Centre of Mass, Chapter Notes, Class 11, Physics (IIT-JEE and AIPMT) PDF Download

Centre of Mass - System of Particles and Rotational Motion, Class 11, Physics

CENTRE OF MASS

1. Centre of Mass :

Every physical system has associated with it a certain point whose motion characterises the motion of the whole system. When the system moves under some external forces, then this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at this point for translational motion. This point is called the centre of mass of the system.

1.1 Centre of Mass of a System of `N' Discrete Particles :

Consider a system of N point masses m₁, m₂, m₃, .................. m_n whose position vectors from origin O are given by ........... respectively. Then the position vector of the centre of mass C of the system is given by.

 ;

where,  is called the moment of mass of particle with respect to origin.

 is the total mass of the system.

Further,  and

So, the cartesian co-ordinates of the COM will be

x_com =

or x_COM =

Similarly, y_COM = and

Note :

• If the origin is taken at the centre of mass then  = 0. hence, the COM is the point about which the sum of "mass moments" of the system is zero.
• If we change the origin then  changes. So  also changes but exact location of center of mass does not change.

1.2 Position of COM of two particles : -

Consider two particles of masses m₁ and m₂ separated by a distance l as shown in figure.

Let us assume that m₁ is placed at origin and m₂ is placed at position (l, 0) and the distance of centre of mass from m₁ & m₂ is r₁ & r₂ respectively.

So x_COM =

r₁ =  =  ...(1)

r₂ =  =  ...(2)

From the above discussion, we see that

r₁ = r₂ =  if m₁ = m₂, i.e., COM lies midway between the two particles of equal masses.

Similarly, r₁ > r₂ if m₁ < m₂ and r₁ < r₂ if m₂ < m₁ i.e., COM is nearer to the particle having larger mass.

From equation (1) & (2)

m₁r₁ = m₂r₂ ...(3)

Centre of mass of two particle system lie on the line joining the centre of mass of two particle system.

Ex.1 Two particle of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre of mass. 

Sol.

Since, both the particles lie on x-axis, the COM will also lie on x-axis. Let the COM is located at x = x, then

r₁ = distance of COM from the particle of mass 1 kg = x

and r₂ = distance of COM from the particle of mass 2 kg = (3 -x)

Using

or

or x = 2 m

thus, the COM of the two particles is located at x = 2m.

Ex.2 Two particle of mass 4 kg & 2kg are located as shown in figure then find out the position of centre of mass. 

Sol. First find out the position of 2 kg mass

x_2kg = 5 cos 37° = 4 m

y_2kg = 5 sin 37° = 3 m

So these system is like two particle system of mass 4 kg and 2kg are located (0, 0) and (4, 3) respectively. then

x_com =  =  =  =

y_com =  =  = 1 m                              

So position of C.O.M is

Ex.3 Two particles of mass 2 kg and 4 kg lie on the same line. If 4 kg is displaced rightwards by 5m then by what distance 2 kg should be move for which centre of mass will remain at the same position. 

Sol. Let us assume that C.O.M. lie at point C and the distance of C from 2kg and 4kg particles are r₁ & r₂ respectively. Then from relation

m₁r₁ = m₂r₂

2r₁ = 4r₂                   ...(i)

Now 4kg is displaced rightwards by 5m then assume 2kg is displaced leftwards by x distance to keep the C.O.M. at rest.

To keep the C.O.M at rest 2 kg displaced 10 m left wards

Aliter : If centre of mass is at rest then we can write

m₁x = m₂y

2 × x = 4 × 5

x = 10 m

Ex.4 Two particles of mass 1 kg and 2 kg lie on the same line. If 2kg is displaced 10m rightwards then by what distance 1kg should displaced so that centre of mass will displaced 2m right wards. 

Sol. Initially let us assume that C.O.M is at point C which is r₁ & r₂ distance apart from mass m₁ & m₂ respectively as shown in figure.

from relation m₁ r₁ = m₂ r₂

⇒ (1) r₁ = 2r₂

Now 2kg is displaced 10 m rightwards then we assume that 1 kg is displaced x m leftward to move the C.O.M 2m rightwards.

So from relation m₁r₁' = m₂r₂'

from eq. (i) & (ii) x = 14m (leftwards)

Ex.5 Three particles of mass 1 kg, 2 kg, and 3 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1m. Find the distance of their centre of mass from A. 

Sol. Assume that 1kg mass is placed at origin as shown in figure.

co-ordinate of A = (0, 0)

co-ordinate of B = (1cos60°,1sin60°) =

co-ordinate of C = (1, 0)

1.3 Centre of Mass of a Continuous Mass Distribution

For continuous mass distribution the centre of mass can be located by replacing summation sign with an integral sign. Proper limits for the integral are chosen according to the situation

 ...(i)

 = M (mass of the body)

here x,y,z in the numerator of the eq. (i) is the coordinate of the centre of mass of the dm mass.

 =

Note : 

• If an object has symmetric mass distribution about x axis then y coordinate of COM is zero and vice-versa

(a) Centre of Mass of a Uniform Rod 

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at
x = L. Mass per unit length of the rod l =

Hence, dm, (the mass of the element dx situated at x = x is) = l dx

The coordinates of the element dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be

x_COM =  =  =

The y-coordinate of COM is

y_COM =  = 0

Similarly, z_COM = 0

i.e., the coordinates of COM of the rod are , i.e, it lies at the centre of the rod.

Ex.6 A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/length) l of the rod varies with the distance x from the origin as l = Rx. Here, R is a positive constant. Find the position of centre of mass of this rod. 

