Chapter - 5_Centre of Mass

CHAPTER-5 CENTRE OF MASS FILL IN THE BLANKS 1. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is ..... (1987; 2M) 2. The magnitude of the force (in newtons) acting on a body varies with time t (in microseconds) as shown in the fig. AB, BC and CD are straight line segment. The magnitude of the total impulse of the force on the body from t = 4 µs to t = 16 µs is ...... N-s (1994; 2M) C

Force (N) →

800 – 600 – 400 – 200 –A 0

B E 2

4

6 8 10 12 14 16 Time (µs) →

TRUE/FALSE 1.

D

F

3.

Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2 m/s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s, the velociy of the centre of mass is 0.75 m/s. (1989; 2M)

A shell is fired from a cannon with a velocity v (m/s) at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (m/s) of the other piece immediately after the explosion is : (1986; 2M) (a) 3 v cos θ (b) 2 v cos θ (c)

3 v cos θ 2

(d)

3 v cos θ 2

OBJECTIVE QUESTIONS Only One option is correct : 1. Two particles A and B initially at rest, move towards each other mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is : (1982; 2M) (a) 3V (b) V (c) 1.5 V (d) zero 2.

A ball hits the floor and rebounds after an inelastic collision. In this case : (1986; 2M) (a) the momentum of the ball just after the collision is the same as that just before the collision. (b) the mechnical energy of the ball remains the same in the collision (c) the total momentum of the ball and the earth is conserved (d) the total energy of the ball and the earth is conserved

4.

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 C; 1M) (a) y = – 5 cm (b) y = + 20 cm (c) y = + 20 cm (d) y = – 20 cm

5.

Two particles of masses m1 and m2 in projectile motion r r have velocities v1 < v2 respectively at time t = 0. They r r collide at time t0 . Their velocities become v ′1 and v ′ 2 at time 2t0 while still moving in air. The value of : r r r r | ( m1v ′1 + m2v ′ 2 ) − ( m1v1 + m2v2 ) | (2001; S)

39

(a) zero

(b) (m1 + m2 )gt0

(c) 2 (m1 + m2 )gt0

(d)

1 (m + m2 )gt0 2 1

6.

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 blocks in the direction of the lighter block. The velocity of the centre of mass is : (2002; S) (a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s

constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then : (1993; 2M) v C

7.

8.

A particle moves in the X-Y plane under the influence of a force such that its linear momentum is r p( t) = A[iˆ cos(kt) − ˆj sin(kt)], where A and k are constants. The angle between the force and the momentum is: (2007; 3M) (a) 0° (b) 30° (c) 45° (d) 90° 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.

Look at the drawing given in the figure which has been drawn with ink of uniform line thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m.

a 10

(b)

a 8

(c)

a 12

(d)

a 3

v (m / K )

v

2v

(d) the maximum compression of the spring is v 2.

m 2K

r r Two balls, having linear moments p1 = pˆi and p 2 = pˆi , undergo a collision in free space. There is no external r r force acting on the balls. Let p'1 and p'2 be their final momenta. The following option figure (s) is (are) NOT ALLOWED for any non-zero value of p, a 1 , a 2 , b 1 , b 2 , c1 and c2 . (2008; 4M) r r (a) p'1 = a1 ˆi+ b1 ˆj+ c1 kˆ p ' 2 = a 2 ˆi + b 2 ˆj

r (b) p ' 1 = c 1 kˆ r (c) p'1 = a1 ˆi+ b1 ˆj+ c1 kˆ

y

r (d) p ' 1 = a 1 ˆi + b1 ˆj

x

r p ' 2 = b 2 ˆj r p' 2 = a2 iˆ+ b2 ˆj− c1 kˆ r p ' 2 = a 2 ˆi + b1 ˆj

SUBJECTIVE QUESTIONS 1.

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 position of the centre of mass of the remaining portion. (1980)

OBJECTIVE QUESTIONS More than one options are correct? 1.

B

A

The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, –a). The y-coordinate of the centre of mass of the ink in this drawing is (2009; M) (a)

L

(a) the kinetic energy of the A-B system, at maximum compression of the spring, is zero (b) the kinetic energy of the A-B system, at maximum compression of the spring, is mv 2 /4 (c) the maximum compression of the spring is

Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A? (2009; M) (a) 4 (b) 3 (c) 2 (d) 1 9.

A

Two blocks A and B each of mass m, are connected by a massless spring of natural length L and spring

42cm 56 cm

40

2.

3.

