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SINGLE CORRECT OPTION TYPE

1. Centre of mass of three particles of masses 1 kg, 2kg and 3 kg lies at the point (1,2,3) and
centre of mass of another system of particles 3kg and 2kg lies at the point (–1, 3, –2). Where
should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the
centre of mass of system?
(A) (0, 0, 0) (B) (1, 3, 2) (C) (–1, 2, 3) (D) (3, 1, 8)

2. The adjacent figure shows an infinite frame of two sides in a gravity free
space. What is the final constant kinetic energy expected of the ball of mass
m projected as shown with an initial velocity v 0. The coefficient of restitution
for the collision between the ball and the frame is e = 0.5. v0

(A)0.5 mv 02 (B) 0.5 mv 02 sin2
(C)0.5 mv 02 cos2 (D) nothing can be said
m
d

3. A bob of mass 10 m is suspended through an inextensible

string of length . When the bob is at rest at the equilibrium
position, two particles of mass ‘m’ each moving with velocity u
making an angle 60 with the string strike and get m
m 60
simultaneously attached to the bob. What is the value of 60
u u
impulsive tension in the string during the impact?

(A) zero. (B) 2 mu (C) mu (D) 12 mu

4. Two particles one of mass m and the other of mass 2 m are m 2m

projected horizontally towards each other from the same level 10 m/s 5 m/s
above the ground with velocities 10 m/s and 5 m/s respectively.
They collide in air and stick to each other. The distance from A
where the combined mass finally land is
60 m

A B
(A) 40 m (B) 20 m (C) 30 m (D) 45m

5. A particle of mass ‘m’ strikes a smooth stationary

m
wedge of mass M with a velocity v0, at an
angle  with horizontal if the collision is v0
perfectly inelastic, the impulse on the wedge M
is smooth floor


Mmv0 Mmv0 cos  Mmv0 sin 

(A)
( M + m) (B)
mv0
(C) ( M + m) (D) ( M + m)

6. A small steel ball A is suspended by an inextensible thread of length l = 1.5 O

m from O. Another identical ball is thrown vertically downwards such that its
surface remains just in contact with thread during downward motion and B
collide elastically with the suspended ball. If the suspended ball just
completes vertical circle after collision, calculate the velocity of the falling
ball just before collision. (g = 10 m s–2)
(A) 15 m/s (B) 5 3 m/s
A
(C) 12.5 m/s (D) 10 3 m/s

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7. Find the position of centre of mass of a uniform disc of radius R from

which a hole of radius r is cut out. The centre of the hole is at a distance
R/2 from the centre of the disc.

Rr 2 Rr 2
(A) towards right of O (B) towards left of O
2(R2 − r 2 ) 2(R2 − r 2 )
2Rr 2 2Rr 2
(C) towards right of O (D) towards left of O
(R 2 + r 2 ) (R 2 + r 2 )

MULTIPLE CORRECT OPTION TYPE

8. The friction coefficient between the horizontal surface and each of 1.0 m/s
the blocks shown in the figure is 0.2. The collision between the
blocks is perfectly elastic. The velocity of 2 kg block in the 2 kg 4 kg
position shown is 1 m/s and the 4 kg block is initially at rest. Take
g = 10m/s2. 16 cm

(A) The velocity of 4 kg mass just after collision is 0.2 m/s

(B) The velocity of 2 kg mass just after collision is 0.2 m/s
(C) The velocity of 4 kg mass just after collision is 0.4 m/s
(D) the separation between them when they finally come to rest is 9 cm

9. A particle of mass m = 2 kg collides head on with another stationary particle of mass M = 3

kg with an initial velocity of 6 m/s. If the particle m stops just after the collision,
(A) the velocity of block M after the collision is 4 m/s
(B) kinetic energy of system before collision is equal to kinetic energy after the collision
(C) the coefficient of restitution for the collision is 2/3
(D) friction of loss of kinetic energy due to the collision is 1/3