Sol. Mass of element dx situated at x = x is

dm = l dx = R x dx

The COM of the element has coordinates (x, 0, 0). Therefore, x-coordinates of COM of the rod will be

The y-coordinates of COM of the rod is  (as y = 0)

Similarly, z_COM = 0

Hence, the centre of mass of the rod lies at

(b) Centre of mass of a Semicircular Ring : 

Figure shows the object (semi circular ring). By observation we can say that the x-coordinate of the centre of mass of the ring is zero as the half ring is symmetrical about y-axis on both sides of the origin. Only we are required to find the y-coordinate of the centre of mass.

To find y_cm we use  ...(i)

Here y is the position of C.O.M. of dm mass

Here for dm we consider an elemental arc of the ring at an angle q from the x-direction of angular width dq. If radius of the ring is R then its y coordinate-will be R sinq, here dm is given as

where l = mass density of semi circular ring.

So from equation ----(i), we have

 ...(ii)

(c) Centre of mass of Semicircular Disc :

Figure shows the half disc of mass M and radius R. Here, we are only required to find the y-coordinate of the centre of mass of this disc as centre of mass will be located on its half vertical diameter. Here to find y_cm, we consider a small elemental ring of mass dm of radius r on the disc (disc can be considered to be made up such thin rings of increasing radii) which will be integrated from 0 to R. Here dm is given as

where s is the mass density of the semi circular disc.

s =

Now the y-coordinate of the element is taken as , (as in previous section, we have derived that the centre of mass of a semi circular ring is concentrated at )

Here y is the position COM of dm mass.

Here y_cm is given as  ⇒

(d) Centre of mass of a Hollow Hemisphere : 

A hollow hemisphere of mass M and radius R. Now we consider an elemental circular strip of angular width dq at an angular distance q from the base of the hemisphere. This strip will have an area.

Its mass dm is given as

Here s is the mass density of a hollow hemisphere

s =

Here y-coordinate of this strip of mass dm can be taken as R sinq. Now we can obtain the centre of mass of the system as.

(e) Centre of mass of a Solid Cone :

A solid cone has mass M, height H and base radius R. Obviously the centre of mass of this cone will lie somewhere on its axis, at a height less than H/2. To locate the centre of mass we consider an elemental disc of width dy and radius r, at a distance y from the apex of the cone. Let the mass of this disc be dm, which can be given as

dm = r × pr² dy

Here r is the mass density of the solid cone

here y_cm can be given as

(f) C.O.M of a solid Hemisphere : -

A hemisphere is of mass density r and radius R To find its centre of mass (only y co-ordinate) we consider an elemental hollow hemispshere of radius r on the solid hemisphere (solid hemisphere can be considered to be made up such hollow hemisphere of increasing radii) which will be integrate from O to R.

Here y Co-ordinate of centre of mass of elemental hollow hemisphere is (0, r/2, 0)

dm = r 2pr² dr

y_CM =  ;  ; y_CM =

(g) Centre of mass of Triangular Plate : 

A triangular plate has mass density s height H and base is 2R. Obviously the centre of mass of this plate will lie some where on its axis at a height less than H/2. To locate the centre of mass we consider an elemental rod of width dy and length 2r at a distance y from the apex of the plate. Let the mass of this rod be dm which can be gives as

dm = s (2r) dy

from the theorem of triangle

⇒ r =

Here Y_CM can be given as

y_CM =  ; y_CM =

y_CM =  ; y_CM =

2. COMBINATION OF STRUCTURE : 

Ex.7 Two circular disc having radius R and mass density s and 2s respectively are placed as shown in figure. Then find out the position of COM of the system. 

Sol. Mass of disc A m_A = spR²

Mass of disc B m_B = 2spR²

Due to symmetry the COM of disc A lie at point O and COM of disc B lie at point O'. So we realize the above

problem in a following way

Centre of mass due to both the disc lie at point C (assume), having distance x from m_A

⇒  ;  ;

So the centre of mass lie in the disc B having distance from O.

Ex.8 Find out the position of centre of mass of the figure shown below. 

Sol.

We divide the above problem in two parts

(i) First find out position of centre of mass of both semicircular plate and rectangular plate separately.

(ii) Then find the position of centre of mass of given structure .

Centre of mass of semicircular disc lie at

⇒

Centre of mass of rectangular plate lie at the centre of plate at point C

⇒ BC = R

⇒

 ;

⇒

⇒ Let us assume COM is at r₁ distance from m_R

⇒  ⇒  Ans. 

3. Cavity Problems : 

If some mass or area is removed from a rigid body then the position of centre of mass of the remaining portion is obtained by assuming that in a remaining part +m & -m mass is there. Further steps are explained by following example.

Ex.9 Find the position of centre of mass of the uniform lamina shown in figure. If the mass density of the lamina is s . 

Sol. We assume that in remaining portion a disc of radius a/2 having mass density s is there then we also include one disc of a/2 radius having -s mass density. So now the problem change in following form

So the centre of mass of both disc A & B lie in their respective centre such as O & O'.

Now

⇒ C.O.M. of the lamina ⇒                             

m_A = s (p a²)

m_B = -s (p) (a/2)² = -s

⇒ c =  =  ;

i.e., C.O.M lie on leftward side from point O.

Ex.10 Find out the position of centre of mass of the uniform lamina as shown in figure. 

Sol. We assume that a disc of radius R having mass density ± s is in the removed section.

Now the problem change in following form

 =

When disc of mass density s and radius R is include than a complete rectangular plate is make having centre of mass at point O. When consider only disc having mass density -s and radius R then C.O.M of this disc lie at point O'

Then the position of C.O.M

=  =  =

i.e., centre of mass lie in the rightwards side from the cavity.

Ex.11 The centre of mass of rigid body always lie inside the body. Is this statement true or false? 

Sol. False.

Ex.12 The centre of mass always lie on the axis of symmetry if it exists. Is this statement true of false? 

Sol. True

Ex.13 If all the particles of a system lie in y-z plane, the x-coordinate of the centre of mass will be zero. Is this statement true or not? 