A body of mass 1 kg initially at rest, explodes and breaks into three fragments of masses in the ratio 1 : 1 : 3. The two pieces of equal mass fly-off perpendicular to each other with a speed of 30 m/s each. What is the velocity of the heavier fragment? (1981; 3M) A Three particles A, B and C of equal mass move with equal speed V along the medians of an equilateral triangle as shown in G fig.They collide at the centroid G of the triangle. B

from the ground with a velocity of 49 m/s. At the same time another identical ball is dropped from a height of 98 m to fall freely along the same path as that followed by the first ball. After some time the two balls collide and stick together and finally fall to the ground. Find the time of flight of the masses. (1985; 8M) L

7.

C

After the collision, A comes to rest, B retraces its path with the speed V. What is the velocity of C? (1982; 2M) 4.

A block of mass M with a semicircular track of radius R, rests on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A (see Fig.). The cylinder slips on the semicircular frictionless track. How far has the block moved when the cylinder reaches the bottom (point B) of the track? How fast is the block moving when he cylinder reaches the bottom of the track?(1983; 7M)

2 . What is the minimum number of collisions after 5 which the amplitude of oscillations becomes less than 60 degrees? (1987; 7M) 8.

An object of mass 5 kg is projected with a velocity of 20 m/s at an angle of 60° to the horizontal. At the highest point of its path the projectile explodes and breaks up into two fragments of masses 1 kg and 4 kg. The fragments separate horizontally after the explosion. The explosion released internal energy such that the kinetic energy of the system at the highest point is doubled. Calculate the separation between the two fragments when they reach the ground. (1990; 8M)

9.

A block A of mass 2m is placed on another block B of mass 4m which in turn is placed on a fixed table. The two blocks have a same length 4d and they are placed as shown in figure. The coefficient of friction (both static and kinetic) between the block B and table is µ. There is no friction between the two blocks. A small object of mass m moving horizontally along a line passing through the centre of mass (CM) of the block B and perpendicular to its face with a speed v collides elastically with the block B at a height d above the table. (1991; 4+4M)

r R

m

M B

5.

Two bodies A and B of masses m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third body C of mass m moves with velocity v0 along the line joining A and B and collides elastically with A as shown in Fig. At a certain instant of time t0 after collision, it is found that the instantaneous velocities of A and B are the same. Further at this instant the compression of the spring is found to be x0 . Determine (i) the common velocity of A and B at time t0 and (ii) the spring contant. (1984; 6M) C

A

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position See figure and released. The ball hits the wall, the coefficient of restitution being

A B

B

4m 2d

m

v

d P

6.

2m

A ball of mass 100 g is projected vertically upwards 41

4d

(a) What is the minimum value of v (call it v0) required to make the block A topple? (b) If v = 2 v0 find the distance (from the point P in the figure) at which the mass m falls on the table after collision (Ignore the role of friction during the collision). [1991; 4+4M] 10. A uniform thin rod of mass M and length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin (0, 0). A slight disturbance at r = 0 causes the lower end to slip on the smooth surface along the positive x-axis, and the rod starts falling. (1993; 1+ 5M) (i) What is the path followed by the centre of mass of the rod during its fall? (ii) Find the equation of the trajectory of a point on the rod located at a distance r from the lower end. What is the shape of the path of this point? 11. A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of larger 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 small sphere reaches the other extreme position. (1996; 3M)

r (a) Find the force F and also the normal force exerted by the table on the wedge during the time ∆t. (b) Let h denote the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due r to the normal force N about the centre of the wedge during the interval ∆t. –2 13. A cylindricalsolidofmass1 0kg 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 particle of uniform density 10–3 kg/m3 . When a dust particles collides with the face of the cylinder, it sticks to its surface. Assuming that the dimensions of the cylinder remains practically unchanged and that the dust sticks only to the front face of the xcoordinates of the front of the cylinder find the xcoordinate of the front of the cylinder at t = 150 s. (1998; 5M)

14. Two blocks of mass 2 kg and M are at rest on an inclined plane and are separated by a distance of 6.0 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 10.0 m/s up the inclined plane. It collides with M, comes back and has a velocity of 1.0 m/s when it reaches its initial position. The other block M after the collision moves 0.5 m up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block M. (1999; 10M) [Take sin θ ≈ tan θ = 0.05 and g = 10 m/s 2 ] M m 6.0

12. A wedge of mass m and triangular cross-section (AB = BC =CA =2R) is moving with a constant velocity − viˆ towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time ∆t r during which the sphere exerts a constant force F on the wedge. (1998; 8M) z

y

A R

x

v

B

C

2kg θ

15. 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 balls 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 carriage, what will be the horizontal velocity of the carriage just after the second impact? (2001; 10M) 42

It is given n = 100, M = 3 kg, m = 0.01kg; b = 2m; a = 1m; g = 10 m/s 2 . (2006; 6M) 19. Two towers AB and CD are situated a distance d apart as shown in figure. AB is 20 m high and CD is 30 m high from the ground. An object of mass m is thrown from the top of AB horizontally with a velocity of 10m/s towards CD.

