Mohammed Asif Name : Roll No. : Topic : Ph : 9391326657, 64606657
WORK – POWER & ENERGY AND CENTRE OF MASS
CONSERVATION OF MOMENTUM, COLLISIONS
SECTION – I
I. This section contains 9 multiple – choice questions numbered 1 to 9. Each question has 4 choice (A), (B), (C), and (D), out of which Only One is correct.
1. A particle of mass m is kept at the edge of a smooth cube of mass M and side L as shown in figure. Cube starts moving at t = 0 with a constant velocity v. Displacement of the center of mass along the horizontal direction from t = 0 to when particle hits the ground is
M m v s m o o t h A) mv L 2L m M v g + + B) mv v 2L m M L g + + C) Mv L 2L m M v g + + D) Mv v 2L m M L g + +
2. There are two masses M and M placed at a distance l apart, Let the centre 1 2
of mass of this system is at a point named C. If M is displaced l towards C 1 1
and M is displaced by l away from C. The distance from C where the new 2 2
centre of mass will be located is A) 2 1 1 2 1 2 M l M l M M + + B) 1 1 2 2 1 2 M l M l M M + + C) 1 1 2 2 1 2 M l M l M M − + C) 2 1 1 2 1 2 M l M l M M − +
3. In a boat of mass 4M and length ‘L’ on a frictionless water surface. Two men A(mass = M) B( mass = 2M) are standing on the opposite ends. Now A
distance 3L/4 relative to boat and meet A. The displacement of he boat on water till A and B meet is
A) 5L 28 B) zero C) L 2 D) 23L 2 4. A paraboloid shaped solid object is formed by
rotating on
parabola y = 2x2 about y – axis as shown in
figure. If the
height of the body is ‘h’ then the position of centre of mass
from origin. (Assume density to be uniform throughout). A) h 3 B) h 4 C) 2h 3 D) 2h 4 O Y h x
5. A pendulum of mass ‘m’ and length ‘1’ is released from
rest in a horizontal position. A nail at a distance ‘d’ below
the pivot causes the mass to move to along the path as
indicated by dotted lines. Find the minimum distance d
in terms of l such that the mass will swing completely
round in the circle shown
m d l A) 2 l B) 34l C) 3 5 l D) 23l 6. A particle of mass m and speed v collides elastically with the end
of a free uniform thin rod of mass M in a horizontal plane as
shown. After the collision, m is stationary. Calculate M. A) m B) 2m C) 3m D) 4m M m v ( T o p v i e w )
7. Two small balls A and B of mass M and 3M respectively hang from the ceiling by ideal strings of equal lengths. The A ball is drawn aside so that it is raised to a height H. If the ball A is released and collides with ball B;
Select the correct answer (s).
A) If collision is perfectly elastic, ball B will rise to a height H/4
B) If the collision is perfectly elastic ball A will rise upto a height H/4
C) If the collision is perfectly inelastic, the combined mass will rise to a height H/16
D) If the collision is perfectly inelastic, the combined
A
mass will rise to a height H/4
8. There are two observers A and B, observer A is fixed to
the smooth plank and observer B is standing on
smooth horizontal ground. The plank is accelerated
with constant acceleration, and C is placed on it then
B mC A
a
s m o o t h
p l a n k s m o o t h
A) Momentum of block C is conserved as seen by observer A B) Momentum of block C is conserved as seen by observer B C) Net force on plank is zero as seen by observer A
D) Net force on plank is non-zero as seen by observer B 9. The figure shows two blocks of mass 2 kg
and 1 kg
accelerating at 4 m/s2 and 8 m/s2
respectively
starting from rest as shown in the figure, then which
of the following are true at any instant (before the collision). 2 k g4 m / s 1 k g 2 8 m 2 / s s m o o t h A) aCM 0r = r B) vCMr ≠ 0r
C) position of centre of mass is constant D) position of centre of mass is not constant
SECTION – II
Assertion – Reason Type
I. This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT – 1
(Assertion) and STATEMENT – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D),
out of which ONLY ONE is correct. NOTE:
-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.
10. A rod is in contact with a wedge which is kept on the ground and can move only along its length between rigid supports as shown. If all the surfaces are smooth and the system is released from rest, then
STATEMENT – 1: - The mechanical energy of the system during the motion will be
conserved.
STATEMENT – 2: - The momentum of the system along horizontal direction is not
conserved.
11. STATEMENT – 1: - The centre of mass of a system of particles is independent of coordinate
system.