Sol. True

4. Motion of centre of Mass and Conservation of Momentum: - 

The position of centre of mass is given by

 ....(1)

Here m₁, m₂, m₃ ..... are the mass in the system and  is the corresponding position vector of m₁, m₂, m₃ respectively

4.1 Velocity of C.O.M of system : 

To find the velocity of centre of mass we differentiate equation (1) with respect to time

⇒

 ...(2)

4.2 Acceleration of centre of mass of the system : -

To find the acceleration of C.O.M we differentiate equation (2)

⇒

 ...(3)

Now (m₁ + m₂ + m₃)  =

The internal forces which the particles exert on one another play absolutely no role in the motion of the centre of mass.

Ex.14 Two particles A and B of mass 1 kg and 2 kg respectively are projected in the directions shown in figure with speeds u_A = 200 m/s and u_B = 50 m/s. Initially they were 90 m apart. Find the maximum height attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant. (g = 10 m/s²) 

Sol.

Case I : If F_net = 0 then we conclude :

(a) The acceleration of centre of mass is zero ()

If a₁, a₂, a₃.... is acceleration of m₁, m₂, m₃ mass in the system then a₁, a₂, a₃ may or may not be zero.

(b) K.E. of the system is not constant it may change due to internal force.

(c) Velocity of centre of mass is constant  but v₁, v₂, v₃ may or may not constant. It may be change due to internal force.

from eq (2)

This is called momentum conservation.

"If resultant external force is zero on the system, then the net momentum of the system must remain constant". 

Case II : When centre of mass is at rest.

(a)  then

 ⇒  = constant.

i.e. ......... may or may not change

Ex.15 A wooden plank of mass 20 kg is resting on a smooth horizontal floor. A man of mass 60 kg starts moving from one end of the plank to the other end. The length of the plank is 10 m. Find the displacement of the plank over the floor when the man reaches the other end of the plank. 

Sol. Here, the system is man +plank. Net force on this system is horizontal direction is zero and initially the centre of mass of the system is at rest. Therefore, the centre of mass does not move in horizontal direction.

Let x be the displacement of the Plank. Assuming the origin, i.e., x = 0 at the position shown in figure.

As we said earlier also, the centre of mass will not move in horizontal direction (x-axis). Therefore, for centre of mass to remain stationary,

x_i = x_f

20x = 60 × (10 -x)

or x =  or x = 7.5 m Ans. 

Ex.16 Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process? 

Sol. Here the system is Mr. Verma + Mr. Mathur + boat. Net force on this system is in horizontal direction is zero and initially the centre of mass of the system is at rest. Therefore the C.O.M does not move in horizontal direction. Let x be the displacement of the boat. Then We can use the concept

Case III : When net force is zero only in one direction. 

Let us assume that F_net in x direction is zero then we conclude

(i) Acceleration of the system in x direction is zero (a_x = 0)

(ii) v_(com)x = constant

i.e., momentum is conserved only in x direction

Ex.17 A man of mass m₁ is standing on a platform of mass m₂ kept on a smooth horizontal surface. The man starts moving on the platform with a velocity v_r relative to the platform. Find the recoil velocity of platform. 

Sol. Absolute velocity of man = v_r -v where v = recoil velocity of platform. Taking the platform and the man a system, net external force on the system in horizontal direction is zero. The linear momentum of the system remains constant. Initially both the man and the platform were at rest.

Hence, 0 = m₁(v_r -v) -m₂v

v =  Ans.

Ex.18 A gun (mass = M) fires a bullet (mass = m) with speed v_r relative to barrel of the gun which is inclined at an angle of 60° with horizontal. The gun is placed over a smooth horizontal surface. Find the recoil speed of gun. 

Sol. Let the recoil speed of gun is v. Taking gun bullet as the system. Net external force on the

system in horizontal direction is zero. Initially the system was at rest. Therefore, applying the principle of conservation of linear momentum in horizontal direction, we get

Mv -m(v_r cos 60° -v) = 0

v =  or v = Ans. 

Ex.19 A particle of mass m is placed at rest on the top of a smooth wedge of mass M, which in turn is placed at rest on a smooth horizontal surface as shown in figure. Then the distance moved by the wedge as the particle reaches the foot of the wedge is : 

Sol. There is no external force in horizontal direction on the wedge block system, So the x-coordinate of the C.O.M of the wedge block system is at rest.

Let us assume that wedge move x when block reaches the ground. We can use the following relation when

x - coordinate of C.O.M is at rest

m₁x₁ = m₂x₂ 

Mx = m (l -x)

Ex.20 A projectile is fired at a speed of 100 m/s at an angle of 37° above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the lighter piece coming to rest. Find the distance from the launching point to the point where the heavier piece lands. 

Sol. Internal force do not effect the motion of the centre of mass, the centre of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is,

x_com =  =

= 960 m

The centre of mass will hit the ground at this position. As the smaller block comes to rest after breaking, it falls down vertically and hits the ground at half of the range, i.e., at x = 480 m. If the heavier block hits the ground at x₂, then

x_com =  ⇒ 960 =  x₂ = 1120 m Ans. 

Ex.21 A shell is fired from a cannon with a speed of 100 m/s at an angle 60° with the horizontal (positive x-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of the other fragment at the time of explosion. 

Sol. As we know in absence of external force the motion of centre of mass of a body remains unaffected. Thus, here the centre of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is v_M = u cos q = 100 × cos 60° = 50 m/s Let v₁ be the speed of the fragment which moves along the negative x-direction and the other fragment has speed v₂,. which must be along positive x-direction. Now from momentum conservation, we have

or 2v = v₂ -v₁ or v₂ = 2v v₁

= (2 × 50) 50 = 150 m/s

Ex.22 A particle of mass 2 m is projected at an angle of 45° with horizontal with a velocity of  After 1 s explosion takes place and the particle is broken into two equal pieces. 

