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BMSAP B.M.SHARMA ACADEMY OF PHYSICS 1 A-PDF PPT TO PDF DEMO: Purchase from www.A-PDF.com to remove the watermark 1 www.a-pdf.com

Center of mass hcv

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Page 1: Center of mass hcv

BMSAPB.M.SHARMA ACADEMY OF PHYSICS

1

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PP

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O P

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: Purchase from

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to remove the w

atermark

1

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Page 2: Center of mass hcv

Consider a gravity- free hall in which a tray of mass M, carrying a cubical block of ice of mass m and edge L, is at rest in the middle . If the ice melts, by what distance does the centre of mass of “ the tray plus the ice”system descend ?

2

Problem 10.Problem 10.

M m

L

2

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Page 3: Center of mass hcv

Problem 12.Problem 12.

Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat.

3

Neglecting friction with water, how far does the boat move in the water during the process ?

Solution.Solution. Let displacement of the boat is xb = X

Mr Verma Mr Mathur(40 kg)

(50 kg) (60 kg)

4m

3

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Page 4: Center of mass hcv

As no external forces are acting on the system the displacement of centre of mass of the system should be zero.

0 0

( )

+ ++ ++ ++ +====

+ ++ ++ ++ +V V M M

cm

b v m

M x M x M xx

M M M

40 50( 2) 60( 2)0

150

x x× + + + −× + + + −× + + + −× + + + −⇒ =⇒ =⇒ =⇒ =

13x m⇒ =⇒ =⇒ =⇒ =

The displacement of Mr. Verma,Xv = (x + 2) m

The displacement of Mr. Mathur,Xm = (x - 2) m

44

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Page 5: Center of mass hcv

Problem 13.Problem 13.

A cart of mass M is at rest on a frictionless horizontal surface and a pendulum bob of mass m hangs from the roof of the cart . The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is L.

5

Find the displacement of the cart during this process.

5

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Page 6: Center of mass hcv

Solution.Solution.

Let displacement of cart

cx x= −= −= −= −Displacement of bob

( )bx L X= −= −= −= −

( ) 0cm systemx ====

. 0cart cart bob bobM x M x+ =+ =+ =+ =

.( ) ( ) 0M x m L X− + − =− + − =− + − =− + − =

mLx

m M====

++++

m

M

L

m

M

L

(-x)

66

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Page 7: Center of mass hcv

Problem 19.Problem 19.A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically. When at a height h from the ground, he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance x from the line of fall.

7

In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond.Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land ?

7

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Page 8: Center of mass hcv

�H

h

hard ground pond

x

88

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Page 9: Center of mass hcv

To avoid the man falling on hard floor the man should be displaced by a horizontal displacement x during which he fall a height h.

Solution.Solution.Considering (man + bag) as a system; No external forces in horizontal direction is acting on man.

H

h

pond

x

99

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Page 10: Center of mass hcv

. 0man bagman bagM V M V→ →→ →→ →→ →

+ =+ =+ =+ =

0b

xM mv

t

− =− =− =− =

The time taken by man to fall a height hT = time to fall H – time to fall (H – h)

2 2( )H H ht

g g

−−−−= == == == =

Hence horizontal velocity of man required

man

xV

t====

As no external force in horizontal direction hence, the linear momentum of ‘man + bag’ system should be conserved (i.e. equal to zero).

10

…(i)

10

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Page 11: Center of mass hcv

.bag

M xv

m t====

[ 2 2( 2)]bag

Mx gv

m H H⇒ =⇒ =⇒ =⇒ =

− −− −− −− −The displacement of centre of mass of “bag + Man”system should be zero.

0man bagmanM x m x→→→→ →→→→

+ =+ =+ =+ =0bagMx mx⇒ − =⇒ − =⇒ − =⇒ − =

bag

Mxx

m⇒ =⇒ =⇒ =⇒ =

1111

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Page 12: Center of mass hcv

Problem 20.Problem 20.

A ball of a mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence 450. The ball is reflected by the plane at equal angle of reflection with the same speed.