STATEMENT – 2: - The centre of mass of system of two particles always divides the line
joining them in the inverse ratio of their masses.
12. STATEMENT – 1: - On the application of impulse, the body will always move along the
direction of the impulse.
STATEMENT – 2: - Change in momentum is always along the direction of impulse.
13. STATEMENT – 1: - Kinetic energy of a system can be increased without applying any
external force on the system.
STATEMENT – 2: - Single external force acting on a particle necessarily changes its
kinetic energy.
SECTION – III
Linked Comprehension Type
I. This section contains 2 paragraphs P14-16 , and P17-19 . Based upon
each paragraph, 3 Multiple – choice questions have to be answered. Each question has 4 choices (A), (B), (C), and (D), out of which Only One is correct.
P14 - 16 : Paragraph for Questions Nos. 14 to 16
A light inextensible thread passes over a small frictionless ideal
pulley. Two blocks of masses m = 1 kg and M = 3 kg respectively
are attached with the thread as shown in the figure. The heavier
block rests on a horizontal surface. A shell of mass 1
m
M A 3 00
kg moving
upward, parallel to incline, with a velocity 10 ms-1
collides at t = 0
and sticks to m. The combined mass travels up the incline to a
distance s, stops for a moment and then moves down the incline.
When the moving mass reaches its initial position, the string becomes taut and the mass M starts to climb up with an initial speed v0. The mass M
moves up to a height h before stopping momentarily. The inclined plane is smooth and along. The acceleration due to gravity is g = 10 m/s2 Answer
the following questions: 14. The value of s is:
A) 1. 5 m B) 2. 0 m C) 2. 5 m D) 1. 0 m 15. The value of v0 is:
A) 1. 0 m/s B) 2. 0 m/s C) 2. 5 m/s D) 5. 0 m/s 16. The value of h is:
A) 0. 5 m B) 1. 0 m C) 2. 0 m D) 2. 5 m
P17 - 19 : Paragraph for Questions Nos. 17 to 19
A circular platform of mass 100 kg is placed symmetrically on a square frame of side 4m having four legs. The radius of the platform is 5m.
17. The maximum permissible mass of a dancer so that the platform does not topple is
A) 33. 3 kg B) 66. 7 kg C) 50 kg D) 75 kg
18. The area on the platform on which any mass of the dancer is permissible is A) 16 m2 B) 78. 5 m2 C) 8 m2 D) 62. 5 m2
19. If five dancers are there on the stage each of mass 40 kg then the maximum number of dancer that can go together at same position along the circumference of the stage with the remaining staying at the centre so that the platform does not topple is
A) 1 B) 2 C) 3 D) 4
SECTION – IV
Matrix Match Type
This section contains 3 questions. Each question contains statement given in two columns which have to be matched. Statement (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 X4 matrix should be follows:
A B C D
p q r s
20. When a body is moving vertically up with constant velocity under the action of a lifting force and air resistance is absent.
21. A ball strikes the ground at an angle α and rebound at an angle β with the vertical as shown in figure. Then match the column I with column II from the
combinations shown: α
β
22. Match the Column I with Column II
Column - I Column - II
A Work done by lifting force (p) Negative B Total work done by all the
forces (q) Positive
C Work done by gravity (r) Zero D If air resistance starts
acting then work done by air resistance if it acts opposite to motion.
(s) Non zero
Column - I Column - II
A Elastic collision (p) tan
(
)
tanα
α ≠ β β
B Perfectly inelastic collision (q) α <β β <, 900
C In – elastic collision (except
perfectly inelastic collision) (r) α = β D Coefficient of restitution (s) β = 900
SECTION – I
I. This section contains 9 multiple – choice questions numbered 1 to 9. Each question has 4 choice (A), (B), (C), and (D), out of which Only One is correct.
1. A ball is thrown up with a certain given velocity at angle θ to the horizontal. The kinetic energy, KE of the ball varies with horizontal displacement x as:
a) b) K E O x K E O x c) d) K E O x K E O x
2. A particle of mass m is released from height ‘h’ on smooth quarter fixed circular wedge. The horizontal surface AB, following the circular path of wedge is rough with coefficient of friction µ between surface and particle. Find the horizontal distance from A where the mass stops.
