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 1g 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 2a 1a 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
+
Wt
(m0 – qt - qt)(v + v)
=
mve
[Wt = g(m0 – qt)t]
[mve = qt(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 v2
m2
m1
v2
B
v1
A
...(i)
m2
m1
v2
B
(1)
v1
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 v2 v1 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 m2v2
m1v1 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 mM
(b)
(d)
2gR
2MgR mM
From COLM,
Mvm mvm From COE
mgR
1 1 MVM2 mv2m 2 2
wmgR Vm
M
m2 2 Vm mVm2 2 m
2MgR Mm
. 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 Mm (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 Mm
its speed Ans. (b) is, wrong. Ans. (c) is, therefore wrong.
M Mm
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
2mv cosq mv mv 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 2R 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 =
zR
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 3 103 N .S
The magnitude of resultant impulse is given by
6.
J=
J12 J 22 103 (15)2 (5 3)2
J=
3 102 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 2MV2 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 M20 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 / 2M20 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 102 kg, , A 104 m2
m m0 + mass of dust collected so far
,
v0 103 m and dust 103 kg / m3
m0 Axdust
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
Axdx 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
VM 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 tan1 e (b)
e tan θ
e tan1 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 ]
4r u
2r 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)
2v2Scos
(B)
2v2Ssin
2v2S sin (C)
2
(D)
2v 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 :
2M mΔE Mm
2MΔE M mm
2M mΔE 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(
(A) moves with a velocity Vm/M horizontal direction
Q.55
Q.56
Q.57
Q.58
(B) moves with a velocity
Vm M
VM m (D) moves with a velocity M m in the
(C) remains at rest horizontal direction. Two persons standing on a floating boat run in succession along its length with a speed 4.2 m/s relative to the boat and dive off from the end. The mass of each man is 80kg and that of boat is 400kg. If the boat was initially at rest, find final velocity of boat. Neglect friction – (A) 0.6 m/s (B) 0.7 m/s (C) 0.1 m/s (D) 1.3 m/s A thin uniform wire is bent to form the two equal sides AB and AC of triangle ABC, where AB = AC = 5cm. The third side BC of length 6 cm is made from uniform wire of twice the density of the first. The distance of centre of mass from A is : (A) (34/11) cm (B) (11/34) cm (C) (34/9) cm (D) (11/45) cm A ball is dropped from a height h on ground. If coefficient of restitution is e, height to which ball goes up after is rebounds for the nth time is (A) h e2n (B) h en (C) e2n/h (D) h/e2n Two particles of mass 1 kg & 0.5 kg are moving in same direction with speed of 2 m/s and 6 m/s respectively on a smooth horizontal surface. speed of centre of mass of system is : (A) 3.33 m/s (B) 4 m/s (C) 5.5 m/s (D) 5 m/s
Q.59 A block of mass M is tied to one end of a massless rope. other end of rope is in hands of a man of mass 2M as shown in figure. The block and the man are resting on a rough wedge of mass M as shown in the figure.
The whole system is resting on a smooth horizontal surface. The man pulls the rope. Pulley is massless and frictionless. What is displacement of wedge when the block meets pulley. (Man does not leave his position during the pull) ? (A) 0.5 m (B) 1 m (C) zero (D) (2/3) Q.60 A t = 0, the positions and velocities of particles are as shown in figure. They are kept on a smooth surface and being mutually attached by gravitational force. Find the position of centre of mass at t = 2s :
(A) X = 5 m Q.61
(B) X = 7 m
(C) X = 3 m
(D) X = 2 m horizontal direction. At highest point in its trajectory it explodes into two pieces of equal mass. One of pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is-
(D) ( 3 2 Q.62 A particle of mass 1 kg moving with a velocity of 5m/s collides elastically with rough ground at with vertical as shown. What can be the minimum coefficient of friction between the particle and the ground
q ////////////////////////////////////////////////// (A) 1
(B) 0.5
(C) 0.75
(D) 0.25
Q.63 Consider the following two statements (a) Linear momentum of the system remains constant (b) Centre of mass of the system remains at rest. (A) a implies b and b implies a (B) a does not imply b and b does not imply a (C) a implies b but b does not imply a (D) b implies a but a does not imply b. Q.64 A particle at rest suddenly disintegrates into two particles of equal masses which start moving The two fragments will(A) Move in the same direction with equal speeds (B) Move in any directions with any speed (C) Move in opposite directions with equal speeds (D) Move in opposite directions with unequal speeds Q.65 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 bullet has moved through a distance 'd' towards the block, the centre of mass of the bullet block system is at a distance of
D d m (a)
Mm
from the block
2 d m DM Mm
Q.66
Q.67
Q.68
Q.69
Q.70
Q.71
Q.72
Q.73
(b)
D d
md MD Mm
from the rifle
(c)
M Mm
from the rifle (d) from the bullet (A) b, c (B) a, d (C) a, b, c (D) a, b A body has its centre of mass at origin. The x-coordinates of particles (A) May be all positive (B) May be all negative (C) May be all non-negative (D) May be +ve for some cases & negative in other cases A train of mass M is moving on a circular track of radius R with constant speed V. length of train is half of the perimeter of track. The linear momentum of train will be – (A) 0 (C) MVR (D) MV 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 : (A) v (B) Zero (C) 2v (D) 3v/2 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 th (A) 1.3 N-s (B) 1.05 N-s (C) 1300 N-s (D) 105 N-s 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 : (A) A/M (B) M/A (C) AM (D) 2A / M A 500 kg boat has an initial speed of 10 m/s as it passes under a bridge. At that instant a 50 kg man jumps straight down into the boat from the bridge. The speed of the boat after the man and boat attain a common speed is (A) 100/11 m/s (B) 10/11 m/s (C) 50/11 m/s (D) 5/11 m/s Consider a system of two identical particles. One of particles is at rest and the other has an acceleration . The centre of mass has an acceleration – (A) zero (B) a / 2 (C) a (D) 2a A man of mass m climbs on a rope of length L suspended below a balloon of mass M. balloon is stationary with respect to ground. If man begins to climb up rope at a speed vrel (relative to rope). In what direction and with what speed (relative to ground) will the balloon move
