Centre of mass, impulse momentum (obj. assignments)

BMSAPB.M.SHARMA ACADEMY OF PHYSICS

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Illustration Illustration 17.17.

A projectile of mass m if fired with an initial velocity v0 at an angle θθθθ to the horizontal. At its highest point, it explodes into two fragments of equal mass. One of the fragments fall vertically with zero initial speed .

y

m

xCMR/2

R

X2

O

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(a)v

g

θθθθ202 sin2

(b)v

g

θθθθ20 sin2

(c)v

g

θθθθ20 sin2

2(d)

v

g

θθθθ203 sin2

2

AnswerAnswer

(b)

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Since the only external force acting on the system is gravitational, the motion of centre of mass of the system (the fragments) follows the same parabolic path as the projectile would have followed if there had been no explosion. Force of explosion is internal, it cannot change the trajectory of the system.

No external force acts on the system in x-direction. Therefore

CM

m x m xx

m m1 1 2 2

1 2

→ →→ →→ →→ →

++++====

++++

m R m xR

m2/ 2

2

× + ×× + ×× + ×× + ×====

x R2 3 /2====

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If we choose origin at position of centre of mass, then

CMx m x m x1 1 2 20→→→→ → →→ →→ →→ →

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

m R m x2( / 2) 0− + × =− + × =− + × =− + × =

x R2 / 2= += += += +

If one of the fragments lands back at the initial position of the projectile,

CM

m m xx R

m m20× + ×× + ×× + ×× + ×

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

x R2 2====

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If the vertical component of velocity of both the fragments, after explosion, is zero, they land at the same time. If one of the fragments is moving downward after the explosion, the other fragment will have upward component of velocity.

In this case the downward moving fragment strikes the ground first. The ground exerts a force on it before the second fragment reaches the ground, that is an external force on the system, therefore our analysis does not apply after this instant.

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A small sphere of radius R is released from rest on the inner surface of a large sphere as shown in figure. There is no friction between any surfaces of contact. When the small sphere reaches the other extreme position,.

7

(a) (b)

(c) (d)

AnswerAnswer

(c)

R R

2R 2R

y

x

M

System

6R

(L,0)R

4M

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y

x

M

System

6R

(L,0)R

4M

Therefore the position of the centre of mass of the system remains fixed in the horizontal direction. When the small sphere moves to left, the larger sphere moves to right so that the centre of mass does not move in the x-direction.

Now, CM

m x m xx

m m1 1 2 2

1 2

→ →→ →→ →→ →→→→→ ∆ + ∆∆ + ∆∆ + ∆∆ + ∆

∆ =∆ =∆ =∆ =++++

m x m x1 1 2 2 0→ →→ →→ →→ →

∆ + ∆ =∆ + ∆ =∆ + ∆ =∆ + ∆ =

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Let centre of large sphere move through x towards right .

M R x Mx( 10 ) 4 0− + + =− + + =− + + =− + + =

x R2====Now coordinates of centre of mass are (L + 2R,0)

Alternatively, CM i

M L M L Rx L R

M M

(4 )( ) ( 5 )( ) ( )

4

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

++++

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x-coordinate of centre of mass is fixed.

CM f

M x M x Rx x R

M M

(4 )( ) ( 5 )( ) ( )

4

+ −+ −+ −+ −= = −= = −= = −= = −

++++

Therefore

CM i CM fx x( ) ( )====

L R x R( ) ( )+ = −+ = −+ = −+ = −

x L R2= += += += +

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A rocket projected towards east explodes at the topmost point of its trajectory, 550 m from the point of projection. One of the fragments is found at a location 550 m east and 120 m north of the launch point. Second fragment is found at a location 550 m east and 65 m south of the launch point.

First two fragments are of equal mass m and third fragment has mass 2m. If all the three fragments struck the ground simultaneously, what is the location of the third fragment ?

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(a)

AnswerAnswer

(a)

1100 m east and 27.5 m south

(b) 1100 m east and 27.5 m north

(c) 1100 m east and 54 m south

(d) 1100 m east and 54 m north

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Solution.Solution.

Forces generated by explosion are internal forces. Motion takes place under the gravitational force even after explosion.

Therefore the centre of mass of the system continues its parabolic trajectory, i.e., centre of mass strikes the ground at the same place where the entire rocket would have done, at a distance R = 1100 m east of the launch point.

Z(m)

X(m)

(East)

(North)Fragment 1

Fragment 2

550O

Fragment 3

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CM

m x m x m xx

m m m1 1 2 2 3 3

1 2 3

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

+ ++ ++ ++ +

m m mxm

m3(550) (550) 2

(1100 )4

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

14

x m3 1650====

CM

m y m y m yy

m m m1 1 2 2 3 3

1 2 3

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

+ ++ ++ ++ +

Similarly,

Z(m)

X(m)

(East)

(North)Fragment 1

Fragment 2

550O

Fragment 3

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m m my3(120) ( 65) 20

4

+ − ++ − ++ − ++ − +====

y m3 27.5= −= −= −= −

Because yCM =0

Therefore location of third fragment is 1650 m east and 27.5 m south of the launch point.

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For Illustration 21For Illustration 21--23.23.

