# conservation of linear momentum formula class 11 derivation physics state or derive the

find conservation of linear momentum formula class 11 derivation physics state or derive the ?

**Newton’s First Law of Motion**

Newton’s first law of motion states that every body continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by an external unbalanced force.

Newton’s Second Law of Motion

Newton’s second law of motion states that 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 in which the force acts. Linear Momentum Newton defined linear momentum as the product of the mass and the velocity of a body.

p = mv

where v is the velocity at a certain instant of time. Differentiating this equation with respect to time, we get

note – 1. Force = slope of momentum–time (p – t) graph.

2. Change in momentum = area under the force–time (F – t) graph.

## Newton’s Third Law of Motion

Newton’s third law of motion states that whenever one body exerts a force on a second body, the second body exerts an equal and opposite force on the first, or, to every action there is an equal and opposite reaction. The action and reaction forces act on different body.

**Law of Conservation of Linear Momentum**

The law of conservation of linear momentum may be stated as ‘when no net external force acts on a system consisting of several particles, the total linear momentum of the system is conserved, the total linear momentum being the vector sum of the linear momentum of each particle in the system’.

Recoil of a Gun The gun and the bullet constitute a two-body system. Before the gun is fired, both the gun and the bullet are at rest. Therefore, the total momentum of the gun-bullet system is zero. After the gun is fired, the bullet moves forward and the gun recoils backwards. Let mb and mg be the masses of the bullet and the gun. If vb and vg are their respective velocities after firing, the total momentum of the gun-bullet system after firing is (mb vb + mg vg). From the law of conservation of momentum, the total momentum after and before the gun is fired must be the same, i.e.

The negative sign indicates that the gun recoils in a direction opposite to that of the bullet. In terms of magnitudes, we have

Impulse Consider a collision between two bodies A and B moving in the same straight line. Let △t be the duration of the collision, i.e. the time for which the bodies were in contact during which time the transfer of momentum took place. We assume that the bodies continue moving in the same straight line after the collision with velocities different from their initial velocities. Impulse of a force is the product of the average force and the time for which the force acts and it is equal to the change in momentum of the body during that time. Impulse is a vector and is measured in kg m s^{–1} or N s.

**Contact Forces**

**(1) Normal Reaction**

The force exerted by one body when placed on the surface of another body is known as contact force. If the two surfaces in contact are perfectly smooth (i.e., frictionless), then the contact force acts only perpendicular (normal) to their surface of contact and is known as normal reaction (R). If a block of mass m is placed on a horizontal frictionless surface [Fig. 3.2 (a)], the normal reaction R = mg. If the block is placed on an inclined plane of inclination α [Fig. 3.2 (b)], the normal reaction R = mg cos α

If there is friction between the surfaces of contact, then the component of the contact force perpendicular to their surface gives the normal reaction and the other component which acts along the tangent to the surface of contact gives the force of friction. The normal reaction, tension and friction are examples of contact force.

**(2) Tension**

The force in a string is called tension (T). If the string is massless, the tension has the same magnitude at all points of the string. Tension in the string always acts away from the body to which it is attached. If the string passes over a frictionless pulley and its ends are attached to two bodies, the tension in the entire string has the same magnitude and its direction is towards its point of contact with the pulley.

**Friction**

Friction is the force which comes into play when one body slides or rolls over the surface of another body and acts in a direction tangential to the surfaces in contact and opposite to the direction of motion of the body. The maximum (or limiting) force of friction when a body just begins to slide over the surface of another body is called the limiting friction. The force of friction just before one body begins to slide over another is called the limiting or static friction (f_{s}). The coefficient of limiting or static friction (µ_{s}) is defined as

where R is the normal reaction, i.e. the normal force pressing the two surfaces together. The force necessary to maintain a body in uniform motion over the surface of another body, after motion has started, is called the kinetic or sliding friction ( f_{k} ). The coefficient of kinetic friction (µ_{k}) is defined as