10. A ball A collides elastically with an another identical ball B at rest, with velocity 10 m/s at an
angle of 30 from the line joining their centres C1 and C2. Select the correct alternative(s)
(A) velocity of ball A after collision is 5 m/s
(B) velocity of ball B after collision is 5 3 m/s
(C) both the balls move at right angles after collision
(D) kinetic energy will not be conserved here, because collision is not head on

11. Two balls A and B having masses 1 kg and 2 kg, moving with speeds 21 m/s and 4 m/s
respectively in the opposite direction, collide head on. After collision A moves with a speed of
1 m/s in the same direction, then the correct statements is (are)
(A) The velocity of B after collision is 6 m/s opposite to its direction of motion before collision.
(B) The coefficient of restitution is 0.2.
(C) The loss of kinetic energy due to collision is 200 J.
(D) The impulse of the force between the two balls is 40 Ns.

12. Three thin rods of each of length L are arranged in an inverted U, as shown L
in the figure. The two rods on the arms of the U each have mass M, the
third rod has mass 3M. 3M
L
M M L

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2L
(A) the center of mass is located at a distance of from the 3M rod
5
L
(B) the center of mass is located at a distance of from the 3M rod
5
(C) the center of mass is equidistant from both the rods of mass M
(D) if another rod of length L and mass 3M is introduced to complete the square, the centre of
mass will shift downwards

13. Two blocks of masses m and M are interconnected by an ideal spring of v0

stiffness k. If m is pushed with a velocity v0 m M

mv 0 Mmv 02
(A) v C = (B) KE of the system (M + m) about CM is equal to
M+m 2(M + m)
Mmv 02
(C) the maximum work done by the spring is equal to –
2(M + m)
Mm
(D) x max = v 0
(M + m)k

Comprehension-1
A long rod of mass 2 kg and length 4m is placed on a smooth horizontal table. Two particles of
masses 2 kg and 1 kg strike the rod simultaneously and strike to the rod after collision as shown.

14. Velocity of centre of mass of the rod after the collision is 2kg 10m/s
(A) 3 m/s (B) 6 m/s
(C) 9 m/s (D) 12 m/s 4m
1kg 10m/s

2kg
15. Angular velocity of rod after collision is
30 19
(A) radian/sec (B) radian/sec (C) 3 radian/sec (D) ( 3 –1)radian/sec
17 17

16. If two particular strike the rod in opposite direction, then after collision, as compared to the
previous situation, so the rod will
(A) rotate faster and translate slower (B) rotate slower and + translate faster
(C) show no change in linear or angular velocity (D) rotate faster, but translate at the same rate

Comprehension-2
A pendulum is suspended from a peg on a vertical wall. The pendulum is L
pulled away from the wall to a horizontal position and released. The ball hits
2
the wall, the coefficient of restitution being . Mass of the bob is m.
5
17. What maximum angle with vertical will the pendulum make after the first collision with the
wall?
−1 2  2  −1 4 −1 1
(A) cos (B) cos −1 1 −  (C) cos (D) cos
5  5 5 5

18. What is the total impulse imparted to the wall before it comes to rest against the wall?
 2   2 
(A) m 2gL (B) m 2gL 1 +  (C) m 2gL( 5 + 2)2 (D) 1m 2gL 1 + 
 5  5

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19. What is the minimum number of collisions after which the amplitude of oscillations becomes
less than 60?
(A) 1 (B) 2 (C) 3 (D) 4

INTEGER TYPE
20. A carpenter has constructed a toy as shown in the figure. If
the density of the sphere is 12 times that of the cone, the
position of the centre of mass of the toy is at a distance of Kr
from O. the value of k is:

21. If the system is released, then the acceleration of the centre of mass of the system
g
is . The value of n is:
n

22. Three spherical bodies each are connected by light inextensible

strings with each other as shown and rest over a smooth v0 = 2 3m / s
horizontal table. At moment t = 0, ball B is imparted a velocity v0
= 2 3 m / s . Calculate the velocity of A when it collides with ball A l B l C
C? 1kg 2kg 1kg

