CENTRE OF MASS-JEE(MAIN)

CENTRE OF MASS  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.

CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES Consider a system of N point masses m1, m2, m3, ................ mn whose    position vectors from origin O are given by r1 , r2 , r3 ,...............  rn respectively. Then the position vector of the centre of mass C of the system is given by.    m1r1  m 2 r2  ........  mn rn   ; rcm = rcm = m1  m 2  ........  m n

1  rcm = M

n



m r

i i

i 1 n

m

i

i 1

n



m r

i i

i 1

 where, m i ri is called the moment of mass of the particle w.r.t O.

  M = 

n

 m i  is the total mass of the system.  i1 



n

Note:

If the origin is taken at the centre of mass then



 m r =0. hence, the COM is the point about which i i

i 1

the sum of “mass moments” of the system is zero.

POSITION OF COM OF T WO PARTICLES Centre of mass of two particles of masses m 1 and m 2 separated by a distance r lies in between the two particles. The distance of centre of mass from any of the particle (r) is inversely proportional to the mass of the particle (m) i.e. r  1/m or

r1 m2 = r2 m1

or

m1r 1 = m 2r 2

or

 m2   m1  r and r =  r 1 =  2 m  m 1  2  m1  m 2

  r 

Here, r 1 = distance of COM from m 1 and r 2 = distance of COM from m 2 From the above discussion, we see that r 1 = r 2 = 1/2 if m 1 = m 2, i.e., COM lies midway between the two particles of equal masses. Similarly, r 1 > r 2 if m 1 < m 2 and r 1 < r 2 if m 2 < m 1, i.e., COM is nearer to the particle having larger mass.

RESONANCE

AIEEE_CENTRE OF MASS - 1

Example 1.

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

Solution : Since, both the particles lies on x-axis, the COM will also lie on x-axis. Let the COM is located at x = x, then r 1 = distance of COM from the particle of mass 1 kg = x

and r 2 = distance of COM from the particle of mass 2 kg = (3 – x) Using

r1 m2 r2 = m1

or

x 2 = or x = 2 m 3x 1

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

Ans.

The position vector of three particles of masses m 1 = 1 kg, m 2 = 2 kg and m 3 = 3 kg are    r1  ( î  4 ˆj  kˆ ) m , r2  ( î  ˆj  kˆ ) m and r3  ( 2î  ˆj  2kˆ ) m respectively. Find the position vector of their centre of mass.

Solution : The position vector of COM of the three particles will be given by     m1r1  m 2 r2  m 3 r3 rCOM  m1  m 2  m 3 Substituting the values, we get  1 (1)( î  4ˆj  kˆ )  (2)(î  ˆj  kˆ )  (3)(2î – ˆj – 2kˆ ) rCOM   ( 3 î  ˆj  kˆ ) m 2 1 2  3

Example 3.

Ans.

Four particles of mass 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and D of a square of side 1 m. Find the position of centre of mass of the particles.

Solution : Assuming D as the origin, DC as x -axis and DA as y-axis, we have m 1 = 1 kg, (x 1, y1) = (0, 1m) m 2 = 2 kg, (x 2, y2) = (1m, 1m) m 3 = 3 kg, (x 3, y3) = (1m, 0) and m 4 = 4 kg, (x 4, y4) = (0, 0) Co-ordinates of their COM are x COM =

m1x1  m 2 x 2  m3m 3  m 4 x 4 m1  m 2  m 3  m 4 =

(1)(0)  2(1)  3(1)  4(0) 5 1 = = m = 0.5 m 1 2  3  4 10 2

RESONANCE


m1y1  m 2 y 2  m 3 y 3  m 4 y 4 m1  m 2  m 3  m 4

Similarly, yCOM =

= 

(1)(1)  2(1)  3(0 )  4(0) 3 = = 0.3 m 1 2  3  4 10

(x COM, yCOM) = (0.5 m, 0.3 m)

Ans.

Thus, position of COM of the four particles is as shown in figure. Example 4.

Consider a two-particle system with the particles having masses m 1 and m 2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as to keep the centre of mass at the same position?

Solution : Consider figure. Suppose the distance of m 1 from the centre of mass C is x1 and that of m 2 from C is x 2. Suppose the mass m 2 is moved through a distance d towards C so as to keep the centre of mass at C.

Then, m 1x 1 = m 2x 2 and m 1(x 1 – d) = m 2 (x 2 – d). Subtracting (ii) from (i) m 1d = m 2 d or,

.........(i) .........(ii)

m1 d = m d, 2

 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

x cm =

 x dm , y  dm

cm

=

 y dm , z  dm

 dm = M (mass of the body)

1  rcm = M

cm

=

 z dm  dm

x = x component of centre of mass of element



 r dm .

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

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 =

M L

M dx L The coordinates of the element dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be Hence, dm, (the mass of the element dx situated at x = x is) =

RESONANCE


L



x COM =

x dm

dx

0

 dm L

 =

0

1 L

=

x=x

x=0

x=L

M  ( x )  dx  L  M L



0

x dx 

L 2

The y-coordinate of COM is

yCOM = Similarly,

 y dm  dm

= 0

z COM = 0

L  i.e., the coordinates of COM of the rod are  , 0, 0  , i.e. it lies at the centre of the rod. 2 

Example 5.

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

Solution :

y

Mass of element dx situated at x = x is dm =  dx = Rx dx The COM of the element has coordinates (x, 0, 0). Therefore, x-coordinate of COM of the rod will be

dx

L

x COM =

x=0

 x dm  dm

=

2

R x dx

( x )(Rx) dx 0

(Rx) dx 0



=

L



0 L

R x dx

 0

The y-coordinate of COM of the rod is yCOM =

Similarly,

=

 x3     3  0 L

 x2     2  0

 y dm  dm

x



2L 3

=0

(as y = 0)

z COM = 0

 2L  Hence, the centre of mass of the rod lies at  ,0, 0  3  

RESONANCE

x=L

L

L

L



x=x

0

Ans.


 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 abnout y-axis on both sides of the origin. Only we are required to find the y-coordinate of the centre of mass. Y

Rd y=Rsin

ycm d 

To find ycm we use

ycm =

1 M

X

 dm y

...(i)

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

M × R d R

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

ycm

1 = M

ycm =



 0

M Rd (R sin) R



R = 

 sin  d 0

2R 

...(ii)

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 ycoordinate of the centre of mass of this disc as centre of mass will be located on its half vertical diameter. Here to find ycm , we consider a small elemental ring of mass dm of radius x 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 dm =

2M R 2

( x)dx Y ycm dx x

X R

Now the y-coordinate of the element is taken as

2x , as in previous section, we have derived that 

the centre of mass of a semi circular ring is concentrated at R

R

Here ycm is given as

RESONANCE

1 ycm = M

2x dm  0



1 = M

4M

 R 0

2

x 2 dx

2R 

ycm =

4R 3


1.

Centre of mass of a uniform rectangular, square or circular plate lies at its centre. Axis of symmetry plane of symmetry.

2.

For a laminar type (2-dimensional) body with uniform negligible thickness the formulae for finding the position of centre of mass are as follows :      m r  m 2 r2  .... A 1t r1  A 2 t r2  .... rCOM  1 1  ( m = At) m1  m 2  .... A 1t  A 2 t  ....

   A r  A 2 r2  .... rCOM  1 1 A 1  A 2  ....

or

Here, A stands for the area, 3.

If some mass of area is removed from a rigid body, then the position of centre of mass of the remaining portion is obtained from the following formulae: (i)

      m1r1  m 2 r2 A 1r1  A 2 r2 rCOM  rCOM  m1  m 2 or A1  A 2

(ii)

x COM 

m1x1  m 2 x 2 A x  A 2x2 x COM  1 1 or m1  m 2 A1  A 2

y COM 

m1y1  m 2 y 2 A y  A 2y2 y COM  1 1 or m1  m 2 A1  A 2

z COM 

m1z1  m 2 z 2 A 1z1  A 2 z 2 or z COM  A  A m1  m 2 1 2

and

   Here, m1, A1, r1 , x1, y1 and z1 are the values for the whole mass while m2, A2, r2 , x 2 , y2 and z2 are the values

for the mass which has been removed. Let us see two examples in support of the above theory. Example 6.

Find the position of centre of mass of the uniform lamina shown in figure. Y

a O

Solution :

X

Here, A 1 = area of complete circle = a 2

a A 2 = area of small circle =    2

2

=

a 2 4

(x 1, y1) = coordinates of centre of mass of large circle = (0, 0)

RESONANCE


and

a  (x 2, y2) = coordinates of centre of mass of small circle =  , 0  2 

Using x COM =

A 1x1  A 2 x 2 A1  A 2

 1 a 2  a      8 4 2 a = = a = – 2 3   6 a   a 2  4 4 

we get x COM

and yCOM = 0 as y1 and y2 both are zero.

 a  Therefore, coordinates of COM of the lamina shown in figure are   , 0   6 

Ans.

 CENTRE OF MASS OF SOME COMMON SYSTEMS 

A system of two point masses m1 r1 = m2 r2 The centre of mass lies closer to the heavier mass.



Rectangular plate (By symmetry)

xc =



b 2

yc =

A triangular plate (By qualitative argument)

at the centroid : yc =



h 3

A semi-circular ring

yc =



L 2

2R 

xc = 0

A semi-circular disc

yc =

4R 3

xc = 0

RESONANCE




A hemispherical shell yc =



3R 8

xc = 0

A circular cone (solid) yc =



xc = 0

A solid hemisphere

yc =



R 2

h 4

A circular cone (hollow)

yc =

h 3

Example 7.

A uniform thin rod is bent in the form of closed loop ABCDEFA as shown in the figure. The ycoordinate of the centre of mass of the system is

(A)

2r 

(B) 

6r 3  2

(C) 

2r 

Ans.

(B)

Solution.

The centre of mass of semicircular ring is at a distance

(D) Zero 2r from its centre. (Let  = mass/ 

length)  Ycm =

2r 4r    2r      6r r   r  r    2r 3  2

r 

MOTION OF CENTRE OF MASS AND CONSERVATION OF MOMENTUM : Velocity of centre of mass of system m1  v cm =

dr1 dr dr dr      m 2 2  m 3 3 .......... ....  m n n m1 v1  m 2 v 2  m 3 v 3 ..........  mn v n dt dt dt dt = M M

Here numerator of the right hand side term is the total momentum of the system i.e., summation of momentum of the individual component (particle) of the system

RESONANCE


Hence velocity of centre of mass of the system is the ratio of momentum of the system to the mass of the system.   PSystem  = M v cm

Acceleration of centre of mass of system     dv dv dv dv m1 1  m 2 2  m 3 3 .......... ....  mn n m1a1  m 2 a 2  m 3 a 3 ..........  mn an  dt dt dt dt = a cm = M M Net force on system Net External Force  Net int ernal Force Net External Force = = M M M (action and reaction both of an internal force must be within the system. Vector summation will cancel all internal forces and hence net internal force on system is zero)    Fext = M a cm  where Fext is the sum of the 'external' forces acting on the system. The internal forces which the =

particles exert on one another play absolutely no role in the motion of the centre of mass. If no external force is acting on a system of particles, the acceleration of centre of mass of the system will be zero. If a c = 0, it implies that v c must be a constant and if v cm is a constant, it implies that the total momentum of the system must remain constant. It leads to the principal of conservation of momentum in absence of external forces.   If Fext  0 then v cm = constant “If resultant external force is zero on the system, then the net momentum of the system must remain constant”.

Motion of COM in a moving system of particles: (1)

COM at rest : If F ext = 0 and V cm = 0, then COM remains at rest. Individual components of the system may move and have non-zero momentum due to mutual forces (internal), but the net momentum of the system remains zero. (i) All the particles of the system are at rest. (ii)

Particles are moving such that their net momentum is zero. example:

(iii)

A bomb at rest suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal & there is no external force on the system for explosion therefore, the COM of the bomb will remain at the original position and the fragment fly such that their net momentum remains zero.

(iv)

Two men standing on a frictionless platform, push each other, then also their net momentum remains zero because the push forces are internal for the two men system.

(v)

A boat floating in a lake, also has net momentum zero if the people on it changes their position, because the friction force required to move the people is internal of the boat system.

(vi) (vii)

Objects initially at rest, if moving under mutual forces (electrostatic or gravitation)also have net momentum zero. A light spring of spring constant k kept compressed between two blocks of masses m 1 and m 2 on a smooth horizontal surface. W hen released, the blocks acquire velocities in opposite directions, such that the net momentum is zero.

(viii)

In a fan, all particles are moving but COM is at rest

RESONANCE


(2)

COM moving with uniform velocity : If F ext = 0, then V cm remains constant therefore, net momentum of the system also remains conserved. Individual components of the system may have variable velocity and momentum due to mutual forces (internal), but the net momentum of the system remains constant and COM continues to move with the initial velocity. (i)

All the particles of the system are moving with same velocity. e.g.: A car moving with uniform speed on a straight road, has its COM moving with a constant velocity.

(ii)

Internal explosions / breaking does not change the motion of COM and net momentum remains conserved. A bomb moving in a straight line suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal & there is no external force on the system for explosion therefore, the COM of the bomb will continue the original motion and the fragment fly such that their net momentum remains conserved.

(iii)

Man jumping from cart or buggy also exert internal forces therefore net momentum of the system and hence, Motion of COM remains conserved.

(iv)

Two moving blocks connected by a light spring on a smooth horizontal surface. If the acting forces is only due to spring then COM will remain in its motion and momentum will remain conserved.

(v) (3)

Particles colliding in absence of external impulsive forces also have their momentum conserved. COM moving with acceleration : If an external force is present then COM continues its original motion as if the external force is acting on it, irrespective of internal forces. Example: Projectile motion : An axe thrown in air at an angle  with the horizontal will perform a complicated m otion of rotation as well as parabolic m otion under the effect of gravitation

H com =

R com

u2 sin2  2g

u2 sin 2 = g

T=

2u sin  g

Rcom Fc

Example: Circular Motion : A rod hinged at an end, rotates, than its COM performs circular motion. The centripetal force (F c) required in the circular motion is assumed to be acting on the COM. 2

 Fc

Fc

mg

Fc mg

mg mg

Fc  mù R COM

RESONANCE


Example 8.

Solution :

A projectile is fired at a speed of 100 m/s at an angle of 37º 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. 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, 2

x COM =

2u sin  cos  = g

2  10 4 

3 4  5 5

10

m

= 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 x 2, then x COM =

960 =

m1x1  m 2 x 2 m1  m 2

(m)( 480)  (3m)( x 2 ) (m  3m)

x 2 = 1120 m

Ans.