Consider two particles of mass m1 = 3 kg, velocity and mass m2 = 5 kg, velocity

sliding on a frictionless surface.2

ˆ(2 / )( )m s i→→→→

υ = −υ = −υ = −υ = −

1ˆ(10 / )m s i

→→→→

υ =υ =υ =υ =

16

(a) i2.50 m/s (b) i−−−− ˆ2.50 m/s

(c) i5.00 m/s (d) i7.50 m/s

AnswerAnswer

(a)

Illustration 21Illustration 21..

Find the velocity of the centre of mass for a fixed observer.

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(a) kg i(22.5 m/s) (b) kg i−−−− ˆ( 22.5 m/s)

(c) kg i(25.0 m/s) (d) kg i(12.5 m/s)

AnswerAnswer

(b)


What is the momentum of each particle in the center of mass reference frame ?

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(a) i(1.25m/s) (b) i(5.00m/s)

(c) i(2.50m/s) (d) i(3.50m/s)

AnswerAnswer

(c)


If a completely inelastic one-dimensional collision takes place, then find the velocity of the composite particle in the ground reference frame.

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Solution.Solution.

(a) Velocity of the centre of mass is given by the expression

Y

X

5.00 kg

2.00 m/s10.00 m/s

3.00 kg

System

m1 m2

VCM

8.00 kg

System

i i

iCM

i

i

mV

Vm

→→→→

====∑∑∑∑

∑∑∑∑

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ˆ ˆ(3.00)(10.00) (5.00)(2.00)( )

3.00 5.00

i i+ −+ −+ −+ −====

++++

ˆ2.50 /m s i====

(b) Velocity of first particle relative to centre of mass

ˆ ˆ(10.00) (2.50)i i= −= −= −= −

1, 1CM CM

→ →→ →→ →→ →→→→→

υ = υ − υυ = υ − υυ = υ − υυ = υ − υ

ˆ(7.50 / )m s i====

Velocity of second particle relative to the centre of mass

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2, 2CM CM

→→→→→ →→ →→ →→ →

υ = υ − υυ = υ − υυ = υ − υυ = υ − υ

ˆ ˆ(2.00)( ) (2.500)i i= − −= − −= − −= − −

ˆ( 4.50 / )m s i= −= −= −= −

The momentum of the first particle in the centre of mass reference frame,

1,ˆ(3.00)(7.50)CMP i

→→→→

====

ˆ(22.5 / )kg m s i====

2,ˆ(5.00)( 4.50)CMP i

→→→→

= −= −= −= −

ˆ( 22.5 / )kg m s i= −= −= −= −

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(c) The total momentum of the system relative to the centre of mass.

, 1, 2,system CM CM CMP P P→ → →→ → →→ → →→ → →

= += += += +

ˆ ˆ(22.5 ) ( 22.5)i i= + −= + −= + −= + −

0 /kg m s====

which shows that the total momentum of the system in the centre of mass reference frame is zero.

(d) The velocity of the composite particle can be obtained from conservation of momentum.

i fP P→ →→ →→ →→ →

====

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ˆ ˆ(3.00)(10.00) (5.00)(2.00)( ) (3.00 5.00)i i V→→→→

+ − = ++ − = ++ − = ++ − = +

ˆ(2.50 / )V m s i→→→→

====

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For Illustration For Illustration 2424--27.27.

A man of mass 80 kg is riding on a trolley of mass 40 kg which is rolling along a level surface at a speed of 2 m/s. He jumps off the back of the trolley so that his speed relative to the ground is 1 m/s in the direction opposite to the motion of the trolley.

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What is the speed of the centre of mass of the man-trolley system before his jumps?

(a) (b)

(c) (d)

2 m/s 3 m/s

4 m/s 2.5 m/s

AnswerAnswer

(a)

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What is the speed of the trolley after the man jumps ?

(a) (b)

(c) (d)

3 m/s 4 m/s

5 m/s 8 m/s

AnswerAnswer

(d)

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What is the speed of the centre of mass of the system after the man hits the ground and comes to rest ?

(a) (b)

(c) (d)

2.76 m/s 5.26 m/s

3.14 m/s 6.23 m/s

AnswerAnswer

(a)

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How much energy did the man expend in jumping?

(a) (b)

(c) (d)

540 J 1080 J

1250 J 460 J

AnswerAnswer

(b)

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Solution.Solution.

m1 m2

v

Initial state

m1

Final state

v’1

v’2

(a) Velocity of CM of man-trolley system before the man jumps,

1 1 2 2

1 2

CM

m m

m m

υ + υυ + υυ + υυ + υυ = = υυ = = υυ = = υυ = = υ

++++1 2[ ]as υ = υ = υυ = υ = υυ = υ = υυ = υ = υ

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before the man touches the ground, there is no

external force, hence v’CM = vCM = v

(b) Denoting the final velocities of man and trolley by v’1 and v’2, we have, from conservation of momentum,

' '

1 1 2 2 1 2( )m m m m− υ + υ = + υ− υ + υ = + υ− υ + υ = + υ− υ + υ = + υ

υ = υ + υ + υ = + + =υ = υ + υ + υ = + + =υ = υ + υ + υ = + + =υ = υ + υ + υ = + + =' '12 1 1

2

80( ) 2 (2 1) 8 /

40

mm s

m

(c) The man comes to rest after hitting the ground, so the speed of the centre of mass of the system is

'" 1 2 2

1 2

0 40 82.67 /

80 40CM

m mm s

m m

× + υ× + υ× + υ× + υ ××××υ = = =υ = = =υ = = =υ = = =

+ ++ ++ ++ +30

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(d) Due to the force of friction exerted by the ground, on the man, the velocity of CM is charged.

f iE KE KE∆ = −∆ = −∆ = −∆ = −

(e) While jumping, the force between man and trolley is internal to the system. It has no influence on the motion of CM.