A
23. In the arrangement shown in Figure, ball and block have the same mass
m = 1 kg each,  = 600 and length l = 2.50 m. Co–efficient of friction 
between block and floor is 0.5. When the ball is released from the position
shown in the figure, it collides with the block and the block stops after
m
moving a distance 2.50 m.
Find coefficient of restitution for collision between the ball and the block.
m

24. Figure shows force verses time during the collision of a 40 gm Force ( N )
tennis ball with a wall. The initial velocity of the ball is 50 m/s
perpendicular to the wall. It rebounds with the same speed, also
perpendicular to the wall. The value of Fmax ( in Newtons ) is 10x,
Fmax
then find the value of x?

time ( sec )
O 0.2 0.4 0.6

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MATRIX MATCH TYPE
25. Match the column – I with column–II.
Column - I Column - I
m

v0 e=1

Momentum is conserved along

(A) (p)
horizontal axis
Fixed Wedge

Momentum is conserved along

(B) (q)
vertical axis

smooth wall
v
= 0
R e =1
(C) v0
(r) Kinetic energy is conserved

v1
m

m
(D) v0 (s) Kinetic energy is not conserved
900
m
v=0
m
v2
Velocity of approach is equal to
(t)
velocity of separation.

26. A particle of mass m is projected with an u

initial velocity u at an angle  with the  H  H2
horizontal. It collides the horizontal smooth A B
surface inelasticlly at B,C,D etc. The 2 C D
t=0 t = T1 t = T1+T2
coefficient of restitution is e. Match the
following
Column – I Column – II
H1
(A) (ratio of successive maximum heights) (p) e −1
H2
AB
(B) (q) e −2
BC
T1
(C) (ratio of successive time of flights) (r) e
T2
tan 
(D) (s) e2
tan 

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KEY

1. D 2. C 3. C 4. C
5. C 6. C 7. B 8. B, C
9. A, C, D 10. A, B, C 11. A, B, C 12. B, C, D
13. A, B, C, D 14. B 15. A 16. A
17. D 18. C 19. D 20. 4
21. 4 22. 3 23. 1 24. 1
25. A → r, t; B → p, s; C → r, t; D → p, q, r, t
26. A → q; B → p; C → p; D → r

SOLUTIONS

1. m1 = 1 + 2 + 3 = 6 kg, (x1, y1, z1) → (1, 2, 3)

m2 = 3 + 2 = 5 kg, (x2, y2, z2) → (–1, 3, –2)
m3 = 5 kg, (x1, y1, z1)?
Given (xcm, ycm, zcm) → (1, 2, 3)
Now apply the formulae
1
component of v0 parallel to the planes remains constant. k = m ( v0 cos  )
2
2.
2
3. J = 2 ( mu cos 600 )
= mu
4. vcm along x − axis = m (10 ) − 2m ( 5 ) = 0
5. J  J cos  = −mv + mv0 cos  = MV
mv0 cos  Mmv0 cos 
( M + m ) v = mv0 cos v= Mv =
M +m M +m
Mmv0 cos 
 J cos  =
M +m
6. C
7. For a disc, mass is proportional to surface area.
So Mass of cut portion: m1 = k  r 2 , Mass of remaining portion: m2 = k  (R2 − r 2 )
R Rr 2
Applying m1x1 = m2 x 2 , where x1 = and solving to get x 2 =
2 2(R2 − r 2 )
8. B, C
9. A, C, D
10. In elastic glancing collision v1 and v2 after collision are
perpendicular to each other
v1 = u,sin 300 = 5 m / s
v2 = u,cos300 = 5 3 u
 = 300