(A) downward,
mv rel mM
Mv rel (B) upwards, m M
mv rel (C) downwards, M
(D) downwards,
(M m) vrel M Q.74 Two particles A and B start moving due to their mutual interaction only. If at any time 't',
v
aA
and
aB
are
v
A and B are their respective velocities, and upto that time W and W their respective accelerations, A B are the work done on A and B respectively by the mutual force, m A and mB are their masses respectively, then which of the following is always correct –
a + aB = 0 m v + mB vB = 0 (C) WA + WB = 0 v + vB = 0 (A) A (B) A A (D) A Q.75 The rotational analogue of force in linear motion is : (A) torque (B) weight (C) moment of inertia (D) angular momentum Q.76 A canon ball is fired with a velocity 200 m/s at an angle 60 0 with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically up with a velocity 100 m/s, the second one falling vertically downwards with a velocity 100 m/s. The third fragment will be moving with a velocity : (A) 100 m/s in the horizontal direction (B) 300 m/s in the horizontal direction (C) 300 m/s in a direction making an angle of 60° with the horizontal
(D) 200 m/s in a direction making an angle of 60° with the horizontal Q.77 A canon shell moving along a straight line bursts in to two parts. Just after the burst one part moves with momentum 20 Ns making an angle 30° with the original line of motion. The minimum momentum of the other part of shell just after the burst is – (A) 0 Ns (B) 5 Ns (C) 10 Ns (D) 17.32 Ns Q.78 A body of mass 1 kg moving in the x-direction, suddenly explodes into two fragments of mass 1/8 kg and 7/8 kgs. An instant later, smaller fragment is 0.14 m above the x-axis. The position of the heavier fragment is (A) 0.02 m above x-axis (B) 0.02 m below x-axis (C) 0.14 m below x-axis (D) 0.14 m above x-axis Q.79 Two particles of equal mass have initial velocities acceleration
2iˆ
ms–1 and
2ˆj ms–1 . First particle has a constant
ˆi ˆj
( ) ms–2 while the acceleration of the second particle is always zero. The centre of mass of the two particles moves in (Aa) Circle (B) Parabola (Cc) Ellipse (D) Straight line Q.80 A bomb travelling in a parabolic path under the effect of gravity explodes in mid air. centre of mass of fragments will: (A) Move vertically upwards & then downwards (B) Move verticall downwards (C) Move in an irregular path (D) Move in parabolic path which unexploded bomb would have travelled. Q.81 Two balls are thrown in air. acceleration of centre of mass of the two balls while in air (neglect air resistance) (A) depends on direction of the motion of the balls (B) depends on the masses of the two balls (C) depends on the speeds of the two balls (D) is equal to g Q.82 A bullet of mass m = 50 gm strikes a sand bag of mass M = 5 kg hanging from a fixed point, with a v
horizontal velocity p . If bullet sticks to the sand bag then the ratio of final & initial kinetic energy of the bullet is: (A) 10–2 (B) 10–3 (C) 10–6 (D) 10–4 Q.83 A solid iron ball A of radius r collides head on with another stationary solid iron ball B of radius 2r. The ratio of their speeds just after the collision (e = 0.5) is : (A) 3 (B) 4 (C) 2 (0) 1 Q.84 Two boys of mass m each are standing in stationary boats also of mass m each. Boy A jumps towards his right with a speed v0 relative to ground just after he has lost contact with boat A, the other boy cuts the string. The relative speed of boats when the first boy has landed in boat B will be B A
(A) Zero
(B) (4/9)v0
(C) (4/ 3)v0
(D) None
ONE/MORE THAN ONE CHOICE MAY BE CORRECT Q.1
A small block of mass 2m initially rests at the bottom of a fixed circular, vertical track, which has a radius of R. The contact surface between the mass and the loop is frictionless. A bullet of mass m strikes block horizontally with initial speed v0 and
remain embedded in the block as the block and the bullet circle the loop. Choose the correct options –
R
m,v0 2m (A) The speed of masses immediately after the impact is v0/3 (B) The min. initial speed of the bullet if the block and the bullet are to
3 5gR
successfully execute a complete ride on the loop is (C) The min. initial speed of the bullet if the block and the bullet are to successfully execute a complete ride
Q.2
on the loop is 3 3gR (D) The speed of the masses immediately after the impact is v0/2 X and Y components of acceleration of C. M. are
(a CM )X (A)
m1m2 g m1 m2
2
m2 (a CM )Y g m1 m2
Q.3
Q.4
Q.5
Q.6
(a CM )X (B)
m1m2 g (m1 m2 )2
m2 (a CM )Y g m1 m2
(C) (D) A projectile is projected at a speed u at an angle into two fragments of mass ratio 1 : 2. The smaller fragments coming to rest. Then : (A) In a given time both parts will cover equal vertical displacement (B) The distance of heavier fragments from the point of projection is 4R/3 where R is horizontal range. (C) The distance of heavier fragments from the point of projection is 5R/4 (D) Linear momentum is conserved only in Y direction. P is the centre of mass of four point masses A, B, C & D which are coplanar but not collinear. (a) P may or may not coincide with one of the point masses. (b) P must lie within the quadrilateral ABCD (c) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C and D three at a time. (d) P must lie on a line joining two of the point A, B, C, D. 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) : (A) Momentum is mv/(M + m) (B) KE is (1/2) Mv2 (C) Momentum is mv (D) KE is m2v2/2(M + m} A solid cone and a solid sphere is arranged as shown in the figure.