However, it changes the total energy of the system by the amount

' 2 ' 2 2

1 1 2 2 1 2

1 1 1( )

2 2 2m m m m

= υ + υ − + υ= υ + υ − + υ= υ + υ − + υ= υ + υ − + υ

2 2 21 1 180 1 40 8 (80 40) 2

2 2 2= × × + × × − + ×= × × + × × − + ×= × × + × × − + ×= × × + × × − + ×

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1080 J====31

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This is equal to the energy expended by the man in jumping.

Conceptual discussion:

If the man jumps with velocity v1 relative to the initial state of trolley the absolute velocity of man is

mG mT TG

→ → →→ → →→ → →→ → →

υ = υ + υυ = υ + υυ = υ + υυ = υ + υ

1| |mGv→→→→

= −υ + υ= −υ + υ= −υ + υ= −υ + υ

The equation of conservation of momentum is

1 1 2 1 2( ) ( )m m m m− −υ + υ + υ = + υ− −υ + υ + υ = + υ− −υ + υ + υ = + υ− −υ + υ + υ = + υ

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Similarly if the man jumps with velocity v1 relative to final state of trolley .

'

1 2| |mGv→→→→

= −υ + υ= −υ + υ= −υ + υ= −υ + υ

The equation of conservation of momentum reduces to

' '

1 1 2 2 2 1 2( ) ( )m m m m− −υ + υ + υ = + υ− −υ + υ + υ = + υ− −υ + υ + υ = + υ− −υ + υ + υ = + υ

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A bead of mass 2m can slide on a smooth rod. A particle of mass m is attached to the bead by a light string of length l. Initially the particle is held horizontally in level with the bead and the string held just taut.

Find the distance through which the bead will move when the string has turned through an angle θθθθ with the horizontal.

θ

(a)

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(a)l + θ+ θ+ θ+ θ(1 cos )

3(b)

l − θ− θ− θ− θ(1 cos )

4

(c)l − θ− θ− θ− θ(1 cos )

3(d)

l − θ− θ− θ− θ2 (1 cos )

3

AnswerAnswer

(c)

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Solution.Solution.

We assign the initial position of bead as origin. There is no external force along the x-axis; hence position of CM will remain unchanged.

2 0( )

3 3CM initial

m ml lX

m

× +× +× +× += == == == =

2 ( cos ) 3 cos( )

3 3CM final

mx m x l x lX

m

+ + θ + θ+ + θ + θ+ + θ + θ+ + θ + θ= == == == =

On equating equations (i) and (ii), we get

3 cos

3 3

l x l+ θ+ θ+ θ+ θ====

…(i)

…(ii)

36

θ

(a)

(1 cos )

3

lx

− θ− θ− θ− θ====36

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Method 2:

CM is initially at rest, in the absence of external forces it will continue to be at rest

11 2 2

1 2

0CM

m x m xx

m m

→ →→ →→ →→ →→→→→ ∆ + ∆∆ + ∆∆ + ∆∆ + ∆

∆ = =∆ = =∆ = =∆ = =++++

1 1 2 2 0m x m x→ →→ →→ →→ →

∆ + ∆ =∆ + ∆ =∆ + ∆ =∆ + ∆ =

where denote absolute displacement of masses m1 and m2.

1 2x and x→ →→ →→ →→ →

∆ ∆∆ ∆∆ ∆∆ ∆

( cos ) 2 0m l l x mx− θ + + =− θ + + =− θ + + =− θ + + =

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(1 cos ) / 3x l= − θ= − θ= − θ= − θ

Conceptual discussion:

A block is released on the convex surface of a hemispherical wedge as shown in figure. We wish to determine the displacement of wedge, when the block reaches the angular position q. There is no external force in the x- direction, so

1 1 2 2 0m x m x→ →→ →→ →→ →

∆ + ∆ =∆ + ∆ =∆ + ∆ =∆ + ∆ =

( sin ) 0m R x Mxθ − − =θ − − =θ − − =θ − − =

sinmRx

m M

θθθθ====

++++

R

M

m

Smoothsurfaceθ

(b)

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An explosive of mass 6 kg is projected at 35 m/s at an angle of 600 with the horizontal. At the top of its flight it explodes, breaking into two parts, one of which has twice the mass of the other.

The two fragments land simultaneously. The lighter fragment lands back at the launch point. Where does the other fragment land. What is the energy of the explosion?

Y

O

H =

XR/2 R/2

Path of m

after explosion2

v sin2 2θ0

v sin22

θ0

2g

R =g

v0

m1 m2

H

v2

y

2g=

(a)

θ

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(a) (b)

(c) (d)

6215 J 5512 J

2525 J 1250 J

AnswerAnswer

(b)

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The internal force of explosion does not change the trajectory of CM.