11. m1u1 + m2 u2 = m1v1 + m2 v2

21 − 7 = 1 + 2v2 v2 = 6 m / s e ( 2i + 5) = ( 6 − 1) e = 0.2
1 1 473 1 1 73
Initial KE = .1.2i 2 + .2.42 = J Find K.E = .1.12 + .2.62 = J
2 2 2 2 2 2

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473 73
KE = − = 200 J J = m1v1 − m1u1 = 2i − 1 = 20 Ns
2 2
12. B, C, D
13. A, B, C, D
14. B
15. A
16. A
17. velocity of the bob just before collision v0 = 2gL L  L
Velocity of the bob just after collision v1 = ev0 5
u12 4 4L
Vertical height raised = = e2 L = L
2g 5 5
1 −1 1
 cos  =   = cos
5 5

18. Total momentum imparted after first collision = mv0(1 + e)

Total momentum imparted after second collision = mev 0(1 + e)
Total momentum imparted after third collision = me 2v0(1 + e)
and so on
total impulse = mv0(1 + e)[1 + e + e2 + ………….]
 2 
 1+
= mv0
(1 + e)
= m 2gL  5  = m 2gL (5 + 2)

1− 2 1 − 2  ( 5 − 2)
 5 
Rationalizing = m 2gL( 5 + 2) 2

19. velocity of rebound after Nth collision = eN v0.

v2
Height raised h = = e2N L L −h  L
2g
L−h
 cos  = = 1 − 1 − e 2N h
L
1
e2 N =  taking log on both sides
2
 2  1  1 
2N log   = log 2 2N log 2 − log 5 = log1 − log 2
 5  2 
 1  3 1
2N log 2 − log10 − log 2 = − log 2 2N  log 2 −  = − log 2
 2  2 2
3 
2N  (0.3) − 0.5  = −0.3 N  3.1  min N required = 4.
2 

20. 4
Sol. Let the density of cone be  then the density of sphere is 12.
1 R3
Mass of cone: m1 = (2R)2 (4R) = 16
3 3
4 3 R3
Mass of sphere: m2 = R (12) = 48
3 3

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16R  4R  48
3
 4  + 3 R (5R)
3
m OQ + m2OQ2 3  
ycm = 1 1 = = 4R
m1 + m2 16R3 48R3
+
3 3
21. 4
3m − m g
Sol. a= g=
3m + m 2
ma + 3m( −a)
acm =
m + 3m
−2ma a g
= =− =−
4m 2 4
22. 3
momentum is conserved 2 3
vertically and horizontally
( )
2 2 3 = 1.vy + 1.vy + 2.vy
vy = 2 3 _________ (1) vy
vy
Mechanical energy is conserved
( 2 )(12 ) = ( 2 ) ( vxn + 3) + ( 2 )( 3)
1 1 1 vx vx
2 2 2
vx = 6 ___________ ( 2)
v = vx2 + v y2 = 3

23. 1
Energy Initially
E = mgl (1 − cos  )
1
E = 10  2.5  = 12.5 __________ (1)
2
Work done against friction
f k d =  mgd = 0.5 110  2.5 f k d = 12.5 __________ ( 2 )
there is no energy loss during collision so e =1

24. 1
(
Impulse = m v f − vi )
F ( 0.4 ) = 0.4 ( 50 + 50 ) 10−3 F = 10 = 10 x x =1

25. A – r,t; B – p,s; C – r,t; D – p,q,r,t

v 2 − v1 0 − u y1
26. e= =
u1 − u 2 u y 2 − 0
 The particle bounces back with a speed which is e times the initial speed along the vertical.
T1 2u y1 / g u y1 uy 1
= = = 1 = = e −1
T2 2u y2 / g u y2 eu y1 e
2
R1 u y1 / 2g  u y1   1 
2 2
−2
= =  =  =e
R 2 u 2y2 / 2g  u y2   e 
tan  u y1 / u x u y1 1
= = = = e −1
tan  u y2 / u x u y2 e

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