# COM,Collisions Master (1)

August 10, 2017 | Author: Raghuram Seshabhattar | Category: Collision, Force, Classical Mechanics, Physical Sciences, Science

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Collisions detailed...

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CONSERVATION OF LINEAR MOMENTUM AND COLLISIONS TOPICS TO BE COVERED Centre of Mass Linear Momentum Conservation of Linear momentum System of Variable Mass Impulse and momentum Collision Types of Collision Line of Impact

CONSERVATION OF LINEAR MOMENTUM CENTRE OF MASS In translational motion each point on a body undergoes the same displacement as any other point as time goes on. In this way the motion of one particle represents the motion of the whole body.

POSITION OF CENTRE OF MASS (a)

System of two particles Consider first a system of two particles m1 and m2 at distances x1 and x2 respectively, from some point origin O. We define a point C, the centre of mass the system, as a distance xcm from the origin O, where xcm is defined by

m1 x1  m2 x2 m1  m2

xcm = ...(1) xcm can be regarded as mass -weighted mean of x1 and x2. (b)

System of many particles (i) If m1, m2 . . . . . , mn, are along a straight line by definition, m1 x1  m 2 x 2  ........  m n x n  m1  m 2  .....  m n

m x m

xcm = where M is total mass of the system.

i

i

m x

i i

i

=

M

...(2)

(ii) If particles do not lie in a straight line but lie in a plane, (suppose x-y plane) the centre of mass C is defined and located by the coordinates xcm and ycm,

where xcm =

ycm =

m x m

m1 x1  m 2 x 2  ......  m n x n  m1  m 2  .....  m n

m1 y1  m 2 y 2 ......  m n y n  m1  m 2  .....  m n

i

m x

i

i i

i

m y m i

i

M

=

m y

i

i i

=

M

...(3)

Y m1 mn

m2

O

x2

x1

X

xn

(iii) If the particles are distributed in space,

m x

m y

i i

i i

M xcm = , ycm = M , zcm = So, position vector of C is given by

r cm = Example-1

 xcmiˆ  ycm ˆj  zcm kˆ =

i i

M

.(4)

m r m r m = M i i

i i

i

A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge of the plate as shown in figure. Find the centre of mass of the remaining portion.

 O2

Solution:

m z

O   O1

A

Let O be the center of circular plate and O1, the center of circular portion removed from the plate. Let O2 be the center of mass of the remaining part. 2

 56    2 Area of original plate =R =  2  = 282 cm2 Area removed from circular part =  r2

...(5)

2

m2 ×

 42  2 2    (21)  cm =  2 

×

m1

×

m Let be the mass per cm2. Then mass of original plate, m = (28)2  mass of the removed part, m1 = (21)2







mass of remaining part, m2 = (28)2  - (21)2 = 343 Now the masses m1 and m2 may be supposed to be concentrated at O1 and O2 respectively. Their combined center of mass is at O. Taking O as origin we have from definition of center of mass,

m1 x1  m 2 x 2 m1  m 2 xcm =

x1 = OO1 = OA - O1A = 28 - 21 = 7 cm x2 = OO2 = ?, xcm = 0

(21)2  7  343  x2 (m1  m2 ) 0= (21)2  7 441 7    9cm 343 343 x = 2

(c)

This means that center of mass of the remaining plate is at a distance 9 cm from the center of given circular plate opposite to the removed portion. Continuous bodies: Lt

xcm =

mi  0

 m x  m

Similarly r cm 

  r dm

 dm

1 M

i i i

 x dm  1 x dm   dm M

 ydm  1 y dm  zdm  1 z dm  M M dm dm ycm =  and zcm = 

 rdm . . . . . (6)

Example-2

The density of a thin rod of length l varies with the distance x from one end as

Solution:

x2 ρ  ρ0 2 l . Find the position of centre of mass of rod. l l  x2   0 2   x ( s  d x )  (dm ) x 0  l  0 x cm  l  l  x2   (dm ) (s  dx ) 0 2   0 0  l  Here, s = area of cross section of rod.

Therefore,

x cm 

3l 4

DISTINCTION B/W CENTER OF MASS AND CENTER OF GRAVITY The position of the center of mass of a system depends only upon the mass and position of each constituent particles,

r CM 

m r m i

i

i

i.e.,

...(1)

The location of G, center of gravity of the system, depends however upon the moment of the gravitational force acting on each particle in the system (about any point, the sum of the moments for all the constituent particles is equal to the moment for the whole system concentrated at G). Hence, if gi is the acceleration vector due to gravity of a particle P, the position vector rG of the center of gravity of the system is given by

rG   mi gi   (ri  mi gi )

...(2)

It is only when the system is in a uniform gravitational field, where the acceleration due to gravity (g) is the same for all particles, that equation (2)

rG  Becomes

m r m

i i

 rCM

i

In this case, therefore the center of gravity and the center of mass coincide. If, however the gravitational field is not uniform and gi is not constant then, in general equation

r (2) cannot be simplified and G

 rCM

.

MOTION OF THE CENTRE OF MASS Assume that the total mass M of the system remains constant with time, then, for our fixed system of particles

M r cm  m1 r 1  m2 r 2  ........  mn r n , where rcm is the position vector identifying the centre of mass in a particular reference frame.

Differentiating this equation W.R.T. time we get

d r cm d r1 dr 2 dr n  m1  m2  ......  m n dt dt dt M dt

or

M

...(7)

m1 v 1  m2 v 2  .....  mn v n

 Vcm =

where v1 is the velocity of the first particle, etc., and drcm /dt (= vcm) is the velocity of the centre of mass. Differentiating equation (7) with respect to time we obtain

d vcm d v1 d v2 d vn  m1  m2  ........  mn dt dt dt M dt

=

m1 a1  m2 a 2  ..........  mn a n

...(8)

,

where a1 is the acceleration of the first particle, etc., and dvcm/dt (= acm) is the acceleration of the centre of mass of system. Now, from Newton's second law, the force F1 acting on the first particle is given by F1 = m1a1. Likewise, F2 = m2a2, etc. We can then write equation (8) as M

 a cm

=

F1  F2  .......  Fn  Finternal  Fexternal

...(9)

Internal forces are forces exerted by the particles on each other. However, from Newton's third law, these internal forces will occur in equal and opposite pairs, so they contribute nothing to the sum.

M acm  F ext This states that the centre of mass of a system of particles moves as through all the mass of the system were concentrated at the centre of mass and all the external forces were applied at that point.

Concept:

Whatever may be the rearrangement of the bodies in a system, due to internal forces (such as one part moving away from the other or an internal explosion taking place, breaking a body into pieces).

(a) If the body was originally at rest, the C.M. will continue to be at rest. (b) If before the change, the body had been moving with a constant velocity, it will continue to move with a constant velocity and In presence of external force if body had been moving with constant acceleration in a particular trajectory, the C.M. will continue to move in the same trajectory, with the same

acceleration as if it had never experienced any explosion only if there is no change in external force. Example-3

In the arrangement shown in figure, mA = 2 kg and mB = 1kg. String is light and inextensible. Find the acceleration of centre of mass of both the

A B

blocks. Neglect friction everywhere. Solution :

Net pulling force on the system is

m

A

 mB g

2  1g  g

or

a A Total mass being pulled is

a

Now,

mA  mB

or 3 kg

B

a

Net pulling force g  T otalmass 3

   mA a A  mBa B 2a   1a  a a COM    mA  mB 1 2 3

g 9 downwards

LINEAR MOMENTUM The momentum of a single particle is a vector v defined as the product of its mass and its velocity v . That is

P =m

 v

From Newton's second law of motion

...(10)

 dv d dP F  ma  m  (mv)  dt dt dt Thus, if m is constant, the rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the direction of that force.