Momentum Conservation : The total linear momentum of a system of particles is equal to the product of the total mass of the   system and the velocity of its centre of mass. P = M v cm

  dP Fext = dt

If

  dP = 0 Fext = 0  dt

;

 P = constant

When the vector sum of the external forces acting on a system is zero, the total linear momentum of the system remains constant.     P + P + P + ...............+ P = constant. 1 2 3 n

Example 9.

A shell is fired from a cannon with a speed of 100 m/s at an angle 60º with the horizontal (positive xdirection). 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.

Solution : As we know in absence of external force the motion of centre of mass of a body remains uneffected. 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 = ucos = 100 ×cos60º = 50 m/s. Let v 1 be the speed of the fragment which moves along the negative x-direction and the other fragment has speed v 2,. which must be along positive x-direction. Now from momentum conservation, we have mv = or

m m v1 + v 2 2 2

or

2v = v 2 – v 1

v 2 = 2v + v 1 = (2 × 50) + 50 = 150 m/s

RESONANCE


Example 10.

A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts moving on the platform with a speed v relative to the platform, with what velocity relative to the ice does the platform recoil ?

Solution : Consider the situation shown in figure. Suppose the man moves at a speed w towards right and the platform recoils at a speed V towards left, both relative to the ice. Hence, the speed of the man relative to the platform is V + w. By the question, w V + w = v, or w = v – V .............(i) Taking the platform and the man to be the system, there is no V external horizontal force on the system. The linear momentum ice of the system remains constant. Initially both the man and the ///////////////////////////////////////////////// platform were at rest. Thus, 0 = MV - mw or, MV = m (v – V) [Using (i)]

mv . Mm 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. or,

Example 11.

V=

Solution : 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. m

m v

M

///////////////////////////////////////////////////////////////////

u

M

///////////////////////////////////////////////////////////

Initially, the system was at rest, thus according to momentum conservation, momentum after jump must be zero, as m (u – v) = M v

mu mM In a free space a rifle of mass M shoots a bullet of mass m at a stationary block of mass M distance D away from it. When the bullet has moved through a distance d towards the block the centre of mass of the bullet-block system is at a distance of : v=

Example 12.

Ans.

md  M D from the rifle Mm

(A)

(D  d) m from the block Mm

(B)

(C)

2 d m  DM from the rifle Mm

(D) (D  d)

M from the bullet Mm

(A,B,D) Bullet of mass(m) Block

Rifle

Solution.

COM x

d D-d-x

D

As;

Mx = m(D – d – x) x=

and

M

m(D  d) from the block Mm

x' = D – d – x =

RESONANCE

(D  d)M from the bullet. Mm AIEEE_CENTRE OF MASS - 12

Example 13.

The centre of mass of two masses m & mmoves by distance distance x and m is kept fixed. The ratio

Ans.

(A) 2 (B)

Solution

(m + m)

 Example 14.

x when mass m is moved by 5

m is m

(B) 4

(C) 1/4

(D) None of these

x = mx + mO 5

m + m = 5 m

;

m = 4 m

;

m =4 m

A uniform disc of mass ‘m’ and radius R is placed on a smooth horizontal floor such that the plane surface of the disc is in contact with the floor. A man of mass m/2 stands on the disc at its periphery. The man starts walking along the periphery of the disc. The size of the man is negligible as compared to the size of the disc. Then the centre of disc.

(A) moves along a circle of radius

R 3

(C) moves along a circle of radius

(C) moves along a circle of radius

R 2

(D) does not move along a circle

2R 3

Ans. Solution.

(A) The centre of mass of man + disc shall always remain at rest. Since the man is always at periphery of disc, the centre of disc shall always be at distance R/3 from centre of mass of two body system. Hence centre of disc moves in circle of radius R/3.

Example 15.

A person P of mass 50 kg stands at the middle of a boat of mass 100 kg moving at a constant velocity 10 m/s with no friction between water and boat and also the engine of the boat is shut off. With what velocity (relative to the boat surface) should the person move so that the boat comes to rest. Neglect friction between water and boat.

Ans. Solution.

(A) 30 m/s towards right (B) 20 m/s towards right (C) 30 m/s towards left (D) 20 m/s towards left (A) Momentum of the system remains conserved as no external force is acting on the system in horizontal direction.  (50 + 100) 10 = 50 × V + 100 × 0  V = 30 m/s towards right, as boat is at rest. VPboat = 30 m/s

Example 16.

Two men of masses 80 kg and 60 kg are standing on a wood plank of mass 100 kg, that has been placed over a smooth surface. If both the men start moving toward each other with speeds 1 m/s and 2 m/s respectively then find the velocity of the plank by which it starts moving.

Solution.

Applying momentum conservation ; (80) 1 + 60 (– 2) = (80 + 60 + 100) v v=

 40 1 =  m/sec. 240 6

RESONANCE


Example 17.

Each of the blocks shown in figure has mass 1 kg. The rear block moves with a speed of 2 m/s towards the front block kept at rest. The spring attached to the front block is light and has a spring constant 50 N/m. Find the maximum compression of the spring. Assume, on a friction less surface

1kg

k=50N/m

1kg

/////////////////////////////////////////////////////////////////

Solution : Maximum compression will take place when the blocks move with equal velocity. As no net external horizontal force acts on the system of the two blocks, the total linear momentum will remain constant. If V is the common speed at maximum compression, we have, (1 kg) (2 m/s) = (1 kg)V + (1 kg)V or, V = 1 m/s. Initial kinetic energy =

1 (1 kg) (2 m/s)2 = 2 J. 2

Final kinetic energy =

1 1 (1 kg) (1m/s)2 + (1 kg) (1 m/s)2 = 1 J 2 2

The kinetic energy lost is stored as the elastic energy in the spring.

Example 18.

Hence,

1 (50 N/m) x 2 = 2J – 1J = 1 J 2

or,

x = 0.2 m.

Figure shows two blocks of masses 5 kg and 2 kg placed on a frictionless surface and connected with a spring. An external kick gives a velocity 14 m/s to the heavier block towards the lighter one. Deduce (a) velocity gained by the centre of mass and (b) the separate velocities of the two blocks with respect to centre of mass just after the kick. 5kg

2kg

//////////////////////////////////////////////////////////////

Solution :

(a)

Velocity of centre of mass is v cm =

(b)

Example 19.

Ans.

5  14  2  0 = 10 m/s 52

Due to kick on 5 kg block, it starts moving with a velocity 14 m/s immediately, but due to inertia 2 kg block remains at rest, at that moment. Thus, velocity of 5 kg block with respect to the centre of mass is v 1 = 14 – 10 = 4 m/s and the velocity of 2 kg block w.r.t. to centre of mass is v 2 = 0 – 10 = –10 m/s

The two blocks A and B of same mass connected to a spring and placed on a smooth surface. They are given velocities (as shown in the figure) when the spring is in its natural length :

(A) (B) (C) (D) (A)

the maximum velocity of B will be 10 m/s the maximum velocity of B will be greater than 10 m/s the spring will have maximum extension when A and B both stop the spring will have maximum extension when both move towards left.

RESONANCE


Solution.

Suppose B moves with a velocity more than 10 m/s a should move at a velocity greater than 5 m/s and increases the overall energy which is not possible since there is no external force acting on the system. Hence B should move with a maximum velocity 10 m/s. Also both A and B can never stop so as to keep the momentum constant. Also both A and B can never move towards left simultaneously for momentum remaining conserved. Hence only (A) is correct.

 IMPULSE 

Impulse of a force F acting on a body for the time interval t = t1 to t = t2is defined as :  t2   dv   F dt m  dt = m dv  = Fdt =  = t1 dt



Also,







      = m( v 2 - v 1 ) = Ä P = change in momentum due to force F  t2   Res = FRes dt = Ä P (impulse - momentum theorem)



t1

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.

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 momenta. Thus, we can write.        F dt  P  Pf  Pi



Important Points : (1) It is a vector quantity. (2) Dimensions = [MLT –1] (3) S unit = kg m/s (4) Direction is along change in momentum. (5) Magnitude is equal to area under the F-t. graph.     (6)  = Fdt = Fav dt = Fav  t (7)

Example 20.

Solution :





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.

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. 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 (50 kg  m / s)  20 = = 250 N. 4s In order to hold the gun, the hero must exert a force of 250 N against the gun.

RESONANCE


 Impulsive force :

Note:

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 NonImpulsive force. Usually colliding forces are impulsive in nature. Since, the application time is very small, hence, very little motion of the particle takes place.

1. 2. 3.

Important points : Gravitational force and spring force are always non-Impulsive. Normal, tension and friction are case dependent. 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

eg.

Ni = Impulsive; Ng = Non-impulsive

N1 Both normals are Impulsive

N2 /////////////////////////////

N1 N1 ////////////////////////////////////////////////////////////////// N3 N2

N1, N3 = Impulsive; N2 = non-impulsive

Both normals are Impulsive

RESONANCE


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 : W hen 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. (a) 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 there. 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. (b)

/////////////////// Both ends of the string attached to movable objects : In this case equal and opposite impulses act on T the two objects, producing equal and opposite T is Impulsive A changes in momentum. The total momentum of the system therefore remains constant, although the B momentum of each individual object is changed in T is non-impulsive the direction of the string. Perpendicular to the string C however, no impulse acts and the momentum of each T is non-impulsive particle in this direction is unchanged. All normal are impulsive but tension T is impulsive only for the ball A

For this example: In case of rod, Tension is always impulsive and in case of spring, Tension is always non-impulsive.

Example 21. 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 : m A

C

m u

m B

/////////////////////////////////////////////////////

(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.

RESONANCE


Solution :

(a)

By Conservation of linear momentum v =

(b)

 T dt 

(c)

 N dt  m 3  u 

(d)

 (N  T) dt =  Ndt   T dt 

mu 3

u

 N dt 

u 3



=

2mu 3 mu 3

2mu 3

 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 : (a)

In a collision, particles may or may not come in physical contact.

(b)

The duration of collision, t is negligible as compared to the usual time intervals of observation of motion.

(c)

In a collision the effect of external non impulsive forces such as gravity are not taken into a account as due to small duration of collision (t) average impulsive force responsible for collision is much larger than external forces acting on the system.

The collision is infact a redistribution of total momentum of the particles. Thus, law of conservation of linear momentum is indispensable in dealing with the phenomenon of collision between particles.

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.

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 : 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. Thus, kinetic energy of system after collision is equal to kinetic energy of system before collision. Thus in addition to the linear momentum, kinetic energy also remains conserved before and after collision.

RESONANCE


(ii) Inelastic collision : In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. 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 after collision is not equal to that of before collision. However, in the absence of external forces, law of conservation of linear momentum still holds good. (iii) Perfectly inelastic : If velocity of separation along the line of impact just after collision becomes zero then the collision is perfectly inelastic. Collision is said to be perfectly inelastic if both the particles stick together after collision and move with same velocity, Note : Actually collision between all real objects are neither perfectly elastic nor perfectly inelastic, its inelastic in nature.

Examples of line of impact and collisions based on line of impact (i)

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

Head on Collision (ii)

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

Oblique Collision (iii)

Ball is falling on a stationary wedge.

Oblique Collision

COEFFICIENT OF RESTITUTION (e) The coefficient of restitution is defined as the ratio of the impulses of reformation and deformation of either body.

RESONANCE


Im pulse of reformatio n e = Im pulse of deformatio n =

 F dt  F dt r

d

Velocity of seperation along line of impact Velocity of approach along line of impact

=

The most general expression for coefficient of restitution is e=

velocity of separation of points of contact along line of impact velocity of approach of point of contact along line of impact

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. Just Before collision u1

Just After collision v1 v2

u2 Line of impact

C

D

B

A

Line of impact C

D

A

B

u1 > u2

v1 < v2 u2

v

u1 ND

m1

ND m2

NR m1

v

v2 v1

NR m2

m1

m2

 Deformation

Reformation

 

Fext = 0 momentum is conserved for the system. m 1u1 + m 2 u2 = (m 1 + m 2)v = m 1v1 + m 2v2



v=

m1u1  m 2u 2 m1v 1  m 2 v 2 = m1  m 2 m1  m 2

.......(1)

Impulse of Deformation : J D = change in momentum of any one body during deformation. = m 2 (v – u2) for m 2 = m 1 (–v + u1) for m 1 Impulse of Reformation : J R = change in momentum of any one body during Reformation. = m 2 (v2 – v) for m 2 = m 1 (v – v1) for m 1  v 2  v1 Velocity of separation along line of impact Impulse of Reformation ( JR )  e= = u u = Velocity of approach along line of impact 1 2 Impulse of Deformation ( J ) D

RESONANCE


Note : e is independent of shape and mass of object but depends on the material. The coefficient of restitution is constant for a pair of materials. (a) e=1  Impulse of Reformation = Impulse of Deformation  Velocity of separation along the LOI = Velocity of approach along the LOI

(b)

(c)

         

e=0

0
Kinetic energy of particles after collision may be equal to that of before collision.

Collision is elastic. Impulse of Reformation = 0 Velocity of separation along the LOI= 0 Kinetic energy of particles after collision is not equal to that of before collision.

Collision is perfectly inelastic . Impulse of Reformation < Impulse of Deformation Velocity of separation along the LOI < Velocity of approach along the LOI Kinetic energy of particles after collision is not equal to that of before collision.

Collision is Inelastic.

Note : In case of contact collisions e is always less than unity.  0e1 Important Point : In case of elastic collision, if rough surface is present then k f < k i (because friction is impulsive) Where, k is Kinetic Energy.

Rough /////////////////////////////////////

A particle ‘B’ moving along the dotted line collides with a rod also in state of motion as shown in the figure. The particle B comes in contact with point C on the rod. To write down the expression for coefficient of restitution e, we first draw the line of impact. Then we resolve the components of velocities of points of contact of both the bodies along line of impact just before and just after collision.