Q.7
Q.8
The centre of mass is : (A) at 3R if m1 = m2 = m and R1 = R2 (B) at 1 2 and R1 = R2 then distance from the centre of solid cone is 11R/3 1 2 (D) at 1.5R if m1 = m2 = m and R1 = R2 In head on elastic collision of two bodies of equal masses (A) the velocities are interchanged (B) the speeds are interchanged (C) the momenta are interchanged (D) the faster body slows down and the slower body speeds up. An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of following cannot be centre of mass of the object.
y
B A (A) (R/3, R/3) (C) (R/4, R/4) Q.9
x (B) (R/3, R/2) (D)
Mass m1 hits and sticks with m2 while sliding horizontally with velocity v along the common line of centres of the three equal masses (m1 = m2 = m3 = m). Initially masses m2 and m3 are stationary and the spring is unstretched. Choose the correct options k v m3 m1 m2
//////////////////////////////////////////////////////////////////////////////////////////////////////// Frictionless (A) the velocities of m1, m2 and m3 immediately after impact are v/2, v/2, 0 (B) the maximum kinetic energy of m3 is 2mv2/9 (C) the minimum kinetic energy of m2 is mv2/72
m v 6k
(D) max. compression of spring is Q.10 A man of mass M is carrying a ball of the mass M/2. The man is initially in the state of rest at a distance D from fixed vertical wall. He throws the ball towards the wall with a velocity V with respect to earth at t = 0.
As a result of throwing, the man also starts moving backwards. The ball rebounds elastically from the wall.
The man finally collects the ball. Assuming friction to be absent. (A) the velocity of the man + ball system after the man has collected the ball is 2V/3 (B) impulse by ball on man is MV/3 (C) impulse by ball on man is MV/6 (D) he catches the ball again at t = 4D/V Q.11 Two cars initially at rest are free to move in the x direction. Car A has mass 4 kg and car B has mass 2 kg. They are tied together, compressing a spring in between them. When the spring holding them together is burned, car A moves off with a speed of 2m/s. Choose the correct options – (A) The speed with which car B leave is 4 m/s (B) Energy stored in the spring before it was burned is 24 J (C) The speed with which car B leave is 2 m/s (D) Energy stored in the spring before it was burned is 12 J Q.12 Choose the incorrect statements (B) CM of a uniform semicircular ring of radiu (D) CM of a hemisphere shell of radius R = R/2 from the centre Q.13 A smooth ball of mass 1 kg is projected with velocity 7 m/s horizontally from a tower of height 3.5m. It collide elastically with wedge of mass 3 kg and inclination 45° kept on ground. Ball collides the wedge at a height of 1m above the ground. (Neglect friction at any contact) Choose the correct options – 1 kg 7m/s
3.5m 3kg 45°1m (A) The velocity of the wedge after collision is 4 m/s.
5 2 m/s
(B) The velocity of the ball after collision is (C) The velocity of the wedge after collision is 7 m/s. (D) The velocity of theball after collision is 14 m/s
Q.14
normal to the surface at the point of contact & rebounds at (A)
(B)
Q.15 In an inelastic collision – (A) the velocity of both the particles may be same after the collision (B) kinetic energy is not conserved (C) linear momentum of the system is conserved (D) velocity of separation will be less than velocity of approach. Q.16 Two particles, P of mass 2m and Q of mass m, are subjected to mutual force of attraction and no other force acts on them. At t = 0, P is at rest at point O and Q is moving away from O with a speed 5u. At a later instant t = T (before any collision has taken place), Q is moving towards O with speed u. Then – (A) momentum of particle P at t = T is zero (B) momentum of particle P at t = T is 6mu 2
–3 mu2 Q.17 Three carts move on a frictionless track with inertias and velocities as shown. The carts collide and stick together after successive collisions. Choose the correct options
m1 = 2kg v1 = 1m/s
m2 = 1kg v2 = 1m/s
A
B
m3 = 2kg v3 = 2m/s
+ve
C
//////////////////////////////////////////////////////////////////////////////////////////////////////// (A) Loss of mechanical energy when B and C stick together is 3 J (B) Magnitude of impulse experienced by A when it sticks to combined mass (B and C) is 12/5 Ns (C) Loss of mechanical energy when B and C stick together is 1.5 J (B) Magnitude of impulse experienced by A when it sticks to combined mass (B and C) is 5/12 Ns Q.18 The simple pendulum A of mass mA B. If the system –
B
q A mA mB
(A) The velocity vB
s 3mAg +
m 2A g mB
2g m 1 A mB
mB mA (C) The velocity vB
2g m 1 A mB
2m 2A g mB
Ag + Q.19 In which of the following cases the centre of mass of a rod is certainly not at its centre? (A) density continuously increments from left to right (B) density continuously decrements from left to right (C) density decreases from left to right upto the centre and then increments (D) density increases from left to right upto the centre & then decreases Q.20 A body is moving towards a finite body which is initially at rest collides with it. In the absence of any external impulsive force, 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. Q.21 The density at one end being twice that of the other end. Choose the correct options– (A) An expression for linear mass density as a function of distance x from end A where linear mass density
x 2L
0 is (B) The position of the centre of mass from end A is 7/9 L (C) An expression for linear mass density as a function of distance x from end A where linear mass density
x L
is (D) The position of the centre of mass from end A is 5/9 L Q.22 In a one-dimensional collision between two particles, their relative velocity is before the collision and after the collision 0
(A)
v1 v2
(C)
| v2 | | v1 | in all cases
if the collision is elastic
(B)
v1 v2
(D)
v1 kv2
Q.23
if the collision is inelastic in all cases, where k lifted
vertically with constant velocity by a force P. 2)
(B) no energy will loss in this process 1 r gL2 + r v2 L (C) work done by force will be 2
1 r gLv 2 (D) Loss in energy 2