As a result of explosion one part comes to rest. Find the maximum height attained by the other part. 

(Take g = 10 m/s²)  

Sol. Applying conservation of linear momentum at the time of collision, or at t = 1 s,

At 1 sec, masses will be at height :

After explosion other mass will further rise to a height :

m

u_y = 20 m/s just after collision.

Total height h = h₁ h₂ = 35 m

Ex.23 A plank of mass 5 kg placed on a frictionless horizontal plane. Further a block of mass 1 kg is placed over the plank. A massless spring of natural length 2m is fixed to the plank by its one end. The other end of spring is compressed by the block by half of spring's natural length. They system is now released from the rest. What is the velocity of the plank when block leaves the plank? (The stiffness contant of spring is 100 N/m)

Sol. Let the velocity of the block and the plank, when the block leaves the spring be u and v respectively.

By conservation of energy  =  +  [M = mass of the plank, m = mass of the block]

⇒ 100 = u² + 5 v²  ...(i)

By conservation of momentum mu + Mv = 0

⇒ u = -5 v ...(ii)

Solving Eqs(i) and (ii)

30v² = 100 ⇒

From this moment until block falls, both plank and block keep their velocity constant.

Thus, when block falls, velocity of plank = Ans. 

Ex.24 Two identical blocks each of mass M = 9 kg are placed on a rough horizontal surface of frictional coefficient m = 0.1. The two blocks are joined by a light spring and block B is in contact with a vertical fixed wall as shown in figure. A bullet of mass m = 1kg and v₀ = 10 m/s hits block A and gets embedded in it. Find the maximum compression of spring. (Spring constant = 240 N/m, g = 10 m/s²) 

Sol.

Ex.25 A flat car of mass M is at rest on a frictionless floor with a child of mass m standing at its edge. If child jumps off from the car towards right with an initial velocity u, with respect to the car, find the velocity of the car after its jump. 

Sol. Let car attains a velocity v, and the net velocity of the child with respect to earth will be u -v, as u is its velocity with respect to car.

Initially, the system was at rest, thus according to momentum conservation, momentum after jump must be zero, as

m(u -v) = M v

Ex.26 A flat car of mass M with a child of mass m is moving with a velocity v₁ on a friction less surface. The child jumps in the direction of motion of car with a velocity u with respect to car. Find the final velocities of the child and that of the car after jump. 

Sol. This case is similar to the previous example, except now the car is moving before jump. Here also no external force is acting on the system in horizontal direction, hence momentum remains conserved in this direction. After jump car attains a velocity v₂ in the same direction, which is less than v₁, due to backward push of the child for jumping. After jump child attains a velocity u + v₂ in the direction of motion of car, with respect to ground.

According to momentum conservation

(M + m) v₁ = Mv₂ + m (u + v₂)

Velocity of car after jump is

Velocity of child after jump is u + v₂ =

Ex.27 Two persons A and B, each of mass m are standing at the two ends of rail-road car of mass M. The person A jumps to the left with a horizontal speed u with respect to the car. There after, the person B jumps to the right, again with the same horizontal speed u with respect to the car. Find the velocity of the car after both the persons have jumped off. 

Sol. Let car attain the velocity v in right ward and velocity of man A with respect to ground is v' then

v' = v -u

from momentum conservation

0 = mv' (M m)v

⇒ m(v -u) (M m)v = 0 ⇒

After wards mass B jumps to the right with the same horizontal speed u with respect to car, than car attain v" velocity from linear momentum conservation.

(M m)v = m(u v") Mv"

Now v" =

Ex.28. A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure. Assuming frictionless surfaces find the velocity of the triangular block when the smaller reaches the bottom end. 

Sol. Let us assume that wedge move leftward with velocity v and block move down ward with velocity u with respect to wedge.

Net force is horizontal direction is zero so momentum is conserved in x direction.

Now velocity of block with respect to ground is

5. Spring block system : 

A light spring of spring constant k and natural length l₀ attached in a compressed condition between two blocks of mass m₁ & m₂ on a smooth horizontal surface as shown in the figure. The spring is initially compressed by a distance x₀.

When system is released the block acquire velocities in opposite direction. Let us assume that the velocities of block m₁ & m₂ is v₁ & v₂ respectively at natural length of the spring and since no external force acts on this system in horizontal direction. Hence the linear momentum remains constant. Then from momentum conservation.

In initial condition there is no external force on the system and both the block is at stationary condition. Therefore centre of mass of the system is at rest. So we can write.

from above figure we can conclude

Due to inertia both the block move further from the position of the natural length of the spring.

Maximum extension occur when both the blocks come to rest. Let us assume that x₁' & x₂' are the extension in the spring from the initial position due to block m₁ & m₂ from natural length

So at maximum extension v₁ = v₂ = 0

Centre of mass is at rest Therefore we can write

Ex.29 A light spring of spring constant k is kept compressed between two blocks of masses m and M on a smooth horizontal surface. When released, the blocks aquirse velocities in opposite directions. The spring loses contact with the blocks when it acquires natural length. If the spring was initially compressed through a distance x, find the final speeds of the two blocks. 

Sol. Consider the two blocks plus the spring to be the system. No external force acts on this system in horizontal direction. Hence, the linear momentum will remain constant. Suppose, the block of mass M moves with a speed v₁ and the other block with a speed v₂ after losing contact with the spring. From conservation of linear momentum in horizontal direction we have

As there is no friction, mechanical energy will remain conserved.

IInd Format : Figure shows two blocks of masses 2m and m are placed on a frictionless surface and connected with a spring. An external kick gives a velocity v₀ m/s to the m mass towards right

Now velocity of centre of mass is

 ⇒ V_com =  =

Due to kick on m mass block is starts moving with a velocity v₀ towards right immediately but due to inertia 2m block remain at rest at that moment. Thus velocity of block A & B with respect to the centre of mass is

v_A =  =  m/sec. (towards right)

v_B = 0 - = - =  (towards left)

Now the following figure shown the condition when centre of mass is rest.

