for (man + platform) system :
2mg – 4T = 2m(a) 2mg – 4 2 mg = 2m (a) [ T = 2 mg ] a = 0 WEIGHING MACHINE
A weighing machine does not measure the weight but measures the force exerted by object on its upper surface.
20. A man of mass 60 Kg is standing on a weighing machine placed on ground. Calculate the reading of machine (g = 10 m/s2).
weighing machine
Sol.For calculating the reading of weighing machine, we draw F.B.D. of man and machine separately.
F.B.D of man N Mg N = Mg F.B.D of man taking mass of man as M weighing machine N N1 Mg F.B.D. of weighing machine
15 Here force exerted by object on upper surface is N
Reading of weighing machine N = Mg
= 60 × 10
N = 600 N.
SPRING BALANCE
It does not measure the weight. It measures the force exerted by the object at the hook. Symbolically, it is represented as shown in figure.
A block of mass ‘m’ is suspended at hook. When spring
balance is in equilibrium, we draw the F.B.D. of mass m for calculating the reading of balance.
m spring balance hook F.B.D. of ‘m’. mg – T = 0 T = mg
Magnitude of T gives the reading of spring balance. 21. A block of mass 20 kg is suspended through two light
spring balances as shown in figure . Calculate the :
(1) reading of spring balance (1). (2) reading of spring balance (2).
Sol.For calculating the reading, first we draw F.B.D.of 20 kg block. F.B.D. 20 kg T 20 g mg – T = 0 T = 20 g = 200 N
Since both the balances are light so, both the scales will read 200 N.
22. (i) A 10 kg block is supported by a cord that runs to a spring scale, which is supported by another cord from the ceiling figure (a). What is the reading on the scale ?
(ii) In figure (b) the block is supported by a cord that runs around a pulley and to a scale. The opposite end of the scale is attached by cord to a wall. What is the reading of the scale.
(iii) In figure (c) the wall has been replaced with a second 10 kg block on the left, and the assembly is stationary. What is the reading on the scale now ?
spring balance hook 10 kg T T (a) T T T
(b)
10kg T T T T 10kg 10kg (c)Sol. In all the three cases the spring balance reads 10 kg. To understand this let us cut a section inside the spring as shown;
16 As each part of the spring is at rest, so F= T. As the
block is stationary, so T= 10g = 100N.
23. A block of mass m is placed on a weighing machine and it is also attached with a spring balance. The whole system is placed in a lift as shown in figure. Lift is moving with constant acceleration 10 m/s2 in upward
direction. The reading of the weighing machine is 500 N and the reading of spring balance is 300 N. Find the actual mass of the block in kg. (Take g = 10 m/s2)
Sol. m = mass of the block
Applying Newton's Law on the block in vertical direction T + N – mg = ma ...(1)
T = 300 N (given reading of spring balance)
N = 500 N (given reading of W.M.) a = 10 m/s2 m = a g N T = 20 300 500 m = 40 kg Ans. 24. Pull is easier than push
Push : Consider a block of mass m placed on rough horizontal surface. The coefficient of static friction between the block and surface is . Let a push force F is applied at an angle with the horizontal.
As the block is in equilibrium along y-axis, so we have
Fy 0; or N = mg + F sin To just move the block along x-axis, we have F cos = N = (mg + F sin ) or F = sin – cos mg ...(i)
Pull : Along y-axis we have ;
Fy 0; N = mg – F sin To just move the block along x-axis, we have F cos = N = (mg – F sin ) or F = sin cos mg . ...(ii)
It is clear from above discussion that pull force is smaller than push force.
25. Discuss the direction of friction in the following cases : (i) A man walks slowly, without change in speed. (ii) A man is going with increasing speed. (iii) When cycle is gaining speed.
(iv) When cycle is slowing down .
Sol. (i) Consider a man walks slowly without acceleration, and both the legs are touching the ground as shown in figure (a). The frictional force on rear leg is in forward direction and on front leg will be on backward direction of motion.