1 1 2 2m x m x====

2

2

4 2

Rx R= × == × == × == × =

Coordinate of

= + = == + = == + = == + = =2

3162

2 2

R Rm R m

m1 m2

(b)

X1X2

CM lies here

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Solution.Solution.

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Solution.Solution.

Just before the explosion the projectile has velocity components vx = v cos θθθθ, vy = 0 and is at the topmost point of its trajectory.

From conservation of momentum along x- axis, we have

' '

1 1 2 2x x xM m mυ = υ + υυ = υ + υυ = υ + υυ = υ + υ

Fragment m1 will land back at the initial launch point if

'

1x xυ = −υυ = −υυ = −υυ = −υTherefore

'' 1 1 1 12 0

2 2 2

x xx x x

M m M m M m

m m m

υ − υ + +υ − υ + +υ − υ + +υ − υ + +υ = = υ = υυ = = υ = υυ = = υ = υυ = = υ = υ

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Time taken by m2 to reach the ground

02 yHt

g g

υυυυ= == == == =

Distance covered during this time

0' 1 12 0

2 2 2

y

xt x

M m M m Rd

m g m

υυυυ + ++ ++ ++ += υ = υ ⋅ == υ = υ ⋅ == υ = υ ⋅ == υ = υ ⋅ =

Coordinate of second fragment,

1

2 22 2 2

M mR R R Mx d R

m m

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

26 (35)sin60

4 9.81= × °= × °= × °= × °

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where m1 = 2 kg, m2 = 4 kg, v0 \= 35 m/s, θθθθ = 300

162 m====

The energy of the explosion is

' 2 ' 2 2

1 1 2 2

1 1 1

2 2 2x x xE m m M∆ = υ + υ − υ∆ = υ + υ − υ∆ = υ + υ − υ∆ = υ + υ − υ

210

2

2x

m M

m= υ= υ= υ= υ

22 2 6(35 cos 30 ) 5512

4J

× ×× ×× ×× ×= ° ≈= ° ≈= ° ≈= ° ≈

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A 75.0 kg man standing in a stationary 50.0 kg boat throws a 10.0 kg anchor horizontally with a speed of 2.00 m/s. How fast fall system (man + boat) recoil?

45

(a) (b)

(c) (d)

2.65 m/s 3.25 m/s

0.55 m/s 1.85 m/s

AnswerAnswer

(d)

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Solution.Solution.

First we assign subscripts as B, man + boat; A anchor; and W, water.

(a) Before

V’AWV’

BW

(b) After

The given velocity 2.00 m/s is the velocity of the anchor w.r.t. boat; we denote is as VAB.

Velocity of anchor w.r.t. water is VAW.

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From law of conservation of momentum,

' 'B BW A AW B BW A AWm V m V m V m V+ = ++ = ++ = ++ = +

Note that we have used all the velocities relative to single frame of reference, i.e.., water.

AB AW BWV V V→ → →→ → →→ → →→ → →

= −= −= −= −

AW AB BWV V V→ → →→ → →→ → →→ → →

= += += += +

' ' 'AW AB BWV V V→ → →→ → →→ → →→ → →

= += += += +

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On substituting expression for and in equation (i), we obtain

'AWV→→→→

AWV→→→→

0 ' ( ' ' )B BW A AB BWm V m V V→ → →→ → →→ → →→ → →

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

' 'ABW AB

B A

mV V

m m

→ →→ →→ →→ → = −= −= −= − ++++

10.0(2.00)

125.0 10.0

= −= −= −= −

++++

0.148 /m s= −= −= −= −

The anchor travels with velocity

' ' 'AW AB BWV V V→ → →→ → →→ → →→ → →

= += += += +

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2.00 ( 0.148)= + −= + −= + −= + −

1.85 /m s====

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A man of mass 60 kg moves on a 6 m long, 240 kg stationary boat, at a speed 1.2 m/s.

50


Determine the velocity of the boat relative to water.

(a) (b)

(c) (d)

-0.96 m/s 0.76 m/s

0.36 m/s 0.48 m/s

AnswerAnswer

(a)

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If the man starts from one end of the boat and ends at the other, find the distance by which the boat is shifted.

(a) (b)

(c) (d)

1.5 m 2.5 m

1.2 m 3.2 m

AnswerAnswer

(c)

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What is the velocity of the boat when the man stops at the other end ?

(a) (b)

(c) (d)

0 1.2 m/s

0.24 m/s 3.6 m/s

AnswerAnswer

(a)

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Determine the velocity of the boat if the man falls out of the boat at the other end while walking.

(a) (b)

(c) (d)

0.36 m/s 0

0.12 m/s 0.24 m/s

AnswerAnswer

(d)

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Solution.Solution.

(a) We assign subscripts as M, man; B, boat and W, water.