Column - I Column - II
A Graph of kinetic energy (on y –
axis) with momentum (on x – axis) (p) B Graph of square root of kinetic
energy (on y – axis) with momentum (on x – axis)
(q)
C Graph of square root of kinetic energy with(1 / momentum) (on x – axis)
(r)
D Graph of kinetic energy with velocity by (on x – axis) (s)
h
A B
m
(Assuming AB to be infinitely large and particle does not falls from it)
A) hµ B) µh C) h D) none of these
3. An object of mass m is tied to a string of length l and a
variable horizontal force is applied on it which is initially
is zero and gradually increases until the string makes an
angle θ with the vertical. Work done by the force F is
θ l
F
A) mgl 1 sin
(
− θ)
B) mgl C) mgl 1 cos(
− θ)
D) mgl 1 cos(
+ θ)
4. A particle moves in a circular path with the decreasing speed. Choose the correct statement
A) Angular momentum remains constant B) Acceleration
( )
ar is towards the centreC) particle moves in a spiral path which decreasing radius D) the direction of angular momentum remains constant
5. A particle moves in the X – Y plane under the influence of a force such that its linear momentum is P tur( ) = A i cos kt$ ( ) − $j sin kt( ), where A and k are constant. The angle between the force and the momentum is:
A) 00 B) 300 C) 450 D) 900
6. Block A in the figure is released from the rest when the
extension in the spring is x0. The maximum
downward
displacement of the block. A) 0 Mg x 2k − B) 0 Mg x 2k + C) 0 2Mg x k − D) 0 2Mg x k + A M
7. Consider the three cases shown in the figure, particle strikes in the vertical plane the perfectly elastic surfaces such that velocity of particle just striking is perpendicular to surface in three cases. In first case time taken by particle to strike to the surface is 2s, in second case time is 4s. Then the time taken by particle to strike the surface in 3rd case is:
θ α C a s e I θ C a s e I I θ C a s e I I I α A) 5 s B) 6 s C) 7 s D) 8 s
8. From the circular disc of radius 4R, two small discs of same radius R, are cut off as shown in the figure. The centre of mass of the new structure will be: A) R
( )
i$ j 5 +$ B)(
$)
3R i j 15 − +$ C) −R5( )
i$+$ D) j −3R14( )
i$+$j 4 R R R y x9. A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1m. Its speed at the bottom is 2 m/s. The work done by frictional force is:
A) 8 J B) - 8 J C) 4 J D) – 4 J
SECTION – II
Assertion – Reason Type
I. This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT – 1
(Assertion) and STATEMENT – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D),
out of which ONLY ONE is correct. NOTE:
-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.
10. STATEMENT – 1: - In a perfectly elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
and
STATEMENT – 2: - In an elastic collision, the linear momentum of the system is conserved.
11. STATEMENT – 1: - A particle of mass m strikes a wedge of mass M horizontally. If collision is perfectly inelastic then particle sticks to the wedge.
m M m
θ
STATEMENT – 2: - In perfectly inelastic collision velocity of both bodies is same along common normal just after collision.
12. STATEMENT – 1: - Power development in a uniform circular motion is always zero.
and
STATEMENT – 2: - Work done in case of a uniform circular motion is zero. 13. STATEMENT – 1: - A shell lying on ground at rest explodes in small
fragments. The centre of mass of fragments may move along a straight line.
STATEMENT – 2: - In explosion the linear momentum of the system always remains conserved.
SECTION – III
Linked Comprehension Type
I. This section contains 2 paragraphs P14-16 , and P17-19 . Based upon
each paragraph, 3 Multiple – choice questions have to be answered. Each question has 4 choices (A), (B), (C), and (D), out of which Only One is correct.
P14 - 16 : Paragraph for Questions Nos. 14 to 16
Two identical beads are attached to free ends of two identical
springs of spring constant k
(
2 3 mg)
3R+
= Initially both
springs make an angle of 600 at the fixed point,
normal
length of each spring is 2R. Where R is the radius of smooth
ring over which bead is sliding Ring is placed on vertical plane
and beads are at symmetry with respect to vertical line as
diameter.
6 0
m m
A) mg
2 B)
3mg
2 C) mg D) Insufficient
data
15. Relative acceleration between two beads at the initial moment
A) g2 vertically away from each other B) g2 horizontally towards each other
C) 2g 3 vertically away from each other B) 2g 3 horizontally towards each other
16. The speed of the bead when spring is at normal length A)
(
2 3 gR)
3 + B)(
2 3 gR)
3 − C) 2gR 3 D) 3gRP17 - 19 : Paragraph for Questions Nos. 17 to 19
In the figure shown ADB and BEF are two fixed smooth circular paths of radius R and R’ respectively. A ball of mass m enters in the tube ADB through point A with minimum velocity to reach point B. From there it moves on other circular path BEF and it is just able to complete the circle.