As a = 0, Fnet = 0 or f1 – f2 = 0 f1 = f2 & N1 = N2. N1 N2 (b) Ground f1 f1 f2 f2 N1 N2
(ii) When man is gaining the speed : The frictional force on rear leg f1 will be greater than frictional force on front leg f2 (fig. b).
acceleration of the man, a = m
f f1 2
17 (iii) When cycle is gaining speed : In this case torque
is applied on the rear wheel of the cycle by the chain-gear system. Because of this the slipping tendency of the point of contact of the rear wheel is backward and so friction acts in forward direction. The slipping tendency of point of contact of front wheel is forward and so friction acts in backward direction. If f1 and f2 are the frictional forces on rear and front wheel, then acceleration of the cycle a =
M f – f1 2
, where M is the mass of the cycle together with rider (fig. a).
N1 N2 f1 f2 (a) N1 N2 f1 f2 (b)
(iv) When cycle is slowing down : When torque is not applied (cycle stops pedaling), the slipping tendency of points of contact of both the wheels are forward, and so friction acts in backward direction (fig. b). If f1 and f2 are the frictional forces on rear and front wheel, then retardation
a = M
f f1 2
26. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in fig.. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding.
50g
50g
Sol. The FBD for the two cases are shown in figure. In Ist case, let the force exerted by the man on the floor is N1. Consider the forces inside the dotted box, we have N1 = T + 50 g.
Block is to be raised without acceleration, so T = 25 g.
N1 = 25 g + 50 g = 75 g = 75 × 9.8 = 735 N
In IInd case, let the force exerted by the man on the floor in N2 . Consider the forces inside the dotted box, we have
N2 = 50 g – T
and T = 25 g N2 = 50 g – 25 g
= 25 g = 25 × 9.8 = 245 N.
As the floor yields to a downward force of 700 N, so the man should adopt mode .
27. Figure shows a weighing machine kept in a lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is 15 kg and 45 kg respectively.
Answer the following questions. Assume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion. (take g = 10 m/s2) (i) Find the mass of the object in kg and the normal force acting on the block due to weighing machine? (ii) Find the acceleration of the lift such that weighing machine shows its true weight ?
Sol. (i)
T + N – Mg = Ma
45 g + 15 g = M(g + a) 450 + 150 = M(10 + 5) M = 40 kg
Normal force is the reaction applied by weighing machine i.e. 15 × 10 = 150 N.
18 (ii) T + N – Mg = Ma 45 g + 40 g = 40(g + a) 450 + 400 = 400 +40 a a = 40 450 = 4 45 m/s2
E X E R C I S E
1. Two blocks of masses 2 kg and 1 kg are placed on a smooth horizontal table in contact with each other. A horizontal force of 3 newton is applied on the first so that the blocks move with a constant acceleration. The force between the blocks would be :
(A) 3 Newton (B) 2 Newton (C) 1 Newton (D) Zero
2. An object of mass 10 kg. moves at a constant speed of 10ms–1. A constant force acts for 4 s on the object and gives it a speed of 2 ms–1 in opposite direction. The force acting on the ob-ject is : (A) – 3N
(B) –30 N (C) 3 N
(D) 30 N
3. A machine gun has a mass 5kg. It fires 50 gram bullets at the rate of 30 bullets per minute at a speed of 400 ms–1. What force is required to keep the gun in position :
(A) 10 N (B) 5 N
(C) 15 N (D) 30N
4. A particle of mass m1 moving with velocity v collides with a mass m2 at rest, then they get embedded. Just after collision, velocity of the system : (A) Increases
(B) Decreases (C) remains constant (D) becomes zero
5. When a car turns on a curved road, you are pushed against one of the doors of the car because of :
(IJSO/Stage-I/2012) (A) inertia
(B) the centripetal force (C) the centrifugal force (D) the frictionaI force
6. What is the reading of the spring balance shown in the figure below? (IJSO/Stage-I/2012)
T T T 0.2kg (A) 0 (B) 2N (C) 4N (D) 6N
7. Two blocks are in contact on a frictionless table. One has mass m and the other 2m.A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be :
(A) Same (B) 1 : 2
(C) 2 : 1 (D) 1 : 3
8. Two forces of 6N and 3N are acting on the two blocks of 2kg and 1kg kept on frictionless floor. What is the force exerted on 2kg block by 1kg block ?:
2kg 1kg
6N
3N
(A)1N (B) 2N
(C) 4N (D) 5N
9. There are two forces on the 2.0 kg box in the overhead view of figure but only one is shown. The second force is nearly : y 30º F = 20 N1 a = 12 m/s2 x (A) –20ˆjN (B) – 20 iˆ + 20ˆj N (C) –32iˆ – 12
3
ˆjN (D) –21iˆ – 16ˆjN10. A dish of mass 10 g is kept horizontally in air by firing bullets of mass 5 g each at the rate of 100 per second. If the bullets rebound with the same speed, what is the velocity with which the bullets are fired :
(A) 0.49 m/s (B) 0.098 m/s (C) 1.47 m/s (D) 1.96 m/s
19 11. A block of metal weighing 2 kg is resting on a
frictionless plank. If struck by a jet releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block will be :
(A) 2.5 m/s2 (B) 5.0 m/s2
(C) 10 m /s2 (D) none of the above
12. A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and between m and M’ is N’ then
(A) N= N’ (B) N > N’
(C) N’> N
(D) cannot be determined
13. STATEMENT-1 : Block A is moving on horizontal surface towards right under action of force. All surface are smooth. At the instant shown the force exerted by block A on block B is equal to net force on block B.