C

P

2m

A

60°

B

16. A particle of mass m, moving in a circular path of radius R with a constant speed v2 is located at point (2R, 0) at time t = 0 and a man starts moving with a velocity v1 along the positive y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. man as a function of time. (2003; 2M)

A

C

m

d

y

D

B

v2

v1

Simultaneously another object of mass 2m is thrown from the top of CD at an angle of 60° to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid air and stick to each other. (1994; 6M) (i) Calculate the distance d between the towers. (ii) Find the position where the object hit the ground.

R (0,0)

m

x

17. Two point masses m1 and m2 are connected by a 20. Three objects A, B and C are kept in a straight line on spring of natural length l0 . The spring is compressed a frictionless horizontal surface. These have masses m, such that the two point masses touch each other and 2m and m, respectively. The object A moves towards then they are fastened by a string. Then the system is B with a speed 9 m/s and makes an elastic collision moved with a velocity v0 along positive x-axis. When with it. Thereafter, B makes completely inelastic collision the system reaches the origin the string breaks (t = 0). with C. All motions occur on the same straight line. The position of the point mass m1 is given by x1 = v0 Find the final speed (in m/s) of the object C. t – A (1 – cos ω t) where A and ω are constants. Find the position of the second block as a function of time. m 2m m Also find the relation between A and l0 . (2003; 4M) 18. There is a rectangular plate of mass M kg of dimensions (a × b). The plate is held in horizontal position by striking n small balls each of mass m per unit area per unit time. These are striking in the shaded all region of the plate. The balls are colliding elastically with velocity v. What is v?

A

B

C

ASSERATION AND REASON

1.

b

a

43

This question contains, statement I (assertion) and statement II (reasons). Statement-I : In an elastic collision between two bodies the relative speed of the bodies after collision is equal to the relative speed before the collison. (2007; 3M) Because : Statement-II : In an elastic collision, the linear momentum of the system is conserved. (a) Statement-I is true, statement -II is true, statementII is a correct explanation for statement-I. (b) statement-I is true, statement-II is true; statementII is NOT a correct explanaion for statement-I. (c) statement-I is true, statement-II is false. (d) statement-I is false, statement-II is true.

2. Statement-I : 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; 3M) Because : Statement-II : The linear momentum of an isolated system remains constant. (a) Statement-I is true, statement -II is true, statementII is a correct explanation for statement-I (b) statement-I is true, statement-II is true; statementII is NOT a correct explanaion for statement-I (c) statement-I is true, statement-II is false. (d) statement-I is false, statement-II is true.

1.

2.

PASSAGE BASED PORBLEM Passage

3.

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 60º 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/s 2 )

The speed of the block at point B immediately after it strikes the second incline is (2008; 4M) (a)

60 m/s

(b)

45 m/s

(c)

30 m/s

(d)

15 m/s

The speed of the block at point C immediately before it leaves the second incline is (2008; 4M) (a)

120 m/s

(b)

105 m/s

(c)

90 m/s

(d)

75 m/s

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 (2008; 4M) (a)

30 m/s

(c) zero A

(b)

15 m/s

(d)

75 m/s

v 60º

B

30º 3m

C

3 3m

ANSWERS FILL IN THE BLANKS 1.

3 2 mv 2

2. 5 × 10–3 N-s

TRUE/FALSE 1. F

OBJECTIVE QUESTION (ONLY ONE OPTION) 1. (d) 8. (c)

2. (c) 9. (a)

3.

(a)

4. (a)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION) 1. (b, d)

2. (a), (d)

44

5. (c)

6. (c)

7. (d)

SUBJECTIVE QUESTIONS 1. 3. 4.

9.

9 cm from centre of bigger circle (leftwards)

2. 10 2 m/s at 45° Velocity of C is V in a direction opposite to velocity of B.

2mV02 V0 5. (i) (ii) 3 3x02

m( R – r ) 2 g (R – r ) ,m M +m M ( M + m) 5 6µgd (b) 6d 3µ 4

(a)

10. (i) Straight line (ii)

6. 6.53 s

13. 105 m

 

16.  − mv2 sin

 2mv  ( 3iˆ − kˆ ),  + mg kˆ 3∆t  3∆ t  14. e = 0.84 M, = 15 kg 15. 12s, 15.75 m/s

4mv

12. (a)

v2  ˆ v   t i + m v2 cos 2 t − v1  ˆj R  R  

8. 44.25 m

x2 y2 + = 1 , ellipse L  r2  –r 2 

2mv

11. (L + 2R, 0)

7. 4

17. x2 = v0t +

(b)

3∆t

h

m  m1 A(1 − cos ? t ), l0 =  1 + 1 A m2 m  2 

18. 10 m/s

ASSERATION AND REASON 1. (b)

2.(d)

PASSAGE BASED PROBLEM 1.