Q.24 A sphere A is released from rest in the position shown and the strikes the block B which is at rest. If e = 0.75 between A and B and µk = 0.5 between B and the support. Choose the correct options –
A 2kg
90° 1.5m 4kg B
(A) the velocity of A just after the impact is
g/6 m/s
49 m (B) the maximum displacement of B after the impact 48
g /12 m / s
(C) the velocity of A just after the impact is
37 m 48
(D) the maximum displacement of B after the impact Q.25 A rod of mass M – m carries an insect of mass m at its bottom end and its top end is connected with a string which passes over a smooth pulley and the other end of the string is connected to a counter mass M. Initially the insect is at
/////////////
M M–m m rest. Choose the correct option(s) – (A) As insect starts moving up relative to rod, the acceleration of centre of mass of the system (insect + rod + counter mass) becomes non-zero. (B) As insect starts moving up relative to rod, tension in the string remains constant and is equal to Mg. (C) An insect starts moving up relative to rod, the tension in the string becomes more than Mg (D) Acceleration of centre of mass of the system (insect + rod + counter mass) is zero when insect moves with constant velocity. Q.26 If the external forces acting on a system have zero resultant, centre of mass (A) must not move (B) must not accelerate (C) may move (D) may accelerate Q.27 In a one dimensional collision between two identical particles A and B, B is stationary and A has momentum p before impact. During its impact B gives impulse J to A. (A) The total momentum of the ‘A plus B’ system is p before and after the impact, and (p-J) during the impact (B) During the impact A gives impulse of magnitude J to B
(C) The coefficient of restitution is
2J 1 p
(D) The coefficient of restitution is
J 1 p
Q.28 A body is thrown vertically upwards from ground with a speed of 10 m/s. If coefficient of restitution of ground, e = 1/2. Choose the correct options – (A) The total distance travelled by the time it almost stops is 20/3 m (B) Time elapsed (after the ball has been thrown) when it is at its subsequent maximum height for the third time is 6.25 sec. (C) The total distance travelled by the time it almost stops is 40/3 m (D) Time elapsed (after the ball has been thrown) when it is at its subsequent maximum height for the third time is 3.25 sec. Q.29 A block moving in air explodes in two parts then just after explosion (A) the total momentum must be conserved (B) the total kinetic energy must be conserved (C) the total momentum must change (D) the total kinetic energy must increase Q.30 A smooth sphere A of mass 0.1 kg is moving with speed 5m/s when it collides head on with another smooth stationary sphere of same radius. If A is brought to rest by the impact and e = 1/2. Choose the correct options – (A) The mass of B is 0.2 kg (B) Speed of B just after impact is 2.5 m/s (C) Magnitude of impulse during collision is 0.5 Ns (D) Speed of B just after impact is 1.5 m/s Q.31 In a game of Carom Board, the Queen (a wooden disc of radius 2cm. and mass 50gm) is placed at the exact center of the horizontal board. The striker is a smooth plastic disc of radius 3cm. and mass 100gm. The board is frictionless. The striker is given an initial velocity u parallel to the sides BC or AD so that it hits the Queen inelastically with coefficient of restitution = 2/3. The impact parameter for the collision is ‘d’ (shown in the figure). Queen rebounds from edge AB of the board inelastically with same coefficient of restitution = 2/3 and enters the hole D following the dotted path shown. The side of the board is L. Choose the correct options –
A
L
B
u L
D
d
C
5 (A) The value of impact parameter ‘d’ is
17
cm
(B) The time which the Queen takes to enter hold D after collision with the striker is
2 (C) The value of impact parameter ‘d’ is
17
153L 40u
cm
153L (D) The time which the Queen takes to enter hold D after collision with the striker is 80u Q.32 In an elastic collision in absence of external force, which of the following is/are correct:
(A) The linear momentum is conserved (B) The potential energy is conserved in collision (C) The final kinetic energy is less than the initial kinetic energy (D) The final kinetic energy is equal to the initial kinetic energy Q.33 A particle of mass m moving with a velocity finally moves with velocity
(2iˆ ˆj)
ˆ (3iˆ 2j) m/s collides with stationary body of mass M and
m/s. Then
ˆ ˆ (A) Impulse received by m m(5i 5j)
(B) Impulse received by
m m(5iˆ ˆj)
M m(5iˆ ˆj)
M m(5iˆ ˆj)
(C) Impulse received by (D) Impulse received by Q.34 A particle of mass m1 (A) elastically collides with another stationary particle (B) of mass m2. Then :
m1 1 m 2 and the particles fly apart in the opposite direction with equal velocities. 2 (A) (B)
m1 1 m2 3
(C)
m1 2 m2 1
and the particles fly apart in the opposite direction with equal velocities.
and the collision angle between the particles is 60° symmetrically.
m1 2 m2 1
(D) and the particles fly apart symmetrically at an angle 90° Q.35 A block of mass m moving with a velocity v0 collides with a stationary block of mass M at the back of which a spring of stiffness k is attached, as shown in figure. Choose the correct alternative(s).
(A) The velocity of the centre of mass is v0/2
1 mM 2 v0 (B) initial kinetic energy of system in the centre of mass frame is 4 M m v0
mM 1 (m M) k
(C) maximum compression in spring is (D) When the spring is in the state of maximum compression, the kinetic energy in the centre of mass frame is zero
ASSERTION AND REASON QUESTIONS (Q.36-Q.45) Note : Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct. (A) Statement-1 is the True, Statement-2 is also True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is the True, Statement-2 is alsoTrue; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement -1 is True, Statement-2 is False. (D) Statement -1 is False, Statement-2 is True. (E) Statement -1 is False, Statement-2 is False. Q.36 Statement 1 : A quick collision between two bodies is more violent than slow collision, even when initial and final velocities are identical.