Note that µ_{k} is always less than µ_{s}. Angle of Friction Angle of friction is the angle between the resultant of the force of limiting friction ( f) and the normal reaction (R). In Fig. 3.3, θ is the angle of friction, which is given by

Angle of Repose Suppose a body is placed on an inclined plane. The angle of inclination is gradually increased until the body just begins to slide along the plane. When this happens the angle of inclination α of the inclined surface with the horizontal is called the angle of repose (see Fig. 3.4). It follows from the figure that

Force of limiting friction ( f ) = mg sin α

Force of normal reaction (R) = mg cos α

### Solving Problems in Mechanics by Free Body Diagram Method

In mechanics, we often have to handle problems which involve a group of bodies connected to one another by strings, pulleys, springs, etc. They exert forces on one another. Furthermore, there are frictional forces and the force of gravity acting on each body in the group. To solve such complicated problems, it is always convenient to choose one body in the group, find the magnitude and the direction of the forces acting on this body by all the remaining bodies in the group. Then we find the resultant of all the forces acting on the body to obtain the net force exerted on it. We then use the laws of motion to determine the dynamics of the body. We apply this procedure to all other bodies in the group one by one. It is useful to draw a separate diagram for each body, showing the directions of the different forces acting on it. Such a diagram is called the free body diagram (F.B.D.) of the body.

**(1) Two masses tied to a string going over a frictionless pulley** Consider two bodies of masses m1 and m2 (m1 > m2) connected by a string which passes over a pulley, as shown in Fig. 3.5(a). When the bodies are released, the heavier one moves downwards and the lighter one moves up.

Net force in the direction of motion of m1 is m_{1}g – T. Therefore, the equation of motion of m_{1} is

Net force in the direction of motion of m_{2} is (T – m_{2}g). Therefore, the equation of motion of m_{2} is

(2) Two masses in contact Figure 3.6(a) shows two blocks of masses m_{1} and m_{2} placed in contact on a horizontal frictionless surface. A force F is applied to mass m_{1}. As a result, the masses move with a common acceleration a. To find a and the contact force on m_{2}, we draw the free body diagrams as shown in Figs. 3.6(b) and (c).

R = normal reaction force between the blocks. From Figs. 3.6(b)and (c), we get

(3) Three masses in contact Figure 3.7(a) shows three blocks of masses m_{1}, m_{2}, and m_{3} placed in contact on a horizontal frictionless surface. A force F is applied to m1. As a result, the three masses move with a common acceleration a. To find a and the contact forces on m2 and m3, we draw the free body diagrams as shown in Figs. 3.7(b) (c) and (d).

R’ = contact force on m_{2} = reaction force between m_{1} and m_{2}

R = contact force on m_{3} = reaction force between m_{2} and m_{3}

It follows from Figs. 3.7(b), (c) and (d) that

**(4) Two masses connnected with a string**

Figure 3.8(a) shows two blocks of masses m_{1} and m_{2} connected with a string and lying on a horizontal frictionless surface. A force F is applied to m2. As a result, the masses move with a common acceleration a. To find a and force exerted on m_{1}, we draw the free body diagrams as shown in Figs. 3.8(b) and (c). T is the tension in the string.

**(5) Three masses connected by strings**

Figure 3.10 (a) shows three blocks of masses m_{1}, m_{2} and m_{3} connected by two strings and placed on a horizontal frictionless surface. A force F is applied to m_{1}. As a result, the blocks move with a common acceleration a. To find a and the forces acting on m_{2} and m_{3}, we draw free body diagrams as shown in Fig. 3.10(b) and (c) and (d). T is the tension in the string between m_{1} and m_{2} and T’ is the tension in the string between m_{2} and m_{3}.

The tension in the string between m1 and m2 is T, which is obtained by adding (i) and (ii).

The tension in the string between m2 and m3 is T’, which from (i) is given by

(6) Two masses connected by a string and suspended from a support Two blocks of masses m_{1} and m_{2} are connected by two strings and suspended from a support as shown in Fig. 3.11(a). Mass m_{2} is pulled down by a force F. The tension T in the string between m1 and m2 and tension T’ in the string between m_{1} and the support can be found from the free body diagrams as shown in Fig. 3.11(b) and (c).