12

Calculate (a) the magnitude of the change in momentum of the

ball(b) the change in the magnitude of the momentum of

the ball.

Solution.Solution.(a) The momentum of the

ball just before strikingv v

mm

45045

0

12

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Page 13: Center of mass hcv

Change in linear momentum, final initialp p p→→→→→→→→ →→→→

∆ = −∆ = −∆ = −∆ = −

0 ˆ2 cos45p mv j∆ =∆ =∆ =∆ =

| | 2p mv⇒ ∆ =⇒ ∆ =⇒ ∆ =⇒ ∆ =

0.14 /p kg m s⇒ ∆ =⇒ ∆ =⇒ ∆ =⇒ ∆ =

| | | |initial finalp p→→→→ →→→→

====(b) As,

Hence, no change in linear momentum of the ball.

0 0ˆ ˆsin45 cos45initialp mv i mv j→→→→

= −= −= −= −0 0ˆ ˆsin45 cos45finalp mv i mv j

→→→→= += += += +

1313

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Page 14: Center of mass hcv

Problem 21.Problem 21.Light in certain cases may be considered as a stream of particles called photons. Each photon has a linear momentum h/λ where h is the Planck’s constant and λ is the wavelength of the light. A beam of light of wavelength λ is incident on the plane mirror at an angle of incidence θ.

14

Calculate the change in the linear momentum of a photon as the beam is reflected by the mirror.Solution.Solution.No change in linear momentum parallel to mirror but there will be change in linear momentum in the direction normal to mirror.

p = λ

hp =

λh

θ θ

14

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Page 15: Center of mass hcv

| | ( cos ) ( cos )p p p∆ = θ − − θ∆ = θ − − θ∆ = θ − − θ∆ = θ − − θ2 cosp= θ= θ= θ= θ

2| | cos

hp∆ = θ∆ = θ∆ = θ∆ = θ

λλλλ

1515

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Page 16: Center of mass hcv

Problem 24.Problem 24.During a heavy rain, hailstones of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstones strike every square meter of a 10 m ×10 m roof perpendicularly in one second and assume that the hailstones do not rebound.

16

Calculate the average force exerted by the falling hailstones on the roof. Density of the hailstones is 900 kg/m3.

Solution.Solution. Force applied by hail stone

| | t lmv mvpF

t t

−−−−∆∆∆∆= == == == =

∆ ∆∆ ∆∆ ∆∆ ∆0

| | l lmv mvF

t t

−−−−= == == == =

∆ ∆∆ ∆∆ ∆∆ ∆

16

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Page 17: Center of mass hcv

34 1

2000 10 10 900 203 2 100

| |1

F

× × × π × ×× × × π × ×× × × π × ×× × × π × ×

×××× ====

| | 1900F N====

1717

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Page 18: Center of mass hcv

Problem 25.Problem 25.A ball of mass m is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls.

18

Find the average force exerted by the ball on the floor during a long time interval.

18

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Page 19: Center of mass hcv

Problem 26.Problem 26.A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. If the car recoils with a speed v backward on the rails, with what velocity is the man approaching the engine ?

19

Solution.Solution.As linear momentum of car + man system in horizontal direction will remain constant (i.e., will be zero)

vm,c

m

Engine

M

v

19

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Page 20: Center of mass hcv

,( ) ( ) 0m cm v v M v− + − =− + − =− + − =− + − =

, ( )m cmv m M v⇒ = +⇒ = +⇒ = +⇒ = +

,

( )m c

m Mv v

m

++++⇒ =⇒ =⇒ =⇒ =

2020

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Page 21: Center of mass hcv

Problem 27.Problem 27.A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50 m where m is the mass of one shell. If the muzzle velocity of the shells is 200 m/s, what is the recoil speed of the car after the second shot ? Neglect friction.

21

Solution.Solution.