If the maximum extension of the spring is x₀ then at this position both the block come to rest condition with respect to COM so from mechanical energy conservation

Put the above value is equation 1

III^rd format : 

Example 

A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring.

We solve the situation in the reference frame of centre of mass. As only F is the external force acting on the system, due to this force, the acceleration of the centre of mass is F/(M + m). Thus with respect to centre of mass there is a Pseudo force on the two masses in opposite direction, the free body diagram of m and M with respect to centre of mass (taking centre of mass at rest) is shown in figure.

Taking centre of mass at rest, if m moves maximum by a distance x₁ and M moves maximum by a distance x₂, then the work done by external forces (including Pseudo force) will be

W = .x₁ +  =

This work is stored in the form of potential energy of the spring as

U =

Thus on equating we get the maximum extension in the spring, as after this instant the spring starts contracting.

 =

x_max = x₁ + x₂ =

Ex.30 Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. 

(a) Find the displacement of the centre of mass at time t

(b) If the extension of the spring is x₀ at time t, find the displacement of the two blocks at this instant.

Sol. (a) The acceleration of the centre of mass is

a_COM =

The displacement of the centre of mass at time t will be

x =  =  Ans. 

(b) Suppose the displacement of the first block is x₁ and that of the second is x₂. Then,

x =

or,

or, x₁ + x₂ =  ...(i)

Further, the extension of the spring is x₁ -x₂. Therefore,

x₁ -x₂ = x₀ ...(ii)

From Eqs. (i) and (ii),

x₁ =  and x₂ = Ans. 

6. Impulse :

Impulse of a force  acting on a body for the time interval t = t₁ to t = t₂ is defined as

change in momentum due to force

Also               (impulse - momentum theorem)

Note :

* Impulse applied to an object in a given time interval can also be calculated from the area under force time (F-t) graph in the same time interval.

6.1 Instantaneous Impulse :

There are many cases when a force acts for such a short time that the effect is instantaneous, e.g., a bat striking a ball. In such cases, although the magnitude of the force and the time for which it acts may each be unknown but the value of their product (i.e., impulse) can be known by measuring the initial and final momentum. Thus, we can write.

Important Points :

(1) It is a vector quantity.

(2) Dimensions = [MLT^-1]

(3) SI unit = kg m/s

(4) Direction is along change in momentum.

(5) Magnitude is equal to area under the F-t. graph.

(6)

(7) It is not a property of a particle, but it is a measure of the degree to which an external force changes the momentum of the particle.

Ex.31 The hero of a stunt film fires 50 g bullets from a machine gun, each at a speed of 1.0 km/s. If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period. 

Sol. The momentum of each bullet

= (0.050 kg) (1000 m/s)

= 50 kg-m/s.

The gun has been imparted this much amount of momentum by each bullet fired. Thus, the rate of change of momentum of the gun

=

= 250 N

In order to hold the gun, the hero must exert a force of 250 N against the gun.

Ex.32 A ball of mass m = 1kg strikes smooth horizontal floor shown in figure. Find out impulse exerted on the floor is : 

Sol. As the ball strike on the surface on impulsive normal force is exerted on the ball as shown in figure.

This normal force can change only the component v_y. So in x direction momentum is conserved. (F_{net x} = 0)

⇒ v' cos 37° = 5 cos 53°

v' =  =

So, v'_y = v' sin37° =  =

Impulse = change in linear momentum in y direction

 = m(v_y -(-v'_y)) =  = 6.25 N-sec

6.2 Impulsive force :

A force, of relatively higher magnitude and acting for relatively shorter time, is called impulsive force. An impulsive force can change the momentum of a body in a finite magnitude in a very short time interval. Impulsive force is a relative term. There is no clear boundary between an impulsive and Non-Impulsive force.

Note : 

* Usually colliding forces are impulsive in nature.

Since, the application time is very small, hence, very little motion of the particle takes place.

Important points : 

1. Gravitational force and spring force are always non-Impulsive.

2. Normal, tension and friction are case dependent.

3. An impulsive force can only be balanced by another impulsive force.

1. Impulsive Normal : In case of collision, normal forces at the surface of collision are always impulsive

e.g. N_i = Impulsive; N_g = Non-impulsive

 Both normals are Impulsive

 N₁, N₃ = Impulsive; N₂ = non-impulsive

 Both normals are Impulsive

2. Impulsive Friction : If the normal between the two objects is impulsive, then the friction between the two will also be impulsive

 Friction at both surfaces is impulsive

Friction due to N₂ is non-impulsive and due to N₃ and N₁ are impulsive

3. Impulsive Tensions : 

When a string jerks, equal and opposite tension act suddenly at each end. Consequently equal and opposite impulses act on the bodies attached with the string in the direction of the string. There are two cases to be considered.

One end of the string is fixed : 

The impulse which acts at the fixed end of the string cannot change the momentum of the fixed object. The object attached to the free end however will undergo a change in momentum in the direction of the string. The momentum remains unchanged in a direction perpendicular to the string where no impulsive forces act.

Both ends of the string attached to movable objects : 

In this case equal and opposite impulses act on the two objects, producing equal and opposite changes in momentum. The total momentum of the system therefore remains constant, although the momentum of each individual object is changed in the direction of the string. Perpendicular to the string however, no impulse acts and the momentum of each particle in this direction is unchanged.

For this example : 

In case of rod, tension is always impulsive and in case of spring, tension is always non-impulsive.

Ex.33 A block of mass m and a pan of equal mass are connected by a string going over a smooth light pulley. Initially the system is at rest when a particle of mass m falls on the pan and sticks to it. If the particle strikes the pan with a speed v, find the speed with which the system moves just after the collision. 