Suppose that instead of a single particle we have a system of n particles with masses m1, m2 . . . etc, and their velocities v1, v2 etc respectively then the total momentum

P= P

+ P

1

2

+..... + P

P

in a particular reference frame is,

n

    m1v1  m2 v 2  ...... mn v n

P

 M V cm =

Also,

   dP dVcm M  Macm  Fext dt dt

\

 Fext

dP = dt

CONSERVATION OF LINEAR MOMENTUM If the sum of the external forces acting on a system is 0. Then,

 Fext or

dP = dt

P

=0

= constant.

So, when the resultant external force acting on the system is zero, total vector momentum of system remains constant. This is called as principle of the conservation of linear momentum. The momentum of the individual particles may change but their sum remain constant if there is no net external force.

Example-4

A man of mass m climbs a rope of length L suspended below a balloon of mass M. balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed v (relative to rope) in upward direction and with what speed (relative to ground)

balloon

M

vb man Solution:

m

vm

with the balloon move? Balloon is stationary No net external force acts on it. Conservation of linear momentum of a system (balloon + man) is valid M vb

 mv m  0

v  m[v

, where

vm

=

v mb  v b

v ]0

mb b M b where vmb = velocity of man relative to the balloon (rope)

vb  

mv mb M m

mv where vmb = v vb = M  m and directed opposite to that of the motion of the man. Example-5

Two identical buggies move one after the other due to inertia (without friction) with the same velocity v0. A man of mass m rides the rear buggy. At a certain moment man jumps into the front buggy with a velocity u relative to his buggy. The mass of each buggy is M. Find the velocities with which the buggies will move lster.

After jumping

v1

M

M

v2

Before jumping

v0

M Solution:

M

v0

Initial momentum of rear buggy = (M + m)v0. The momentum of man when he jumps = m(v1 + u), where v1 is the velocity of buggy as he jumps. By conservation of linear momentum (M + m)v0 = Mv1 + m(v1 + u) v1(M + m) = (M + m)v0 - mu m u v =v - M m 1

0

Initial momentum of front buggy = Mv0 Mv0 + m(v1 + u) = (M + m)v2

mu   u  u   (M  m)v2  v0  M  m  Mv0 + m  Mu    v0    ( M  m)v2 M m Mv + m  0

mMu  ( M  m)v2 (M + m)v0 + M  m

mMu 2 v2 = v0 + M  m 

SYSTEM OF VARIABLE MASS Till now we have studied system with constant mass. Now, here, we will study about the system which gains or loses mass during its motion, e.g. in case of Rocket propulsion, its motion depends upon the constant ejection of fuel from it. Let us choose system of mass M (as shown in figure 1) whose centre of mass is moving with velocity as seen from a particular reference frame. An external force ext acts on the system.

y

t

y

M v cm x

O

t+t M-M M u v+v cm cm

O

(1)

x

(2)

At a time  t later the configuration has changed to that shown in figure 2. A mass DM has been ejected from the system, its centre of mass moving with velocity as seen by our observer. The mass of the body is reduced to M - DM and the velocity of the centre of mass of the system is changed to v +  v.

Fext = dP/dt P t

Fext 

{ for finite time interval  t}

\

Pf  Pi =

t

[ M  M )(v  v )  Mu ]  [ Mv ] t v M M  [u  (v  v )] t t =

Fext 

as Dt 0; M t

dv v t approaches dt dM - dt

approaches and is negligible as compared to v.

dv dM  Fext  (u  v ) dt dt \ (as M is decreasing with time) dv dM M  Fext  urel F  Fthrust dt dt = ext M

where

urel

Example-6

, is the relative velocity of the ejected mass with respect to main body.

A rocket of initial mass m0 (including shell and fuel) is fired vertically at time t = 0. The fuel is consumed at a constant rate q = dm/dt and is expelled at a constant speed u relative to the rocket. Derive an expression for the magnitude of the velocity of the rocket at time t,

neglecting the resistance of the air and variation of acceleration due to

v

Solution:

the gravity(g). At time t, the mass of the rocket shell and remaining fuel is m = m0 - qt, and the velocity is v. During the time interval  t, a mass of fuel  m = q  t is expelled with a speed u relative to the rocket. Denoting by v the e absolute velocity of expelled fuel, we apply principle of impulse and we write

(m0 – qt)v v

+

Wt

(m0 – qt - qt)(v + v)

=

mve

[Wt = g(m0 – qt)t]

[mve = qt(u – v)] (m0 - qt)v - g(m0 - qt)  t = (m0 - qt - q  t)(v +  v) - q  t(u - v) Dividing throughout by t and letting t approach zero we obtain

dv  qu -g(m0 - qt) = (m0 - qt) dt Separating variables and integrating from t = 0, v = 0 to t = t, v = v

 du    g dt m  qt  dv =  0 v = [u ln(m0

t gt] 0 - qt) -

 m0     gt m  qt  v = u ln  0

IMPULSE AND MOMENTUM We know that force is related to

F

dP dt

momentum as

Fdt  d P

We can find the change in momentum of the body during a collision (fromi P to f P ) by integrating over the time of collision and assuming that the force during collision has a constant direction,

; in which the subscripts i (= initial) and f(= final) refer to the times before and after the collision. The integral of a force over the time interval during which the forces acts is called the impulse of the force. The impulse of this force, , is represented in magnitude by area under the force time curve.

COLLISIONS In a collision a relatively large force acts on each colliding particle for a relatively short time.

CONSERVATION OF MONMENTUM DURING COLLISIONS Consider now a collision between two particles, such as those of masses m1 and m2, shown in figure. During the brief collision these particles exert large forces on one another. At any instant F1 is the force exerted on particle 1 by particle 2 and F2 is the force exerted on particle 2 by particle 1. By Newton's third law these forces at any instant are equal in magnitude but oppositely directed.

m1 m2 F1

F2

1

1

The change in momentum of particle 1 results in from the collision is

tf

p1

 F dt  F t 1

1

=

ti

in which F 1 is the average value of the force F1 during the time interval of the collision Dt = tf - ti. The change in momentum of particles 2 resulting from the collision is tf

p 2

 F dt  F t 2

2

ti

= in which is the average value of the force F2 during the time interval of the collision Dt = tf - ti. If no other forces act on the particles, then Dp1 and Dp2 gives the total change in momentum for each particle. But we have found that at each instant

F1

= - F 2 , and therefore

p1  p2 If we consider the two particles as an isolated system, total momentum of system is

P  p1  p2 , And the total change in momentum of the system as a result of the collision is zero that is,

P  p1  p2  0

p 1  p2 

constant Hence, if there are no external forces acting on the system, the total momentum of the system is not changed by the collision. The impulsive forces acting during the collision are internal forces which have no effect on the total momentum of the system.

TYPES OF COLLISION: Collision b/w two bodies may be classified in two ways: 1.

Elastic collision and inelastic collisionn.

Collision is said to be elastic if both the bodies come to their original shape and size after the collision, i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies, otherwise, it is called an inelastic collisionn. Thus in addition to the linear momentum, kinetic energy also remains conserved before and after collision. 2.

Head on collision or oblique collision.

If the directions of the velocity of colliding objects are along the line of action of the impulses, acting at instant of collision then it is called as head-on or direct collision. Otherwise impact is said to be oblique or indirect or eccentric.

NEWTON’S LAW OF RESTITUTION: Experimental evidence suggests that the ratio of relative speed of separation to relative speed of approach is constant for two given set of objects.