Statement 2 : The rate of change of momentum determines that the force is small or large. Q.37 Statement 1 : If collision occurs between two elastic bodies their kinetic energy decreases during the time of collision. Statement 2 : During collision intermolecular space decreases and hence elastic potential energy increases. Q.38 Statement 1 : If some portion of a disc is taken out the centre of mass shifts to the opposite side of the removed portion. Statement 2 : The position of centre of mass depends on the position of individual masses only. Q.39 Statement 1 : In a two body collision, the momenta of the particle are equal and opposite to one another, before as well as after the collision when measured in the center of mass frame. Statement 2 : The momentum of the system is zero from the centre of mass frame. Q.40 Statement 1 :Two identical bodies undergo an elastic collisionn. Speed of both balls before collision is u.
Q.41 Q.42
Q.43
Q.44
The maximum speed of any ball after collision can be u 2 . Statement 2 : In oblique collision component of velocity along tangent will remains unchanged. Statement 1 : When you lean behind over the hind legs of the chair, the chair falls back after a certain angle. Statement 2 : Centre of mass lying outside the system makes the system unstable. Statement 1 : The centre of mass of a proton and an electron in a stationary hydrogen atom, related from their respective positions remains at rest. Statement 2 : The centre of mass remains at rest, if no external force is applied. Statement 1 : Two balls make a head on collision w.r.t. one frame. Coefficient of restitution is measured to be e. Then coefficient of restitution is same w.r.t. all frames. Statement 2 : Coefficient of restitution is equal to one for perfectly elastic collisionn. Statement 1 : If no external force acts on a system of particles, then the centre of mass will not move in any direction. Statement 2 : If net external force is zero, then the linear momentum of the system remains constant.
M
m
Q.45 Statement 1 : The centre of mass shifts upwards, as man moves upwards. Statement 2 : The centre of mass accelerates under the action of net force.
MATCH THE COLUMN TYPE QUESTIONS (Q.46-Q.50) Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II.
////////// m u m
//////////
spring
m 2u m
Q.46 Two pan of sa me man are connected by string. Two ball of same moving with speed u and 2u collide with the pan and sticks with pan after collision. Now match the column I and II. Column I Column II
(A) For left pan (B) For right pan (C) For left ball (D) For right ball
(p) Impulse by normal reaction (q) Impulse by tension (r) Impulse by spring (s) Impulse by weight
Q.47 Two blocks of masses 3 kg and 6 kg are connected by an ideal spring and are placed on a frictionless horizontal surface. The 3 kg block is imparted a speed of 2 m/s towards left. (consider left as positive
2m/s
6kg 3kg //////////////////////////////////////////////////
direction) Column I
Column II
2 (A) When the velocity of 3 kg block is 3 m/s 2 (B) When the speed of 3 kg block is 3 m/s (C) When the speed of 3 kg block is minimum (D) When the speed of 6 kg block is maximum each other
(p) Velocity of centre of mass is
2 3 m/s
(q) Deformation of the spring is zero (r) Deformation of the spring is maximum (s) Both the blocks are at rest with respect to
Q.48 Two balls of masses m1 and m2 are moving towards each other with speeds u1 and u2. They collide head on and their speeds are v1 and v2 after collision. (m1 = 8kg, m2 = 2 kg, u2 = 3 m/s)
u1 m1
u2
m2
Column I (A) The speed u1 (in m/s) so that both balls move in same direction if coefficient of restitution is e = 0.5 (B) The speed u1 (in m/s) so that both maximum energy is transformed to m2 (assume elastic collision) (C) Coefficient of restitution if m2 stops after collision and u1 = 0 .5m/s (D) If collision is inelastic and u1 = 3 m/s, the loss of kinetic energy (in J) after collision may be
Column II (p) 1/14 (q) 1/8 (r) 2 (s) 4
Q.49 In the figure shown, when the persons A and B exchange their positions, then match the Column I and
Column II.
Column I (A) the distance moved by the centre of mass of the system is (B) the distance moved by the plank is (C) the distance moved by A with respect to ground is (D) the distance moved by B with respect to ground is
Column II (p) 20 cm (q) 1.8 m (r) zero (s) 2.2 m
Q.50 In each situation of column I, a system involving two bodies is given. All strings and pulleys are light and friction is absent everywhere. Initially each body of every system is at rest. Consider the system in all situations of column I from rest till any collision occurs. Then match the statements in column-I with the corresponding results in column II. Column I Column II (A) block plus wedge system is placed over (p) shifts towards right smooth horizontal surface. After the system is released from rest, the centre of mass of system.
m M (B) The string connecting both the blocks of mass m is horizontal. Left block is placed over smooth horizontal table as shwn. After two block system is released from rest, the centre of
m
(q) shifts downwards
m
mass of system (C) The block and monkey have same mass. The monkey starts climbing up rope. After the monkey starts climbing up, the centre of mass
of monkey + block system.
(r) shifts upwards
m
(D) Both block of mass m are initially at rest. The left block is given initial velocity u downwards. Then, the centre of mass of two block system
(s) does not shift
afterwards.
m
INTEGER TYPE QUESTIONS (Q.51-61) Q.51
Q.52
Q.53 Q.54 Q.55
Q.56 Q.57 Q.58 Q.59
Answer to each of the questions is a single/double/triple digits integer. A 3 kg mass moving at a speed of 15 ms-1 collides with a 6 kg object initially at rest. They stick together. Find the velocity (in m/s) of the combination after the collision. Two particles A and B lighter particle having mass m, are released from infinity. They move towards each other under their mutual force of attraction. If their speeds are v and 2v respectively and K.E. of the system is xmv2. Find the value of x. A block of mass m moving at a speed v collides with another block of mass 2m at rest. The lighter block comes to rest after collision. The coefficient of restitution is 1/A. Find the value of A. A particle is dropped on a surface from a height of 20 m. If the coefficient of restitution is 0.5, the total distance covered by the particle before coming to rest is (100/A) m. Find the value of A. A ball of mass m is dropped onto a floor from a certain height. collision is perfectly elastic & the ball rebounds to the same height and again falls. The average force exerted by the ball on the floor during long time interval is xmg. Find the value of x. Sphere of mass m1 = 3 kg impinges with a velocity of 7 m/s directly on another sphere of mass m2 = 5 kg. The velocities after impact are in the ratio 2 : 3. The coefficient of restitution is 1/A. Find the value of A. In the above question, loss of energy is (in joule). Two solid spheres of radii 6 cm and 12 cm and masses 36 g & 18 g respectively touch each other. The position of centre of mass is x cm along OO' from the sphere of smaller radius. Find the value of x. A particle A of mass m moving on a smooth horizontal surface collides with a stationary particle B of mass 2m directly, situated at a distance d from a wall. The coefficient of restitution between A and B and between B and the wall is e = 1/4. The distance from the wall where they collide again is 3d/x. Find the value of x. ( Assume that the entire motion takes place along a straight line perpendicular to the wall).