Then

v 2 x  v 1x e = u u 1x 2x

Collision in one dimension (Head on) m1

m1

m2 u2

u1 (a) Before Collision

u1 > u2

v 2  v1 e = u u 1 2



RESONANCE

m2 v2

v1 (b) After Collision

v2 > v1

(u1 – u2)e = (v2 – v1)


By momentum conservation, m 1u 1 + m 2u 2 = m 1v 1 + m 2v 2 v2 = v1 + e(u1 – u2) and

v1 =

m1u1  m 2u 2  m 2 e(u1  u 2 ) m1  m 2

v2 =

m1u1  m 2u 2  m1e(u1  u 2 ) m1  m 2

Special Case : (1) e=0  v1 = v2  for perfectly inelastic collision, both the bodies, move with same vel. after collision. (2)

e=1 and m 1 = m 2 = m, we get v1 = u2 and v2 = u1 i.e., when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities., e.g.

v1=0 m

2m/s

m

Before Collision

(3)

m 1 >> m 2 m 1 + m 2  m 1 and

m2 0 m1

 v1 = u1 No change and v2 = u1 + e(u1 – u2) Now If e = 1 v2 = 2u1 – u2

Example 22.

Two identical balls are approaching towards each other on a straight line with velocity 2 m/s and 4 m/s respectively. Find the final velocities, after elastic collision between them. m

2m/s

4m/s

m

Solution : The two velocities will be exchanged and the final motion is reverse of initial motion for both. 4m/s

RESONANCE

m

m

2m/s


Example 23.

Three balls A, B and C of same mass ‘m’ are placed on a frictionless horizontal plane in a straight line as shown. Ball A is moved with velocity u towards the middle ball B. If all the collisions are elastic then, find the final velocities of all the balls. m m m u A B C //////////////////////////////////////////

Solution : A collides elastically with B and comes to rest but B starts moving with velocity u m m m u A B C //////////////////////////////////////////

After a while B collides elastically with C and comes to rest but C starts moving with velocity u m m m u A B C //////////////////////////////////////////

 Final velocities VA = 0; VB = 0 and VC = u Example 24.

Ans.

Four identical balls A, B, C and D are placed in a line on a frictionless horizontal surface. A and D are moved with same speed ‘u’ towards the middle as shown. Assuming elastic collisions, find the final velocities. u u A B C D /////////////////////////////////////////////////////

Solution : A and D collides elastically with B and C respectively and come to rest but B and C starts moving with velocity u towards each other as shown u u A B C D /////////////////////////////////////////////////////

B and C collides elastically and exchange their velocities to move in opposite directions u u A B C D /////////////////////////////////////////////////////

Now, B and C collides elastically with A and D respectively and come to rest but A and D starts moving with velocity u away from each other as shown u A B C D ///////////////////////////////////////////////////// u

 Final velocities VA = u (  ); VB = 0; VC = 0 and VD = u (  ) Example 25.

Ans.

Two particles of mass m and 2m moving in opposite directions on a frictionless surface collide elastically with velocity v and 2v respectively. Find their velocities after collision, also find the fraction of kinetic energy lost by the colliding particles.

m

2v

v

2m

Solution : Let the final velocities of m and 2m be v 1 and v 2 respectively as shown in the figure:

m By conservation of momentum: m(2v) + 2m(–v) = m(v 1) + 2m (v 2) or 0 = mv 1 + 2mv 2 or v 1 + 2v 2 = 0

RESONANCE

2m

v2

.........(1) AIEEE_CENTRE OF MASS - 23

and since the collision is elastic: v 2 – v 1 = 2v –(–v) or v 2 – v 1 = 3v Solving the above two equations, we get, v 2 = v and v 1 = –2v Ans.

.........(2)

i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2v in the direction shown in figure: 2m

m

v

The collision was elastic therefore, no kinetic energy is lost, KE loss = KE i - KE f or,

Example 26.

1 1 1  1   m( 2v ) 2  ( 2m)(  v ) 2    m( 2v ) 2  ( 2m)v 2  = 0 2 2 2 2    

On a frictionless surface, a ball of mass m moving at a speed v makes a head on collision with an identical ball at rest. The kinetic energy of the balls after the collision is 3/4th of the original. Find the coefficient of restitution.

Solution : As we have seen in the above discussion, that under the given conditions :

By using conservation of linear momentum and equation of e, we get,  1 e  v 1'    v and  2 

Kf 

Given that

 1 e  v2'  v  2 

3 Ki 4

or

1 1 3  1 mv 2    mv 1’ 2 + mv 2’ 2 =  2 2 4 2

or

e =

Substituting the value, we get

 1 e     2  Example 27.

2

 1 e   +   2 

2

=

3 4

1 2

Ans.

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 surfaces, find the final velocities of the blocks. 2m/s 2kg

10m/s

very heavy object

/////////////////////////////////////////////////////////////////////

Solution : Let v1 and v2 be the final velocities of 2kg block and heavy object respectively then, v1 = u1 + 1 (u1 – u2) = 2u1 – u2 = –14 m/s v2 = –2m/s

RESONANCE


Example 28.

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

Solution : 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. Example 29.

Two balls of masses 2 kg and 4 kg are moved towards each other with velocities 4 m/s and 2 m/s respectively on a frictionless surface. After colliding the 2 kg ball returns back with velocity 2m/s. Just before collision

2kg

4m/s 2m/s

Just after collision 2m/s

4kg

////////////////////////////////////////////

2kg

4kg

v2

////////////////////////////////////////

Then find: (a) velocity of 4 kg ball after collision (b) coefficient of restitution e. (c) Impulse of deformation JD. (d) Maximum potential energy of deformation. (e) Impulse of reformation JR. Solution : (a)

(b) (c)

(d)

By momentum conservation, 2(4) – 4(2) = 2(–2) + 4(v2)



v2 = 1 m/s

velocity of separation 1  ( 2) 3  = 0.5 e = velocity of approach = 4  ( 2 ) 6 At maximum deformed state, by conservation of momentum, common velocity is v = 0. JD = m 1(v – u1) = m2(v – u2) = 2(0 – 4) = –8 N -s = 4(0 – 2) = – 8 N - s or = 4(0 – 2) = – 8 N - s Potential energy at maximum deformed state U = loss in kinetic energy during deformation. or

1 1 1 2 2 U =  m1u1  m 2u 2  – (m1 + m 2)v2 2 2  2 1 1 1 2 2 =  2( 4)  4( 2)  – (2 + 4) (0)2 2 2  2

or

RESONANCE

U = 24 Joule AIEEE_CENTRE OF MASS - 25

(e)

JR = m 1(v1 – v) = m2 (v – v2) = 2 (–2 – 0) = –4 N-s = 4(0 – 1) = –4 N-s

or or

JR e = J D

 JR = eJD

= (0.5) (–8)

= –4 N-s

 C ollisio n in tw o dimension (o blique) 1.

A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual particles do 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)

Example 30.

A ball of mass m hits a floor with a speed v0 making an angle of incidence a with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball.

Solution : The component of velocity v 0 along common tangential direction v 0 sin  will remain unchanged. Let v be the component along common normal direction after collision. Applying, Relative speed of separation = e (Relative speed of approach) along common normal direction, we get v = ev 0 cos 

v (= ev0cos) v'  v0sin

Thus, after collision components of velocity v’ are v 0 sin  and ev 0 cos 

Note :



v '  ( v 0 sin  )2  (ev 0 cos  )2

and

v 0 sin  tan  = ev cos  0

or

tan  =

tan  e

Ans.

For elastic collision, e = 1  v’ = v 0 and

RESONANCE

Ans.

 =  AIEEE_CENTRE OF MASS - 26

Example 31.

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.

Solution : 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 

becomes zero after collision, while that of 2 becomes v cos . W hile 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 below. Component along common tangent direction

Ball

1 2

Before collision v sin  

After collision v sin  

Component along common normal direction Before collision v cos  

After collision  v cos 

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

 VARIABLE MASS SYSTEM :  If a mass is added or ejected from a system, at rate  kg/s and relative velocity v rel (w.r.t. the system),  then the force exerted by this mass on the system has magnitude  v rel .  Thrust Force ( Ft )

   dm  Ft  v rel    dt   Suppose at some moment t = t mass of a body is m and its velocity is v . After some time at t =   t + dt its mass becomes (m – dm) and velocity becomes v  dv . The mass dm is ejected with    relative velocity v r . Absolute velocity of mass ‘dm’ is therefore ( v + v r ). If no external forces are

acting on the system, the linear momentum of the system will remain conserved, or   Pi  Pf      or m v = (m – dm) ( v + d v ) + dm ( v + v r )        or m v = m v + md v – (dm) v – (dm) (d v ) + (dm) v + v r dm

RESONANCE


 The term (dm) (d v ) is too small and can be neglected.    dm   dv    m   = vr     md v = – v r dm or dt  dt   

1. 2.

Here,

  dv   m  = thrust force F t dt  

and

–

 

dm = rate at which mass is ejecting dt

or

   dm  Ft  v r    dt 

Problems related to variable mass can be solved in following four steps Make a list of all the forces acting on the main mass and apply them on it.

  dm   Apply an additional thrust force Ft on the mass, the magnitude of which is v r   dt  and direc   tion is given by the direction of v r in case the mass is increasing and otherwise the direction of  – v r if it is decreasing.

3.

Find net force on the mass and apply   dv Fnet  m (m = mass at the particular instant) dt

4.

Integrate it with proper limits to find velocity at any time t.

 Ro cket pro pu lsion : Let m 0 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.

  dm   be the mass of the gas ejected per unit time and v r the exhaust velocity of the Further, let   dt    dm   and v r are kept constant throughout the journey of gases with respect to rocket. Usually   dt 

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

  dm   Ft = vr   dt 

(upwards)

2. 3.

W eight of the rocket Net force on the rocket

W = mg F net = F t – W

(downwards) (upwards)

RESONANCE


or

4.

  dm   –mg F net = v r   dt 

Net acceleration of the rocket

or

vr dv = m dt

or

dv =



u

F m

  dm    –g  dt 

vr m

 dm

dv  v r



m

v

or

a=

m0

– g dt

 dm –g m

 m0   Thus, v = u – gt + v r n   m 

t



dt 0

...(i)

dm  dm   is upwards, as v r is downwards and Note : 1. F t = v r   is negative. dt  dt 

 m0  . 2. If gravity is ignored and initial velocity of the rocket u = 0, Eq. (i) reduces to v = v r ln   m 

Example 32.

A rocket, with an initial mass of 1000 kg, is launched vertically upwards from rest under gravity. The rocket burns fuel at the rate of 10 kg per second. The burnt matter is ejected vertically downwards with a speed of 2000 ms –1 relative to the rocket. If burning stopsafter one minute. Find the maximum velocity of the rocket. (Take g as at 10 ms –2)

Solution : Using the velocity equation m  v = u – gt + v r ln  0   m  Here u = 0, t = 60s, g = 10 m/s 2, v r = 2000 m/s, m 0 = 1000 kg and m = 1000 – 10 × 60 = 400 kg  1000   v = 0 – 600 + 2000 ln   400  or v = 2000 ln 2.5 – 600 The maximum velocity of the rocket is 200(10 ln 2.5 – 3) = 1232.6 ms –1

W e get

Example 33. A flat car of mass m 0 starts moving to the right due to a constant horizontal force F. Sand spills on the flat car from a stationary hopper. The rate of loading is constant and equal to  kg/s. Find the time dependence of the velocity and the acceleration of the flat car in the process of loading. The friction is negligibly small. Solution :

Ans.

 m0

F

Initial velocity of the flat car is zero. Let v be its velocity at time t and m its mass at that instant. Then

RESONANCE


At t = 0, v = 0 and m = m 0 at t = t, v = v and m = m 0 + t Here, v r = v (backwards) dm =  dt

Ft = vr



dm = v dt

(backwards)

Net force on the flat car at time t is F net = F – F t dv = F – v dt

or

m

or

(m 0 + t)



–



 m  t   F    = n  0 n    F  v   m0 



F m0  t = F  v m0

Solution :

 or

or



0

dv = F  v

t



0

dt m 0  t

1 1 [n (F – v)] 0v = [n (m 0 + t)] 0t  

=

or

v=

Ft m0  t

Ans.

dv = acceleration of flat car at time t dt

F  v m

Ft  F  m  0  t a =  m 0  t  

Example 34.

v

dv =F – v dt

From Eq. (i),

or

....(i)

     

or

a=

Fm 0 (m 0  t ) 2

Ans.

A cart loaded with sand moves along a horizontal floor due to a constant force F coinciding in direction with the cart’s velocity vector. In the process sand spills through a hole in the bottom with a constant rate kg/s. Find the acceleration and velocity of the cart at the moment t, if at the initial moment t = 0 the cart with loaded sand had the mass m 0 and its velocity was equal to zero. Friction is to be neglected. In this problem the sand spills through a hole in the bottom of the cart. Hence, the relative velocity of the sand v r will be zero because it will acquire the same velocity as that of the cart at the moment. vr = 0 v m F dm    as Ft  v r  Thus, F t = 0 dt   v and the net force will be F only. F net = F  dv   = F m  dt 

But here

RESONANCE

....(i)

m = m 0 – t



(m 0 – t)

dv =F dt AIEEE_CENTRE OF MASS - 30

t

v

dv =

or





v=

0



F dt  t

0 m0

F  n ( m0   t )  0t 

or

v=

F  m 0  ln   m 0  t 

Ans.

From eq. (i), acceleration of the cart a =

dv F = dt m

or

F a = m  t 0

Ans.

 LINEAR MOMENTUM CONSERVATION IN PRESENCE OF EXTERNAL FORCE. 

 dP Fext = dt

Note:



  Fext dt = dP



  dP = Fext )mpulsive dt



 If Fext )mpulsive = 0



 dP = 0

or

 P is constant

Momentum is conserved if the external force present is non-impulsive. eg. gravitation or spring force

Example 35.

Two balls are moving towards each other on a vertical line collides with each other as shown. Find their velocities just after collision.

2kg 3m/s 4m/s 4kg

4m/s

Solution :

Let the final velocity of 4 kg ball just after collision be v. Since, external force is gravitational which is non - impulsive, hence, linear momentum will be conserved. Applying linear momentum conservation: 2(–3) + 4(4) = 2(4) + 4(v) 1 or v = m/s 2

RESONANCE

2kg

v 4kg AIEEE_CENTRE OF MASS - 31

/////////////////////

Example 36. A bullet of mass 50g is fired from below into the bob of mass 450g of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of 1.8 m. Find the speed of the bullet. Take g = 10 m/s 2.

v

Solution :

Let the speed of the bullet be v. Let the common velocity of the bullet and the bob, after the bullet is embedded into the bob, is V. By the principle of conservation of the linear momentum, (0.05 kg) v v V = 0.45 kg  0.05 kg = 10 The string becomes loose and the bob will go up with a deceleration of g = 10 m/s2. As it comes to rest at a height of 1.8 m, using the equation v2 = u2 + 2ax, 1.8 m =

Problem 1.