Re lative speed of separation e Re lative speed of approach The ratio e is called the coefficient of restitution and is constant for two particular objects. In general, 0£e£1 e = 0, for completely inelastic collision, as both objects stick together. e = 1, for an elastic collisionn.

HEAD-ON COLLISION Consider two spheres A and B of mass m1 and m2, which are moving in the same straight line and to the right with known velocities v1 and v2 as shown in figure. If v2 is larger than v1, particle B will eventually strike the sphere A. Under the impact, the two spheres will deform and at the end of the period of deformation, they will have the same velocity u as shown in figure. A period of restitution will then place, at the end of which, depending upon the magnitude of the impact forces and upon the materials involved, the two spheres either will have regained their original shape or will stay permanently deformed. The purpose here is to determine the velocities v’1 and v’2 of the spheres at the end of the period of restitution as shown in figure. (i)

For elastic collision Considering first the two spheres as a single system, we note that there is no impulsive, external force. Applying law of conservation of linear momentum (COLM). m1v1 + m2v2 = m1 v1 + m2 v2

m2

m1

v2

B

v1

A

...(i)

m2

m1

v2

B

(1)

v1

A (2)

In an elastic collision kinetic energy before and after collision is also gets conserved. Hence,

1 1 1 1 m1v12  m2 v22  m1v'12  m2 v' 22 2 2 2 2 Solving eqs. (i) and (ii) for

v'1 and

v' 2

v'1 and v'2 , we get

 m1  m2   2m 2   v1   v2 m  m m  m 2  2   1 = 1  m2  m1   2m1   v2   v1  m1  m2   m1  m2 

=

SPECIAL CASES : 1.

...(ii)

If m1 = m2, then from eqs. (iii) and (iv), we can see that v1¢ = v2 and v2¢ = v1

...(iii)

...(iv)

i.e., when two particles of equal mass collide elastically and the collision is head on, then they exchange their velocities e.g. 2.

If m1 >> m2 and v1 = 0.

m2 0 m1

3.

Then With these two substitutions We get the following two results: v1¢ » 0 and v2¢ » -v2 i.e., the particle of mass m1 remains while the particle of mass m2 bounces back with same speed v2. If m2 >> m1 and v1 = 0 With the substitution and v1 = 0, we get the results v1¢ » 2v2 and v2¢ » v2 i.e., the mass m1 moves with velocity 2v2 while the velocity of mass m2 remains unchanged.

(ii)

For Inelastic Collision The kinetic energy of particles no longer remains conserved. Suppose the velocities of two particles of mass m1 and m2 before collision are v1 and v2 in the directions shown in figure. Let v1¢ and v2¢ be the velocities after collision. Applying the law of COLM. m1v1 + m2v2 = m1v¢1 + m2v¢2 ...(i) Applying Newton's Law of Restitution, separation speed = e(approach speed) v¢1 - v¢2 = e(v2 - v1) ...(ii) Solving eqs. (i) and (ii), we get

and

 m1  em2   m  em2   v1   2 v 2 m1  m2  m1  m2    v1 1 =  m1  em2   m  em2   v1   2 v2  m1  m2   m1  m2  v21 =

...(iii)

...(iv)

SPECIAL CASES: 1.

If collision is elastic, i.e. e = 1, then

 m1  m2   2m 2   v1   v 2 m1  m2  m1  m2    v1’ = and which are same as eqs. (iii) and (iv). 2.

If collision is perfectly inelastic, i.e., e = 0, then

m1v1  m2 v2 m1  m2 = v, (say) V’1 = v2’ =

 m2  m1   2m1   v2   v1 m1  m2  m1  m2    v2’ =

3.

If m1 = m2 and v1 = 0, then

1 e  1 e  v1 '   v2 and v2 '   v2  2   2 

LINE OF IMPACT It is important to know the line of impact during the collision. line of impact is line along which the impulsive force act on the bodies. To find it draw tangent at point of contact of two bodies. Draw a normal to tangent at the point. This normal line is known to be line t v2 v1 A

Line of impact

B

n

v2

v1

of impact.

OBLIQUE COLLISION : Let us now consider the case when the velocities of the two colliding spheres are not directed along the line of impact as shown in figure. As already discussed the impact is said to be oblique. Since velocities v¢1 and v¢2 of the particles after impact are unknown in direction and magnitude, their determinaion will require the use of four independent equations. We choose as coordinate axes the n-axis along the line of impact, i.e. along the common normal to the surfaces in contact, and the t-axis along their common tangent. Assuming that the sphere are perfectly smooth and frictionless, we observe that the only impulses exerted on the sphere during the impact are due to internal forces directed along the line of impact i.e. along the n axis. It follows that t

t

t m2v2

m1v1 A

A

B n

m1v1

m2v2

+

B n

=

A

B n

(i)

(ii)

(iii)

The component along the t axis of the momentum of each particle, considered separately, is conserved; hence the t component of the velocity of each particle remains unchanged throughout. We can write. (v1)t = (v¢1)t; (v2)t = (v¢2)t The component along the n axis of the total momentum of the two particles gets conserved. We write. m1 (v1)n + m2(v2)n = m1(v¢1)n + m2(v¢2)n The component along the n axis of the relative velocity of the two particles after impact is obtained by multiplying the n component of their relative velocity before impact by the coefficient of restitution. (v¢2)n - (v¢1)n = e[(v1)n - (v2)n] We have thus obtained four independent equations, which can be solved for the components of the velocities of A and B after impact.

Note: Definition of coefficient of restitution can be applied along common normal direction in the case of oblique collisions .

Example-7

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

q

q v

Solution:

v

Suppose the angle of reflection is

q and the speed after

collision is v . It is an oblique impact. Resolving the velocity v along the normal and tangent, the components

θ

are v cos θ and v sin . Similarly, resolving the velocity after reflection along the normal and along the tangent the components are - v cos q and Since there is no tangential action,

v sin q

.

v sin q = v sin q Applying Newton's law for collision,

(- v cos q - 0) = -e(v cos q - 0) v cos q = ev cos q From equations (i) and (ii),

...(i)

...(ii)

2

= v2 sin2 q + e2 v2cos2

q

= and tan q =

 tan q    e  -1 

= tan Example-8

Solution:

.

A ball of mass m, moving with a velocity v along X-axis, strike another ball of mass 2m kept at rest. The first ball comes to rest after collision and the other breaks into two equal pieces. One of the pieces starts moving along Y-axis with a speed v1. What will be velocity of the other piece ? The total linear momentum of the balls before the collison is mv along the X-axis. After the collision, momentum of the first ball = 0, momentum of the first piece = mv1 along gthe Y-axis and momentum of the second piece = mv2 along its direction of motiion where v2 is the speed of the second piece. These three should add to mv along the x-axis, which is the intiial momentum of the system. v1 Y Y

v

v 2m

m

X

m

q

v2 Taking components along the X-axis,

mv2 cos q  mv

.....(1)

& taking components along Y-axis

mv2 sin q  mv1 From (1) and (2),

tan q  v1 / v v2 

v  v12  v 2 cos q

.....(2)

X

OBJECTIVE PROBLEMS (SOLVED) 1.

Solution:

A small cube of mass m slides down a circular path of radius R cut into a larger block of mass M, as shown in figure. M rests on table, and both blocks move without friction. blocks are initially at rest, & m starts from top of path. velocity v of cube as it leaves the block is

(a)

2mgR M

(c)

2mgR mM

(b)

(d)

2gR

2MgR mM

From COLM,

Mvm  mvm From COE

mgR 

1 1 MVM2  mv2m 2 2

 wmgR   Vm 

M

m2 2 Vm  mVm2 2 m

2MgR Mm

. Ans. (d)

2.