1.2 /MB m s→→→→

υ =υ =υ =υ =

MB MW BW

→ →→ →→ →→ →→→→→

υ = υ − υυ = υ − υυ = υ − υυ = υ − υ

MW MB BW

→ →→ →→ →→ →→→→→

υ = υ + υυ = υ + υυ = υ + υυ = υ + υ

If the boat moves with velocity V relative to water,

1.2 /Mw m s V→→→→

υ = − +υ = − +υ = − +υ = − +

From law of conservation of momentum

' 'M MW B BW M MW B BWM V M V M V M V+ = ++ = ++ = ++ = +

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Initially the boat and the man were at rest. Therefore

0 0 ' 'M MW B BWM V M V+ = ++ = ++ = ++ = +

0 (60)( 1.2 ) (240)( )V V= − + += − + += − + += − + +

0.24 /V m s====

1.2 0.24MWV = − += − += − += − +

0.96 /m s= −= −= −= −

We have assumed flow of water to be negligible. So velocity of boat and man relative to water may be considered to be their absolute velocities, i.e., velocity observed by a fixed observer.

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(b) Displacement of man relative to boat xMB = 6 m

65

1.2MB

MB

xs

V= = == = == = == = =

Time taken by the man to reach the other end of the boat

Distance travelled by boat,

(0.24)(5) 1.2BW BWx V t m= × = == × = == × = == × = =

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(c) Initially momentum of the system is zero, therefore that boat should stop as the man stops, to make final momentum zero.

' '0 M W B BM V M V= += += += +

System

(d) Our system consists of boat and man, therefore we ma y use equation (i).

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(60)( 1.2 ) (120)V V= − + += − + += − + += − + +

Thus the boat will continue to move to the right at 0.24 m/s.

0.24 /V m s====

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A smooth wedge of mass M rests on a smooth horizontal surface. A block of mass m is projected from its lowermost point with velocity v0 .

What is the maximum height reached by the block?

M

m v0

(a) (b)

V

Initial position

Final position

V = Vx

V = Vx

V = 0yf

Vy

h

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(a)M

g m M

υυυυ ++++

20 (b)

M

g m M

υυυυ ++++

20

2

(c)M

g m M

υυυυ ++++

202

(d)M

g m M

υυυυ ++++

203

2

AnswerAnswer

(b)

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Solution.Solution.

At the instant the block breaks contact with the wedge, they have common x-component of velocity. In addition, the block has a vertical component of velocity.

Due to this vertical component the block rises upwards till the vertical component of velocity vanishes.

From momentum conservation along x-axis,

0 ( )m m M Vυ = +υ = +υ = +υ = +

0

( )

mV

m M

υυυυ====

++++

…(i)

…(ii)

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2 2

0

1 1( )

2 2m m M V mghυ = + +υ = + +υ = + +υ = + +

From energy conservation between initial and final positions of block,

22 2

0 0

1 1

2 2

mm mgh

m M

υ = υ +υ = υ +υ = υ +υ = υ + ++++

2

0

2

Mh

g m M

υυυυ ==== ++++

…(iii)

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A wedge of mass m with its upper surface hemispherical in shape, as shown in figure (a), rests on a smooth horizontal surface near the wall.

A small block of mass m2 slides without friction on the hemispherical surface of the wedge.

What is the maximum velocity attained by the wedge?

r

m2

m1

(a) (b)

A N

N

N’

N’C

B

2gr

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(a)m gr

m m++++2

1 2

2(b)

m gr

m m++++2

1 2

2

(c)m gr

m m++++2

1 2

4 2(d)

m gr

m m++++2

1 2

2 2

AnswerAnswer

(d)

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Solution.Solution.

As long as the block moves from A to B, the reaction on the wedge presses it to the wall. When the block reaches the lowermost positions, its velocity from energy conservation is 2grυ =υ =υ =υ =

When the block moves along the right half of the wedge, during its upward journey as well as downward journey the reaction of the block on the wedge is towards right as shown in figure (b). Therefore during the entire motion of the block from B to C and C to B, the wedge is accelerated towards right. 65

(b)

A N

N

N’

N’C

B

2gr

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Thus maximum velocity is attained by the wedge at the instant when the block passes point B during the return journey.

i fP P====

From conservation of momentum, at the instant when the wedge is separated from the wall,

2 1 1 2 22m gr m m= υ + υ= υ + υ= υ + υ= υ + υ

From energy conservation,

i fE E====

2 2

1 1 2 22

2 2

m mm gr

υ υυ υυ υυ υ= += += += +

…(i)

…(ii)

66

(b)

A N

N

N’

N’C

B

2gr

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1 20, 2grυ = υ =υ = υ =υ = υ =υ = υ =

On solving equations (i) and (ii) simultaneously, we obtain two solutions

21

1 2

22 ,

mgr

m mυ =υ =υ =υ =

++++2 1

2

1 2

2m m

grm m

−−−−υ =υ =υ =υ =

++++

The first solution corresponds to the instant when the block reaches for the first time at point B. At this instant the block moves with velocity v2 and the wedge is at rest.

The second solution corresponds to the instant when the block has the maximum velocity

21 max

1 2

2 2( )

m gr

m mυ =υ =υ =υ =

++++

and

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For IllustrationFor Illustration 3737--40.40.

A cricket ball of mass 0.145 kg moving with a velocity of 20.0 m/s horizontally, is struck by a bat so that it travels with 50.0 m/s in the opposite direction

68

IllustrationIllustration 37.37.

What is the impulse of the bat on the ball?

(a) i−−−− ˆ( 10.15) kg m/s (b) i−−−− ˆ( 20.30) kg m/s

(c) i−−−− ˆ( 5.25) kg m/s (d) i−−−− ˆ( 30.20) kg m/s

AnswerAnswer

(a)

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IllustrationIllustration 38.38.