STATEMENT-2 : From Newtons’s third law, the force
exerted by block A on B is equal in magnitude to force exerted block B on A
(A) statement-1 is true, Statement 2 is true, statement-2 is correct explanation for statement-1.
(B) statement-1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True
14. A certain force applied to a body A gives it an acceleration of 10 ms–2 . The same force applied to body B gives it
an acceleration of 15 ms–2 . If the two bodies are joined
together and same force is applied to the combination, the acceleration will be :
(IJSO/Stage-I/2011) (A) 6 ms–2 (B) 25 ms–2
(C) 12.5 ms–2 (D) 9 ms–2
15. Four blocks are kept in a row on a smooth horizontal table with their centres of mass collinear as shown in the figure. An external force of 60 N is applied from left on the 7 kg block to push all of them along the table. The forces exerted by them are :(IAO/Sr./Stage-I/2008)
7 kg 5 kg 2 kg 1 kg 60N
P Q R S
(A) 32 N by P on Q (B) 28 N by Q on P (C) 12 N by Q on R (D) 4 N by S on R
16. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle with the vertical in equilibrium,then the tension in the string AB is :
(A) F sin (B) F /sin (C) F cos (D) F / cos
17. In the system shown in the figure, the acceleration of the 1kg mass and the tension in the string connecting between A and B is : (A) g 4 downward, 8g 7 (B) g 4 upward, g 7 (C) g 7 downward, 6 7 g (D) g 2 upward, g 18. A body of mass 8 kg is hanging from another body of
mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s–2. The tension T
1
and T2 will be respectively :(Use g =9.8 m/s2)
(A) 200 N, 80 N (B) 220 N, 90 N (C) 240 N, 96 N (D) 260 N, 96 N 19. Two masses M1 and M2 are attached to the ends of a
light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination and . If M2 > M1 then the
acceleration of block M2 down the inclined will be :
(A) 2 1 2 M (sin ) M M g (B) 1 1 2 M g(sin ) M M (C) 2 1 1 2 M M sin M sin M g (D) Zero
20 20. Three masses of 1 kg, 6 kg and 3 kg are connected to
each other by threads and are placed on table as shown in figure. What is the acceleration with which the system is moving ? Take g = 10 m s–2:
(A) Zero (B) 1 ms–2
(C) 2 m s–2 (D) 3 m s–2
21. The pulley arrangements shown in figure are identical the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force F= 2 mg, where g is acceleration due to gravity. The acceleration of mass in case I is :
(A) Zero
(B) More than that in case II (C) Less than that in case II (D) Equal to that in case II
22. A 50 kg person stands on a 25 kg platform. He pulls massless rope which is attached to the platform via the frictionless, massless pulleys as shown in the figure. The platform moves upwards at a steady velocity if the force with which the person pulls the rope is :
(A) 500 N (B) 250 N
(C) 25 N (D) 50 N
23. Figure shows four blocks that are being pulled along a smooth horizontal surface. The mssses of the blocks and tension in one cord are given. The pulling force F is :
4kg 3kg 2kg 1kg 60º F 30N (A) 50 N (B) 100 N (C) 125 N (D) 200 N
24. A10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground. The magnitude of the least acceleration the monkey must have if it is to lift the package off the ground is :
(A) 4.9 m/s2 (B) 5.5 m/s2
(C) 9.8 m/s2 (D) none of these
25. Two blocks, each of mass M, are connected by a massless string, which passes over a smooth massless pulley. Forces F act on the blocks as shown. The tension in the string is :
(A) Mg (B) 2 Mg
(C) Mg + F (D) none of these 26. Two blocks of mass m each is connected with the
string which passes over fixed pulley, as shown in figure. The force exerted by the string on the pulley P is :
(A) mg (B) 2 mg
21 27. One end of a massless rope, which passes over a
massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that rope can bear is 360 N, with what minimum safe acceleration (in m/s2) can a monkey of 60 kg move
down on the rope : P
C
(A) 16 (B) 6
(C) 4 (D) 8