(b)

2. (b)

3. (c)

SOLUTIONS  v  = mv2 + m    2

FILL IN THE BLANKS 1.

From conservation of linear momentum magnitude of r r P3 should be 2mv in a direction opposite of P12 r r (resultant of P1 and P2 ). ∴

Speed of particle of mass 2m ( v′) =

P=mv 2

P3 v = 2m 2

= 2.

2

3 2 mv 2

∫

Impulse = Fdt = area under F-t graph ∴ Total impulse from t = 4 µs to t = 16 µs = Area EBCD = Area of trapezium EBCF +Area of triangle FCD

P12 = 2 mv

1 1 (200 + 800)2 × 10–6 + × 800 × 10 × 10 –6 2 2 = 5 × 10–3 N-s =

P1= mv P3

∴ Total energy released is,

1  1 E = 2  mv 2  + ( 2m ) v'2 2  2

TRUE FALSE 1.

Since, net force on the system is zero. Velocity of centre of mass will remain constant.

OBJECTIVE QUESTIONS (ONLY ONE OPTION) r 1. Net force on system is zero. So ∆ VCOM = 0 and 45

because initially centre of mass is at rest. Therefore, centre of mass always remains at rest. ∴ (d) 2.

3.

4.

θ=

In an inelastic collision only momentum of the system remain conserved. Some energy is always lost in the form of heat, sound etc. ∴ (c)

yCM

∴

meet m2 at A(at θ = ∴ 9.

8.

(c)

ycm =

6m( 0) + m ( a) + m( a) + m (0) + m( − a) a = 6m + m + m + m + m 10

(– a, a ) (a , a )

(0, 0) (0, –a)

r r vCM = m1v1 + m2v2 m1 + m2

∴

10 ×14 + 4 × 0 10 × 14 = = 10 m/s 10 + 14 14

x

(a)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION) v2 = 0 B

1.

m2 = 4kg

r r dP F= = − kAsin(kt)iˆ − kA cos(kt) ˆj dt r P = A cos(kt) iˆ − A sin(kt) ˆj r r Since, F .P = 0 r r ∴ Angle between F and P should be 90°. ∴ (c)

After collisions between C and A, C stops while A moves with speed of C i.e. v [in-head on elastic collision, two equal masses exchange their velocities]. At maximum compression, A and B will move with same speed v/2 (from conservation of linear momentum). Let x be the maximum compression in this position. ∴ KE of A-B system at maximum compression

=

1  v (2m )   2 2

2

or, Kmax = mv 2 /4 From conservation fo mechanical energy in two positions shown in figure.

Let first collision be at an angle θ, ∴

2π anticlockwise from OC) 3

y

v1 =14m/s

7.

2π anticlockwise from OB) 3

Now, again m1 will move back with speed v and

= 2 (m1 + m2 ) gt 0 ∴ (c)

m 1 = 10kg

C

(again at θ =

y2 = – 5cm (a) r r r r | ( m1v ′1 + m2v′ 2) − (m1v1 + m2v2 ) |

=

θ

After first collision at B, m2 will move back with speed v and make collision with m1 at C.

= | m1 (v1 − 2gt0 ) + m2 ( v2 − 2 gt0 ) − m1v1 − m2 v2 |

6.

O

B

or ∴ 5.

2v θ

Before explosion, particle was moving along x-axis i.e., it has no y-component of velocity. Therefore, the centre of mass will not move in y-direction or we can say yCM = 0. Now,

m1 A m2

v

Let v' be the velocity of second fragment. From conservation of linear momentum, 2m (v cos θ) = mv' – m (v cosθ) ∴ v'= 3 v cos θ ∴ (a)

m 3m ×15 + × y2 m1 y1 + m2 y2 4 = ⇒0= 4 m1 + m2 m

2π = 120 ° 3

or,

θr (2π − θ) r = v 2v

1 2 1 1 mv = mv 2 + kx2 2 4 2 1 2 1 kx = mv 2 2 4

2θ = 2π − θ 46

∴ ∴ 2.

move along horizontal we should have Mx = m (R – r – x)

m x =v 2k (b, d)

m(R − r) M+m (ii) Let V1 be the speed of m towards right and v2 the speed of M towards left. From conservation fo linear momentum. mv 1 = Mv2 ...(1) From conservation of mechanical energy x=

r r Initial momentum of the system p1 + p2 = 0 r r ∴ Final momentum p'1 + p' 2 = 0 should also be zero. Option (b) is allowed because if we put c1 = −c2 ≠ 0 r r p ' 1 + p ' 2 will be zero. Similary, we can check other

1 2 1 mv1 + Mv22 2 2 Solving these two equations, we get mg (R – r) =

options. ∴ correct options are (a) and (d).