Sol. Let the required speed is V.

Further, let J₁ = impulse between particle and pan

and J₂ = impulse imparted to the block and the pan by the string

Using, Impulse = change in momentum

For particle J₁ = mv -mV ...(i)

For pan J₁ -J₂ = mV ...(ii)

For block J₂ = mV ...(iii)

Solving, these three equation, we get Ans. 

Alternative solution : 

Applying conservation of linear momentum along the string ;

mv = 3mV

we get, V = Ans. 

Ex.34 Two identical block A and B, connected by a massless string are placed on a frictionless horizontal plane. A bullet having same mass, moving with speed u strikes block B from behind as shown. If the bullet gets embedded into the block B then find : 

(A) The velocity of A, B, C after collision

(B) Impulse on A due to tension in the string

(C) Impulse on C due to normal force of collision.

(D) Impulse on B due to normal force of collision.

Sol. Let us assume that all the three are move with velocity v

Take rightward direction is ve. Then first we write impulse equation on bullet

Now impulse equation on block B

 ...(2)

Impulse equation on block A

 ...(3)

(a) Add eq. (1), (2), (3) then

0 = 3mv -mu ⇒ v =

(b) Impulse on A due to Tension in the string from eq. (3)

(c) Impulse on C due to normal force of collision

from eq. (1)  =

(d) Impulse on B due to normal force of collision

from eq. (2)

 =

7. Coefficient of Restitution (e) 

The coefficient of restitution is defined as the ratio of the impulses of reformation and deformation of either body.

e =  =

Example for calculation of e : 

Two smooth balls A and B approaching each other such that their centres are moving along line CD in absence of external impulsive force. The velocities of A and B just before collision be u₁ and u₂ respectively. The velocities of A and B just after collision be v₁ and v₂ respectively.

e =

Note : Coefficient of restitution is a factor between two colliding bodies which is depends on the material of the body but independent of shape.

We can say e is a factor which relates deformation and reformation of the body.

Ex.35 If a body falls normally on a surface from height h, what will be the height regained after collision if coefficient of restitution is e? 

Sol. 

If a body falls from height h, from equations of motion we know that it will hit the ground with a velocity say u =  which is also the velocity of approach here.

Now if after collision it regains a height h₁ then again by equations of motion v = which is also the velocity of separation. So, by definition of e,

e =  or h₁ = e²h 

Ex.36 A block of mass 2 kg is pushed towards a very heavy object moving with 2 m/s closer to the block (as shown). Assuming elastic collision and frictionless surface, find the final velocities of the blocks.  

Sol. Let v₁ and v₂ be the final velocities of 2kg block and heavy object respectively then,

Ex.37 A ball is moving with velocity 2 m/s towards a heavy wall moving towards the ball with speed 1 m/s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediately after the collision. 

Sol. The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e = 1).

separation speed = approach speed

or v -1 = 2 + 1

or v = 4 m/s Ans. 

Ex.38 A ball is dropped from a height h on to a floor. If in each collision its speed becomes e times of its striking value (a) find the time taken by ball to stop rebounding (b) find the total change in momentum in this time (c) find the average force exerted by the ball on the floor using results of part (a) and (b). 

Sol. (a) When the ball is dropped from a height h, time taken by it to reach the ground will be

t₀ =  and its speed v₀ =

Now after collision its speed will becomes e times, i.e., v₁ = ev₀ =  and so, it will take time to go up till its speed becomes zero = (v₁/g). The same time it will take to come down. So total time between I and II collision will be t₁ = 2v₁/g. Similarly, total time between II and III collision t₂ = 2v₂/g.

So total time of motion

(b) Change in momentum in I collision

= mv₁ -(-mv₀) = m (v₁ + v₀)

Change in momentum in II collision = m(v₂+v₁)

Change in momentum in nth collision = m(v_n + v_n-1)

Adding these all total change in momentum

(C) Now as  so, F_av =

Substituting the value of DT and Dp from Eqns. (1) and (2)

F_av =  ×  = mg                                                   . ..(3)

7.1 Line of Motion 

The line passing through the centre of the body along the direction of resultant velocity.

7.2 Line of Impact 

The line passing through the common normal to the surfaces in contact during impact is called line of impact. The force during collision acts along this line on both the bodies.

Direction of Line of impact can be determined by :

(a) Geometry of colliding objects like spheres, discs, wedge etc.

(b) Direction of change of momentum.

If one particle is stationary before the collision then the line of impact will be along its motion after collision.

Examples of line of impact  

(i) Two balls A and B are approaching each other such that their centres are moving along line CD.

ii) Two balls A and B are approaching each other such that their centre are moving along dotted lines as shown in figure.

(iii) Ball is falling on a stationary wedge.

Note : In previous discussed examples line of motion is same as line of impact. But in problems in which line of impact and line of motion is different then e will be

e =

Ex.40 A ball of mass m hits a floor with a speed v making an angle of incident q with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball. 

Sol. Suppose the angle of reflection is q' and the speed after the collision is v' (shown figure)

The floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives

Ex.41 A ball is projected from the ground at some angle with horizontal. Coefficient of restitution between the ball and the ground is e. Let a, b and c be the ratio of times of flight, horizontal range and maximum height in two successive paths. Find a, b and c in terms of e? 

Sol. Let us assume that ball is projected with speed u at an angle q with the horizontal.

Then Before first collision with the ground.

Time fo flight

Horizontal range

Maximum Height H_max =  ...(1)

After striking the ground the component u_y is change into e u_y, so

Time of flight T' = ,

 ...(2)

from eq (1) & (2)

Now

 ;  = c

Ex.42 A ball is projected from the ground with speed u at an angle a with horizontal. It collides with a wall at a distance a from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall. 

Sol. A ball is a projected with speed u at an angle a with horizontal. It collides at a distance a with a wall parallel to y-axis as shown in figure.