A block of metal weighing 2 kg is resting on a frictionless plane. It gets struck by jet releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block is

Block 2kg

(a)

5 3 m/s2

(c)

25 8

25 (b) 4 m/s2

m/s2

V

(d)

5 2

m/s2

m f dm  5 Kg 2  a   2.5m / s 2 s m dt

Solution:

Force Ans. (d)

3.

A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. fraction of incident energy transferred to the heavier ball is

(a) (c)

Solution:

n 1 n

n 2 (b) (1  n)

2n (1  n) 2

4n (1  n) 2 . (d)

Let the velocities of the two balls, after collision, be v1 and v2 in the direction of u - which we take as positive momentum conservation mu + 0 = m v1 + nm v2 e = 1 v2 – v1 = u Solving these equations, we have

v2 

2u n 1

fraction of energy transferred 

KE of ball of mass nm Incident KE 2

1  2u  (nm)   2  n 1  4n 1  mu 2 (n  1) 2 2 Ans. (d) 4.

An insect of mass m is initially at one end of the stick of length L and mass M, which rests on a smooth horizontal floor. coefficient of friction b/w the insect and stick is k. The minimum time in which the insect can reach the other end of the stick is t.

 kmg   t (a) the centre of mass of the plank has velocity magnitudes  M  with respect to horizontal floor at time t.

t

2LM k(M  m)g

(b) (c) the magnitude of the linear momentum of the insect at time ‘t’ is (kmgt) w.r.t. horizontal floor (d) all the above Solution:

m kg Acceleration of insect and the stick will be kg and M respectively, but in opposite direction. Therefore, relative acceleration  m  kg 1    M 

5.

2 M 1  m 2 kg 1   t  t  kg(M  m) 2  M

Ans. (b) A strip of wood of mass M and length l is placed on a smooth horizontal surface. An insect of mass m starts at one end of the strip and walks to the other end in time t, moving with a constant speed

 (a) the speed of the insect as seen from the ground is

t

l Mm   (b) the speed of the insect as seen from the ground is t  M  m 

 m    (c) the speed of the insect as seen from the ground is t  M  m  2 1 l m  M   2 t (d) the total kinetic energy of the system is Solution:

As the insect moves from one end to the other, the strip moves in the opposite direction. The insect, therefore, gets to cover a distance less than and its speed <

t Ans. (a)

 The distance covered by the insect w.r.t ground

M t Mm

its speed Ans. (b) is, wrong. Ans. (c) is, therefore wrong.

M Mm

2

1  M  1 m   m   M  The total KE (seen from ground frame) 2  t M  m  2  t M  m  1   mM(M  m) 1       (M  m)   2 t . 2  t  (M  m) Ans. (d) is wrong. 2

6.

2

2

A ball strikes on the ground at an angle of 30º from the

vertical and rebound at an angle of from the vertical as shown in the figure. If the coefficient of restitution between ball and ground is 1/ 3 , then is

 30º r

e

1 3

(a) 30º (b) 45º (c) 60º (d) 15º Solution: Velocity along the plane before collision is equal to after collision v sin 30° Velocity perpendicular to plane after collision is equal to e v cos 30° vsin 30 1 tan    ev cos 30 e. 3 = 1 Therefore,

  45

Ans. (b) 7.

A ball of mass 1 kg is dropped on a horizontal ground from a height of 128 meter. If coefficient of restitution between ball and ground is 1/2, then height gained by ball just after 3rd impact from the ground is (a) 64 meter (b) 32 meter (c) 16 meter (d) 2 meter v v  Solution: eV 1 / 2 V Veloicty of ball just before collision is

2gh

e 2gh

Velocity of ball just after collision is Height attained by ball after Ist collision is [at maximum height veloicty is zero]

h  e 2 h ; 2 2  Therefore, after 2nd collision h  (e ) h

Therefore, after 3rd collision Ans. (d) 8.

h  (e2 )3 h

The linear momentum P of a particle varies with time as P = a + bt2, where a and b are constants. The net force acting on the particle is (a) zero (b) constant (c) proportional to t (d) proportional to t2

F Solution:

dp  2bt dt

Ans. (c) 9.

A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact.If half the kinetic energy is lost by impact, what is the value of the coefficient of restitution ? 1 (a) 2 2 (b) 1

1

3 2

2 (c) (d) Solution: Balls exchange there velocities in the case of a perfectly elastic collisionn. Therefore, e = 1. Ans. (b) 10.

A shell is fired from a cannon with a velocity v (m/s) at an q angle with the horizontal direction. At highest point in its trajectory it explodes into two pieces of equal mas. One of the pieces retraces its path to the cannon and the speed (m/s) of the other piece immediately after the explosion is 3 v cos q (a) (b) 2 v cos q 3 v cos q (c) 2

Solution: Let

v

(d) be the velocity of second fragment. From conservation of linear momentum

2mv cosq  mv  mv cos q Therefore,

Ans(a).

3 v cos q 2

v  3v cos q

SUBJECTIVE PROBLEMS (SOLVED) 1. Solution:

Locate the centre of mass of a uniform semicircular rod of radius R and linear density λ kg/m. From the symmetry of the body we see that the CM must lie along the y axis, so xCM = 0. In this case it is convenient to express the mass element in terms of the angle θ , measured in radians. The element, which subtends an angle d at the origin, has a length R d

λ d. Its y coordinate is y = R sin

mass dm = R

Therefore, yCM =

θ

θ

and a

.

ydm M

yCM =

1 R 2 2R 2 2   R sin q d q  [  cos q ]  0 M 0 M M

The total mass of the ring is M = 2. Solution:

π R λ ; Therefore, y

CM

2R =

 .

Find the centre of mass of a uniform solid hemisphere of radius R and mass M with centre of sphere at origin and the flat of the hemisphere in the x, y plane. Let the center of the sphere be the origin and let the flat of the hemisphere lie in the x, y plane as shown. By symmetry x  y  0. Consider the hemisphere divided into a series of slices parallel to x, y plane. Each slice is of thickness dz The slice between z and (z + dz) is a disk of radius, r =. R  z Let r be the constant density of the uniform sphere. 2

2

Mass of the slice, dm =

  r 2  dz    R 2  z2  dz

z

r

R

y

x R

 z dm 0

The

z

value is obtained by

z=

M

R

  ( R

2

z  z 3 )dz

0

M

=

  R 2 z 2

 M  2 =

zR

z 4    4  z 0

 R4 R4      2 4   z M

R 4 z 4M

2  V    R 3  3  Since M = z

R 4 3  R 2  8 4  R 3  3 

3    0, 0, R  8  . Hence center of mass has positive coordinates as  3.

The balloon, the light rope and the monkey shown in figure are at rest in the air. If monkey reaches top of rope, by what distance does balloon descend ? Mass of the balloon – M, mass of the monkey = m and the length of the rope M L

m

ascended by the monkey = L. Solution:

Mx = m(L - x) x=

mL M+ m

4.

A bullet of mass m strikes a block of mass M connected to a light spring of stiffness k, with a speed v0 and gets embedded into mass M. Find the loss of K.E. of the system

just after impact.

Solution:

The process of impact of bullet and block is transient. Within a very short time of impact, the compression of the spring is negligible. Therefore the corresponding spring force is negligible. Even though it is external to the system (M + m), we can conserve its momentum just before and after the impact (impact force is internal). Conservation of linear momentum of bullet plus block just after and before impact yields mV0

(M + m)V = mV0

V = M m

where V = common velocity of block & a bullet. Therefore the loss of K. E. of the system

1 1 mV02  ( M  m)V 2 2  KE = 2 mV0 Putting V = m  M we obtain,

5.