28. Which figure represents the correct F.B.D. of rod of mass m as shown in figure :
(A) (B)
(C) (D) None of these
29. Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is :
(A) 15 kg (B) kg
2 15
(C) 5 kg
(D) Infinitely large (or not possible)
30. In the figure, the blocks A, B and C of mass each have acceleration a1 ,a2 and a3 respectively . F1 and F2 are external forces of magnitudes 2 mg and mg respectively then which of the following relations is correct :
(A) a1 = a2 = a3 (B) a1 > a2 > a3 (C) a1 = a2 , a2 > a3 (D) a1 > a2 , a2= a3
31. A weight is supported by two strings 1.3 and 2.0 m long fastened to two points on a horizontal beam 2.0 m apart. The depth of this weight below the beam is : (IAO/Jr./Stage-I/2007)
(A) 1.0 m (B) 1.23 m
(C) 0.77 m (D) 0.89 m
32. A fully loaded elevator has a mass of 6000 kg. The tension in the cable as the elevator is accelerated downward with an acceleration of 2ms–2 is (Take g = I0
ms–2)
(A) 7·2 × 104 N (B) 4.8 × 104 N
(C) 6 × 104 N (D) 1.2 × 104 N
33. A light string goes over a frictionless pulley. At its one end hangs a mass of 2 kg and at the other end hangs a mass of 6 kg. Both the masses are supported by hands to keep them at rest. When the masses are released, they being to move and the string gets taut. (Take g = 10 ms–2) The tension in the string during the
motion of the masses is :
(A) 60 N (B) 30 N
(C) 20 N (D) 40 N
34. In the given figure. What is the reading of the spring balance:
(A) 10 N (B) 20 N
(C) 5 N (D) Zero
35. Two bodies of masses M1 and M2 are connected to each other through a light spring as shown in figure. If we push mass M1 with force F and cause acceleration a1 in mass M1 what will be the acceleration in M2 ?
(A) F/M2 (B) F/(M1 + M2) (C) a1 (D) (F–M1a1)/M2
36. A spring balance is attached to 2 kg trolley and is used to pull the trolly along a flat surface as shown in the fig. The reading on the spring balance remains at 10 kg during the motion. The acceleration of the trolly is (Use g= 9.8 m–2) :
(A) 4.9 ms–2 (B) 9.8 ms–2
22 37. A body of mass 32 kg is suspended by a spring balance
from the roof of a vertically operating lift and going downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is :
(A) Decreasing at 20 m & increasing at 50 m (B) Increasing at 20 m & decreasing at 50 m
(C) Continuously decreasing at a constant rate throughout the journey
(D) Continuously increasing at constant rate throughout the journey
38. A ship of mass 3 × 107 kg initially at rest is pulled by a
force of 5 × 104 N through a distance of 3m. Assume
that the resistance due to water is negligible, the speed of the ship is :
(A) 1.5 m/s (B) 60 m/s
(C) 0.1 m/s (D) 5 m/s
39. When a horse pulls a cart, the force needed to move the horse in forward direction is the force exerted by : (A) The cart on the horse
(B) The ground on the horse (C) The ground on the cart (D) The horse on the ground
40. A 2.5 kg block is initially at rest on a horizontal surface. A 6.0 N horizontal force and a vertical force Pare applied to the block as shown in figure. The coefficient of static friction for the block and surface is 0.4. The magnitude of friction force when P = 9N : (g = 10 m/s2)
(A) 6.0 N (B) 6.4 N
(C) 9.0 N (D) zero
41. The upper half of an inclined plane with inclination is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is : (A) 2 tan (B) tan
(C) 2 sin (D) 2 cos
42. Minimum force required to pull the lower block is (take g = 10 m/s2) :
(A) 1 N (B) 5 N
(C) 7 N (D) 10 N
43. N bullets each of mass m are fired with a velocity v m/ s at the rate of n bullets per sec., upon a wall. If the bullets are completely stopped by the wall, the reaction offered by the wall to the bullets is :
(A) N m v / n (B) n m v (C) n N v / m (D) n v m / N
44. A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be , then the stopping distance is : (A) mg 2 P (B) 2 mg P2 (C) g m 2 P 2 (D)
2
m
g
P
2 2
45. What is the maximum value of the force F such that the block shown in the arrangement, does not move :