( v / 10) 2 2  10 m / s 2

or,

v = 60 m/s.

Three particles of masses 0.5 kg, 1.0 kg and 1.5 kg are placed at the three corners of a right angled triangle of sides 3.0 cm, 4.0 cm and 5.0 cm as shown in figure. Locate the centre of mass of the system.

Solution :

1.5 kg (C) y

taking x and y axes as shown. coordinates of body A = (0,0) coordinates of body B = (4,0) coordinates of body C = (0,3)

5cm 3cm

m A x A  mB xB  MCrC x - coordinate of c.m. = m A  mB  mC =

0.5  0  1.0  4  1.5  0 0.5  1.0  1.5

=

4 cm = cm = 1.33cm 3 kg

similarly y - wordinates of c.m. =

0.5 kg (A)

4cm

x 1.0 kg (B)

0.5  0  1.0  0  1.5  3 4.5 = = 1.5 cm 0.5  1.0  1.5 3

So, certre of mass is 1.33 cm right and 1.5 cm above particle A. Problem 2.

A block A (mass = 4M) is placed on the top of a wedge B of base length l (mass = 20 M) as shown in figure. When the system is released from rest. Find the distance moved by the wedge B till the block A reaches at lowest end of wedge. As sume all surfaces are frictionless.

RESONANCE


Solution :

Initial position of centre of mass

X BMB  X A MA 5 XB   XB .20 M  .4M = = 6 MB  MB 24M

=

Final position of centre of mass 5( XB  x )  x ( XB  x )20M  4Mx = 6 24M

=

since there is no horizontal force on system centre of mass initially = centre of mass finally. 5XB +  = 5XB + 5x + x  = 6x

x

 6

Problem 3.

An isolated particle of mass m is moving in a horizontal xy plane, along x-axis. At a certain height above ground, it suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. Find the position of heavier fragment at this instant.

Solution :

As particle is moving along x-axis, so, y-coordinate of COM is zero. YM M = YM 4

M   + Y3M 4 4

 3M     4 

M 0 × M = 15   + Y3M 4 4



 3M     4 

Y3M  5cm 4 Problem 4.

A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg and m kg. The fragments of masses 1 kg and 2 kg fly off with speeds 12 m/s along x-axis and 8 m/s along yaxis respectively. If m kg flies off with speed 40 m/s then find the total mass of the shell.

Solution :

As initial velocity = 0, Initial momentum = (1 + 2 + m) × 0 = 0 



Finally, let velocity of M = V . We know | V | = 40 m/s. Initial momentum = final momentum .  0 = 1 × 12 î + 2 × 8 ˆj + m V



|V | =

(12)2  (16 )2 1 = m m2





V =–

(12î  16ˆj) m

(12)2  (16) 2 = 40 {given}

(12)2  (16 )2 = 0.5 kg 40 Total mass = 1 + 2 + 0.5 = 3.5 kg m=

Problem 5.

A block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equal parts continuing in the same direction. If one of the parts moves at 30 m/s, with what speed does the second part move and what is the fractional change in the kinetic energy of the system.

Solution :

RESONANCE


Applying momentum conservation ;

m m V+ × 30 2 2 So, V = 10 m/s m × 20 =

V + 15 2

initial kinetic energy =

1 m × (20)2 = 200 m 2

final kinetic energy =

1 m 1 m . . (10)2 + × (30)2 = 25 m + 225 m = 250 m 2 2 2 2

fractional change in kinetic energy = Problem 6.

20 =



1 ( final K. E)  (initial K. E ) 250m  200m = = initial K.E 200m 4

A block at rest explodes into three equal parts. Two parts starts moving along X and Y axes respectively with equal speeds of 10 m/s. Find the initial velocity of the third part.

Solution : Let total mass = 3 m,

initial linear momentum = 3m × 0 

Let velocity of third part = V Using conservation of linear momentum :  m × 10 î + m × 10 ˆj + m V = 0  So, V = (– 10 î – 10 ˆj ) m/sec. 

|V | = Problem 7.

(10)2  (10)2 = 10 2 , making angle 135o below x-axis

Blocks A and B have masses 40 kg and 60 kg respectively. They are placed on a smooth surface and the spring connected between them is stretched by 1.5m. If they are released from rest, determine the speeds of both blocks at the instant the spring becomes unstretched.

Solution :

Let, both block start moving with velocity V1 and V2 as shown in figure Since no horizontal force on system so, applying momentum conservation 0 = 40 V1 – 60 V2

2 V1  3 V2

........(1)

Now applying energy conservation, Loss in potential energy = gain in kinetic energy

1 2 1 1 kx = m1V12 + mV2 2 2 2 2 2 1 1 1 × 600 × (1.5)2 = × 40 × V12 + × 60 × V22 2 2 2 Solving euation (1) and (2) we get, V1 = 4.5 m/s, V2 = 3 m/s. Problem 8.

.......(2)

Find the mass of the rocket as a function of time, if it moves with a constant accleration a, in absence of external forces. The gas escaps with a constant velocity u relative to the rocket and its initial mass was m 0.

RESONANCE


Solution :

 – dm   Using, Fnet = Vrel   dt  dm dt Fnet = ma Solving equation (1) and (2)

Fnet = – u

ma = – u m



mo

dm m

.......(1) ......(2)

dm dt t

=

– adt u o



m  at / u mo = e

Problem 9.

n

m at = mo u

m  m0 e

 atu

Ans.

A ball is approaching to ground with speed u. If the coefficient of restitution is e then find out:

m

(a) the velocity just after collision.

u

(b) the impulse exerted by the normal due to ground on the ball.

/////////////////////////

Solution :

e=

velocity of separation v = u velocity of approah

(a) velocity after collision = V = eu ........(1) (b) Impulse exerted by the normal due to ground on the ball = change in momentum of ball. = {final momentum} – {initial momentum} = {m v} – {– mu} = mv + mu = m {u + eu} = mu {1 + e} Ans.

RESONANCE


PART - I : OBJECTIVE QUESTIONS Hkkx - I : oLrqfu"B iz'u ¼OBJECTIVE QUESTIONS½ SECTION (A) : CALCULATION OF CENTRE OF MASS nz O;eku d sUnz d h x.kuk A-1. The centre of mass of a body : oLrqd k nzO;eku d sUnz: (1) Lies always at the geometrical centre (ges 'kk T;kferh d sUnzij fLFkr gksrk gSaA) (2) Lies always inside the body (ges 'kk oLrqd sv Unj gksrk gS) (3) Lies always outside the body (ges 'kk oLrqd sckgj gksrk gS) (4*) Lies within or outside the body (oLrqd sv Unj ;k ckgj gks rk gSA) Sol.

Centre of mass is a point which can lie within or outside the body.

nzO;eku d sUnzog fcUnqgSt ksfd oLrqd sv anj ;k ckgj gksl d rk gSA A-2.

A uniform solid cone of height 40 cm is shown in figure. The distance of centre of mass of the cone from point B (centre of the base) is : ,d l eku Bksl 'kad qft l d h Å ¡p kbZ40 l seh- gS] fp=kkuql kj fn[kk;k x;k gSA fcUnqB l snzO;eku d sUnzd h nwjh

¼v k/kkj d sd sUnzl s½ gksxh &

(1) 20 cm Sol.

Centre of mass are rcm =

nzO;eku d sUnzrcm = A-3.

Sol.

(2) 10/3 cm

(3) 20/3 cm

(4*) 10 cm

h 40 = = 10 cm 4 4

h 40 = = 10 cm 4 4

A body has its centre of mass at the origin. The x-coordinates of the particles (1) may be all positive (2) may be all negative (3) must be all non-negative (4*) may be positive for some particles and negative in other particles ,d oLrqd k nzO;eku d sUnzewy &fcUnqij gSA d .kksad sx–funsZ'kkad & (1) lkjs/kukRed gksld rsgS A (2) lkjs_ .kkRed gksld rsgS A (3) lkjsv_ .kkRed gks usghpkfg,A (4*) d q N d .kksad sfy ;s/kukRed o d qN d .kksad sfy ;s_ self explaintory

RESONANCE

.kkRed gksld rsgSA

NIT_CENTRE OF MASS - 36

SECTION (B) : MOTION OF CENTRE OF MASS B-1.

nzO;eku d sUnz d h xfr

A bomb travelling in a parabolic path under the effect of gravity, explodes in mid air. The centre of mass of fragments will:

,d ce xq: Roh; izHkko esaijoy ; iFk ij xfr d jrk gS]chp gok esafoLQ ksfVr gkst krk gSA Vqd M+ksd k nzO;eku d sUnz (1) Move vertically upwards and then downwards

igy sl h/ksÅ ij d h v ksj fQ j uhpsd h xfr d jsxk (2) Move vertically downwards

l h/ksuhpsd h v kSj xfr d jsxkA (3) Move in irregular path

v l eku iFk ij xfr d jsxkA (4*) Move in the parabolic path which the unexploded bomb would have travelled.

ml ijoy ; iFk ij xfr d jsxk ft l ij ce Q Vusl sigy sd j jgk gksrk gSA B-2.

If a ball is thrown upwards from the surface of earth and during upward motion :

v xj i`Foh d h l rg l sxsan Å ij d h v ksj Q sad h t krh gS] rksÅ ij d h v ksj xfr d snkSjku – (1) The earth remains stationary while the ball moves upwards

i`Foh : d h gqbZjgrh gS] t cfd xsan Å ij d h v ksj xfr d jrh gSA (2) The ball remains stationary while the earth moves downwards

xasn : d h jgrh gSt cfd i`Foh uhpsd h v ksj xfr d jrh gS (3) The ball and earth both moves towards each other

xsan rFkk i`Foh nksuksa,d nwl jsd h v kSj xfr d jrsgSA (4*) The ball and earth both move away from each other

xsan rFkk i`Foh nksuksa,d nwl jsl snwj d h v ksj xfr d jrsgSA Sol.

In abscence of external force both move away from each other to keep the centre of mass at rest.

cká cy d h v uqifLFkfr esanksuksa,d nwl jsl snwj d h v ksj xfr d jsxsaft l l snzO;eku d sUnzfojke ij jgsA B-3.

Internal forces can change :

vkUrfjd cy ifjofrZr d j ld rk gS&

Sol.

(1) the linear momentum but not the kinetic energy of the system. (2*) the kinetic energy but not the linear momentum of the system. (3) linear momentum as well as kinetic energy of the system. (4) neither the linear momentum nor the kinetic energy of the system. (1) fud k; d k jS f[kd laosx fd Urqxfrt Å t kZugha (2*) fud k; d h xfrt Å t kZfd UrqjS f[kd laosx ugha (3) fud k; d k jS f[kd laosx o lkFk ghxfrt Å t kZHkh (4) fud k; d k u rksjS f[kd laosx]u gh xfrt Å t kZ Internal forces canot change velocity but can do work.

v karfjd cy osx d ksifjofrZr ugh d jrsgSay sfd u d k;Zd j l d rsgSA B-4.

If the external forces acting on a system have zero resultant, the centre of mass (1) must not move (2) must accelerate (3*) may move (4) may accelerate

;fn fd lh fud k; ij y x jgsckº; cy ksad k ifj.kkeh 'kwU; gks]nzO;eku d sUnz& (1) xfr ughad js xkA (2) Rofjr gks xkA (3*) xfr d j ld rk gS A Sol.

(4) Rofjr gksld rkgS A If initial velocity of system is not zero then centre of mass moves with constant velocity.

;fn fud k; d k izkjfEHkd osx 'kwU; ugh gSrksnzO;eku d sUnzfu;r osx l sxfr d j l d rk gSA B-5.

Two balls are thrown in air. The acceleration of the centre of mass of the two balls while in air (neglect air resistance) (1) depends on the direction of the motion of the balls (2) depends on the masses of the two balls (3) depends on the speeds of the two balls (4*) is equal to g

nksxsans,d lkFk ok;qesaQ sad h x;h gaSA nksuksaxsanksad k nzO;eku d sUnzd k Roj.k t c osok;qesagS& (1) xs anksd h xfr d h fn'kk ij fuHkZj d jrk gSA (2) nks uksaxsanksd snzO;ekuksaij fuHkZj d jrk gSA (3) nks uksaxsanksad h pky ksaij fuHkZj d jrk gSA (4) g d scjkcj gS A Sol.

acm =

m1g  m 2 g m1  m 2 = g

RESONANCE


B-6.

Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speed of 2m/s and 6m/s respectively on a smooth horizontal surface. The speed of centre of mass of the system is : nksnzO;eku 1 kg rFkk 0.5 kg ,d gh fn'kk esa2 m/sec rFkk 6 m/sec l sfpd usl rg ij xfr d j jgsgSA l ewg d s

nzO;eku d sUnzd k osx gksxk& (1*)

10 m/s 3

(2)

10 m/s 7

(3)

11 m/s 2

(4)

12 m/s 3

1 6 10 2  m / sec . 1  1/ 2 3

1 2  Sol.

vcm =

B-7.

Two particles having mass ratio n : 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is : nksd .k ft ud k nzO;eku d k v uqikr n : 1 gSos,d gYd h v forkU; jLl h l st qM +sgq, gSat ksfpd uh f?kjuh l sxqt jrh

gSA v xj fud k; d ksN ksM +fn;k t krk gSrksl ewg d snzO;eku d sUnzd k Roj.k gksxkA 2

(1) (n – Sol.

a=

1)2

2

 n  1  g (2)   n  1

g

 n  1  g (3*)   n  1

 n  1 g (4)   n  1

(nm – m) g nm  m

(n – 1) g (n  1) a1 = a2 = a

=

acm =

acm =

nma1 – ma 2 (n – 1) a = (nm  m) (n  1) (n – 1) 2 (n  1) 2

g.

SECTION (C) : CONSERVATION OF LINEAR MOMENTUM C-1.

Sol.

js[kh; l aosx l aj{k.k

Two particles A and B initially at rest move towards each other under a mutual force of attraction. The speed of centre of mass at the instant when the speed of A is v and the speed of B is 2v is : [JEE - 89] izkjEHk esafLFkj A o B fi.M ijLij v kd "kZ.k cy l s,d nwl jsd h v ksj xfr d jrsgSA t c A d h pky v o B d h pky 2 v gSrksnz O;eku d sUnzd h pky gksxh \ [JEE - 89] (1) v (2*) Zero 'kw U; (3) 2 v (4) 3 v /2 Net external force is zero so net momentum will remain zero.

d qy cká cy 'kwU; gSA bl fy , d qy l aosx 'kwU; gksxkA C-2.