Solution:

MmV02 KE = . 2( M  m)

A bullet of mass 10–3 kg strikes an obstacle and moves at angle 60° to its original direction. If its speed also changes from 20 m/s to 10m/s during collision. Find the magnitude of impulse acting on the bullet. m = 10–3 kg Consider components of impulse along initial velocity J1 = 10–3[–10cos 60° – (–20)] J1 = 15 10–3 N.S Similarly impulse perpendicular to initial velocity we have

J2 = 10-3[10 sin60 - 0] =

5 3  103 N .S

The magnitude of resultant impulse is given by

6.

J=

J12  J 22  103 (15)2  (5 3)2

J=

3 102 N .S.

A particle of mass M is attached with a string of length l and is released from the position as shown in the figure. When string becomes vertical particle strikes with the block of mass M as shown in the figure. Calculate maximum compression in the spring if collision between particle and block is perfectly elastic

M

l

k

M

M

smooth Solution:

V=

2gl

Velocity of left block just after collison When compression is maximum then velocity of block will be same

MV  2MV 

V 2

Using conservation of energy

1 2 1 1 kx  MV2  2MV2 2 2 2

=

M MV 2 2gl 2 4

1 2 M2gl Mgl kx = Mgl = 2 2× 2 2 x=

7.

Mgl k

A system of two blocks A and B are connected by an inextensible massless string as shown. The pulley is massless and frictionless.

A bullet of mass 'm' moving with a velocity 'u' as shown hits the

u m m B A

Solution:

3m block 'B' and gets embedded into it. Find the impulse imparted by tension force to the system. Let velocity of B and A after collision has magnitude v. At the time of collision, tension = T Impulse provided by tension =

 Tdt

Consider change of momentum of (B + bullet) mu Consider change of momentum A

 Tdt

 Tdt = 3mv

= 2mv

...(i)

...(ii)

from (i) and (ii), mu = 5mv v = u/5

 u  3mu   T dt  5 . Hence, Impulse = 3mv = 3m  5  8.

Solution:

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 parts moves at 30 m/s, with what speed does second part move and what is the fractional change in the kinetic energy. There is no external force on the block. Internal forces break block in two parts. The linear momentum of the block before the on block. Internal forces break block in two parts. of the two parts after the break. As all the velocities are in same direction

M  20 m / s  

M 30 m / s   M v 2 2

where v is the speed of theother part. From this equation v = 10 m/s. The change in kinetic energy is 1M 30 m / s 2  1 M 10 m / s 2  1 M20 m / s 2 2 2 2 2 2

M m2  m2   450  0  400 2   50 2 M 2 s  s   m2  M 50 2   s  1  2 Hence, the fractional change in the kinetic energy 1 / 2M20 m / s  4

9.

A cylindrical solid of mass 10–2 kg and cross-sectional area 10–4 m2 is moving parallel to its axis (the X-axis) with a uniform speed of 103 m/s in the positive direction. At t = 0, its front face passes the plane x = 0. The region to the right of this plane is filled with stationary dust particles of uniform density 10–3 kg/m3. When a dust particle collides with the face of cylinder, it sticks to THE surface. Assuming that the dimensions of the cylinder remains practically unchanged, and that dust sticks only to front face of the cylinder, find x-coordinate of front of the cylinder at t = 150 s.

Solution:

Given

m0  102 kg, , A  104 m2

m  m0 + mass of dust collected so far

,

v0  103 m and dust    103 kg / m3

m0 Axdust

m  m0  Ax x A

v0

m0

v

m

x

x=0

x=0

At t = 0

At t = t

The linear momentum at t = 0 is

P0  m0 v0

x

P  mv  (mv  Ax)v

t 0 and momentum at t = t is According to law of conservation of linear momentum

P0  Pt Therfore,

m0 v0  ( m0  Ax)v or m0 v0  ( m0  Ax)

m or

0

dx dt

 Axdx  m0 v0dt or  ( m0  Ax) dx  m0 v0  dt x

150

0

0

x

or

 x2  150  m0 x  A   m0 v0 t 0 2 0 

Hence,

x2 m0 x  A  150m0 v0 2

Solving this quadratic equation and substituting the values of positive value of x as 105 m. Therefore,

10.

x  105 m

A freight car is moving on smooth horizontal track without any external force. Rain is falling with a velocity u m/s at an angle with the vertical. Rain drops are collected in the car at the rate of m kg/s. If initial mass of the car is m0 and velocity v0 then

m0 , A,  and v0 , we get

find its velocity after time(t)

usinq q ucosq

u

v Solution:

After time t mass of the car with water is mt = (m0 + mt)kg. Let at that momentum speed of the car be v.

mt

dv dm  Fext  vrel dt dt

  dv    (m0 + mt)  dt  = 0 + (u sin v

dv

t

 (u sin q  v)    m

v0

0

q

dt

  0  t    

+ v)m

  m0   t       u sin q  v       ln    m0 /    u sin q  v0      ln

m0 u sin q  v  u sin q  v0 (m0  t )

 m0  m0 v0  t  m0 v0   1  v  u sin q   m  t  (m0  t )  m0  t  (m0  t) v = u sin θ  0

.

JEE MAIN MODEL SECTION - I Level – I 1.

A dog weighing 5 kg is standing on a flat boat so that it is 10 m from the shore. The dog walks 4 m on the boat towards the shore and then halts. boat weighs 20 kg & one can assume that there is no friction between it and the water. How far is the dog from shore at the end of this time? (a) 3.2 m (b) 0.8 m (c) 10 m (d) 6.8 m

2.

A system consists of two identicals particles. One particle is at rest and the other particle has an acceleration ‘a’. The centre of mass of the system has an acceleration of (a) 2a (b) a

a (c) 2 3.

(d)

a 4

ˆ ˆ ˆ Two bodies of mass 10 kg and 2 kg are moving with velocities 2i  7 j  3k and

 10iˆ  35ˆj  3kˆ m/s respectively. The velocity of their centre of mass is

4.

ˆ (a) 2i m/s (b) 2kˆ m/s ˆ ˆ ˆ ˆ ˆ (c) (2 j  2k ) m/s (d) (2i  2 j  2k ) m/s A rope thrown over a pulley has a ladder with a man of mass m on one of its ends and a counterbalancing mass M on its other end. The man climbs with a velocity vr, relative to the ladder. Ignoring the masses of pulley and the rope as well as the friction on the pulley axis, velocity of centre of mass of this system is m vr (a) M

m vr (b) 2M

M vr (c) m

2M vr (d) m

5. Two balls are dropped from same height h on a smooth plane and the other on a rough plane having same inclination q with horizontal. Both the planes have same coefficient of restitution. If range and time of flight of first and second balls are R1,

h

h R1 T1 & R2, T2 respectively, then

T1 q

Smooth plane

(a) T1 = T2, R1 = R2 (b) T1 < T2 , R1 < R2 (c) T1 = T2, R1 > R2 (d) T1 > T2, R1 > R2

R2

T2 q

Rough plane

6.

A projectile is fired with initial momentum p at an angle 45° from a point P as shown in figure. Neglecting air resistance, the magnitude of change in momentum between

45°

leaving P and arriving at Q is P

7.

Q

(a) p/2 (b) 2p (c) p (d) 2p Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of an equilateral triangle of side b as shown in the figure. The co-ordinates of the centre of mass are y

3kg

2kg

1kg

x

 7b 3 3 b  0, ,  12 12   (a)

 3 3 b 7b  , ,0  12 12   (b)

 7b 3 3 b  ,0  , 12 12   (c)

 7b 3 3 b   ,0,  12 12  (d) 

8.