Taking Gun and car as a system no external force are acting on the system in horizontal direction at any time.

v1

gun vs,g

=200 m/s

21

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Page 22: Center of mass hcv

After IInd shot 20049 48

49m m− × = −− × = −− × = −− × = −

2

200200

49v m

+ −+ −+ −+ −

2

1 1200

49 48v

= += += += +

149 .200mv m− =− =− =− =

1

200/

49v m s====

After Ist shot

2222

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Page 23: Center of mass hcv

Problem 28.Problem 28.Two persons each of mass m are standing at the two extremes of a railroad car of mass M resting on a smooth track. The person on left jumps to the left with a horizontal speed u with respect to the state of the car before the jump.

23

Thereafter, the other person jumps to the right, again with the same horizontal speed u with respect to the state of the car before his jump. Find the velocity of the car after both the persons have jumped off.

23

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Page 24: Center of mass hcv

Solution.Solution.Considering both man and car as a system.

Initially the linear momentum of the system is zero.

1 10 ( ) ( )m v u M m v= − + += − + += − + += − + +

12

muv

M m====

++++

Now, taking II man and car as a system

Initially linear momentum of the system is

…(i)

v1

2424

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Page 25: Center of mass hcv

As, pinitial = pfinal

1( ) ( )( 2 )

initial

mup m M v m M

M m= + = += + = += + = += + = +

++++

2 2( )finalp m u v Mv= + += + += + += + +

…(ii)

…(iii)

Now taking II man and car as a system

Initially linear momentum of the system isv

2

2525

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Page 26: Center of mass hcv

2 2( ) ( )( 2 )

mum M m u v Mv

M m+ = + ++ = + ++ = + ++ = + +

++++2

2( 2 )( )

m uv

M m m M⇒ = −⇒ = −⇒ = −⇒ = −

+ ++ ++ ++ +

Hence, the car will move with velocity2

( 2 )( )

m u

M m m M+ ++ ++ ++ +towards left.

2626

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Page 27: Center of mass hcv

Problem 29.Problem 29.Figure shows a small block of mass m which is started with a speed v on the horizontal part of the bigger block of mass M placed on a horizontal floor. The curved part of the surface shown is semicircular. All the surfaces are frictionless.

27

Find the speed of the bigger block when the smaller block reaches the point A of the surface.

Av

m

27

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Page 28: Center of mass hcv

Solution.Solution.Considering bigger block and smaller block as a system.The system has no force in horizontal direction.As surface of M is smooth i.e. no friction between M & m. Hence the bigger block will start moving when m reaches at the semicircular track.

When “m” reaches at A, the ‘m’ block will move in vertical direction and in horizontal direction both M and m will move with same velocity (v).

A vm

A

v’

v

2828

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Page 29: Center of mass hcv

initialp mv====

( )finalp m M v= += += += +

initial finalp p⇒ =⇒ =⇒ =⇒ = mvv

M m⇒ =⇒ =⇒ =⇒ =

++++

Considering the linear momentum of the system in horizontal direction.

2929

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Page 30: Center of mass hcv

Problem 30.Problem 30.In a typical Indian Bugghi (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which one person can sit. A bugghi of mass 200 kg is moving at a speed of 10 km/hr. As it overtakes a school boy walking at a speed of 4 km/hr, the boy sits on the wooden plate.

30

If the mass of the boy is 25 kg, what will be the new velocity of the bugghi ?Solution.Solution.Considering the ‘Bugghi + School boy’ a system

200 10 25 4initialp = × + ×= × + ×= × + ×= × + ×

(200 25)finalp V= ×= ×= ×= ×

30

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Page 31: Center of mass hcv

As, pinitial = pfinal

200 10 25 4 225 V× + × =× + × =× + × =× + × =28

/3

V km hr⇒ =⇒ =⇒ =⇒ =

3131

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Page 32: Center of mass hcv

Problem 33.Problem 33.Consider a head- on collision between two particles of masses m1 and m2. The initial speeds of the particles are u1 and u2 in the same direction. The collision starts at t = 0 and the particles interact for a time interval ∆t. During the collision, the speed of the first particle varies as

32

Find the speed of the second particle as a function of time during the collision.