Q.60 inelastically so that its vertical component of the velocity becomes zero and horizontal component remains unchanged. It moves along the smooth horizontal surface. fter first collision. Take g = 10 m/s2 Q.61 A long platform moves with a constant acceleration a = g horizontally. A particle is projected from the rear e particle makes collision with the platform and retraces the path. The particle makes three collisions with the platform all collisions perfectly elastic)
PASSAGE BASED QUESTIONS Passage 1- (Q.62-Q.64) projected across the table with velocity u.
Q.62 Find the speed with which the second particle begins to move if the direction of u is along BA
u 2
u 13 (B) 8
u 3 (C) 4
u 2
u 13 (B) 8
u 3 (C) 4
u 2
u 13 (B) 8
u 3 (C) 4
u 3 (A) (D) 2 Q.63 Find the speed with which the second particle begins to move if the direction of u is at an angle of 120° with AB – u 3 (A) (D) 2 Q.64 Find the speed with which the second particle begins to move if the direction of u is perpendicular to AB –
(A)
u 3 (D) 2
Passage 2- (Q.65-Q.67) Two equal spheres of mass m' are suspended by vertical strings so that they are in contact with their centers at same level. A third equal spheres of mass m falls vertically and strikes elastically the other two simultaneously so that their centres at the instant of impact form an equilateral triangle in a vertical plane. If
///////////////////////////// u m v m'
u is velocity of m just before impact.
m' v
Q.65 Find the velocity of m' just after impact
(A)
2 3u 7
(B)
5u 7
(C)
2u 7
(D)
6u 7 Q.66 Find the velocity of m just after impact
2 3u 7
(A) (B) Q.67 Find the impulse of tension of the strings
6 mv (A) 7
5u 7
1 mv (B) 7
(C)
2u 7
(C)
5 2 mv mv 7 (D) 7
6u (D) 7
Passage 3- (Q.68-Q.70) Collision is the transfer of momentum due to only the internal forces between the particles taking part in collision. When exchange of a momentum takes place between two physical bodies due to their mutual interactive force, it is defined as collision between two bodies. Two bodies move in different directions interact each other at the point of intersection of their line of motion and the reaction due to their physical contact is the interaction force which is the cause of transfer of momentum from one body to another. Collision may be either elastic or inelastic. In case of elastic collision
momentum and K. E. are both conserved but in case of inelastic collision only momentum is conserved and K. E. is not conserved.
r Q.68 Two particles having the position 1
ˆ (3iˆ 5j) m and
ˆ r2 (5iˆ 3j) m move with velocities
ˆ m/s V1 (4iˆ 3j) and V2 (aiˆ 7ˆj) m / s . If the particles collide, then value of a must be : (A) 8 (B) 6 (C) 4 Q.69 A block of mass m is moved towards a movable wedge of mass M = Km and height h with velocity u (All the surface are smooth). If the block
(D) 2
just reaches the top of the wedge, the value of u is :
(A)
2gh 2gh (1 K ) K
(B)
2ghK 1 K 1 2gh 1 K
(C) (D) Q.70 Two equal spheres A and B lie on a smooth horizontal circular groove at opposite ends of a diameter. A projected along the groove and at the end of time t impinges on B. If e is coefficient of restitution, second impact will occur after time (A) 2t (B) 2et (C) 2t/e (D) 2t/e2 *Passage 4- (Q.71-Q.73) In a billiards game a stationary red ball is aimed at a white ball that had been given a speed of 10 m/s. After the collision, the red ball moves at an angle of 30° with the original direction of motion of the white ball. Assuming the collision to be elastic and the two balls have same mass find Q.71 The speed of the red ball
5 2 m/s
2 5 m/s
3 5 m/s
(A) (B) (C) Q.72 The speed of the white ball (A) 3 m/s (B) 5 m/s (C) 4 m/s m/s Q.73 The direction of motion of the white ball with its original direction of motion – (A) 60° (B) 54° (C) 66° Passage 5- (Q.74-Q.76) Fig. shows a uniform square plate from which four identical squares at the corners can be removed. If the centre of mass of the complete plate is
(D)
5 3 m/s (D) 5.5
(D) 50°
y 1
4
2 II I O III IV
x 3
at the origin. Q.74 Where will be the centre of mass after the removal of square 1 – (A) In first quadrant (B) In the second quadrant (C) In the third quadrant (D) In the fourth quadrant Q.75 In above question, where will be the centre of mass after the removal of square 1 and 2 – (A) On the y-axis, below the origin. (B) On the y-axis, above the origin (C) In the third quadrant. (D) In the fourth quadrant. Q.76 In above question, where will be the centre of mass after the removal of squares 1, 2 & 3 – (A) In the third quadrant (B) In the fourth quadrant (C) In the first quadrant (D) At the origin Passage 6- (Q.77-Q.79) Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on the first block pulling it away from the other as shown in figure. Q.77 The displacement of the centre of mass at time t is –
m
k
m Ft 2 3m
Ft 2 2m
F Ft 2 4m
Ft 2 m
(A) (B) (C) (D) Q.78 If the extension of the spring is x0 at time t, then the displacement of the first block at this instant is –
ö 1 æFt 2 + x0 ÷ ç 2 è 2m ø
-
ö 1 æFt 2 + x0 ÷ ç 2 è 2m ø
ö 1 æFt 2 - x0 ÷ ç 2 è 2m ø
æFt 2 ö ç 2m + x 0 ÷ è ø
(A) (B) (C) (D) Q.79 If the extension of the spring is x0 at time t, then the displacement of the second block at this instant is –
(A)
æFt 2 ö ç 2m - x 0 ÷ è ø
(B)
ö 1 æFt 2 + x0 ÷ ç 2 è 2m ø
(C)
ö 1 æ2Ft 2 - x0 ÷ ç 2è m ø
(D)
ö 1 æFt 2 x 0÷ 2 çè 2m ø Passage 7- (Q.80-Q.82) Two particles of masses m1 and m2 separated by a distance L from each other and are released from their initial state of rest. Only gravitational force acts between m1 and m2
x cm
m1x1 m2 x 2 dx dx , v1 1 , v2 2 m1 m2 dt dt
vcm
m1v1 m2 v2 m1 m2
.