SUBJECTIVE QUESTIONS 1.

r Rcom =

π ( 42)

v2 = m

2

π[(56)2 –(42) 2 ]

× 7 ( − iˆ) 5.

= 9 cm 2.

uur From conservation of linear momentum P3 should be uuur uur uur equal and opposite to P12 (resultant of P1 and P2 ). So, let v' be velocity of third fragment, then (3m)v' =

...(2)

2 g( R − r ) M ( M + m)

(i) Collision between A and C is elastic and mass of both the blocks is same. Therefore, they will exchange their velocities i.e., C will come to rest and A will be moving will velocity vo . Let V be the common velocity of A and B, then from conservation of linear momentum, we have At rest

2mv v0

∴

v' =

2 v 3

C

v0 A

B

(a)

(b) C

45° 45°

→ P1 = mv

→ P3

or

mA v0 = (mA + mB ) V mv 0 = (m + 2m) V

or

V=

v0 3

1 1 2 1 m Av0 = (m A + mB )V 2 + kx02 2 2 2

2 × 30 3

2

= 10 2 m/s This velocity is at 45° as shown in figure.

4.

B

(ii) Further, from conservation of energy we have,

Here, v =30 m/s (given)

3.

A (c)

→ P2=mv

v' =

V

V

→ P12 = 2 mv

∴

A

C

Before collision net momentum of the system was zero. No external force is acting on the system. Hence momentum after collision should also be zero. A has come to rest. Therefore B and C should have equal and opposite momentum or velocity of C should be V in opposite direction of velocity of B.

6.

or

1 2 1  V0  2 1 mv0 = (3m)   + kx0 2 2 2  3 

or

1 2 1 2 kx0 = mv 0 3 2

or

k=

For collision of both the balls,

49 t −

(i) The centre of mass of M + m in this case will not move in horizontal direction. For centre of mass not to 47

2mv 20 3x 02

1 1 × 9. 8t 2 = 98 − × 9. 8t 2 2 2

B

∴

t = 2 sec., At this instant height from ground

1 2 = 49 t − × 9. 8t = 78. 4 m 2 and using conservation of linear momentum, we get, m1v1 − m2v 2 = 2mv ,

S = ut +

1 2 at 2

8.

⇒

− 78. 4 = 4.9t − 4.9t 2 ⇒ t = 4.53 sec. Total time of flight = 4.53 + 2 = 6.53 sec. 7.

As shown in figure initially when the bob is at A, its potential energy is mgl. When the bob is released it strikes the wall at B. If v be the velocity with which the bob strikes the wall, then l

O

n > 3 (here n is an integer)

Let v1 and v2 be the velocities after explosion in the directions shown in figure. From conservation of linear momentum, we have 5 (20 cos 60°) = 4v1 – 1 × v2 or 4v1 – v2 = 50 ...(1) Further it is given that, kinetic energy after explosion becomes two times. Therefore,

or

1 2 mv or v = 2 gl ...(1) 2 Speed of the bob after rebounding (first time) mgl =

Here, t =

..(2)

The speed after second rebound is v2 = e2 (2 gl)

∴ 9.

In general after n rebounds, the speed of the bob is vn = en (2 gl) ..(3) Let the bob rises to a height h after n rebounds. Applying the law of conservation of energy, we have

vn2 e 2n .2 gl 2n h = 2g = 2g = e .l

∴

 2  h=    5

at

? = 60 °,

u sin θ 20sin60 ° T = = = 1.77 s g 9.8 2

x = 25 × 1.77 = 44.25 m

If v1 and v2 are the velocities of object of mass m and block of mass 4m, just after collision then by conservation of momentum, mv = mv 1 + 4mv 2 , i.e. v = v1 + 4v2 ...(1) Further, as collision is elastic

v2 = 0 or v2 = Therefore,

n

4 .l =   l 5

...(2)

1 2 1 1 mv = mv12 + 4mv22 , i.e. v 2 = v12 + 4v22 ...(2) 2 2 2 Solving, these two equations we get either

1 2 mvn = mgh 2 or

4v12 + v22 =1000

Solving Eqs. (1) and (2), we have v1 = 15 m/s, v2 = 10 m/s or v1 = 5 m/s and v2 = – 30 m/s In both the cases relative velocity of separation in horizontal direction is 25 m/s. ∴ x = 25t = distance between them when they strike the ground

C

v1 = e 2 gl

n

1 2 = 2  × 5 × (20cos60°)  2 

B

h = l(1− cos 60 °) =

∴

A

h

2n

or

l 1 4 4   l < ⇒  < 2 2 5 5 

1 1 × 4 × v12 + ×1 × v22 2 2

θn

Similarly,

? < 60° ⇒ h < n

Solving this, we get v = 4 . 9 m/sec ∴

l 2

∴

...(4)