If the contact time between the bat and the ball is 1.00 × 10-3 s, what is the average force on the ball ?

(a) i− ×− ×− ×− × 4 ˆ( 5.24 10 N) (b) i− ×− ×− ×− × 4 ˆ( 1.02 10 N)

(c) i− ×− ×− ×− × 4 ˆ( 3.02 10 N) (d) i− ×− ×− ×− × 3 ˆ( 1.02 10 N)

AnswerAnswer

(b)

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Compare the average force of the bat on the ball with the weight of the ball.

(a) (b)

(c) (d)

3.16 × 103 N 7.17 × 103 N

5.24 × 103 N 2.16 × 103 N

AnswerAnswer

(b)

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If the ball leaves the bat in vertical direction at speed 40.0 m/s, what is the impulse of the bat on the ball ?

(a) (b)

(c) (d)

3.48 kg m/s 2.64 kg m/s

6.48 kg m/s 3.64 kg m/s

AnswerAnswer

(c)

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Solution.Solution.

(a)

i fP J P→ → →→ → →→ → →→ → →

+ =+ =+ =+ =

ˆ ˆ(0.145)(20.0) (0.145)(50.0)( )i J i→→→→

+ = −+ = −+ = −+ = −

ˆ( 10.15)J i→→→→

= −= −= −= −

72

P1

Rdt

P1

Rdt =+

Initial momentum + impulse = final momentum

From impulse-momentum equation

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(b) .avgJ F t→→→→→→→→

= ∆= ∆= ∆= ∆

which shows that the average force exerted by bat on the bat is much larger than the weight.

73

4

3

ˆ( 10.15) ˆ( 1.02 10 )1.00 10

avg

iF i

→→→→

−−−−

−−−−= = − ×= = − ×= = − ×= = − ×

××××

43(1.02 10 )

7.17 10(0.145)(9.81)

avgF

mg

××××= = ×= = ×= = ×= = ×(c)

P1

P1

=

J

I

+

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(d) ˆ ˆ2m i J m j→→→→

υ + = υυ + = υυ + = υυ + = υ

The magnitude of impulse

74

ˆ ˆ( 2 )J m i j→→→→

= υ − += υ − += υ − += υ − +

ˆ ˆ(0.145)(20.0)( 2 ) /i j kg m s= − += − += − += − +

2 2 1/2[( 1) 2 ]m= υ − += υ − += υ − += υ − +

5m= υ= υ= υ= υ

6.48 /kg m s====

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After falling from rest through a height h a body of mass m begins to raise a body of mass M (M>m) connected to it through a pulley.

Mg

(a) Before the string tautens

B

mgNA

a

A

a

mg

Mg

T

T

(b) After the string tautens

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(a)m h

M m g++++

2 2(b)

m h

M m g−−−−

2 2

(c)m h

M m g−−−−

2(d)

m h

M m g−−−−

2

AnswerAnswer

(b)


Determine the time it will take for the body of mass Mto return to its original position.

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(a)M

M m++++(b)

M

M m−−−−

(c)M

M m++++

2(d)

M

M m++++2( )

AnswerAnswer

(a)


Find the fraction of kinetic energy lost when the body of mass M is jerked into motion

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Solution.Solution.

The speed of the body B just before the string becomes taut is . When the string is jerked, large impulsive reactions are generated in the string. At this moment effect of gravity is negligible.

m M m( ) 'υ = + υυ = + υυ = + υυ = + υ

So momentum of the system is conserved at this instant. Let v’ be the common speed of the two bodies after they are jerked into motion. From conservation of momentum, we have

m

M m'υ = υυ = υυ = υυ = υ

++++

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The acceleration of the system is

F mg Mg M m a( )= − = += − = += − = += − = +∑∑∑∑

M ma g

M m

−−−−= −= −= −= −

++++

The acceleration is negative, opposite to v’

Let the system return to original position at time t.

t at 21

0 '2

= υ += υ += υ += υ +

m ht

a M m g

2 ' 2 2υυυυ= − == − == − == − =

−−−−

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(b) The frictional loss of kinetic energy is

m M mM

M mm

2 2

2

1 1( ) '

2 21

2

υ − + υυ − + υυ − + υυ − + υ====

++++υυυυ

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An elastic collision takes place between two masses m1 and m2, moving on a frictionless surface, as shown in figure.

The spring constant is k = 600 N/m.

m1 m2

v = 4 m/s1i

v = 2.5 m/s2i

k

m 1.60 kg1= m 1.60 kg1=

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What is the velocity of the block 2 at the instant block 1 is moving to the right with a velocity 3.00 m/s ?

(a) (b)

(c) (d)

1.50 m/s -2.62 m/s

4.64 m/s -1.74 m/s

AnswerAnswer

(d)

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What is compression of the spring at that instant?

(a) (b)

(c) (d)

0.52 m 0.4 m

0.173 m 0.84 m

AnswerAnswer

(c)

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Solution.Solution.