Two particles A and B, initially at rest move towards themselves under a mutual force of attraction. At instant when speed of A is v and the speed of B is 2v, the speed of centre of mass is (a) Zero (b) v (c) 1.5 v (d) 3v

9.

A shell is fired from a gun with a muzzle velocity u m/sec at an angle q with the horizontal. At top of the trajectory shell explodes into two fragments P and Q of equal mass. If speed of fragment P immediately after explosion becomes zero, where does the centre of mass of fragments hit the ground ?

(a)

u 2sin 2q g

u 2sin 2 2q 2g (c)

u 2sin 2q g (b) u sin q g (d)

10.

In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance 1.12 × 10–10 m. The distance of the centre of mass from the carbon atom is (a) 0.48 × 10–10 m (b) 0.51 × 10–10 m (c) 0.56 × 10–10 m (d) 0.64 × 10–10 m

11.

Four particles of masses m1 = 2m, m2 = 4m, m3 = m and m4 are placed at four corners of a square. What should be the value of m4 so that the centres of mass of all the four particles are exactly at the

m4

m3

centre of the square ? m1 (a) 2 m (c) 6 m

m2 (b) 8 m (d) Can never be at the centre of the square

12.

If the KE of a body becomes four times of its initial value, then the new momentum will be : (a) three times its initial value (b) four times its initial value (c) twice its initial value (d) unchanged.

13.

A particle of mass M is moving in a horizontal circle of radius R with uniform speed V. When it travels from one point to a diametrically opposite point its (a) Momentum does not change (b) Momentum changes by 2MV (c) KE changes by MV2 (d) KE changes by (1/4)MV2 A surface is hit elastically and normally by n balls per unit time, all the balls having the same mass m and moving with the same velocity u. force on the surface is (a) mnu2 (b) 2mnu (c) (1/2)mnu2 (d) 2mnu2 A lead ball strikes a wall and falls down. A tennis ball having the same mass and same velocity strikes the same wall and bounces back. Which is the correct statement ? (a) tennis ball suffers a greater change in momentum (b) lead ball suffers greater change in momentum (c) Both balls suffers equal change in momentum (d) momentum of lead ball is greater than that of tennis ball K.E. of a body of mass m & momentum p is given by (a) mp (b) (p2/2m) (c) p2m (d) (m2/2p) Two bodies of masses mA and mB have equal K.E. The ratio of their momenta is

14.

15.

16.

17.

(a) mA : mB

(b) mB : mA

mA2 : mB2

mA : mB

18.

19.

20.

(c) (d) A bomb of mass 9 kg explodes into two pieces of masses 3 kg & 6 kg. velocity of mass 3 kg is 16 m/s. The K.E. of mass 6 kg in joule is (a) 96 (b) 384 (c) 192 (d) 768 A body of mass 1 kg initially at rest, explodes & breaks into three fragments of masses in the ratio 1 : 1 : 3. two pieces of equal mass fly off perpendicular to each other with a speed of 15 m/s each. The speed of the heavier fragment is (a) 5 2 m/s (b) 45 m/s (c) 5 m/s (d) 15 m/s A cart of mass M is tied to one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M, the entire system is on a smooth horizontal surface. The man is at x = 0 and the cart at x = 10 m. If the man pulls the cart by a rope, the man and the cart will meet at the point (a) x = 0 (b) x = 5 m (c) x = 10 m (d) They will never meet

Level – II 1.

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 ms-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is : (a) 30 ms-1 (b) 20 ms-1 (c) 10 ms-1 (d) 5 ms-1

2.

A particle of mass m is projected from the ground at an angle momentum when it is at highest point of its track. (a) mu sin

3.

4.

θ

θ

(b) 2 mu sin

from vertical. The change in

θ

(c) mu cos θ (d) 2 mu cos θ . The position of centre of mass of a system consisting of two particles of masses m1 and m2 separated by a distance L apart from m1 (a)

m2L m1  m 2

(c)

m2L m1  m 2

(b)

m1 L m1  m 2

(d)

m1 L m1  m 2

A block of mass moving with speed collides with another block of mass 5m at rest. The lighter block comes to rest after the collision. The coefficient of restitution is : (a) 1/5 (b) 4/5 (c) 1/ (d) 1/25.

5.

Lower surface of a plank is rough and lies over a rough horizontal surface. Upper surface of the plank is smooth and has a smooth hemisphere placed over it through a light string as shown. After the string is burnt trajectory of C.M of the sphere is

C

(a) circle (c) straight line 6.

(b) ellipse (d) none of these

A loaded spring gun of mass M fires a shot of mass m with a velocity V at an angle of elevation . The gun is initially at rest on a horizontal frictionless surface. Just after firing, the centre of mass of the gun-shot system :

V (a) moves with a velocity

m M

Vm (b) moves with a velocity M cos

θ

in the horizontal direction

(c) remains at rest

VM  m  (d) moves with a velocity M  m  in the horizontal direction. 7.

A steel ball strikes a steel plate placed on a horizontal surface at an angle θ with the vertical. If the co-efficient of restitution is e, the angle at which the rebound will take place is :

(a)

(c)

θ

 tan θ  tan1    e  (b)

e tan θ

 e  tan1    tan θ  (d) .

8.

The magnitude of the momentum of a particle varying with time is shown in figure. The variation of force acting on the particle is shown

P

t0

as :

t

2t0

(a)

9.

(b)

(c) (d) . -1 A 1 kg ball, moving at 12 ms , collides head-on with a 2 kg ball moving in the opposite direction at 24 ms-1.If the coefficient of restitution is (a) 60 J (c) 240 J

10.

2 3

, then the energy lost in the collision is : (b) 120 J (d) 480 J.

A ball of mass m falls vertically from a height h and collides with a block of equal mass m moving horizontally with a velocity v on a surface. coefficient of kinetic friction b/w the block and the surface is 0.2, while the coefficient of restitution (e) between the ball and the block is 0.5. Three is no friction acting between the ball and the block. The velocity of the block decreases by

m m (a)

0.5 2gh

v

k=0.2 (b) 0

0.1 2gh

(c)

(d)

0.3 2gh

SECTION - I Level – I ) 1. 3. 5. 7. 9. 11. 13 15. 17. 19.

(d) (b) (c) (c) (b) (d) (b) (a) (c) (a)

2. 4. 6. 8. 10. 12. 14. 16. 18. 20.

(c) (b) (b) (a) (d) (c) (b) (b) (c) (b)

2. 4. 6. 8. 10.

(a) (a) (c) (c) (d)

Level – II 1. 3. 5. 7. 9.