A R D B E C F R
17. The minimum velocity at A is given to the ball is
A) 4Rg B) 2Rg C) 3Rg D) 5Rg
18. The ratio of radius of two circular paths ADB and BEF is A) RR ,= 23 B) RR ,= 32 C) R , 2
R = 5 D)
R , 5 R = 2 19. The normal reaction at point E is
A) 4 mg B) 5 mg C) 6 mg D) 7 mg
SECTION – IV
Matrix Match Type
This section contains 3 questions. Each question contains statement given in two columns which have to be matched. Statement (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in
A B
matrix should be follows:
20.
21. Consider the situation in which a block is kept on smooth hemisphere. The block is a top of
hemisphere.
Column - I Column - II
A Potential energy (p) When no external or internal non conservative forces are present.
B Kinetic energy (q) Only for conservative forces. C Mechanical energy (r) Work done by all the forces in
the absence of internal non- conservative forces.
D Power developed zero. (s) Force is always perpendicular to velocity of object.
Column - I Column - II A
R = 1 m
m
Block of mass ‘m’ given a slight push rightwards and smooth hemisphere fixed to ground.
(p) Normal reaction on block is mg/4 at angle cos-1 (3/4) with vertical.
B 2 m
R
Block of mass ‘2m’ given a slight push at top of smooth and fixed hemisphere.
(q) Normal reaction of block is zero at angle cos-1 (2/3) with vertical.
C
R = 1 m
m
a = 2 0 / 9 m2 / s
Block of mass ‘m’ at top of smooth hemisphere moving with acceleration 20/9 m/s2.
(r) Normal reaction on block is zero at angle cos-1 (4/5) with vertical.
D
R = 1 m
m v = 4 g r
A block placed at bottom of smooth and fixed sphere pushed with velocity 4gr
(s) Normal reaction is 2mg at height ‘R’ from the ground.
22.
Column - I Column - II
A From natural length to extension ‘l’. (The work done by spring) (p)
2
1/ 2 kl
−
B From natural length to
Compression ‘l’. (The work done by spring)
(q)
2
1/ 2 kl
+
C From compressed length ‘l’ to extended length l. (The work done by spring)
(r)
2
kl D From compressed length ‘l’ to
natural length. (The work done by spring)
(s)
KEY
PAPER – 1
PAPER - 2
1. (C) 1. (C) 2. (B) 2. (B) 3. (A) 3. (C) 4. (C) 4. (D) 5. (C) 5. (D) 6. (D) 6. (A) 7. (A, B, C) 7. (B) 8. (B, C, D) 8. (D) 9. (A, C) 9. (B) 10. (B) 10. (B) 11. (B) 11. (D) 12. (D) 12. (A) 13. (C) 13. (C) 14. (C) 14. (C) 15. (B) 15. (D) 16. (A) 16. (C) 17. (B) 17. (B) 18. (A) 18. (B) 19. (C) 19. (C) 20. (A – q. s), (B – r), (C – p, s), (D – p, s) 20. (A – q), (B – r), (C – p), (D – s) 21. (A – r), (B – s), (C – q), (D – p) 21. (A – p, q),(B – q),(C – r),(D – q, s) 22. (A – s), (B – q), (C – r), (D – s) 22. (A – p), (B – p), (C – s), (D – q)HINTS & SOLUTION
PAPER – I
1. (C)
1. Initially particle will remain stationary till time t ,1
Here 1
L t
V
=
Centre of mass velocity CM
MV V
m M
= +
After this time particle will fall down vertically and reaches ground in time t2 2 2h t g =
(
)
CM CM 1 2 x = V t +t 2. (B) 5g l d( − ) 2. cm 1 1 2 2 1 2 M x M x x M M ∆ + ∆ ∆ = + 3. (A) 3. 2M 3L x M L x 4Mx 4 4 − = + + 4. (C) 4. cm y dm y dV h dV dm =∫
=∫
∫
( )
2 dV= πx dy 5. (C)5. Velocity at lowest position is 5g l d
( )
− 6. (D)7. (A, B, C)
7. Velocity of A just before collision ν = 2gH
Let ν1 and ν2 are the velocities of M and 3M after the collision.