Let v_x and v_y be the components of its velocity along x and y-directions at the time of impact with wall. Coefficient of restitution between the ball and the wall is e.

Component of its velocity along y-direction (common tangent) v_y will remain unchanged while component of its velocity along x-direction (common normal) v_x will becomes ev_x is opposite direction.

*Further, since v_y does not change due to collision, the time of flight (time taken by the ball to return to the same level) and maximum height attained by the ball remain same as it would had been in the absence of collision with the wall. Thus,

From O A B, R = a = u cos a . t_OAB

from BCO, R = a = eucosa. t_BCO

T = t_OAB + t_BCO

or  =  or  = -

or  e =

or e = Ans.

Ex.43 To test the manufactured properties of 10 N steel balls, each ball is released from rest as shown and strikes a 45° inclined surface. If the coefficient of restitution is to be e = 0.8. determine the distance s to where the ball must strike the horizontal plane at A. At what speed does the ball strike at A? (g = 9.8 m/s²) 

Sol. v₀ =  =  = 5.42 m/s

Component of velocity parallel and perpendicular to plane at the time of collision.

v₁ = v₂ =  = 3.83 m/sec.

Component parallel to plane (v₁) remains unchanged, while component perpendicular to plane becomes ev₂, where

ev₂ = 0.8 × 3.83 = 3.0 m/s

Component of velocity in horizontal direction after collision

v_x =  =  = 4.83 m/s

While component of velocity in vertical direction after collision.

v_y =  = = 0.59 m/s

Let t be the time, the particle takes from point C to A, then

1.0 = 0.59 t +  × 9.8 × t^{2 ;} t = 0.4 sec

Solving this we get,

DA = v_xt = (4.83)(0.4) = 1.93 m

S = DA -DE = 1.93 -1.0

S = 0.93 m

v_yA = v_yc + gt = (0.59) + (9.8) (0.4) = 4.51 m/s

v_xA = v_xC = 4.83 m/s

v_A =  = 6.6 m/s

Ex.44 A ball of mass m = 1 kg falling vertically with a velocity v₀ = 2m/s strikes a wedge of mass M = 2kg kept on a smooth, horizontal surface as shown in figure. The coefficient of restitution between the ball and the wedge is e = . Find the velocity of the wedge and the ball immediately after collision. 

Sol. Given M = 2kg and m = 1kg

Let, J be the impulse between ball and wedge during collision and v₁, v₂ and v₃ be the components of velocity of the wedge and the ball in horizontal and vertical directions respectively.

Applying impulse = change in momentum

we get J sin 30° = Mv₁ = mv₂

or  = 2v₁ = v₂

J cos 30° = m(v₃ + v₀) ...(i)

or  = (v₃ + 2) ...(ii)

Applying, relative speed of separation = e

(relative speed of approach) in common normal direction, we get

Thus, velocities of wedge

and ball are v₁ = m/s

and v₂ = in horizontal direction as shown in figure.

8. Collision or Impact 

Collision is an event in which an impulsive force acts between two or more bodies for a short time, which results in change of their velocities.

Note : 

• In a collision, particles may or may not come in physical contact.
• The duration of collision, Dt is negligible as compared to the usual time intervals of observation of motion.
• In a collision the effect of external non impulsive forces such as gravity are not taken into account as due to small duration of collision (Dt) average impulsive force responsible for collision is much larger than external forces acting on the system.

The collision is in fact a redistribution of total momentum of the particle :

Thus law of conservation of linear momentum is indepensible in dealing with the phenomenon of collision between particles. Consider a situation shown in figure.

Two balls of masses m₁ and m₂ are moving with velocities v₁ and v₂ (<v₁) along the same straight line in a smooth horizontal surface. Now let us see what happens during the collision between two particles.

figure (a) : Balls of mass m₁ is behind m₂. Since v₁ > v₂, the balls will collide after some time.

figure (b) : During collision both the balls are a little bit deformed. Due to deformation two equal and opposite normal forces act on both the balls. These forces decreases the velocity of m₁ and increase the velocity of m₂

figure (c): Now velocity of ball m₁ is decrease from v₁ to v₁' and velocity of ball m₂ is increase from v₂ to v₂'. But still v₁' > v₂' so both the ball are continuously deformed.

figure(d) : Contact surface of both the balls are deformed till the velocity of both the balls become equal. So at maximum deformation velocities of both the blocks are equal

at maximum deformation

figure(e) : Normal force is still in the direction shown in figure i.e. velocity of m₁ is further decreased and that of m₂ increased. Now both the balls starts to regain their original shape and size.

figure (f) : These two forces redistributes their linear momentum in such a manner that both the blocks are separated from one another, Velocity of ball m₂ becomes more than the velocity of block m₁ i.e., v₂>v₁

The collision is said to be elastic if both the blocks regain their original form, The collision is said to be inelastic. If the deformation is permanent, and the blocks move together with same velocity after the collision, the collision is said to be perfectly inelastic.

8.1 Classification of collisions 

(a) On the basis of line of impact 

(i) Head-on collision : If the velocities of the colliding particles are along the same line before and after the collision.

(ii) Oblique collision : If the velocities of the colliding particles are along different lines before and after the collision.

(b) On the basis of energy : 

(i) Elastic collision : 

(a) In an elastic collision, the colliding particles regain their shape and size completely after collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies.

(b) Thus, kinetic energy of system after collision is equal to kinetic energy of system before collision.

(c) e = 1

(d) Due to F_net on the system is zero linear momentum remains conserved.

(ii) Inelastic collision : 

(a) In an inelastic collision, the colliding particles do not regain their shape and size completely after collision.

(b)  Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles no longer remains conserved.

(c) However, in the absence of external forces, law of conservation of linear momentum still holds good.