2 v 5

v2 = 2 v 5

Substituing in Eq. (1) v1 = 3 v 5

l 2

when v2 = 0, v1 = v, but it is physically unacceptable. (a) Now, after collision the block B will start moving with velociy v2 to the right. Since, there is no friction 48

Therefore, CM will fall vertically downwards towards negative y-axis i.e. the path of CM is a straight line. (ii) Refer figure (b). We have to find the trajectory of a point P (x, y) at a distance r from end B. CB = L/2, OB = (L/2) cos θ, MB = r cos θ ∴ x = OB – MB = cos θ {(L/2 – r)}

between blocks A and B the upper block A wil stay at its positions and will topple if B moves a distance s such that s > 2d ...(3) However, the motion of B is retarded by frictional force f = µ (4m + 2m)g between table and its lower surface. So, the distance moved by B till it stops

 6µmg  0 = – 2  s i.e., s = 3µg  4m  Substituting this value of s in Eq. (3), we find that for toppling of A v22

v22 > 6µgd ⇒

y ...(2) r Squaring and adding Eqs. (1) and (2), we get or,

sin +

5 6µgd 2 or,

speed.

2d g

cos2 θ

x2 =

{( L /2) 2 – r 2 }

+

y2 r2

x2 y2 + =1 {( L / 2) − r }2 r 2

...(3)

This is an equation of an ellipse. Hence, path of point P is an ellipse whose equation is given by (3).

3 v1 = v = 3 6 µgd 5 and as time taken by it to fall down

2h = g

sin θ =

2θ

(b) if v = 2v0 = 5 6µgd , the object will rebound with

t=

...(1)

Similarly, y = r sin θ

2 v > 6µgd ⇒ v > 5 6µgd 5 2

v min = v 0 =

or

x {( L / R) − r}

cos θ =

or

v22

[as h = d]

11. Since, all the surfaces are smooth, no external force is acting on the system in horizontal direction. Therefore, the centre of mass of the system in horizontal direction remains stationary.

The horizontal distance moved by it to the left of P in this time

y

y CC 1 2= 5R (in both cases)

x = v1t = 6d 3µ Note : (a) Toppling will take place if line of action of weight does not pass through the base area in contact. (b) v1 and v2 can obtained by using the equations of head on elastic collision r r r v1 = 2v cm − u1 r r r v 2 = 2vcm − u 2

c1 (L,0)

C

x1 =

y (a)

M θ

x (x,0)

(L + 5R,0)

Final

m1 x1 + m2 x2 m1 + m2

( 4M )( L ) + M ( L + 5 R) = (L + R ) ...(1) 4M + M Let (x, 0) be the coordinates of the centre of large sphere in final position. Then, x-coordinate of CM finally will be =

( 4M )( x) + M ( x − 5R) = ( x − R) ...(2) 4M + M Equating Eqs. (1) and (2), we have x = L + 2R Therefore, coordinates of large sphere, when the smaller sphere reaches the other extreme position, are (L + 2R, 0) x1 =

(O, L/2 sin θ)

O

c1

x- coordinate of CM initially will be given by

P (x, y) B

c2

c 2 = (x – 5R, 0)

A

C

x

Initial

10. (i) Since, only two forces are acting on the rod, its weight Mg (vertically downwards) and a normal reaction N at point of contact B (vertically upwards). No horizontal force is acting on the rod (surface is smooth). A

c2

B

x (b)

49

12. (a) (i) Since, the collision is elastic and sphere is fixed,

Magnitude of torque of N about CM = magnitude of torque of F about CM = F.h

the wedge will return with velocity of viˆ z

–v i

y

+v i

x

Fixed

F

 4mv  r h | t N |=    3 ∆t  13. Given : m0 = 10–2 kg, A = 10–4 m2, v0 = 103 m/s

A

m0

v

vo

m

F sin 30°

Now, linear impulse in x-direction = change in momentum in x-direction. ∴ (F cos 30°)∆t=mv – (– mv) = 2mv ∴

F= F=

∴ or

x

x=0 At t = 0

At t = t

ρ dust = ρ = 10 –3 kg/m3 m = m0 + mass of dust collected so far = m0 + Ax ρ dust or m = m0 + Ax ρ