(c) (a) (c) (b) (c)

JEE ADVANCED ONLY ONE OPTION IS CORRECT Q.1

A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. fraction of incident energy transferred to the heavier ball is : n

Q.2

Q.3

n (1  n) 2

2n (1  n) 2

4n (1  n) 2

(A) 1  n (B) (C) (D) A ball of mass m moving with a certain velocity collides against a stationary ball of mass m. two balls stick with each otherduring collision. If E be initial kinetic energy, then the loss of kinetic energy in the collision is (A) E (B) E/2 (C) E/3 (D) E/4 Six identical balls are lined up along a straight frictionless groove. Two similar balls moving with speed v along the groove collide with this row on extreme left side end. Then : (A) One ball from the right end will move on with speed 2v, all the other remains at rest

Q.4

(B) Two balls from extreme right will move on with speed v each and the remaining balls will be at rest (C) All te balls will start moving to right with speed v/8 each (D) All the six balls originally at rest will move on with speed u/6 each and the two incident balls will come to rest. A sphere of mass m moving with a constant velocity u hits another stationary sphere of same mass. If e is coefficient of restitution, then ratio of velocities of two spheres (First sphere/Second sphere) after collision will be :

Q.5

 e  1  

1  e   

1 e   

 e 1  

(A)  1  e  (B)  1  e  (C)  e  1  (D) .  e  1  Two spheres A and B of equal mass are free to move on a smooth horizontal surface. A and B move towards each other with velocity vectors

ˆi

aiˆ  bjˆ and ciˆ  djˆ respectively and collide when the line joining their

centres is parallel to . After impact A and B have velocity vectors and respectively. coefficient of restitution b/w the spheres is e (< 1). (A) b = q (B) c = r (C) a + c = p + r (D) ea = p Q.6

A child of mass 4 kg jumps from cart B to cart A and then immediately back to the cart B. The mass of each cart is 20kg and they are initially at rest. In both cases the child jumps at 6m/s relative the cart. If the cart moves along the same line with negligible friction with the final velocities of v B and vA respectively. Find

A

B

the ratio of 6vB and 5vA.

Q.7

(A) 0.5 (B) 0.75 (C) 0.25 (D) 1 A ball rolls off a horizontal table with velocity v0 = 5 m/s . The ball bounces elastically from a vertical wall at a horizontal distance D = 8 m from the table, as shown in figure. The ball then strikes the floor a distance

v0=5m/s h=20m

x0

x0 from the table (g = 10 m/s2). The value of x0 is : (A) 6 m Q.8

(B) 4 m

D

(C) 5 m

A smooth sphere is moving on a horizontal surface with velocity vector

ˆj

(D) 7 m

3iˆ  ˆj

immediately before it hits a

vertical wall. The wall is parallel to the vector and the coefficient of restitution between the wall and sphere is 1/3. The velocity vector of the sphere after it hits the wall is : (A) Q.9

ˆi  ˆj

(B)

3iˆ  ˆj / 3

(C)

ˆi  ˆj

(D)

ˆi  ˆj

A ball of mass m collides perpendicularly on a smooth stationary wedge of mass M. If the coefficient

m v0

of restitution of collision is e, the velocity of the wedge after collision is :

(1  e) m v 0

emv 0

q

M

(1  e) m v 0 sin q

M v 0e

M (A) (B) M  m (C) M  m Q.10 Two masses A and B connected with an inextensible string of

(D)

M  msin 2 q

v m/s along the ground perpendicular to line AB as shown in figure. Find the tension in string during their subsequent motion.

B m 

A 2m 2mv 2 (A) 3 2

mv

v mv 2 (B) 3

2mv 2

(C) (D) Q.11 Four particles of mass 5, 3, 2, 4 kg are at the points (1, 6), (–1, 5), (2, –3), (–1, – 4). Find the coordinates of their centre of mass. (A) (1/7, 23/14) (B) (1/4, 23/14) (C) (1/7, 13/14) (D) (2/7, 13/14) Q.12 A semicircular portion of radius 'r' is cut from a uniform rectangular plate as shown

C

O

in the figure. The distance of centre of mass C of remaining plate from O is :

(A)

2r 3 

2r 4

3r (B) 2(4  )

2r 3(4  )

(C) (D) Q.13 Bullets of mass 109 each are fired from a machine gun at rate of 60 bullets/minute. The muzzle velocity of bullets is 100 m/s. The thrust force due to firing bullets experienced by the person holding the gun stationary is (A) 2N (B) 3N (C) 5N (D) 1N

Q.14 A body of mass M (fig) with a small disc of mass m placed on it rests on a smooth horizontal plane. disc is set in motion in horizontal direction with velocity v. To what height (relative to the initial level) will the disc

m

v M

rise after breaking off body M? friction is assumed to be absent.

mv 2 2g  M  m 

mv 2 2mv 2 mv 2 3g  M  m 3g  M  m g M  m (A) (B)  (C) (D) Q.15 If a ball is thrown upwards from the surface of earth: (A) earth remains stationary while ball moves upwards (B) ball remains stationary while earth moves downwards (C) ball and earth both moves towards each other (D) ball and earth both move away from each other Q.16 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

m m R remains in vertical plane.

(C) R/2 (D) R Q.17 A spaceship is moving with constant speed v0 in gravity free space along + Y-axis suddenly shoots out one third of its part with speed 2v0 along + X-axis. Find the speed of the remaining part. 3 v0 2

13 v0 4

13 v0 2

5 v0 2

(A) (B) (C) (D) Q.18 Three particles A, B and C each of the same mass m and connected by strings AB and BC of length a lie at rest with the string in a straight line on a smooth horizontal table. B is projected with a speed v0 at right angles to AB. Find the speed of particles A relative to C at the moment it collides with particle C. v0

B

A

1 3

C

4 v0

3

v0

2 3

v0

2 5

v0

(A) (B) (C) (D) Q.19 A ball is dropped from a height of 1m. The coefficient of restitution between the ground and the ball is 1/3. The height to which the ball will rebound after two collisions with ground is (A) (1/81) m (B) (1/27) m (C) (1/36) m (D) (1/45) m Q.20 A moving body with a mass m1 strikes a stationary body of mass m2. The masses m1 and m2 should be in the ratio m1/m2 so as to decrease the velocity of the first body 1.5 times assuming a perfectly elastic impact. Then the ratio m1/m2 is (A) 1/25 (B) 1/5 (C) 5 (D) 25

Q.21 Two small bodies of masses 'm' and '2m' are placed in a fixed smooth horizontal circular hollow tube of mean radius 'r' as shown. The mass 'm' is moving with speed 'u' and the mass '2m' is stationary. After their first collision, the time elapsed for next collision is : [ coefficient of restitution e = 1/2 ]

4r u

2r u

3 r (C) u

12  r u

(A) (B) (D) Q.22 Two balls of equal masses are projected upward simultaneously, one from the ground with speed 50 m/s and other from a 40 m high tower with initial speed 30 m/s. Find the maximum height attained by their centre of mass. (A) 50m (B) 25m (C) 100m (D) 75m Q.23 A small steel ball A is suspended by an inextensible thread vertically downwards such that its surface remain just in contact with thread during downward motion and collides elastically with the suspended ball. If the suspended ball just completes vertical circle after collision, calculate the velocity (in cm/s) of the falling ball just before collision. (g O

B

A

= 10 m/s2) (A) 625

(B) 1250

(C) 500

(D) 750

Q.24 Two particles of mass 1 kg & 0.5 kg are moving in same direction with speed of 2m/s and 6m/s respectively on a smooth horizontal surface. speed of centre of mass of system is : (A) 10/3 m/s (B) 10/7 m/s (C) 11/2 m/s (D) 12/3 m/s Q.25 A bullet of mass m strikes an obstruction and deviates off at 60° to its original direction. If its speed is also changed from u to v, find the magnitude of the impulse acting on the bullet. (A)

m u 2  uv  v2

(B)

m u 2  2uv  v2

(C)

m u 2  uv  v2

m u 2  uv  2v2 (C) R/2

(D) R

(D)

Q.26 All the particles of a body are situated at a distance R from origin. distance of centre of mass of the body from the origin is : (A) = R (B) (C) > R Q.27 In the arrangement shown in the figure, mA = 2 kg and mB = 1 kg. String is light and inextensible. Find the acceleration of centre of

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

A B

mass of both the blocks. Neglect friction everywhere. (A) g/9 upwards (B) g/9 downwards (C) g/7 downwards (D) g/7 upwards

Q.28 Water flows through a pipe bent at an angl by water on the bend of the pipe of area of cross section S ? (A)