(d) (Energy loss)_{Perfectly Inelastic} > (Energy loss)_{Partial Inelastic}

(e) 0 < e < 1

(iii) Perfectly Inelastic collision : 

(i)  In this the colliding bodies do not return to their original shape and size after collision i.e. both the particles stick together after collision and moving with same velocity

(ii) But due to F_net of the system is zero linear momentum remains conserved.

(iii) Total energy is conserved.

(iv) Initial kinetic energy > Final K.E. Energy

(v) Loss in kinetic energy goes to the deformation potential energy

(vi) e = 0

8.2 Value of Velocities after collision : 

Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e.

e =

⇒ (u₁ -u₂)e = (v₂ -v₁) ...(i)

By momentum conservation

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂  ...(ii)

v₂ = v₁ e(u₁ -u₂) ...(iii)

from above equation

v₁ =  ...(iii)

v₂ =  ...(iv) 

Special cases : 

i.e when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities.

8.3 Collision in two dimension (oblique) : 

1. A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual particles change along common normal direction. If mass of the colliding particles remain constant during collision, then we can say that linear velocity of the individual particles change during collision in this direction.

2. No component of impulse act along common tangent direction. Hence, linear momentum or linear velocity of individual particles (if mass is constant) remain unchanged along this direction.

3. Net impulse on both the particles is zero during collision. Hence, net momentum of both the particles remain conserved before and after collision in any direction.

4. Definition of coefficient of restitution can be applied along common normal direction, i.e., along common normal direction we can apply Relative speed of separation = e (relative speed of approach)

Ex.45 A ball of mass m makes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after collision. 

Sol. In head on elastic collision between two particles, they exchange their velocities. In this case, the component of ball 1 along common normal direction, v cos q becomes zero after collision, while

that of 2 becomes v cos q. While the components along common tangent direction of both the particles remain unchanged. Thus, the components along common tangent and common normal direction of both the balls in tabular form are given a head :

From the above table and figure, we see that both the balls move at right angles after collision with velocities v sin q and v cos q.

Note : When two identical bodies have an oblique elastic collision, with one body at rest before collision, then the two bodies will go in directions.

Ex.46 Two spheres are moving towards each other. Both have same radius but their masses are 2kg and 4kg. If the velocities are 4m/s and 2m/s respectively and coefficient of restitution is e = 1/3, find.

(a) The common velocity along the line of impact.

(b) Final velocities along line of impact.

(c) Impulse of deformation.

(d) impulse of reformation

(e) Maximum potential energy of deformation

(f) Loss in kinetic energy due to collision.

Sol. In DABC sinq =  =  =

or q = 30°

(a) By conservation of momentum along line of impact.

2(4 cos 30°) -4(2cos30°) = (2 + 4)v

or v = 0 (common velocity along LOI)

(b) Let v₁ and v₂ be the final velocity of A and B respectively then, by conservation of momentum along line of impact, 2(4 cos 30°) -4(2cos30°) = 2(v₁) + 4(v₂)

or 0 = v₁ + 2v₂ ........(1)

By coefficient of restitution,

e =

or  =  or v₂ -v₁ =  ...(2)

from the above two equations,

v₁ =  and v₂=

(c) J₀ = m₁(v -u₁) = 2 (0 -4 cos 30°) = -4 N-s

(d) J_R = eJ₀ = (-4) = -

(e) Maximum potential energy of deformation is equal to loss in kinetic energy during deformation upto maximum deformed state,

or U = 18 Joule

(f) Loss in kinetic energy

9. Variable Mass

In our discussion of the conservation linear momentum, we have so far dealt with systems whose system whose mass remains constant. We now consider those mass is variable, i.e., those in which mass enters or leaves the system. A typical case is that of the rocket from which hot gases keep on escaping thereby continuously decreasing its mass.

In such problem you have nothing to do but apply a thrust force  to the main mass in addition to the all other force acting on it. This thrust force is given by,

Here  is the velocity of the mass gained or mass ejected relative to the main mass. In case of rocket this is sometimes called the exhaust velocity of the gases.  is the rate at which mass is increasing or decreasing.

The expression for the thrust force can be derived from the conservation of linear momentum in the absence of any external forces on a system as follows :

Suppose at some moment t = t mass of a body is m and its velocity is . After some time at t = t dt its mass becomes (m -dm) and velocity becomes . The mass dm is ejected with relative velocity . Absolute velocity of mass `dm' is therefore . If no external forces are acting on the system, the linear momentum of the system will remain conserved,

Here,  = thrust force  and = rate at which mass is ejecting

Problems related to variable mass can be solved in following three steps  

1. Make a list of all the forces acting on the main mass and apply them on it.

2. Apply an additional thrust force  on the mass, the magnitude of which is  and direction is given by the direction of  in case the mass is increasing and otherwise the direction of  if it is decreasing.

3. Find net force on the mass and apply

(m = mass at that particular instant)

9.1 Rocket Propulsion 

Let m₀ be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that moment. Initially let us suppose that the velocity of the rocket is u.

Further, let  be the mass of the gas ejected per unit time and v_r the exhaust velocity of the gases. Usually  and v_r are kept constant throughout the journey of the rocket. Now, let us write few equations which can be used in the problems of rocket propulsion. At time t = t

1. Thrust force on the rocket

                     (upwards)

2. Weight of the rocket

W = mg                              (downwards)

3. Net force on the rocket

F_net = F_t -W                           (upwards)

or

Note :

Ex.47 A uniform chain of mass per unit length l begins to fall with a velocity v on the table. Find the thrust force exerted by the chain on the table. 

Sol. Let us assume that the mass of the chain is m and length l.

We assume that after time t, x length of the chain has fallen on the table. Then the speed of the upper part of the chain is  as shown in figure.

Now its time t + dt, length of chain has fallen on the table is v dt. Then the mass of chain has fallen on the table is