4mv 2mv = ∆t cos30° 3∆t

and

4mv

3∆t r F = F cos 30°iˆ − ( F sin 30 °) kˆ

The linear momentum at t = 0 is P0 = m0v0 and momentum at t = t is Pt = mv = (m0 + Ax ρ )v From law of conservation of momentum P0 = Pt

r  2mv   2mv   kˆ F =  iˆ −    ∆t   3∆t 

(ii) Taking the equilibrium of wedge in vertical N

C mg

∴

m0v0 = (m0 + Ax ρ ) v (but v =

∴

m0v0 = (m0 + Ax ρ=)

or

(m0 + A ρ x)dx = m0v0 dt 2

or

F sin 30°

0

z- direction during collision. N = mg + F sin 30° N = mg +

∫

dx dt

150

(m0 + A ρ x)dx =

∫ 0

m0v0dt

x

⇒

2mv 3∆t

r  2mv  ˆ k N =  mg +  3 ∆t   (b) For rotational equilibrium of wedge [about (CM)] N

C mg

F

 x2  150 m0 x + A?  = [m0v0t ]0 2   0

x2 = 150 m0v0 2 Solving this quadratic equation and substituting the values of m0, A, ρ and v0, we get, x = 105m.

or in vector form

h

dx ) dt

Hence, m0 x + Aρ

14. Let v1 = velocity of block 2 kg just before collision. v2 = velocity of block 2 kg just after collision. and v3 = velocity of block M just after collision. Applying work energy theorem, (change in kinetic energy = work done by all the forces at different stages as shown in figure.

50

6

m

v1

V1

M /s m 10 θ h1= 6 sin θ

0.5 V3

2kg

2kg

θ

θ

Figure I. For 2 kg block between starting point and the moment just before collision ∆KE = Wfriction + Wgravity

θ

h= 2 0.5 sin θ

(i) Coefficient of restitution (e)

Relativevelocityof separation = Relativevelocityof approach

1 m{v12 − (10)2} = −6µmg cosθ − 6mg sin θ 2

=

Given, µ = 0.25, sin ? ≈ tan ? = 0. 05,

v2 + v3 v1

5 + 1.73 = 0.84 8 or e = 0.84 (ii) Applying conservation of linear momentum before and after collision 2v1 = Mv3 – 2v2 =

∴

cos ? ≈ 1

∴

v12 − 100 = −2[ 6 × 0. 25× 10× 1 + 6 × 10× 0.05] v12 = 100 − 36 = 64

∴

M

v1 = 8 m/s

Figure 2. For 2 kg block, just after collision and at the initial point ∆KE = Wfriction + Wgravity

∴

M=

2(v1 + v2 ) 2(8 + 5) 26 = = v3 1.73 1.73

M ≈ 15 kg 15. (i) Velocity of the ball relative to ground = 100 m/s y (vertical)

uy u = 100m/s

1m/s

v2 30°

v = 5 3m/s

1 m[(1) 2 − v22 )] = −6µmg cos ? + 6mg sin ? 2 or

1 − v22 = −24

∴

v 22 = 25, or

y (horizontal) ax = 0 and 2 a y= – g = – 10 m/s C 120m A

Horizontal component of its velocity.

v2 = 5 m/s

u x =u cos 30° = 50 3 m/s and vertical component of its velocity, u y = u sin 30° = 50 m/s Vertical displacement of the ball when it strikes the carriage is –120 m or

Figure. 3. For block M, Just after collision and at the point of stop, ∆KE = Wfriction + Wgravity

1 1 2 2 S y = u y t + a y t ⇒ –120 = (50 t) +   (–10) t 2 2

1 M [ 0 − v32 ] = − (0.5)(µ)( M ) g cos ? − 0.5Mg sin ? 2

⇒ t2 – 10t – 24 = 0 ∴ t = 12 s or – 2 s Ignoring the negative time, we have ∴ t0 = 12 s

v32 = ( 0. 5 × 0. 25× 10 ×1) + 0.5 × 10 × 0. 05 v32 = 2 × 1.5 = 3 v3 = 1. 75 m/s

ux

(ii) When it strikes the carriage, its horizontal component of velocity is still 50 3 m/s. It stickes to the carriage. 51

Let V2 be the velocity of (carriage + ball) system after collision. Then, applying conservation of linear momentum in horizontal direction :

or

∴

and

1× 50 3 = (1 + 9)V2

r v v   v p =  − v2 sin 2 tiˆ + v2 cos 2 tˆj  R R   r v m = v1 ˆj

∴ Linear momentum of particle w.r.t man as a function of time is r r r p pm = m( v p − vm )

∴ V2 = 5 3 m/s The second ball is fired when the first ball strikes the carriage i.e. after 12 s. In these 12 s the car will move forward a distance of 12V1 or 60 3 m. The second ball also takes 12 s to travel a vertical displacement of – 120 m. This ball will strike the carriage only when the carriage also covers the same

 v   v   = m  − v 2 sin 2 t iˆ +  v 2 cos 2 t − v1  ˆj  R   R   

distance of 60 3 m in these 12s. This is possible only 17. (i) when resistive forces are zero because velocity of car (V1) = Velocity of carriage after first collision (V2) =

x1 = v0 t – A (1 – cos ωt)