1 1 1

1( ) ( )t u u

tυ = + υ −υ = + υ −υ = + υ −υ = + υ −

∆∆∆∆

32

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Page 33: Center of mass hcv

Problem 34.Problem 34.A bullet of mass m moving at a speed v hits a ball of mass M kept at rest. A small part having mass ‘m’breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed v1 in the direction of the bullet.

33

Find the velocity of the bullet after the collision.

33

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Page 34: Center of mass hcv

Problem 38.Problem 38.Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which Bcatches. Then B rolls the ball toward A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform.

34

(a) Find the speed of A after he rolls the ball for the first time.

(b) Find the speed of A after he catches the ball for the first time.

(c) Find the speed of A and B after the ball has made 5 round trips and is held by A.

(d) How many times can A roll the ball ?

(e) Where is the centre of mass of the system “ A + B + ball” at the end of the nth trip ?

34

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Page 35: Center of mass hcv

35

p = 20kg m/sA

20 kg m/s

A

20kg m/s

B

p =

40kg m/s

A,1 20kg m/s 40kg m/s

Solution.Solution.After every throw ‘A’ or ‘B’ gain the momentum 20 kg/m/s in the direction opposite to the motion of ball.

Ist round

p = 20kg m/sA

m = p = 20 kg m/sball ballv ball

BA

35

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Page 36: Center of mass hcv

At the end of each round both the man receive a momentum 40 kg m/s.

(a) pA = 20 = 40 vA1

/2

Av m s====

p = 60kg m/sA

20 kg m/s

A

40kg m/s

B

20kg m/s

p =

80kg m/s

A,2

20kg m/s 60kg m/s 20kg m/s p =60kg m/sA

Second round

3636

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Page 37: Center of mass hcv

(c) At the end of 5 round trip the linear momentum of (ball + ‘A) system

,5 40 5 200 /Ap kg m s= × == × == × == × =

,5200 (40 4) Av= += += += +

,5

50/

11Av m s====

(d) ‘A’ can catch the ball untill the velocity of ‘A’ is less than 5 m/s. After 5 round when ‘A’ through the ball i.e., 6th time the velocity of ‘A’ become greater than 5 m/s hence A can through the ball only 6 times.

(b) pA,1 = 40 = (40+4) vA,1

,1

40 10/

44 11Av m s= == == == =

3737

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Page 38: Center of mass hcv

Let the centre of mass is at a distance x cm from the initial position of A.

(40 40 4) 40 0 4 0 40cmx d+ + = × + × ++ + = × + × ++ + = × + × ++ + = × + × +

10

21cmx m==== From initial position of A.

(e) As no external force is acting on ‘A’, ‘B’ and ‘ball’system hence the position of centre of mass of system should remain constant always.

3838

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Page 39: Center of mass hcv

39

Problem 41.Problem 41.A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m.

Find the maximum compression of the spring .

1.0 m/s

2 kg 2 kg

39

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Page 40: Center of mass hcv

Problem 42.Problem 42.A bullet of mass 20 g travelling horizontally with a speed of 500 m/s passes through a wooden block of mass 10. 0 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest.Find the friction coefficient between the block and the surface.

40

500 m/s

40

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Page 41: Center of mass hcv

Problem 43.Problem 43.A projectile is fired with a speed u at an angle θ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field ?

4141

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Page 42: Center of mass hcv

Problem 46.Problem 46.A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact.

Find the maximum extension of the spring.

Take g = 10 m/s2.

42

p v

42

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Page 43: Center of mass hcv

Problem 49.Problem 49.Two masses m1 and m2 are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance x0 when the system is released from rest.Find the distance moved by the two masses before they again come to rest.

4343

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Page 44: Center of mass hcv

Problem 50.Problem 50.Two blocks of masses m1 and m2 are connected by a spring of spring constant k. The block of mass m2 is given a sharp impulse so that it acquires a velocity v0towards right.

Find (a) the velocity of the centre of mass,

(b) the maximum elongation that the spring will suffer.