(a) From momentum conservation, we obtain

1 1 2 2 1 1 2 2i i f fm m m mυ υ + υ = υ + υυ υ + υ = υ + υυ υ + υ = υ + υυ υ + υ = υ + υ

2(1.60)(4.00) (2.10)( 2.50) (1.60)(3.00) (2.10) f+ − = + υ+ − = + υ+ − = + υ+ − = + υ

2 1.74 /f m sυ = −υ = −υ = −υ = −

Negative sign implies that block 2 is still continuing in the same direction.

(b) Because no friction or non-conservation force acts on the system, we can use conservation of energy equation. We obtain

2 2 2 2 2

1 1 2 2 1 1 2 2

1 1 1 1 1

2 2 2 2 2i i f fm m m m kxυ + υ = υ + υυ + υ = υ + υυ + υ = υ + υυ + υ = υ + υ

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On substituting numerical values, we obtain

0.173x m====

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Consider a one-dimensional elastic collision between a given incoming body 1 and body 2, initially at rest. How would you choose the mass of B in comparison to the mass of A in order that B should recoil with

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(a) m m<<<<<<<<2 1 (b) m m>>>>>>>>2 1

(c) m m====2 1 (d)m m

mm

++++<<<<<<<<

21 2

2

1

( )

AnswerAnswer

(a)


greatest speed

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(a) m m<<<<<<<<2 1 (b) m m>>>>>>>>2 1

(c) m m====2 1 (d)m m

mm

++++<<<<<<<<

21 2

2

1

( )

AnswerAnswer

(b)


greatest momentum

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Solution.Solution.

Since the collision is elastic energy as well as momentum are conserved.

1 1 1 1 2 2i f fm m mυ = υ + υυ = υ + υυ = υ + υυ = υ + υ

21 1 2

1

i f f

m

mυ = υ = υυ = υ = υυ = υ = υυ = υ = υ

For the sake of simplicity we take a parameter

2

1

mk

m====

…(i)

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Equation (i) reduces to

1 1 2i f fkυ = υ + υυ = υ + υυ = υ + υυ = υ + υ

2 2 2

1 1 1 1 2 2

1 1 1

2 2 2i f fm m mυ = υ + υυ = υ + υυ = υ + υυ = υ + υ

2 2 2

1 1 2i f fkυ = υ + υυ = υ + υυ = υ + υυ = υ + υ

Now we substitute the expression for v1f in equation (ii) to obtain

…(ii)

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12

2

1i

fk

υυυυυ =υ =υ =υ =

++++

(a) Particle 2 will recoil with maximum speed when k is minimum, i.e..,

2 10k or m m→ <<→ <<→ <<→ <<

(b) Momentum of particle 2 is

1 12 2 2

2

1i

f

kmp m

k

υυυυ= υ == υ == υ == υ =

++++

12

1 (1/ )A im

k

υυυυ====

++++

Particle 2 will recoil with maximum momentum when denominator is minimum, which is possible if

2 1k or m m→ ∞ >>→ ∞ >>→ ∞ >>→ ∞ >>91

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(c) Kinetic energy of particle 2 is

2

2 2 2

1

2fk m= υ= υ= υ= υ

2

11

21

2 1ikmk

υυυυ ====

++++

2

1 12

2

(1 )im k

k

υυυυ====

++++

12

4( )

(1 ) 4

KE k

k k====

− +− +− +− +

Particle 2 will recoil with maximum kinetic energy when denominator is minimum,

k = 1 or mA = mB

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A smooth ball is dropped from a height h on a smooth incline, as shown in figure. After collision the velocity of the ball is directed horizontally

93


Find the coefficient of restitution.

(a) (b)

(c) (d)

cot2 θθθθ sin2 θθθθ

tan2 θθθθ cos2 θθθθ

AnswerAnswer

(c)

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(a) m gh θθθθ2 2 tan (b) m gh θθθθ2 2 sin

(c) m gh θθθθ2 2 cot (d) m gh θθθθ2 2 cos

AnswerAnswer

(d)


If the collision is elastic, what is the impulse on the ball ?

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v’

h

θ

Solution.Solution.

(a) Normal axis and tangential axis are shown in the figure. Reaction of the incline is along n-axis and in the absence of friction there is no force along t-axis; therefore velocity along t-axis remains unchanged,

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cos sinuυ θ = θυ θ = θυ θ = θυ θ = θ

From the definition of coefficient of restitution,

sin

cose

u

υ θυ θυ θυ θ= −= −= −= −

− θ− θ− θ− θ

coteuυ = θυ = θυ = θυ = θ

θ

n-axis

v sin θ

v cos θ

u cos θ u sin θ

t - axis

θ

θ

v

u

j

i

(a)

From equations (i) and (iii),

( cot )cos sineu uθ θ = θθ θ = θθ θ = θθ θ = θ

2tane = θ= θ= θ= θ

…(i)

…(ii)

…(iii)

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(b) When the collision is elastic, the component of velocity along n-axis is reversed in direction. Therefore the change in velocity

θ

n-axis

v

u

(b)

θ

θ

2 cos∆υ = υ θ∆υ = υ θ∆υ = υ θ∆υ = υ θ

There is no change in velocity along the t-axis, therefore no impulse along t-axis.

m= ∆υ= ∆υ= ∆υ= ∆υ

2 cosm= υ θ= υ θ= υ θ= υ θ

Velocity of the ball when it strikes the plane

2gh====

Thus impulse 2 2 cosm gh= θ= θ= θ= θ97

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An inelastic ball is projected with velocity, at an angle a to the horizontal, towards a

wall distant d from the point of projection. 0 ghυ =υ =υ =υ =

After collision the ball returns to the point of projection. What is the coefficient of restitution?

v0

α

O d Velocity after impact

Velocity before impact

vy

v cos = v0 xα

ev cos = v0 xα

vy

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(a)d

h dα −α −α −α −

2

( sin 2 )(b)

d

h dα +α +α +α +( sin 2 )

(c)d

h dα −α −α −α −( sin 2 )(d)

d

h dα +α +α +α +

2

( sin 2 )

AnswerAnswer

(c)

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0 cos

d====

υ αυ αυ αυ α

Solution.Solution.