5 3 m/s. Hence, at the time of second collision : Let V be the by velocity of carriage after second collision. Conservation of linear momentum in horizontal direction gives

(As there is no external force, so COM will move with constant velocity v0)

1kg

5 3m/s

Before collision

∴

m x2 = v0t + 1 A(1–cos ωt ) m2

(ii)

a1 =

100 3 m/s or V = 15.75 m/s 11

or,

16. Angular speed of particle about centre of the circle

= – ω2 A cos ωt

v2

∆p b  = n ×  a ×  × (2mv) ∆t 2  Equating the torque about hinge side, we have,

18. F =

v1

(0, 0)

t=0

b 3b b  n ×  a ×  × (2mv) × = Mg × 2 4 2  Substituting the given values we get, v = 10 m/s

x

R

r v p = ( −v2 sin ?iˆ + v 2 cos ?jˆ)

 m1  + 1 A (1 - cos ω t), (cos ω t = 0) l0 =   m2  m  l 0 =  1 + 1 A  m2 

y θ

2

Thus, the relation between l0 and A is,

v θ = ωt = 2 t R

v2

d 2 x1

m1 x2 – x1 = m A (1 – cos ωt) + A (1 – cos ωt) 2

11 V = (1) (50 3 ) + (10) (5 3 ) = 100 3

v ω= 2 , R

m1 x1 + m2 x 2 = v0 t m1 + m2

dt The separation x2 – x1 between the two blocks will be equal to l0 and a 1 = 0 or cos ωt = 0

µx = 50 3 m/s

∴ V =

xCM =

19. Acceleration of A and C both is 9.8 m/s 2 downwards

52

∴

d 2 = (d – d 1) = (17.32 – 11.55)m = 5.77 m Therefore, position from B is d 1 i.e., 11.55 m and from D is d 2 or 5.77 m.

C θ 60° →

→

→

C =V

–V

A

X 10m

→

VC

V CA

Y →

θ vA

A

20. Let after collision velocity of block A and B be vA and vB respectively. r r r v B = (1 + e)vCOM − euB

E 30m

d

 9m  = (1 + 1)  − 0 = 6 m/sec  3m  r r r vC = (1+ e) vCOM − euC

20m

 2m × 6  = (1)  − 0 = 4 m/sec  3m  D

B

ASSERATION AND REASON

Let A be the origin, so

1.

1 1 y1 = − gt 2 & y2 − 10 = −5 3t − gt 2 2 2 For collision y1 = y2 ⇒

10 = 5 3t t=

2 3

sec

d = 10t + 5t =

30

= 17 .32 m 3 Horizontal (or x) component of initial linear momentum of projectile m and 2m, PAx + PCx = mv A − 2m ( vC cos 60 °) = 0 i.e., x-component of momentum of combined mass after collision will also be zero i.e., the combined mass will have momentum or velocity in vertical or y-direction only. ⇒

In case of elastic collision, coefficient of restitution e = 1. or relative speed of approach = relative speed of separation. ∴ (b)

2. If a force is applied at centre of→ F mass of a rigid body, its torque aboutcentreofmasswillbezero butaccelerationwillbenon-zero. Hence, velocity will change. ∴

C

(d)

PASSAGE BASED PROBLEM 1.

Between A and B, height fallen by block

h1 = 3 tan 60 º = 3 m ∴

v1 = 2 gh1 = 2 × 10 × 3 = 60 ms −1

vcx= vcos60° c C V0 60° A

vax= vA P

B

t=

2 s 3

F d1

In perfectly inelastic collision, component of v 1 perpendicular to BC will become zero, while component of v1 parallel to BC will remain unchanged. ∴ speed of block B immediately after it strikes the second incline is, v2 = component of v1 along BC

D d2

Hence, the combined mass will fall at point F just below the point of collision P. Here,

d 1 = VAxt = (10) ×

 3 −1   = v1 cos 30º = ( 60 ) 2  = ( 45 ) ms  

2 3

= 11.55 m 53

∴ 2.

correct option is (b).

Height fallen by the block from B to C

h2 = 3 3 tan30º = 3m Let v3 be the speed of block, at point C, just before it leaves the second incline, then : h3 =

∴ 3.

v 22 + 2 gh

2

= 45+ 2×3 = 105 ms−1 correct option is (b).

In elastic collistion, component of v1 parallel to BC will remain unchanged, while component perpendicular to BC will remain unchanged in magnitude but its direction will be reversed.

 3   v11 = v1 cos 30º = ( 60 ) 2  =   v⊥

45 ms −1

1 −1  = 15 ms 2  

= v1 sin 30º = ( 60 )

Now vertical component of velocity of block : ∴

v = v⊥ cos 30 º− v11 cos 60 º = 0 (c)

54