2 1 1 2 2 1 M M 3M alsoe ev 0 ν − ν ν = ν + ν = − ⇒ ν −ν = −ν 8. (B, C, D)
8. With respect to ground C will be stationary, and w. r. t. plank it is accelerating.
9. CM CM 2 4 8 1 a 0 0 2 1 × − × = = ⇒ ν =
+ (both the masses starts from rest)
10. (B) 10. FBD of the system N1 W1 W 2 N2 11. (B) 12. (D) 13. (C) 14. (C)
14. Momentum of the two remains conserved during the collision 1 10 2× = ×ν ⇒ ν = 5m/ s
2
1 2 2gh, hereh S 2 × ν = = 2 15. (B)
15. When string becomes taut impulse will act. For combined mass − = ν −J 2 0 10
For 3 Kg mass J 3= ν0
16. (A)
16. On writing energy conservation (mechanical) after jerk, we get h.
2 2 1 2 1 3 2 g h 3 g h 0 0 2× ν + × ν = − × × + × ×2 2 17. (B) 17. M 5 2 100 M × = + 18. (A)
18. Area = Area of square frame. 19. (C) 19. n 40 5100 200× + × = 2 20. A – (Q, S) ; B – (R) ; C – (PS) ; D – (PS) 21. A – r, B – s, C –q, D – p ( )
( )
( ) ( ) A→ s ; B→ q ; C→ r ; D→ sHINTS & SOLUTION
PAPER – II
1. (C) 2. (B) 3. (C) 4. (D) 5. (D) 6. (A) Sol.(
0)
2 20 1 1 mgx k 2x x kx 2 2 = + − On solving 0 mg x x 2k = − 7. (B)Sol. See figure
α α C a s e I C as e I I 4 se c 6 s e c 2 se c 4 se c 8. (D) Sol.
(
)
(
)
2 2 1 1 2 2 3 3 cm 2 2 2 1 2 3 0 16R 3R R 0 x A x A x A 3R X A A A 16 R R R 14 π − π − − − = = = − − − π − π − π Similarly cm 3R y 14 = −( )
cm cm cm 3R S X i y j i j 14 = + = − + ur $ $ $ $ 9. (B)Sol. Wnon conservative− = ∆ME=(ME)f −(ME)i 2 1mv mgh 2 1 1 4 1 10 1 8 J 2 = − = × × − × × = − 10. (B) 11. (D)
11. When e = 0 velocity of separation along common normal zero, but there may be relative
velocity along common tangent. 12. (A)
13. (C)
13. There may be an impulsive force due to ground therefore c. m may have an upward
velocity, there fore linear momentum in vertical direction will not be conserved. 14. (C) 14. N mg sin 30= 0 + ∆K. x.cos 300
(
2 3)
(
)
mg . R 2 3 3mg 2 3R 2 mg mg mg 2 2 + = + − = + = 15. (D)15. Acceleration of both beeds in horizontal direction towards each other g .
3
=
16. (C)
16. Mechanical energy conservation
2 mgR 2 1K . x 0 0 1mv 2 ∆ + + = − 2 +2
(
)
x R 2 3 2gR v 3 ∆ = − = 17. (B)17. K.E. at point A = P.E. at point D
2 A A 1mv mgR 2 v 2gR = = 18. (B) 18. (KE)B +(PE)B =(KE)E
(
)
(
)
1 1 m 2gR mgR ' m 5gR ' ; velocity at E to just at F 5gR ' 2 + = 2 = On solving R 3 R '= 2 19. (C) 19. N mg mvE2 R ' − = On solving N = 6 mg21. A→
( )
p,q ; B→( )
q ; C→( )r ; D→( )
q,s 21. ( )A mg cos N mv2 ... 1( ) R θ− =(
)
( ) ( ) ( )( )
( ) ( )(
)
(
)
( ) 2 2 2 2 1 mv mgR 1 cos ... 2 2 2 N 0... 3 cos 3 3 3 3 When cos N mg 2g 1 m 4 4 4 3mg 2g mg m 4 4 4 B q mv C mg cos N ma sin ... 1 R 1 mv mg 1 cos R masin R 2 Rmv 2mgR 1 cos 2maR sin ... 2 mg cos masin N 2mg 1 = − θ = ⇒ θ= θ = ∴ = × − − = − = θ− − θ = = − θ + θ = − θ + θ θ − θ = +
(
−)
1cos 2ma sin R 4 N 0 cos 5 − θ + θ = ⇒ θ = 22. A→