Gm1m2 m1 / m1 2 x Let x = x2 – x1. At t = 0, x1 = 0, x2 = L, x = L, v1 = 0 and v2 = 0. a1 = acceleration of . Let S1 be the frame fixed to mass m1 . Clearly S1 is non-inertial frame . Frame S fixed to O in an inertial frame . Q.80 Choose the incorrect statement
(A)
m1x1 m2 x 2 m1 0 m2 L m1 m2 m1 m2
(B) m1v1 + m2v2 = 0
dx dt
(C) v2 = v1 +
(D)v1= –
m1 dx m m dt 1 2
Q.81 By Newton’s 2nd law , equation of motion for mass m2 in the S1 frame is :
m2 (A)
d2 x dt
2
m2 (C)
Gm2 (m1 m2 ) x
d2 x dt
2
m2
2
(B)
Gm1m2 x
m2
2
(D)
d2 x dt
2
d2 x dt
2
Gm1 (m1 m2 ) x2
Gm1m2 x2
dx 1 1 v2 v1 v velocity 2G L Q.82 x = x2 – x1 , dt of m2 with respect to m1. Let value of v2 is (A) m1m2 these
(B) m2
(C) (m1 + m2
(D) None of
Passage 8- (Q.83-Q.85) A horizontal frictionless string is threaded through a bead of mass m. The string is pulled between two vertical opposite sides of a cart of mass M, as shown in figure. The length of the cart is L and the radius of the bead, r, is very small in comparison with L (r <<
L). Initially, the bead is a the right edge of the cart. The cart is struck and as a result, it moves with velocity v0. When the bead collides with the cart’s walls the collisions are always completely elastic. Q.83 The velocity of the center of mass of the cart and the bead is :
mv0 (A) m M
Mv0 (B) m M
Mv0 (C) M m
Q.84 The first collision of the bead with the cart’s wall take place at the time t 1 is :
mv0 (D) M m
L v0
L v cm
2L (A) (B) (C) vcm Q.85 The velocity of the bead after the first collision, in the centre of mass frame:
2Mv0 (A) M m
(B)
Mv0 Mm
2mv0 (C) m M
(D)
2mMv0 2 2 (D) m M
2L v0
Other Engineering Exams Questions 1.
Three identical spheres of mass 1 kg each are placed on the corners of an equilateral triangle of side 2 m, with origin at one corner. The vector giving centre of mass of the system is
ˆ ˆ (a) i 0.29 j
ˆ ˆ
2.
ˆ ˆ (b) i 0.33 j
(c) i j (d) i 0.58 j A cracker is thrown into air with a velocity of 10 m/s at an angle of 45° with the vertical. When it is at a height of (1/2) m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground ? (g = 10 m/s2)
ˆ
1 (a) 4 5 ms
(c) 5 4 ms
1
1 (b) 2 5 ms
(d) 10 ms–1
ˆ
3.
If the system is released, then the acceleration of the centre of mass
m of the system is
g 4
4.
5.
3m
g (b) 2
(a) (c) g (d) 2g A body of mass M at rest explodes into three pieces, two of which of mass (M/4) each are thrown off in perpendicular directions with velocities of 3 m/s and 4 m/s respectively. third piece will be thrown off with a velocity of (a) 1.5 m/s (b) 2 m/s (c) 2.5 m/s (d) 3 m/s A bullet of mass ‘a’ and velocity ‘b’ is fired into a large block of wood of mass ‘c’ and gets embedded in it. The final velocity of the system is
cb (a) ( a b ) ab (c) ( a c )
b ( a b) (b) c
b (a c) (d) a 6. Two balls each of mass 0.25 kg are moving towards each other in a straight line, one at 3 m/s the other at 1 m/s collide. The balls stick together after the collision. The magnitude of the final velocity of the combined mass is (a) 4 m/s (b) 2 m/s (c) 1 m/s (d) (1/2) m/s 7. A body of mass 2 kg moving with a velocity of 3 m/s collides head on with a body of mass 1 kg moving in opposite direction with a velocity of 4 m/s. After collision the two bodies stick together and move with a common velocity (a) (1/4) m/s (c) (2/3) m/s
(b) (1/3) m/s (d) (3/4) m/s
8.
A bomb travelling in a parabolic path under the effect of gravity & explodes in mid air. centre of mass of fragments will (a) Move vertically upwards & then downwards (b) Move vertically downwards (c) Move in irregular path (d) Move in the parabolic path the unexploded bomb would have travelled
9.
A shell explodes and many pieces fly off in different directions. The following is conserved : (a) Kinetic energy (b) Momentum (c) Neither momentum nor KE (d) Momentum and KE.
10.