If the KE of a body becomes four times its initial value, then the new momentum will be more than its initial momentum by;

vxj fd l hoLrqd hxfrt Å t kZizkjfEHkd l spkj xquhgkst krhgSrksu;k l aosx izkjfEHkd l aosx l sfd rukT;knk gksxkA (1) 50% Sol.

k2 4  k1

Then

(2*) 100% 2

 v2     4  v1 



(3) 125%

(4) 150%

v2 2 v1

 p 2 – p1   mv 2 – mv 1  v    100 =  2 – 1  100 = 100% rc  p   100 =  mv 1 1      v1 

RESONANCE


C-3.

A man of mass 'm' climbs on a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed vrel (relative to rope). In what direction and with what speed (relative to ground) will the balloon move? Mv rel mv rel (1*) downwards, (2) upwards, m M mM mv rel (M  m)v rel (3) downwards, (4) downwards, M M ,d m nzO;eku d k v kneh L y EckbZd h jLl h ij p<+jgk gS] jLl h M nzO;eku d sxqCckjsl sy Vd h gqbZgSaA t ehu d s l kis{k xqCckjk fLFkj gSA vxj vknehjLl h ij mij d hvksj vrel (jLl h d sl kis{k) osx l sp<+uk'kq: d j nsrk gSrksfd l fn'kk esa(t ehu d sl kis{k) rFkk fd l osx l sxqCckjk xfr d jsxkA

mv rel mM mv rel (3) uhps ] M we have ge t kursgSVm + Vb = Vrel  by conservation of linear momentum (1*)

Sol.

uhps]

Mv rel mM (M  m)v rel (4) uhps M Vm = Vrel – Vb

(2)

Å ij,

js[kh; l aosx l j{k.k l s mVm – MVb = 0 So v r% m (Vrel – Vb) – MVb = 0 

Vb =

mv rel mM

SECTION (D) : SPRING - MASS SYSTEM D-1.

fLçax nzO;eku fud k;

In the figure shown the change in magnitude of momentum of the block when it comes to its initial position if the maximum compression of the spring is x 0 will be :

fp=kkuql kj Cy kWd d sl osax esaifjorZu gksxk t c Cy kWd nqckjk izkjfEHkd v oLFkk esav krk gS;fn fLizax d k v f/kd re l EihM+u x0 gSA

(1*) 2 k m x 0 D-2.

(2)

k m x0

(3) zero

(4) none of these bues al sd ksbZugh

Two masses are connected by a spring as shown in the figure. One of the masses was given velocity v = 2 k, as shown in figure where 'k' is the spring constant. Then maximum extension in the spring will be nksnz O;eku fp=kkuql kj fLizax lsca/ksgq;sgS aA ,d nz O;eku d ks2k os x ns rsgSt S l kfp=kesan'kkZ;kx;kgSt gkak fLiza x fu;rkad

gSA rksfLizax esavf/kd re f[kap ko gksxk

(1) 2 m

RESONANCE

(2) m

(3*)

2m k

(4)

3m k


2

Sol.

v 1 1 1 by energy conservation mv2 = (2m)   + kx 2 2 2 2 2 2

Å t kZl ja{k.k l s x=



v 1 1 1 mv2 = (2m)   + kx 2 2 2 2 2

2mK

SECTION (E) : IMPULSE

vkos x

E-1.

A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for t = 0.01 sec., the impulse of the impact force is : ,d 50 gm d h xsan 10 ehVj Å ¡p kbZl sN ksM +h t krh gSA ;g okfil 75 izfr'kr xfrt Å t kZd h gkfu d sckn mN y rh gSA v xj ;g /kjrh d sl kFk t = 0.01 l sd .M , d sfy , Li'kZd jrh gSrksVDd j d scy d k v kosx gksxkA (1) 1.3 N–s (2*) 1.05 N-s (3) 1300 N–s (4) 105 N–s

Sol.

v1 = k2 =

2gh =

2  10  10 = 10 2

1 1 k  v22 = v12 4 1 4

v1 = 5 2 2 |P| = |–mv2 – (mv1)| = m|–v2 – v1|  v2 =

15  10 –2 3 |P| = 50 × × × = 2 10 2 2 J = P = 1.05N-s. 10–3

E-2.

The area of F-t curve is A, where 'F' is the force on one mass due to the other. If one of the colliding bodies of mass M is at rest initially, its speed just after the collision is : F-t oØ d k {ks =kQ y A gSt gk¡'F' ,d nzO;eku ij nwl jsd sd kj.k cy gSA vxj Vd jkusoky h oLrqv ksaesal sM nzO;eku

oky h izkjEHk esafLFkj Fkh rksVd jkusd sckn bl d k osx D;k gksxkA (1*) A/M Sol..

Area of F-t curve = A = Impulse. Impulse vkos ax = dP = A = mv – 0 v=

E-3.

(2) M/A

(3) AM F-t

(4)

2A M

vkjs[k d k {ks=kQ y = A = vkosx

A . M

The given figure shows a plot of the time dependent force Fx acting on a particle in motion along the xaxis. What is the total impulse delivered by this force to the particle from time t = 0 to t = 2second? fn;kx;k fp=kx-v{k d svuqfn'k xfr d jrsgq;s,d d .kij d k;Zd jusoky sl e; ij fuHkZj cy Fx(t) d kxzkQ iznf'kZr d jrk gSA l e; t = 0 l st = 2 sec rd bl cy }kjk d .k d ksfn;k x;k d qy v kosx D;k gS\

(1) 0

RESONANCE

(2) 1 kg-m/s

(3*) 2 kg-m/s

(4) 3 kg-m/s


Sol.

Impulse (v kos x) = F dt



= Area under curve (oØ = E-4.

l sifjc) {ks=kQ y )

1 (2) (2) = 2 kg-m/sec. 2

A mass of 100g strikes the wall with speed 5m/s at an angle as shown in figure and it rebounds with the same speed. If the contact time is 2 × 10–3 sec., what is the force applied on the mass by the wall : 100g nz O;eku d k d ksbZd .k 5m/s d h pky lsfd lh nhokj lsfp=k esan'kkZ;sx, d ks.k ij Vd jkrk gS]rFkk mlh pky ls okil y kSV vkrk gSA ;fn laid Zle; 2 × 10–3 sec gksrksd .k }kjk nhokj ij y xk, x, cy d k eku gS

60º 60º 100g (1) 250 3 to right (1) 250 3 Sol.

Fx =

nka;hvksj

(3*) 250 3 N to left

(4) 250 N to left

(2) 250 N nk;hvks j

(3*) 250 3 N cka ;hvks j

(4) 250 N cka ;hvksj

Px (Pf x – Pi x ) –mV sin 60º–(mV sin 60º )    –250 3 N = 250 3 N towards left cka ;h rjQ t t 2  10 – 3

SECTION (F) : COLLISION F-1.

(2) 250 N to right

VDd j

A block moving in air explodes in two parts then just after explosion (1*) the total momentum must be conserved (2) the total kinetic energy of two parts must be same as that of block before explosion. (3) the total momentum must change (4) the total kinetic energy must not be increased

gok esaxfr'khy ,d xqVd k nksHkkxksaesaVwV t krk gSrFkk Vqd M+savy x&vy x gkst krsgSa]rksfoLQ ksV d srqjUr ckn & (1) d q y laosx lajf{kr jgsxkA (2) nks uksaHkkxksad h d qy xfrt Å t kZogh jgsxh t ksfoLQ ksV d sigy sxqVd sd h Fkh (3) d q y laosx ifjofrZr gkst k;sxkA (4) d q y xfrt Å t kZughac<+uh pkfg,A F-2.

In head on elastic collision of two bodies of equal masses, it is not possible : (1) the velocities are interchanged (2) the speeds are interchanged (3) the momenta are interchanged (4*) the faster body speeds up and the slower body slows down

leku nzO;eku oky h nksoLrqv ksad h lEeq[k çR;kLFk VDd j esa];g lEHko ughagS& & (1) os x ijLij ifjofrZr gkst krsgSA (2) pky s aijLij ifjofrZr gkst krh gSA (3) la osx ijLij ifjofrZr gkst krsgSA (4*) rhoz xkeh oLrqrst gkst k, rFkk /kheh oLrq/kheh py saA F-3.

A bullet of mass m = 50 gm strikes a sand bag of mass M = 5 kg hanging from a fixed point, with a  horizontal velocity v p . If bullet sticks to the sand bag then the ratio of final & initial kinetic energy of the bullet is (approximately) : ,d 50 gm d h xksy h ,d jsr d sy Vd sgq,

cSx l s(ft l d k nzO;eku 5 kg gSa) {ksfrt osx v l sVd jkrh gSA v xj cwy sV jsr d scSx d sfpid t krh gSrksxksy h d h v fUre rFkk izkjfEHkd xfrt Å t kZd k v uqikr D;k gksxk ¼y xHkx½( 1 ) 10–2

RESONANCE

(2) 10–3

(3) 10–6

(4*) 10–4


Sol.

0.5 × vp + m × 0 = 5.05 v

vf 0.05  v = = 10–2 5 i 1 m( v f )2 2  = (10–2)2 = 10–4. 1 2 m( v i ) 2 F-4.

There are hundred identical sliders equally spaced on a frictionless track as shown in the figure. Initially all the sliders are at rest. Slider 1 is pushed with velocity v towards slider 2. In a collision the sliders stick together. The final velocity of the set of hundred stucked sliders will be : fp=kkuql kj 100 Cy kWd ,d nwl jsl sl eku nwjh ij ?k"k.kZjfgr l rg ij mifLFkr gSA izkjEHkeaasl HkhfLFkj gSA igy sCy kWd d ksv osx l snwl jsd h rjQ /kDd k fn;k t krk gSA rFkk nksuksa,d nwl jsd sfpid t krsgSrksl Hkh fpid s100 Cy kWd d k

v fUre osx D;k gksxk \

v v (2*) (3) zero (4) v 99 100 by conservation of linear momentum Pi = Pf  mv = (100 m) u  js[kh; l aosx l j{k.k l sPi = Pf  mv = (100 m) u  u = v/100

(1) Sol.

F-5.

u = v/100

A massive ball moving with speed v collides head-on with a tiny ball at rest having a mass very less than the mass of the first ball. If the collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to: ,d Hkkjh xsan t ksv osx l s,d N ksVh xsan ¼ft l d k nzO;eku cgqr&cgqr d e gSA½ l sl Eeq[k izR;kLFk VDd j d jrh gSrks

VDd j d sckn nwl jh xsan d k osx y xHkx gksxkA Sol.

(1) v (2*) 2v (3) v/2 Velocity of heavy mass donot change after collison

(4) .

VDd j d sckn Hkkjh oLrqd k osx ifjofrZr ugh gksrk gSA v 2 – v1 u2 – u1 = – e = – 1 F-6.

Sol.

F-7.



v2 – v = –1 0–v



v2 = 2v

A ball of mass 'm', moving with uniform speed, collides elastically with another stationary ball. The incident ball will lose maximum kinetic energy when the mass of the stationary ball is ,d 'm' nzO;eku d h xsan fu;r osx l sxfr d j jgh gSA ,d v U; fLFkj xsUn l sizR;kLFk VDd j d jrh gSA v xj vkifrr xsan d h v f/kd re xfrt Å t kZd h gkfu gksrh gSrksfLFkj xsan d k nzO;eku gksxkA [REE - 96] (1*) m (2) 2m (3) 4m (4) infinity vuUr If mass ;fn nz O;eku = m first ball will stop iz Fke xsan : d t k,xh  v = 0 so vr% k.e. = 0 (min U;w ure) In other cases there will be some kinetic energy nw l jsizd j.k esaleku gksxh (K.E. can't be negative _ .kkRed ughgksld rh) 100 J. 3 If before collision the masses are moving in the opposite direction, then their velocity of approach before the collision is :

During the head on collision of two masses 1 kg and 2 kg the maximum energy of deformation is

100 3

nksnzO;ekuksa1kg. rFkk 2kg d h ‘'kh"kkZfHkeq[k VDd j d snkSjku l aihMu d h egRre Å t kZ

t wy gSA ;fn VDd j d s

igy snzO;eku foijhr fn'kk esaxfreku gSa] rc mud k VDd j l sigy sl kehI; osx gS& (1*) 10 m/sec.

RESONANCE

(2) 5 m/sec.

(3) 20 m/sec.

(4) 10 2 m/sec.


Sol.

1 m1 m 2 100 U = 2 (m  m ) (V1 – V2)2 = 3 1 2 2m .m 100 (V1 – V2)2 × 2 (m  2m) = 3 putting m = 1 kg j[kusij: (V1 – V2) = 10 m/sec.

AlternateSolution: When deformation is maximum both the particles are moving with same velocity . So applying momentum conservation.

t c nksuksad .k leku osx lsxfr d jsaxsrc laihM+u vf/kd re gksxkA vr%laosx lj{k.k ls m1v1 + m2v2 = m1v 1’ + m2v1’ v1’ =

m1v 1  m 2 v 2 m1  m 2

Applying energy conservation: Å t kZlj{k.kls

1 1 1 m1v12 + m2v22 = (m 1 + m 2) (v1´ )2 + U deformation 2 2 2  U deformation=

F-8.

m1m 2 100 2 m1  m 2  × (v1 – v2) = 3 v1 – v2 = 10m/sec.

A block A of mass m moving with a velocity ' v ' along a frictionless horizontal track and a blocks of mass m/2 moving with 2 v collides with block elastically. Final speed of the block A is : ,d m nzO;eku d k Cy kWd A, v ' osx l s?k"kZ.k jfgr iFk ij py jgk gSv kSj ,d m/2 nzO;eku d k ,d Cy kWd B, 2 v osx l spy rsgq;sCy kWd A,l sizR;kLFk : i l sVd jkrh gSrksCy kWd A d k v fUre pky gksxhA

(1) Sol.

1 2

5v 3

(2*) v

(3)

2v 3

Let the velocities of plank and body of mass

(4) none of these bues al sd ksbZugha

m 2

m nzO;eku d h oLrqfp=kkuql kj VDd j d sckn 2 move with speed v1 and v2 after collision as shown. Ø e'k%osx v1 o v2 lsxfr d jrh gSA From conservation of momentum.

ekuk Iy kad vkSj

l aosx l j{k.k l s m m 2v = mv1 + v 2 2 2 or 2v1 + v2 = 0 From equation of coefficient of restitution. mv –

..........(1)

iz R;koLFkku xq.kka d v 2  v1 v  2v Solving 1 and 2 we get 1 o 2 d ksgy d jusij v1 = –v e=1=

RESONANCE



v2 – v1 = 3v

..........(2)


F-9.