**(7) Two blocks connected by a string passing over a frictionless pulley fixed at the edge of a horizontal table**

Consider a block of mass m1 lying on a frictionless table connected through a pulley to another block of mass m2 hanging vertically (Fig. 3.12). When the system is released, let acceleration of the blocks be a. From free body diagrams, the equations of motion of m1 and m2 are

If the table top is frictionless, the blocks will move even if m2 < m1.

If m is the coefficient of friction between block m1 and the table, the force of friction is

**(8) Two blocks connected by a string passing over a frictionless pulley fixed at the top of an inclined plane**

Let T be the tension in the string. Since the pulley is frictionless, the tension is the same throughout the string (Fig. 3.14). There are the following two cases: (a) Mass m1 moving up along the incline with acceleration a [Fig. 3.14]

The equations of motion of m1 and m2

If µ is the coefficient of friction between m1 and the inclined plane, the frictional force f = µR = µm_{1}g cosθ will act down the plane because the block m1 is moving up the plane. In this case, Eq. (i) is replaced by

(b) Mass m1 moves down the incline with acceleration a In this case, we get m_{1}g sin q – T = m_{1}θ and T – m_{2}g = m_{12}a which give

(9) Two blocks connected by a string passing over a frictionless pulley fixed at the top of a double inclined plane Let the block of mass m_{1} move up along the inclined plane of angle of inclination θ1, and the block of mass m_{2} move down the inclined plane of angle of inclination θ2 (Fig. 3.15). Let T be the tension in the string. Then, for m_{1} and m_{2}, we have

(10) One blocks are placed on top of the another A block of mass m_{1} is placed on another block of mass m_{2}, which is lying on a horizontal frictionless surface. The coefficient of friction between the blocks is µ.

Case 1: The maximum force that can be applied on the lower block so that the upper block does not slip [Fig. 3.16(a)]

F_{max} = maximum value of force F so that block m_{1} does not slip of block m_{2}

f = frictional force on m_{1} due to m_{2}

= µR = µm_{1}g

Due to friction, m_{2} will try to drag m_{1} to the right. Hence frictional force f acts towards left. From F.B.D. of m_{1},

(a) If F > F_{max }, m_{1} will begin to slide on m_{2} and then their accelerations will be different

(b) If F < F_{max}, m_{1} and m_{2} move together without any relative motion between them.

Case 2: The maximum force that can be applied on the upper block so that it does not slip on the lower block.

F_{max} = maximum value of force F so that block m_{1} just begins to slide on block m_{2} Block m_{1} tries to drag block m_{2} toward right due to frictional force f = µR = µm_{1}g. The frictional force exerted by m_{1} on m_{2} will be towards right. R’ = normal reaction on m_{2} by the horizontal surface. If a is the acceleration of blocks towards right, from F.B.D. of m_{1} we have

(a) If F < F_{max}, the blocks move together without any relative motion.

(b) If F > F_{max}, the blocks slide relative to each other and then their accelerations are different.

## हिंदी माध्यम नोट्स

**Class 6 **

Hindi social science science maths English

**Class 7**

Hindi social science science maths English

**Class 8**

Hindi social science science maths English

**Class 9 **

Hindi social science science Maths English

**Class 10**

Hindi Social science science Maths English

**Class 11 **

Hindi sociology physics physical education maths english economics geography History

chemistry business studies biology accountancy political science

**Class 12 **

Hindi physics physical education maths english economics

chemistry business studies biology accountancy Political science History sociology

## English medium Notes

**Class 6 **

Hindi social science science maths English

**Class 7**

Hindi social science science maths English

**Class 8**

Hindi social science science maths English

**Class 9 **

Hindi social science science Maths English

**Class 10**

Hindi Social science science Maths English

**Class 11 **

Hindi physics physical education maths entrepreneurship english economics

chemistry business studies biology accountancy

**Class 12 **

Hindi physics physical education maths entrepreneurship english economics