Time taken to reach the wall

Time taken to return to the point of projection after impact

0 cos

d

e====

υ αυ αυ αυ α

Note that x-component of velocity after impact is.

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Total time of flight

0 0

1

cos cos cos

d d d e

e egh

++++ = + == + == + == + = υ α υ αυ α υ αυ α υ αυ α υ α α α α α

There is no change in the vertical component of the velocity after impact, therefore total time of flight remains unchanged.

02 sin 2 singhT

g g

υ αυ αυ αυ α αααα= == == == =

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2 sin1

cos

ghd e

e ggh

αααα++++ ====

α α α α

1 sin 2e h

e d

+ α+ α+ α+ α ====

( sin 2 )

de

h d====

α −α −α −α −

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A ball is projected with velocity v0 at an angle ααααto the horizontal, towards a smooth wall which approaches the ball with velocity u. After collision the ball retraces its path to the point of projection.

What is the time t taken by the ball from instant of projection to point of impact?

u

v0

v cos 0 α

vsin

0

α

α

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(a)u

g u

υ α + υ αυ α + υ αυ α + υ αυ α + υ α

υ α +υ α +υ α +υ α +0 0

0

( cos 2 ) sin

( cos )

(b)u

g u

υ α + υ αυ α + υ αυ α + υ αυ α + υ α

υ α −υ α −υ α −υ α −0 0

0

( cos 2 ) sin

( cos )

AnswerAnswer

(a)

(c)u

g u

υ α − υ αυ α − υ αυ α − υ αυ α − υ α

υ α +υ α +υ α +υ α +0 0

0

( cos 2 ) sin

( cos )

(d)u

g u

υ α − υ αυ α − υ αυ α − υ αυ α − υ α

υ α −υ α −υ α −υ α −0 0

0

( cos 2 ) sin

( cos )

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Solution.Solution.

Since the wall is smooth, the vertical component of velocity will not change, as explained earlier, the tangential component remains unchanged.

Collision is elastic, therefore coefficient of restitution is 1.

Re

Re

lative velocity of separatione

lative velocity of approach= −= −= −= −

0

1( cos )

V

u= −= −= −= −

υ α +υ α +υ α +υ α +

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Thus relative velocity of separation

0( cos )V u= − υ α += − υ α += − υ α += − υ α +

Velocity relative to ground 0( cos )u u= − υ α + += − υ α + += − υ α + += − υ α + +

Time of flight, depends on the vertical component of velocity which is unchanged.

106

02 sinT

g

υ αυ αυ αυ α====

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Let time taken before impact be t, the distance covered before and after impact the same.

0 0cos ( )( cos 2 )t T t uυ α = − υ α +υ α = − υ α +υ α = − υ α +υ α = − υ α +

0 0 0( cos cos 2 ) ( cos 2 )t u Tυ α + υ α + = υ α + υυ α + υ α + = υ α + υυ α + υ α + = υ α + υυ α + υ α + = υ α + υ

0 0

0

( cos 2 ) sin

( cos )

ut

g u

υ α + υ αυ α + υ αυ α + υ αυ α + υ α====

υ α +υ α +υ α +υ α +

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A ball moving with a velocity v strikes a wall moving toward the wall with a velocity u. An elastic impact occurs. Determine the velocity of the ball after the impact. What is the cause of the change in the kinetic energy of the ball ? Consider the mass of the wall to be infinitely great.

108

v u

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(a) m v u u−−−−2 ( ) (b) m v u u++++2 ( )

(c) m v u u++++2 ( 2 ) (d) m v u u++++2 (2 )

AnswerAnswer

(b)

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Solution.Solution.

110

Velocity of ball relative to wall before collision is (v + u). After elastic collision the velocity of ball relative to wall will be -(v + u). The velocity of the ball relative to ground will be

( ) ( 2 )v u u v u− + − = − +− + − = − +− + − = − +− + − = − +

Kinetic energy before impact 21

2mv====

Kinetic energy after impact 21( 2 )

2m v u= += += += +

The change in kinetic energy is equal to 2mu(u + v).

Now we calculate the work of reaction forces acting on the ball during the impact. Let the collision continue for t seconds. Assume the reaction force to be constant (the result does not depend on this assumption). Since the impact changes the momentum by 2m(v + u), the force of reaction is

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2 ( )m v uF

t

++++====

The work of this force is

2 ( )( ) 2 ( )

m v u utW Fs F ut m v u u

t

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

So we can see that this work is equal to change in kinetic energy.

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Education

Centre of mass, impulse momentum (obj. assignments)