In a collision between two solid spheres, velocity of separation along the line of impact (assume no external forces act on the system of two spheres during impact) :

nksBksl xksy ksad h VDd j esa] VDd j js[kk d sv uqfn'k nwj t kusd k osx ¼;g ekfu;sfd nksxksy ksad sfud k; ij VDd j d snkSjku d ksbZckº; cy d k;Zugh d jrk gS½ (1*) cannot be greater than velocity of approach

ikl v kusd sosx l sT;knk ughagksl d rk (2) cannot be less than velocity of approach

ikl v kusd sosx l sd e ughagksl d rk (3) cannot be equal to velocity of approach

ikl v kusd sosx d scjkcj ugh gksl d rk (4) none of these (bues al sd ksbZugha) Sol.

| velocity of separation | e = | velocity of approach | For elastic collision e = 1 |Velocity of separation| = |velocity of approach| For inelastic collision e < 1 So |velocity of separation| < |velocity of approach| | nw j

gksusd k lkisf{kd osx | | ut nhd vkusd k lkis f{kd osx | izR;kLFk VDd j d sfy , e = 1 | nw j gksusd k l kisf{kd osx | = | ut nhd v kusd k l kisf{kd osx | v izR;kLFk VDd j d sfy , e < 1 v r% | nwj gksusd k l kisf{kd osx | < | ut nhd v kusd k l kisf{kd osx |

gy %

e=

F-10.

In the figure shown the block A collides head on with another block B at rest. Mass of B is twice the mass of A. The block A stops after collision. The co-efficient of restitution is : fp=kkuql kj Cy kWd A, nwl jsfLFkj Cy kWd B l sl Eeq[k VDd j d jrk gSA B d k nzO;eku A l snqxuk gSA VDd j d sckn A

Cy kd : d t krk gSA izR;kLFkrk xq.kkad gS%

(1*) 0.5 Sol.

v1 =  

F-11.

(2) 1

(3) 0.25

(4) it is not possible

;g l EHko ughagSA

(m1 – em 2 )u1 m 2 (1  e)u 2 (m – e2m)u1 2m(1  e)  0   =0 m1  m 2 m1  m 2 = m  2m m  2m 0 = m – e2m e = 1/2

A particle of mass m moves with velocity v0 = 20 m/sec towards a wall that is moving with velocity v = 5 m/sec. If the particle collides with the wall elastically, the speed of the particle just after the collision is : m nz O;eku d k ,d d .k v0 = 20 m/sec l sxfr d jrk gqv k 5 m/sec l sxfr d jrh nhokj l sfp=kkuql kj Vd jkrk gS

rksçR;kLFk VDd j d sckn d .k d k osx D;k gksxkA

(1*) 30 m/s

RESONANCE

(2) 20 m/s

(3) 25 m/s

(4) 22 m/s


Sol.

Sol. V2 = Z0 Vol. of Sep = Vel of approach ( elastic) nwj t kusd k osx = fud V vkusd k osx ( izR;kLFk)  20 + 5 = V – 5  V = 30 m/s Ans.

F-12.

A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of super-ball to be v0 horizontally, the average force being exerted by the super-ball on one wall is : ,d l qij xsan]nksn`<+nhokjksat ksd nwjh ij gS]d se/; vkxsihN sxfr d jrh gqbZçR;kLFk : i l smN y rh gSA xq: Roh; cy ux.; gSrFkk xsan d k {ksfrt osx v0 gSrksizR;sd nhokj ij l qij xsan }kjk y xk;k x;k cy gksxkA (1)

Sol.

1 mv 02 2 d

(2*)

mv 02 d

(3)

2d t = v (time for succeesive collision Ø ekxr 0

2mv 02 d

(4)

4mv 02 d

VDd jksad schp le;)

N × t = dP = mv0 – (–mv0)

2d N × v = 2mv0 0 N= F-13.

mv 02 d

In the figure shown the block A collides head on with another block B at rest. Mass of B is twice the mass of A. The block A stops after collision. The co-efficient of restitution is : fp=kkuql kj Cy kWd A, nwl jsfLFkj Cy kWd B l sl Eeq[k VDd j d jrk gSA B d k nzO;eku A l snqxuk gSA VDd j d sckn A

Cy kd : d t krk gSA izR;kLFkrk xq.kkad gS%

(1*) 0.5 F-14.

(2) 1

(3) 0.25

(4) it is not possible

;g l EHko ughagSA

A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass. If e is the coefficient of restitution, then ratio of speed of the first sphere to the speed of the second sphere after collision will be : ,d m nzO;eku d k xksy k fu;r osx l sxfr d jrk gqv k l eku nzO;eku d sfLFkj xksy sl sVd jkrk gSA v xj e izR;kLFkrk

xq.kkad gSrksVDd j d sckn izFke xksy sd sosx ,oanwl jsxksy sd sosx v uqikr D;k gksxkA

Sol.

 1 e   (1*)   1 e  mu = mv1 + mv2 u = v1 + v2

 1 e   (2)   1 e  .......(i) .......(i)

v 2 – v1 =e u

......(ii)

as solving have gy

RESONANCE

v1

 e  1  (3)   e  1

 e  1  (4)   e  1

 1– e 

d jusij v =  1  e  .   2


SECTION (G) : VARIABLE MASS G-1.

If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate of combustion of the fuel will be : ;fn 300 m/s, osx lsxSl ksad ksckgj fud ky jgsjkd sV d s}kjk xSal ksaij cy 210 N gSrksbZa/ku d st y usd h nj gksxh : (1) 10.7 kg/sec

Sol.

ifjorZu'khy nzO;eku

F=

(2) 0.07 kg/sec

(3) 1.4 kg/sec

(4*) 0.7 kg/sec

dm dt

210 = 300 ×

dm dt



dm = 0.7 kg/s. dt

PART - II : ASSERTION / REASONING Hkkx - II : d Fku@d kj.k (ASSERTION/REASONING) 1.

STATEMENT-1 : A sphere of mass m moving with speed u undergoes a perfectly elastic head on collision with another sphere of heavier mass M at rest (M > m), then direction of velocity of sphere of mass m is reversed due to collision [no external force acts on system of two spheres] STATEMENT-2 : During a collision of spheres of unequal masses, the heavier mass exerts more force on lighter mass in comparison to the force which lighter mass exerts on heavier mass. oDrO;-1 : ,d xksy k ft l d k nzO;eku m gSrFkk bl d h pky u gSrFkk ;g nwl jsHkkjh xksy sl siw.kZr;k izR;kLFk l Eeq[k VDd j d jrk gSA Hkkjh xksy k izkjEHk esafojke esagSrFkk Hkkjh xksy sd k nzO;eku M gSt gk¡(M > m) gSA VDd j d sd kj.k m nz O;eku d sxksy sd sosx d h fn'kk foijhr gkst krh gS[nksuksaxksy ksd sfud k; ij d ksbZckº; cy d k;Zjr ugh gS] oDrO;-2 : v l eku nzO;eku d sxksy ksad h VDd j d snkSjku Hkkjh oky k xksy k gYd soky sxksy sij]gYd soky sxksy s}kjk

Hkkjh oky sxksy sij y xscy d h rqy uk esav f/kd cy y xkrk gSA

Sol. 2.

(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3*) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True (1) oDrO;-1 lR; gS ] oDrO;-2 lR; gS; oDrO;-2, oDrO;-1 d k lgh Li"Vhd j.k gSA (2) oDrO;-1 lR; gS ] oDrO;-2 lR; gS; oDrO;-2, oDrO;-1 d k lgh Li"Vhd j.k ughagSA (3*) oDrO;-1 lR; gS ]oDrO;-2 vlR; gS; (4) oDrO;-1 vlR; gS ]oDrO;-2 lR; gS Statement-2 contradicts Newton's third law and hence is false. oDrO;-2 U;wVu d srhljsfu;e d k fojks/k d jrk gSrFkk blfy ;svlR; gSA STATEMENT-1 : In a perfectly inelastic collision between two spheres, velocity of both spheres just after the collision are not always equal. STATEMENT-2 : For two spheres undergoing collision, component of velocities of both spheres along line of impact just after the collision will be equal if the collision is perfectly inelastic. The component of velocity of each sphere perpendicular to line of impact remains unchanged due to the impact. oDrO; 1 : nksxksy kasd schp ,d iw.kZv izR;kLFk VDd j d sfy ,] VDd j d srqjUr ckn nksuksaxksy sad sosx ges'kk cjkcj

ugh gksaxsA oDrO; 2 : VDd j d jrsgq;snksxksy sad sfy ,]VDd j d srqjUr ckn VDd j d h js[kk d sv uqfn'k nksuksaxksy ksad sosxksad s ?kVd cjkcj gksaxs;fn VDd j iw.kZr;k v izR;kLFk gSA VDd j d h js[kk d sy Ecor~izR;sd xksy sd sosx d k ?kVd VDd j d sd kj.k v ifjofrZr jgrk gSA (1*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True (1*) oDrO;-1 l R; gS ] oDrO;-2 l R; gS; oDrO;-2, oDrO;-1 d k l gh Li"Vhd j.k gSA

oDrO;-1 l R; gS] oDrO;-2 l R; gS; oDrO;-2, oDrO;-1 d k l gh Li"Vhd j.k ughagSA (3) oDrO;-1 l R; gS ] oDrO;-2 v l R; gS; (4) oDrO;-1 v l R; gS ] oDrO;-2 l R; gS (2)

RESONANCE


Sol.

(Easy)

From statement-2, if the component of relative velocity normal to line of impact is non-zero, they shall not have same velocity after collision. Hence statement-2 is correct explanation of statement-1. (Easy) oDrO;-2 l s , ;fn VDd j d h js [kk d sy Ecor~l kisf{kd osx d k ?kVd v 'kwU; gS]rksVDd j d sckn mud sl eku osx ugh gksaxsA bl fy ;soDrO;-2 ] oDrO;-1 d k l gh Li"Vhd j.k gSA

3.

Statement 1 : If the mass of the colliding particles remains constant, then the linear velocity of the individual particles change during collision along common normal direction. Statement 2 : A pair of equal and opposite impulses act along common normal direction. (1*) Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1. (2) Both statements 1 and 2 are true but statement 2 is not correct explanation of statement 1. (3) Statement 1 is true but statement 2 is false (4) Both statements 1 and 2 are false. oDrO; 1 % v xj VDd j d j jgsad .kksad k nzO;eku fu;r jgsrksVDd j d snkSjku izR;sd d .k d k v iuk jsf[kd osx

mHk;fu"B v fHky Ec fn'kk d sv uqfn'k ifjofrZr gksrk gSA oDrO; 2 % l eku rFkk foijhr v kosx ;qXe mHk;fu"B v fHky Ec fn'kk d sv uqfn'k d k;Zd jrk gSA (1*) nks uksaoDrO; l R; gSrFkk oDrO; 2] oDrO; 1 d h l gh O;k[;k d jrk gSA (2) nks uksoDrO; l R; gSijUrqoDrO; 2] oDrO; 1 d h l gh O;k[;k ugh d jrk gSA (3) oDrO; 1 l R; gSrFkk oDrO; 2 v l R; gS A (4) oDrO; 1 v l R; gSfd UrqoDrO; 2 l R; gS A Ans. (1)

PART - I : OBJECTIVE QUESTIONS Hkkx - I : oLrqfu"B iz'u ¼OBJECTIVE QUESTIONS½ Single choice type

,d y fod Yih izd kj 1.

All the particles of a body are situated at a distance R from the origin. The distance of the centre of mass of the body from the origin is fd lh oLrqd sleLr d .k ewy fcUnqlsR nwjh ij fLFkr gSA oLrqd snzO;eku d sUnzd h ewy fcUnqlsnwjh & [HCV - I] (1) = R (2*)  R (3) > R (4)  R

2.

A uniform thin rod of mass M and Length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin (0,0). A slight disturbance at t = 0 causes the lower end to slip on the smooth surface along the positive x-axis, and the rod starts falling. The acceleration vector of centre of mass of the rod during its fall is : [JEE - 93] [ R is reaction from surface] ,d l e: i iry h M nzO;eku ,oaL y EckbZd h N M+y-v {k d sv uqfn'k fpd us{kSfrt ry ij Å /okZ/kj [kM+h gSA bl d k fupy kfl jk ewy fcUnq(0,0) ij gSy ?kqfo{kksHk l sgh bl d k fupy k fl jk /kukRed x-v{k d h vksj fQ l y rk gsrFkk N M+

fxjuk izkjEHk d jrh gSfxjrsl e; bl d snzO;eku d sUnzd k Roj.k l fn'k gksxkA [ R l rg l sizfrfØ ;k cy gSA]    Mg  R (1*) a CM  M

   Mg  R (2) a CM  M

   (3) a CM  Mg  R

(4) None of these

buesalsd ksbZ

ughAa Sol.

Equation of motion xfr    Mg  R  Ma cm    Mg  R so v r% a cm = M

RESONANCE

d sl ehd j.k l s


3.

A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. The distance over which the skater will move back (the coefficient of friction between the skater and the ice is ) : ,d Ld sVj (nzO;eku m) cQ Zij [kM+k gS]M nzO;eku d k iRFkj v osx l s{ksfrt fn'kk esaQ sd rk gSA Ld sVj ihN sd h v ksj ft ruh nwjh r; d jsxk] oksgksxh\ (cQ ZrFkk v kneh d schp ?k"kZ.k xq.kkad gS) :

Mv 2

M2 v 2 (1) 2 mg

(2)

Pi = 0 Pf = MV – mV1

...(i) ....(ii)

MV – mV1 = 0



M2 v 2 (3*)

2 m 2 g

2 m 2 g

M2 v 2 (4)

2 m 2 2 g

Sol.

4.

using 

02 = v12 – 2ax v12 = 2gx



 MV     m 



x=

v1 =

M V.. m

2

= 2g x.

M2 V 2 2m2g

In a vertical plane inside a smooth hollow thin tube a block of same mass as that of tube is released as shown in figure. When it is slightly disturbed it moves towards right. By the time the block reaches the right end of the tube then the displacement of the tube will be (where ‘R’ is mean radius of tube). Assume that the tube remains in vertical plane.