2v2Scos 

(B)

2v2Ssin 

2v2S sin (C)

 2

(D)

2v Ssin  cos  2

Q.29 A block of mass 2 kg slides along a frictionless table with a speed of 10m/sec. Directly in front of it and moving in the same direction is a block of mass 5 kg moving at 3 m/sec. A massless spring of spring constant k = 1120 N/m is attached to the back side of 5 kg mass as shown in figure. When the blocks collide the maximum compression in the spring (if the spring does not bend) will be

(A) 0.25 m (B) 0.4 m (C) 0.33 m (D) 1.12 m Q.30 A body moving towards a finite body at rest collides with it it is possible that (a) both the bodies come to rest (B) both bodies move after collision (c) the moving body comes to rest & stationary body starts moving (d) the stationary body remains stationary, moving body changes its velocity (A) a, b only (B) c, d only (C) a, c, d (D) a, b, c, d Q.31 A ball hits a floor and rebounds after an inelastic collisionn. In this case (A) momentum of ball just after the collision is same as that just before the collision (B) mechanical energy of ball remains the same during the collision (C) total momentum of ball and the earth is conserved (D) None of these

f Q.32 When two particles of masses m1 and m2 are moving under the action of their internal forces 1 and m1 m 2 m1  m 2

under the force

f1

f2 –

(B) Their relative motion is the same as that of one of the particles with its mass replaced by the reduced mass of the system, with the other particle remaining at rest. (C) Their relative motion can be obtained by assuming one of the particles to have an infinite mass and by replacing the mass of the second particle by the reduced mass of the system, the force between them of the same as before (D) Both the particles move with uniform velocity. Q.33 An electron of mass m moving with a velocity v collides head on with an atom of mass M. As a result of the collision a certain fixed amount of energy is stored internally in the atom. The minimum initial velocity possessed by the electron is :

2M  mΔE Mm

2MΔE M  mm

2M  mΔE Mm

(A) (B) (C) (D) None of these Q.34 A hand ball falls on the ground and rebounds elastically along the same line peratining to motion. Then (A) The linear momentum is conserved (B) The linear momentum is not conserved, the loss in momentum being dissipated as heat in the ball and the ground (C) During the collision the full kinetic energy of the ball is converted into potential energy and then is completely converted to kinetic energy of the ball (D) None of these Q.35 Choose the correct statement – (A) The relative velocity of two particles in a head on collision is unchanged both in magnitude and direction (B) The linear momentum is conserved but not the kinetic energy in elastic collisionn. (C) A quick collision between two bodies is more violent than a slow collision even though the initial and final velocities are identical (D) None of these Q.36 A particle of mass m makes a head-on elastic collision with a particle of mass 2m initially at rest. The velocity of the first particle before and after collision is given to be u1 and v1. Then which of the following statements is true in respect of this collision ? (A) For all values of u1, v1 will always be less than u1 in magnitude and |v1| = u1/3 (B) The fractional loss in kinetic energy of the first particle is 9/8 (C) The gain in kinetic energy of the second particle is (8/9)th of the initial kinetic energy of the first particle. (D) There is a net loss in the energy of the two particles in the collision. Q.37 Hail storms are observed to strike the surface of the frozen lake at 30° with the vertical and rebound at 60° with the vertical. Assume contact to be smooth, the coefficient of restitution is : (A) e  1/ 3 (B) e = 1/3 (C) e  3 (D) e = 3 Q.38 If the external force acting on a system has zero resultant the centre of mass (A) Must not move (B) Must not accelerate (C) May accelerate (D) None of these Q.39 Which one of the following statements does not hold good when two balls of masses m1 and m2 undergo elastic collision (A) When m1 < m2 and m2 at rest, there will be maximum transfer of momentum. (B) When m1 > m2 and m2 at rest, after collision ball of mass m2 moves with four times the velocity of m1. (C) When m1 = m2 and m2 at rest, there will be maximum transfer of K. E. (D) When collision is oblique & m2 at rest with m1 = m2 after collision the ball moves in opposite directions.

(4iˆ - ˆj)

Q.40 A small sphere of mass m = 1 kg is moving with a velocity m/s. It hits a fixed smooth wall and rebound with velocity m/s. The coefficient of restitution between the sphere and the wall is n/16. Find value of n. (A) 6 (B) 7 (C) 4 (D) 9 Q.41 A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving northward with the same speed v. two particles coalesce on collision. The new particle of mass 2m will move in the north-eastern direction with a velocity

(A) –v/2 (B) v/2 (C) Q.42 Internal forces can change (A) linear momentum but not kinetic energy of the system. (B) kinetic energy but not linear momentum of the system. (C) linear momentum as well as the KE of system. (D) neither linear momentum, nor kinetic energy of system. Q.43 Two particles of equal mass have initial velocities

v/ 2

2iˆ m / s

and

(D) None

ˆ /s 2jm . First particle has an acceleration (

ˆi  ˆj

) m/s2, while the acceleration of the second particle is zero. The centre of mass of the two particles moves in : (Aa) circle (B) parabola (Cc) ellipse (D) straight line –1=

Q.44 Two colliding particles of masses m1 and m2 moving with v1 and v2 + m2

–1

m1–1

and is the coefficient of restitution)

(A) Loss of kinetic energy of 2 2 energy of 1/ 2  v1  v2  1 

1/ 2  v1  v2 

2 (1–

 )2

(B) Loss of potential ) (v1 + v2)

 ) (v1 + v2)

Q.45 A bullet of mass 0.01 kg, travelling at a speed of 500 ms –1, strikes a block of mass 2 kg, which is suspended by a string of length 5 m, & emerges out. block rises by a vertical distance of 0.1 m. The speed of the bullet after it emerges from the block is (A) 55 ms–1 (B) 110 ms–1 (C) 220 ms–1 (D) 440 ms–1 Q.46 An anti-air craft shell has been fired and it has kinetic energy K. It explodes in mid air. The sum of kinetic energies of particles is(A) K (B) < K (C) > K (D) 0 Q.47 A radioactive nucleus initially at rest decays by emitting an electron and an anti-neutrino at right angles to one another. The momentum of the electron is 3.2 × 10–23 kg-ms–1 and that of anti-neutrino is 6.4 × 10–23 kgms–1. The direction of recoiling nucleus with that of e – is(A) tan–1 (1/2) (B) tan–1 (2) (C) tan–1 – tan–1 2 Q.48 A neutron travelling with a velocity v and kinetic energy E collides perfectly elastically head on with the nucleus of an atom of mass number A at rest. The fraction of the total energy retained by neutron is :

2 A  1I F G HA J K

2 A  1I F G H J K (C) A  1

2 A  1I F G J HA K (B)

(A) Q.49 Figure shows an irregular wedge of mass m placed on a smooth horizontal surface. Part BC is rough. What minimum velocity should be imparted to a small block of

2 A  1I F G H J K (D) A  1

H wedge same mass m so that it may reach point B – (A) (C)

2 gH 2 g (H  h)

(B) (D)

2gH

gh

m

m

B

C b

Q.50 In the above question, the velocity of wedge when the block comes to rest on part BC is –

gh

g (H  h)

2 gH

(A) (B) (C) (D) None of these Q.51 The diagram shows the velocities just before collision of two smooth spheres of equal radius and mass. The

impact is perfectly elastic. The velocities just after impact are :

(A)

(B)

(C) (D) Q.52 Choose incorrect one. If no external force acts on a system : (A) Velocity of centre of mass remains constant (B) Velocity of centre of mass is not constant (C) Velocity of centre of mass may be zero (D) Acceleration of centre of mass is zero. Q.53 A system consists of mass M and m(