Two equal balls are in contact on a table and are in equilibrium. A third equal ball collides with them simultaneously, symmetrically and remains at rest after impact. coefficient of restitution is (a) 2/3 (b) 3/2 (c) 1/3 (d) 1/2
11.
A 3 kg bomb explodes into 3 equal pieces A, B and C. A files with a speed of 40 m/s and B with a speed of 30 m/s making an angle of 90° with the direction of A. The angle made by the direction of C with that of A is
12.
4 cos 1 5 (a)
4 cos 1 5 (b)
3 cos 1 5 (c)
3 cos 1 5 (d)
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 3
1 (c) 2
(d)
3 2
13.
A body of mass 3 kg is acted on by a force which varies as shown in the graph. The momentum acquired is given by 10 Force/N 0
14.
6 2 4 Time/s
(a) Zero (b) 5 N-s (c) 30 N-s (d) 50 N-s Which of the following is not an inelastic collision ? (a) A man jumps on a cart (c) Collision of two glass balls
15.
(b) A bullet embedded in a block (d) None of the above
In the given figure the position-time graph of a particle of mass 0.1 kg is shown. Linear x(m)
6 4 2 momentum at t = 2 sec is 0
1
2
t (second)
3
(a) Zero
(b) – 0.2 kg m sec–1
(c) 0.1 kg m sec–1
(d) – 0.4 kg m sec–1
Other Engg Examination Questions 1. 3. 5. 7. 9. 11. 13 15.
(d) (a) (c) (c) (b) (b) (d) (a)’
2. 4. 6. 8. 10. 12. 14.
(a) (c) (c) (d) (a) (c) (c)
PREVIOUS IIT QUESTIONS Q.1
Q.2
Q.3
Q.4
A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of large and small spheres are 4M and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the co-ordinates of the centre of the larger sphere when the smaller sphere reaches the
other extreme position. [1996] An isolated particle of mass m is moving in horizontal plane (x-y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at : [1997] (A) y = –5cm (B) y = + 20 cm (C) y = + 5 cm (D) y = – 20 cm Two blocks of mass 2 kg and M are at rest on an inclined plane and are separated by a distance of 6 m as shown. The coefficient of friction between each block and the inclined plane is 0.25. The 2 kg block is given a velocity of 10 m/s up the inclined plane. It collides with M, comes back and has a velocity of 1 m/s when it reaches its initial position. The other block M after the collision moves 0.5m up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block M.
2 [Take sin q tan q 0.05 and g 10m / s ] [1999] Two particles of masses m1 and m2 in projectile motion have velocities v1 < v2 respectively at time t = 0.
They collide at time t0. Their velocities become
Q.5
v1' and v '2 at time 2t
0
while still moving in air. The value
| (m1v1' m2 v'2 ) (m1v1 m2 v2 ) | is of (A) zero (B) (m1 + m2)gt0 (C) 2(m1 + m2)gt0 (D) 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 m/s to the heavier block in the direction of
the lighter block. The velocity of the centre of mass is : [2002] (A) 30 m/s (B) 20 m/s (C) 10 m/s Q.6
(D) 5 m/s
A car P is moving with a uniform speed of 5 3 m / s towards a carriage of mass 9 kg at rest kept on the rails at a point B as shown in figure. The height AC is 120 m. Cannon balls of 1 kg are fired from the car with an initial velocity 100 m/s at an angle 30° with the horizontal. The first cannon ball hits the stationary carriage after a time t0 and sticks to it. Determine t0. At t0, the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second ball also hits and sticks to
the carr the second impact?
iage, what will be the horizontal velocity of the carriage just after [2001]
Q.7
STATEMENT - 1 : If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. [2007] because STATEMENT - 2 : The linear momentum of an isolated system remains constant. (A)Statement- 1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1 (B) Statement -1 is True, Statement -2 is True ; Statement-2 is NOT a correct explanation for Statement - 1 (C) Statement - 1 is True, Statement- 2 is False (D) Statement -1 is False, Statement -2 is True
Q.8
The balls, having linear momenta
p1 pi
p
and
p2 pi , undergo a collision in free space. There is no
p
external force acting on the balls. Let 1 and 2 be their final momenta. The following option (s) is (are) NOT ALLOWED for any non-zero value of p, a1, a2, b1, b2, c1 and c2. [2008]
(A)
p1 a1ˆi b1ˆj c1kˆ p2 a 2 ˆi b2 ˆj
p1 c1kˆ p c2 kˆ (B) 2
(C)
p1 a1ˆi b1ˆj c1kˆ p2 a 2 ˆi b2 ˆj c1kˆ
(D)
p1 a1ˆi b1ˆj p a ˆi b ˆj 2
2
1
Paragraph for Question Nos. 9 to 11 [2008] A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
A
M v 60° B
30° 3m
3 3m
C
Q.9
The speed of the block at point B immediately after it strikes the second incline is –
A 3m 60° B v 3m
30° 30° 3m
(A)
60
C
3 3m
m/s
(B)
45
m/s
(C)
30
m/s
(D)
15
m/s
Q.10 The speed of the block at point C, immediately before it leaves the second incline is (A) 120 m/s (B) 105 m/s (C) 90 m/s (D) 75 m/s Q.11 If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is – (A) 30 m/s (B) 15 m/s (C) 0 (D) 15 m/s Q.12 Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that
v
A
2v
at A, these two particles will again reach the point A? (A) 4 (B) 3 (C) 2 (D) 1 [2009] Q.13 If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that (A) linear momentum of the system does not change in time (B) kinetic energy of the system does not changes in time (C) angular momentum of the system does not change in time (D) potential energy of the system does not change in time [2009]
ANSWERS (1) (5) (8) (12)
(L + 2R, 0) (C) AD (C)
(2) (A) (6) t0 = 12 sec, (9) (B) (13) (A)
(3) (4) (B) (7) (C) (10) (B) (14) 4 m/sec.
(11) (C) (15) (A)