Å /okZ/kj ry esa[kks[ky h ufy d k esal eku nzO;eku d k Cy kWd fp=kkuql kj N ksM +k t krk gSt c ;g ufy d k d s,d fl js ij igq¡p rk gSrksufy d k d k foLFkkiu gksxk (‘R’ ufy d k d h v kSl r f=kT;k gS) ekuksufy d k Å /okZ/kj jgrh gSA

2R 4R R (2) (3*) (4) R   2 Let the tube displaced by x towards left, then block will be displaced by (R – x) towards right ; ekuk ufy d k ck;h rjQ x foLFkkfir gksrh gSA rc Cy kWd nk;h rjQ (R – x) foLFkkfir gksxkA

(1) Sol.

mx = m (R – x)

RESONANCE



x=

R 2


5.

A stationary body explodes into two fragments of masses m 1 and m 2. If momentum of one fragment is p, the minimum energy of explosion is ,d fLFkj oLrqm 1 rFkk m 2 nzO;eku easVwV t krh gSA v xj ,d Hkkx d k l aosx p gSrksfo[k.Mu d h Å t kZgksxh (1)

Sol.

p2 2(m1  m 2 )

use F.E. =

p2

(2)

(3*)

2 m1m 2

p 2 (m1  m 2 ) 2m1m 2

(4)

p2 2(m1  m 2 )

m1v1 = m2v2 =P

1 1 mv12 + m2v22 2 2 2

 P   P  1  + 1 m   = m1  2 2 2  m 2   m1 

2

2 1 P (m 2  m1 ) = . m1m 2 2

6.

A train of mass M is moving on a circular track of radius ' R ' with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the train will be

,d Vªsu ft ld knzO;eku M gS,d leku osx V lso`Ùkkd kj iFkft ld hf=kT;k' R ' gSeasxfr d j jghgSA Vªsu d hy EckbZ Vªsd d hifjf/k d hvk/kh gSA Vªsu d k js[kh; laosx gksxkA (1) 'kw U; 0 Sol.

(2*)

2M V 

(3) MVR

(4) MV

If we treat the train as a ring of mass 'M' then its COM will be at a distance

2R from the centre of the 

circle. Velocity of centre of mass is : 2R 

;fn Vªsu d ks'M' nzO;eku d k oy ; ekusrksml d k nzO;eku d sUnz

nwjh ij gksxkA nzO;eku d sUnzd k osx

VCM = RCM . 2R  V  .   R

=

2R . 



VCM =



The linear momentum of the train =

=

(  =

V ) R

2MV 2V  MVCM =   As the linear momentum of any system = MVCM 2MV 

Ans.

fd l h Hkh fud k; d k js[kh; l aosx = MVCM vr%Vª su

7.

d k js[kh; laosx =

2MV 

Ans.

Two particles approach each other with different velocities. After collision, one of the particles has a  momentum p in their center of mass frame. In the same frame, the momentum of the other particle is [REE - 98]

nksd .k ,d nwl jsd h v ksj fHkUu&fHkUu osx l sv k jgsgSaA VDd j d sckn muesal s,d d .k d k mud snzO;eku d sUnzrU=k  esal aosx p gSA l eku rU=k esav U; d .k d k l aosx gksxkA [REE - 98] Sol.

  (1) 0 (2*) – p (3) – p /2 Net momentum in centre of mass frame should be zero

 (4) – 2 p

nzO;eku d sUnzd sfunsZ'k ra=k esad qy l aosx 'kwU; gksxk     so v r% P  P' = 0  P'  –P

RESONANCE


8.

A particle of mass ‘m’ and velocity ‘ v ’ collides oblique elastically with a stationary particle of mass ‘m’. The angle between the velocity vectors of the two particles after the collision is : ,d ‘m’ nzO;eku d k d .k t ksfd ‘ v ’ osx l sxfr d j jgk gS,d ‘m’ nzO;eku d sfLFkj d .k l sfr;Zd izR;kLFk VDd j d jrk gSA VDd j d sckn nksuksad .kksad sosx l fn'kksad se/; d ks.k gksxkA [REE - 97] (1) 45° (2) 30° (3*) 90° (4) None of these bues al sd ksbZugha

Sol.



be fore collision VDd j l siw oZ So angle between velocity vectors is 90º v r% osx l fn'kksad se/; d ks.k 90º gSA 9.

after collision VDd j

d sckn

Two homogenous spheres A and B of masses m and 2m having radii 2a and a respectively are placed in touch. The distance of centre of mass from first sphere is : nkslekax (homogenous) xksy sA rFkk B ft ud sØ e'k%nzO;eku m rFkk 2m o f=kT;k 2a rFkk a gS,d nwl jsd sikl j[ks

x;sgSA igy sxksy slsnzO;eku d sUnzd h nwjh Kkr d jksA Sol.

10.

(1) a (2*) 2a we have m 1 r1 = m 2 r2  mr = 2m (3a – r) 

(3) 3a

(4) none of these bues alsd kbsZugha

r = 2a

A non–uniform thin rod of length L is placed along x-axis as such its one of ends at the origin. The linear mass density of rod is  = 0x. The distance of centre of mass of rod from the origin is : L y EckbZd h vleku NM+ft ld k js [kh; ?kuRo  = 0x gSA ft lsbl rjg j[kk x;k gSfd bld k ,d fljk ewy fcUnqij gSA x ewy fcUnqlsy EckbZgSA ewy fcUnqlsnzO;eku d sUnzd h nwjh Kkr d jksA (1) L/2 (2*) 2L/3 (3) L/4 (4) L/5

Sol.

L

dm x

x cm



(dx )x

     dm  dx



0x

2

dx

0 L





2L 3

0 xdx

0

11.

A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The centre of mass :

cUn ckWDlsesa,d xsan bl rjg j[khgqbZgSfd ;g xsan ckWDl d hnhokjksalsVd jkrhjgrhgSA ckWDlsd ksfpd uh{kSfrt lrg ij j[kk x;k gSA nzO;eku d sUnz

Sol.

(1) of the box remains constant (2*) of the box plus the ball system remains constant (3) of the ball remains constant (4) of the ball relative to the box remains constant (1) ckW Dl d kfu;r jgsxkA (2*) xs an rFkk ckWDl fud k; d k fu;r jgsxkA (3) xs an d k fu;r jgsxkA (4) xs an d k ckWDl d slkis{k fu;r jgsxkA Net external force on box plus ball system is zero.

xsan v kSj ckDl fud k; d qy cká cy 'kwU; gSA

RESONANCE


12.

A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of plank is M/3, the distance that the plank moves relative to the ground is : M nz O;eku d k,d O;fDr L y EckbZd sr[rst ksfd ?k"kZ .kjfgr lrg ij j[kkgSd s,d fljsij [kM+kgS A O;fDr r[rsd snwl js fljsd hvksj py uk'kq: d jrkgSA ;fn r[rsd knzO;eku M/3 gksrkst c O;fDr nwl jsfljsij igq¡p rkgSrksr[rs}kjkr;

nwjhgks xhA (1*) 3L/4 Sol.

(3) 4L/5

(4) L/3

m (–x) = 0 3

m (L – x) + 4 mx 3

mL = 13.

(2) L/4

Two blocks A and B are connected by a massless string (shown in figure) A force of 30 N is applied on block B. The distance travelled by centre of mass in 2s starting from rest is : nksCy kWd A rFkkB fp=kkuql kj nzO;ekujfgr jLlhlst qM+sgq, gSA F = 30 N d kcy B CykW d ij yxk;kt krkgSrksxfr d jus d s2s i'pkr~nzO;eku d sUnzd s}kjk r; nwjh gksxh :

B

A 10kg

20kg

F=30N

Smooth(?k"kZ .kjfgr) (1) 1m

Sol.

acm =

(2*) 2m

(3) 3m

(4) none of these

buesalsd ksbZugha

30 = 1 ms2 (10  20)

S = 0 (2) +

1 (1) (2)2 2

=2m 14.

The motion of the centre of mass of a system of two particles is unaffected by their internal forces :

nksd .k fud k; d k nzO;eku d sUnzchp d svkUrfjd cy ksalsvçHkkfor jgrk gSrks

Sol.

(1*) irrespective of the actual directions of the internal forces (2) only if they are along the line joining the particles (3) only if they are at right angles to the line joining the particles (4) only if they are obliquely inclined to the line joining the particles. (1*) vkUrfjd cy d hfd lhHkhfn'kkd sfy ,A (2) d s oy t c cy nksauksd kst ksM +usoky h js[kk esay xsA (3) d s oy t c cy nksauksd kst ksM +usoky h js[kk d sy Ecor~y xsA (4) d s oy t c cy nksauksd kst ksM +usoky h js[kk lsfr;Zd : i lsy saxsA vector sum of internal forces on system is zero.

fud k; ij v karfjd cy ksad sl fn'k ;ksx 'kwU; gksrk gSA 15.

Sol.

Two bodies of masses m and 4m are moving with equal linear momentum. The ratio of their kinetic energies is : m nz O;eku 4m nzO;eku d h nksoLrqleku laosx lsxfr d j jgh gS]rksbud h xfrt Å t kZv ksad k vuqikr gksxkA (1) 1 : 4 (2*) 4 : 1 (3) 1 : 1 (4) 1 : 2 2 k1 1 P /(m) k 2 = 2 P 2 / 2( 4m) = 4/1

RESONANCE


16.

If the momentum of a body increases by 20%, the percentage increase in its kinetic energy is equal to : ;fn laosx 20% c<+k fn;k t k, rksoLrqd h xfrt Å t kZesaçfr'kr ifjorZu (c<+uk) gksxk & (1*) 44

Sol.

we have



(2) 88

(3) 66

(4) 20

p 2 – p1  0.2 p1 p2  1.2 p1

 k 2 – k1  k    100   2 – 1  100 so  k k 1    1 

 p 22  p2   =  2 – 1 × 100 (since k = ) 2m  p1  = ((1.2)2 – 1) × 100 = 44 % 17.

Two observers are situated in different inertial reference frames. Then :

nksçs{kd vy x&vy x t M+Ro funsZ'k rU=k esafLFkr gSrks

Sol.

(1*) the momentum of a body by both observers may be same (2) the momentum of a body measured by both observers must be same (3) the kinetic energy measured by both observes must be same (4) none of the above (1*) nks auksçs{kd d svuql kj oLrqd k laosx cjkcj gksld rk gSA (2) nks auksçs{kd d svuql kj oLrqd klaosx cjkcj gh gksxkA (3) nks auksçs{kd d svuql kj oLrqd h xfrt Å t kZcjkcj gh gksxkA (4) bues alsd ksbZugha When velocity of both frames are same then momentum will be same else it will be diffrent according to diffrent observers

t c nksuksfunsZ'k rU=kksd k osx l eku gS] rc l aosx l eku gksxk v U;kFkk v y x v y x iz{ksd d sl kis{k ;g v y x v y x gksxkA 18.

Sol.

A man is sitting in a moving train, then : ,d O;fDr xfr'khy js y xkM+h (train) esacSBk gqv k gS% (1) his momentum must not be zero (2) his kinetic energy is zero (3) his kinetic energy is not zero (4*) his kinetic energy may be zero (1) bld kla osx 'kwU; ughagksxkA (2) bld hxfrt Å t kZ'kw U; gksxhA (3) bld hxfrt Å t kZ'kw U; ughagksxhA (4*) bld h xfrt Å t kZ'kw U; gksld rh gSA Velocity of man w.r.t. train is zero so kinetic energy of man w.r.t. train = 0

jsy xkMh d sl kis{k O;fDr d k osx 'kwU; gSv r% jsy xkMh d sl kis{k O;fDr d h xfrt Å t kZ'kwU; gksxh 19.

A bomb dropped from an aeroplane explodes in air. Its total :

ok;q;ku lsfxjk;k x;k ce gok esafo?kfVr gkst krk gSrksd qy

Sol.

(1) momentum decreases (3*) kinetic energy increases (1) la os x ?kVsxkA (3*) xfrt Å t kZc<+ sxhA Kinetic energy increase due to internal energy.

(2) momentum increases (4) kinetic energy decreases (2) la osx c<+sxkA (4) xfrt Å t kZ?kVs xhA

v kUrfjd Å t kZd sd kj.k xfrt Å t kZc
RESONANCE


20.

Sol. 21.

Two blocks of masses m1 and m2 are connected by a massless spring and placed on smooth surface. The spring initially stretched and released. Then : nksCykW d m1 rFkkm2 d ksnz O;ekujfgr fLça x d hlgk;rklst ks Md+ j fpd uhlrg ij j[kkt krkgS A fLça x d ksçkjEHkes a[kha pd j NksM +nsrsgSA rc : (1) the momentum of each particle remains constant separately (2) the momentums of each body are equal (3) the magnitude of momentums of each body are equal to each other (4) the mechanical energy of system remains constant (5*) both (3) and (4) are correct (1) çR;s d d .k d k vy x&vy x laosx fu;r jgsxkA (2) nks auksd .k d klaosx leku gksxkA (3) nks auksd .kksad klaosx d kifjek.kleku gksxkA (4) fud k; d h;ka f=kd Å t kZlajf{kr jgsxkA (5*) (3) rFkk(4) lghgS A       p1  p 2   p1  p 2  0 p1  – p 2 A shell is fired from a cannon with a velocity v at an angle  with the horizontal direction. At the highest point in its path, it explodes into two equal pieces, one retraces its path to the cannon and the speed of the other piece immediately after the explosion is : ,d xkys sd ksrks i ls{kfSrt ls d ks .kij v os x lsnkxkt krkgS A bld siFkd smPpre fcUnqij ;g nkscjkcj Hkkxkases afo[kf.Mr

gkst krk gSA ,d Hkkx blh iFk d ksrksi d h rjQ vuql fjr d jrk gSrksnwl jsHkkx d h pky VDd j d srqjUr ckn gksxh & (1*) 3 cos Sol.

3 (3)    cos 2

(2) 2 cos

(4)

3  cos 2

P1 = Pf

m m  (–V cos + V) 2 2 V = 3V cos  mV cos  =

22.

The centre of mass of the shaded portion of the disc is : (The mass is uniformly distributed in the shaded portion) : pd rh d sN k;kafd r gq;sHkkx d k nzO;eku d sUnzgksxk %(N k;kafd r Hkkx esanzO;eku l e: i forfjr gS) (1*)

R to the left of A 20

(A l sck;hav ks j

R (2) to the left of A ( A 12

Sol.

l sck;hav ksj

R 20

R 12

(3)

R to the right of A 20

(4)

R R to the right of A (A l snk;hav ks j 12 12

(A l snk;hav ks j

A1 = R2

A2 =

R2 16

x1 = 0

x2 =

3R 4

ij) A

ij)

R 20

ij)

ij)

R2 3R  16 4 – R 2 20 R  R2 – 16

0–

xcen =

RESONANCE


23.