F 60º 1 2 3 m = 3kg (A) 20 N (B) 10 N (C) 12N (D) 15 N
46. A bock of mass 5 kg is held against wall by applying a horizontal force of 100N. If the coefficient of friction between the block and the wall is 0.5, the frictional force acting on the block is : (g =9.8 m/s2)
5kg 100N
(A)100 N (B) 50 N
(C) 49 N (D) 24.9 N
47. A heavy roller is being pulled along a rough road as shown in the figure. The frictional force at the point of contact is : (IAO/Jr./Stage-I/2007)
F
(A) parallel to F (B) opposite to F (C) perpendicular to F (D) zero
48. When a motor car of mass 1500 kg is pushed on a road by two persons, it moves with a small uniform velocity. On the other hand if this car is pushed on the same road by three persons, it moves with an acceleration of 0.2 m/s2. Assume that each person is
producing the same muscular force. Then, the force of friction between the tyres of the car and the surface of the road is : (IAO/Jr./Stage-I/2009)
(A) 300 N (B) 600 N
23 49. A block of mass M is at rest on a plane surface inclined
at an angle to the horizontal The magnitude of force
exerted by the plane on the block is : (A) Mg cos (B) Mg sin
(C) Mg tan (D) Mg
50. A block of mass M rests on a rough horizontal table. A steadily increasing horizontal force is applied such that the block starts to slide on the table without toppling. The force is continued even after sliding has started. Assume the coefficients of static and kinetic friction between the table and the block to be equal. The cor-rect representation of the variation of the frictional forces, ƒ, exerted by the table on the block with time t is
given by :
(A) (B)
(C) (D)
51. A small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed under her pushing force (F) which is obliquely upward as shown . Then
(A) The resultant of the pushing force (F), weight of the toy, normal force by the ground on the toy and the frictional force is zero.
(B) The normal force by the ground is equal and oppo-site to the weight of the toy.
(C) The pushing force (F) of the child is balanced by the equal and opposite frictional force
(D) The pushing force (F) of the child is balanced by the total internal force in the toy generated due to deformation
52. On a horizontal frictional frozen lake, a girl (36 kg) and a box (9kg) are connected to each other by means of a rope. Initially they are 20 m apart. The girl exerts a horizontal force on the box, pulling it towards her. How far has the girl travelled when she meets the box ? (A) 10 m
(B) Since there is no friction, the girl will not move (C) 16 m
(D) 4m
53. Which of the following does NOT involve friction ? (IJSO/Stage-I/2011) (A) Writing on a paper using a pencil
(B) Turning a car to the left on a horizontal road. (C) A car at rest parked on a sloping ground (D) Motion of a satellite around the earth.
54. In the two cases shown below, the coefficient of kinetic friction between the block and the surface is the same, and both the blocks are moving with the same uniform speed. Then, (IAO/Sr./Stage-I/2008) F1 F2 (A) F1 = F2 (B) F1 < F2 (C) F1 > F2 (D) F1 = 2F2 if sin = Mg/4F2
55. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration
‘a’ 3:2. The value of ‘a’ is : (g = acceleration, due to
gravity)
(A) (3/2)g (B) g
(C) (2/3) g (D) g/3
56. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in time t1 when elevator is stationary and in time t2 if it is moving uniformly. Then
(A) t1 = t2 (B) t1 > t2 (C) t1 < t2
(D) t1 < t2 or t1 > t2 depending
57. STATEMENT-1 : A man standing in a lift which is moving upward, will feel his weight to be greater than when the lift was at rest.
STATEMENT-2 : If the acceleration of the lift is ‘a’ upward
then the man of mass m shall feel his weight to be equal to normal reaction (N) exerted by the lift given N = m(g+a) (where g is acceleration due to gravity (A) 1 is true, Statement 2 is true, statement-2 is correct explanation for statement 1.
(B) 1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True