44

m1 m2

v0

44

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Page 45: Center of mass hcv

Problem 51.Problem 51.Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force F instead of any impulse.Find the maximum elongation that the spring will suffer and the distances moved by the two blocks in the process.

4545

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Page 46: Center of mass hcv

Problem 52.Problem 52.Consider the situation of the previous problem. Suppose the block of mass m1 is pulled by a constant force F1 and the other block is pulled by a constant force F2.

Find the maximum elongation that the spring will suffer.

4646

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Page 47: Center of mass hcv

Problem 54.Problem 54.The track shown in figure is frictionless. The block B of mass 2m is lying at rest and the block A of mass m is pushed along the track with some speed. The collision between A and B is perfectly elastic.With what velocity should the block A be started to get the sleeping man awakened ?

47

A

Bh h

47

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Page 48: Center of mass hcv

Problem 55.Problem 55.A bullet of mass 10 g moving horizontally at a speed of 50√7 m/s strikes a block of mass 490 g kept on a frictionless track as shown in the figure. The bullet remains inside the block and the system proceeds towards the semicircular track of radius 0.2 m.Where will the block strike the horizontal part after leaving the semicircular track ?

48

0.2 m

48

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Page 49: Center of mass hcv

Problem 56.Problem 56.Two balls having masses m and 2m are fastened to two light strings of some length l. The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic.(a) Find the velocities of the balls rise after the collision ?

(b) How high will the balls rise after the collision ?

49

m 2m

1 0 1

49

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Page 50: Center of mass hcv

Problem 57.Problem 57.A uniform chain of mass M and length L is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest.

Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.

5050

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Page 51: Center of mass hcv

Problem 58.Problem 58.The blocks shown in figure have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest.

Find the distance travelled by B if

(a) the collision is perfectly elastic and

(b) the collision is perfectly inelastic.

Take g = 10 m/s2.

51

A10 m/s

B

51

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Page 52: Center of mass hcv

Problem 59.Problem 59.The friction coefficient between the horizontal surface and each of the blocks shown in figure is 0.20. The collision between the blocks is perfectly elastic.Find the separation between the two blocks when they come to rest

Take = g = 10 m/s2.

52

2 kg10 m/s

2 kg

16 cm

52

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Page 53: Center of mass hcv

Problem 60.Problem 60.A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure. Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

53

m

M

h

α

53

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Page 54: Center of mass hcv

Problem 61.Problem 61.Figure shows a small body of mass m placed over n larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume that all the surfaces are frictionless.

(a) Find the speed of the larger block when the smaller block is sliding on the vertical part.

(b) Find the speed of the smaller mass when it breaks off the larger mass at height h.

(c) Find the maximum height (from the ground) that the smaller mass ascends.

(d) Show that the smaller mass will again land on the bigger one.

5454

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Page 55: Center of mass hcv

Find the distance traversed by the bigger block during the time when the smaller block was in its flight under gravity .

55

mM

h

55

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Page 56: Center of mass hcv

Problem 62.Problem 62.A small block of superdense material has a mass of 3 ×1024 kg. It is situated at a height h (much smaller than the earth’s radius) from where it falls on the earth’s surface.Find its speed when its height from the earth’s surface has reduced to h/2. The mass of the earth is 6 × 1024

kg.

5656

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Page 57: Center of mass hcv

Problem 63.Problem 63.A body of mass m makes an elastic collision with another identical body at rest. Show that if the collision is not head- on, the bodies go at right angle to each other after the collision.

5757

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Page 58: Center of mass hcv

Problem 64.Problem 64.A small particles travelling with a velocity v collides elastically with a spherical body of equal mass and of radius r initially kept at rest. The centre of this spherical body is located a distance p(< r) away from the direction of motion of the particle .

Find the final velocities of the two particles.

[Hint : the force acts along the normal to the sphere through the contact. Treat the collision as one-dimensional for this direction. In the tangential direction no force acts and the velocities do not change]

58

vrp

58

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