A semicircular portion of radius ‘r’ is cut from a uniform rectangular plate as shown in figure. The distance of centre of mass 'C' of remaining plate, from point ‘O’ is : ,d v ) Zo`rh; Hkkx ft l d h f=kT;k r gSd ksfp=kkuql kj ,d vk;rkd kj Iy sV l scuk;k x;k gSA fcUnqO l scph gqbZIy sV d snzO;eku d sUnzC d h nwjh gksxhA

(1) Sol.

3r (2) 2 ( 4  )

2r (3  )

(3)

2r ( 4  )

2r (4*) 3 ( 4  )

A1 = 2r × r = 2r2 A2 =

r 2 2

x1 =

r 2

x2 =

4r 3

r r 2 r –  2 2 3 r 2 2 2r – 2

2r 2 

xcm =

2  r 3 1 –  2r  3   = 4 –  3 [ 4 – ]   r2    2 

24.

In an elastic collision in absence of external force, which of the following is/are correct : [REE - 95]

,d izR;kLFk VDd j esa] ckg~; cy d h v uqifLFkfr esafuEu esal sd kSul k@l sd Fku l R; gSaA (1) The linear momentum is not conserved

js[kh; l aosx l ajf{kr ughajgrk gSA (2) The potential energy is conserved in collision

VDd j esafLFkfrt Å t kZl ajf{kr jgrh gSA (3) The final kinetic energy is less than the initial kinetic energy

v fUre xfrt Å t kZizkjfEHkd xfrt Å t kZl sd e gksrh gSA (4*) The final kinetic energy is equal to the initial kinetic energy

v fUre xfrt Å t kZizkjfEHkd xfrt Å t kZd scjkcj gksrh gSA 25.

A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined system (bag + bullet) : ,d M nzO;eku d k FkSy k ,d y Ecs/kkxsl sy Vd k gSrFkk ,d m nzO;eku d h xksy h v {kSfrt osx l sv kd j FkSy sesa /kal t krh gSA rksFkSy k rFkk xksy h d sl a;qDr fud k; d sfy , — (1) Momentum is mMv/(M + m) (2) KE is (1/2) Mv2 (3*) Momentum is mv (4) KE is m 2v2/(M + m) (1) l a osx mMv/(M + m) gksxkA (2) xfrt Å t kZ(1/2) Mv2 gks xhA 2 2 (3*) l a osx mv gksxkA (4) xfrt Å t kZm v /(M + m) gks xhA

RESONANCE


Sol.

by conservation of linear momentum pi = pf js[kh; laosx lehd j.k lspi = pf 

mv = (m + M) u

so energy of system =

vr%fud k; d h Å t kZ= 26.



u=

mv (m  M)

1 (m + M) × u2 = m 2v2/2(M + m) 2

1 (m + M) × u2 = m 2v2/2(M + m) 2

A shell explodes in a region of negligible gravitational field, giving out n fragments of equal mass m. Then its total [REE - 97] ,d xksy k,d ux.; xq: Roh; {ks=kesafoLQ kksfVr gksrk gSA rFkk l eku nzO;eku m d sn Vqd M+siznku d jrk gSA rc bl d h

d qy & (1) Kinetic energy is smaller than that before the explosion

xfrt Å t kZfoLQ ksfVr gksusd sigy soky h xfrt Å t kZl sd e gksxhA (2) Kinetic energy is equal to the before the explosion

xfrt Å t kZfoLQ ksfVr gksusd sigy soky h xfrt Å t kZd scjkcj gksxhA (3) Momentum and kinetic energy depend on n l aosx rFkk xfrt Å t kZn ij fuHkZj d jsxhA (4*) Momentum is equal to that before the explosion.

l aosx foLQ ksfVr gksusd sigy soky sl aosx d sl eku gksxkA More than one choice type

cgqfod Yih izd kj 27.*

A system of particles has its centre of mass at the origin. The x-coordinates of all the particles (A) may be positive (B) may be negative (C*) may be non-negative (D*) may be non-positive ,d d .k fud k; d k nzO;eku d sUnzewy &fcUnqij gSA lHkh d .kksad sx–funsZ'kkad & (A) /kukRed gksld rsgS A (B) _ .kkRed gksld rsgS A (C*) v&_ .kkRed gksld rsgS A (D*) v&/kukRed gksld rsgS A

28.*

In which of the following cases the centre of mass of a system is certainly not at its centre ? (A*) A rod whose density continuously increases from left to right (B*) A rod whose density continuously decreases from left to right (C) A rod whose density decreases from left to right upto the centre and then increases (D) A rod whose density increases from left to right upto the centre and then decreases

fuEu eslsfd u fLFkfr;ksad sfy ;sfud k; d k nzO;eku d sUnzfuf'pr : i lsbld sd sUnzij ughagksxk & (A) ,d NM ft ld k ?kuRo cka ;hlsnka;h vksj fujarj c<+rk jgsA (B) ,d NM ft ld k?kuRo cka ;h lsnka;h vksj fujUraj d e gksrk jgsaA (C) ,d NM ft ld k ?kuRo cka ;h lsnka;h vksj e/; fcUnqrd d e gksrk jgsrRi'pkr~c<+sA (D) ,d NM ft ld k ?kuRo cka ;h lsnka;h vksj e/; fcUnqrd c<+rk jgsrRi'pkr~d e gksrk jgsA 29.*

If the net external force acting on a system is zero, then the centre of mass (A) must not move (B*) must not accelerate (C*) may move (D) may accelerate

;fn fd lh fud k; ij y x jgsd qy ckº; cy ksad k ifj.kkeh 'kwU; gks]rksnzO;eku d sUnz& (A) xfr ughad js xkA (B) Rofjr ughagks xkA (C) xfr d j ld rk gS A (D) Rofjr gksld rkgS A

RESONANCE


30.*

In an elastic collision in absence of external force, which of the following is/are correct : [REE - 95]

,d izR;kLFk VDd j esa] ckg~; cy d h v uqifLFkfr esafuEu esal sd kSul k@l sd Fku l R; gSaA (A*) The linear momentum is conserved

js[kh; l aosx l ajf{kr jgrk gSA (B) The potential energy is conserved in collision

VDd j esafLFkfrt Å t kZl ajf{kr jgrh gSA (C) The final kinetic energy is less than the initial kinetic energy

v fUre xfrt Å t kZ] izkjfEHkd xfrt Å t kZl sd e gksrh gSA (D*) The final kinetic energy is equal to the initial kinetic energy

v fUre xfrt Å t kZ, izkjfEHkd xfrt Å t kZd scjkcj gksrh gSA 31.*

A small ball collides with a heavy ball initially at rest. In the absence of any external impulsive force, it is possible that (A) Both the balls come to rest (B*) Both the balls move after collision (C*) The moving ball comes to rest and the stationary ball starts moving (D) The stationary ball remains stationary, the moving ball changes its velocity.

fojkekoLFkkesaj[kh,d Hkkjhxsa n ls,d NksVhxsa n Vd jkrhgSA fd lhHkhcká vkosxhcy d hvuqifLFkfr es a;g lEHko gSfd (A) nks uksaxsansfojkekoLFkkesavkt k;saA (B) VDd j d si'pkr~nks uksaxasnsxfr'khy gks (C) xfr'khy xs an fLFkj gkst k;srFkk fLFkj xsan xfr çkjEHk d j nsaA (D) fLFkj xs an fLFkj jgs]xfr'khy xsan d k osx ifjofrZr gkst k;sA 32.*

A block moving in air explodes in two parts then just after explosion (neglect change in momentum duet to gravity) (A*) The total momentum of two parts must be equal to the momentum of the block before explosion. (B) The total kinetic energy of two parts must be equal as that of block before explosion. (C) The total momentum must change (D*) The total kinetic energy must increase

gokes axfr'khy ,d xq Vd knksHkkxks aesafoLQ ksfVr gkst krkgSrksfoLQ ksV d srq jUr ckn ¼xq: Ro d sd kj.kla os x ifjorZ u ux.; gS½ & (A) nks uksHkkxksad k d qy laosx]foLQ ksV d sigy sCy kWd d slaosx d scjkcj gksxk (B) nks uksaHkkxksad h d qy xfrt Å t kZogh jgsxh t ksfoLQ ksV d sigy sxqVd sd h Fkh (C) d q y laosx vko';d : i lsifjofrZr gkst k;sxkA (D) d q y xfrt Å t kZvko';d : i lsc<+t k,sxhA 33.*

Two bodies of same mass collide head on elastically then (A*) Their velocities are interchanged (B*) Their speeds are interchanged (C*) Their momenta are interchanged (D*) The faster body slows down and the slower body speeds up.

leku nzO;eku oky h nksoLrqv ksad h lEeq[k çR;kLFk VDd j esa& (A) bud sos x ijLij ifjofrZr gkst krsgSA (B) bud h pky s aijLij ifjofrZr gkst krh gSA (C) bud sla osx ijLij ifjofrZr gkst krsgSA (D) rhoz xkehoLrq/khehgkst krhgSrFkk/khehoLrqrs t gkst krhgS A

RESONANCE


PART - II : COMPREHENSION Hkkx - II : v uqPN sn (COMPREHENSION) Comprehension # 1 v uqPN sn # 1 A 3kg block ‘A ‘ moving with 4 m/sec on a smooth table collides inelastically and head on with an 8kg block ‘B’ moving with speed 1.5 m/sec towards ‘A ‘. Given e = 1/2 fpd usest ij 4 m/sec osx lspy rk gqv k ,d 3kg nzO;eku d k Cy kWd ‘A ‘ , 1.5 m/sec osx lsfoifjr fn'kk esaA d h vksj xfr'khy 8kg nzO;eku d sCy kWd d slkFk lEeq[k vçR;kLFk VDd j d jrk gSA fn;k gS: e = 1/2 1.

Final velocities of both the blocks

nksuksaCy kWd d k vfUre osx (1*) VA = 2m/s, VB = (3) VA = 2m/s, VB = 2.

3 m/s 4

(2) VA = 1m/s, VB =

1 m/s 4

3 m/s 4

3 m / s , VB = 2m/s 4

(3) VA =

The impulse of reformation and deformation

l: i.krFkkfo: i.kd kvkosx (1) 12Ns, 6 Ns 3.

(2) 4Ns, 12 Ns

(3*) 6Ns, 12 Ns

(4) 12Ns, 4Ns

(3) 30J

(4*) 33J

The maximum potential energy of deformation

fo: i.kd hvf/kd re fLFkfrt Å t kZ (1) 34J 4.

(2) 35J

Find out loss in kinetic energy of system.

fud k; d h xfrt Å t kZesagkfu Kkr d jksA (1) ) Ans.

89 J 4

(1) VA = 2m/s, VB =

RESONANCE

(2*) )

3 m/s , 4

99 J 4

(3)

(2) 6Ns, 12 Ns,

79 J 4 (3) 33J

(4) zero (4)

99 J 4


Sol.

using momentum conservation

la osx laj{k.kls e=

3 × 4 – 8 × 1.5 = 3V1 + 8V2 12 – 12 = 3 V1 + 8V2  3 V1 + 8V2 = 0

....(1)

coffecient of restitution iz R;koLFkku

xq .kka d

V2 – V1 =

1 2

V2 – V1 1 = 4  1 .5 2

5 .5 2

....(2)

....(2)  5 .5



reat in (i) 3 × 1 + 8  2  V1  = 0   3 × 1 + 22 + 8 × 1 = 0 

V2 = –

 V1 = –

22 = – 2 m/sec 11

3 3 3 V1 = – × (2) = m/sec 8 8 4

(2) applying momentum conservation eqn. la osx lja{k.kls m1 V1 + m2V2 = (m1 + m2)V  V = 0 so | PD | = | m1 ( V – u1) | = | m1 u1| = 3 × 4 = 12 Ns | JR | = |e. JD | = 6 Ns (3) P.E =

1 1 1 mu1 2 + m2V2 – (m1 + m2) V2. 2 2 2

=

1 1 × 3 × u2 + × 8 × (1.5)2 – 0 = 335 2 2

2 1 99   3  ( 2)2  1  8   3   (4) K = Ki – Kf = 33 –  2 2 4   = 4 J.   

Comprehension–2 A smooth ball 'A' moving with velocity 'V' collides with another smooth identical ball at rest. After collision both the balls move with same speed with angle between their velocities 60°. No external force acts on the system of balls. ,d fpd uh xsan 'A' pky 'V' l snwl jh fpd uh ,d l eku xsan l sVd jkrh gSt ksfd igy sfojke esaiM+h FkhA VDd j d s ckn nksuksaxsansl eku pky l spy rh gSo vc mud sosxksad se/; d ks.k 60° gSA xsanksad sfud k; ij d ksbZckº; cy d k;Zjr

ughagksrk gSA

5.

The speed of each ball after the collision is

VDd j d sckn izR;sd xsan d h pky & V V (2) 3 2 From conservation of momentum

(1) Sol.

V (3*)

3

2V (4)

3

l aosx l aj{k.k l s mv = mv' cos30° + mv' cos30° v

v

 v' = 2 cos 30  3

RESONANCE


6.

If the kinetic energy lost is fully converted to heat then heat produced is

;fn xfrt Å t kZesagqbZgkfu iw.kZr;k Å "ek esaifjofrZr gksrh gSrksmRiUu Å "ek gksxh & (1) Sol.

1 mV2 3

Loss in kinetic energy =

7.

(2)

1 mv2 2

(3) 0

(4*)

1 mV 2 6

xfrt Å t kZesagkfu 2

–2×

1  v  1   mv 2 m 2  3  6

The value of coefficient of restitution is izR;ku;u xq.kkad (coefficient of restitution) d k eku

gS&

1 1 (3) (4) 0 3 3 Initially B was at rest, therefore line of impact is along final velocity of B. xsan B izkjEHk esafojke esaFkh v r% v kosx fØ ;k js[kk xsan B d sv fUre osx d sv uqfn'k gksxhA

(1) 1 Sol.

2 mV2 3

e=

(2*)

1 v 2 3 v ' v ' cos 60 1   v cos 30 3 3 v 2

RESONANCE


CENTRE OF MASS-JEE(MAIN)

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