Solved Examples of System

7/30/2019 Solved Examples of System

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Solved examples

Question 1. A body fall from height H.if t1 is time taken for covering first half height and

t2 be time taken for second half.Which of these relation is true for t1 and t2a. t1 > t2

b. t1 < t2

c t1=t2

d Depends on the mass of the body

Question 2 .A 1 kg ball moving at 12 m/s collides head on with 2 kg ball moving with 24 m/s in

opposite direction.What are the velocities after collision if e=2/3?

a. v1=-28 m/s,v2=-4 m/s

b. v1=-4 m/s,v2=-28 m/s

c. v1=28 m/s,v2=4 m/sd. v1=4 m/s,v2=28 m/s

Question 3.A system consisting of two objects has a total momentum of (18 kgm/sec)i and its center

of mass has the velocity of (3 m/s)i.One of the object has the mass 4 kg and velocity (1.5 m/s)i.The

mass and velocity of the other objects area. 2 kg, (6 m/s)i

b. 2 kg, (-6 m/s)i

c. 2 kg, (3 m/s)i

d. 2 kg, (-3 m/s)i

Question 4.A moving bullet hits a solid target resting on a frictionless surface and get embeded in

it.What is conserved in it?

a. Momentum Alone

b KE alone

c. Both Momentum and KE

d. Neither KE nor momentum

Question 5. A stationary body of mass 3 kg explodes into three equal parts.Two of the pieces fly off

at right angles to each other with the velocities 2i m/s and 3j m/s.If the explosion takes place in 10-3

sec.find out the average force action on the third piece in N

a.(-2i-3j)103

b. (2i+3j)103

c (2i-3j)10-3

d. none of these

Question 6.A bullet of mass m is fired horizontally with a velocity u on a wooden block of Mass M

suspended from a support and get embeded into it.The KE of th wooden + block system after the

collisson

a.m2u2/2(M+m)b.mu

2/2


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c. (m+M)u2/2

d. mMu2/2(M+m)

Question 7.A body of Mass M and having momentum p is moving on rough horizontal surface.If it is

stopped in distance s.Find the value of coefficient of friction

a.p2/2M

2gs

b. p/2Mgs

c. p2/2Mgs

d. p/2M2gs

Question 8. Which of these is true of a conservative force?

a. Workdone between two points is independent of the path

b. Workdone in a closed loop is zero

c. if the workdone by the conservative is positive,its potential energy increases

d. None of the these

Question 9. A simple pendulum consists of a mass attached to a light string l. if the system is

oscillating through small angles which of the following is true

a.The freqiency is independent of the acceleration due to gravity g

b.The period depends on the amplitude of the ocsillation

c.the period is independent of mass m

d. the period is independent of lenght l

Question 10. A body of mass m is dropped from a certain height.it has velocity v1 when it is at a

height h1 above the ground.it has velocity v2 when it is at a height h2 above the ground.which of the

following is true

a.v12-v2

2=2g(h1-h2)

b.v12-v2

2=2g(h2-h1)

c. v1-v2=2g(h2-h1)d. v1-v2=2g(h1-h2)

Question 11.A pendulum has a length l.Its bob is pulled aside from its equilibrium position throughany angle and then released.The speed of the bob when its passes through it equalibrium position a.&radic:2gl

b. &radic:2gl(1-cos)c.&radic:2glcosd.&radic:2gl(1-sin

Question 12.A rockets works on the principle of conservation of

a. Linear momentum

b.mass

c.energyd. angular momentum


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Question 13.A flat car of weight W roll without resistance along on a horizontal track .Intially the car

together with weight w is moving to the right with speed v.What invcrement of the velocity car will

obtain if man runs with speed u reltaive to the floor of the car and jumps off at the left?a.wu/w+W

b. Wu/W+w

c. (W+w)u/w

d. none of the above

Question 14.A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficent

speed to reach the top of the incline.The plane is 3 mlong and is inclined at 20.Coefficent of friction

between the package and the inclined plane is .40. what minimum intial KE must the boy suply to the

package given as sin20=.342 cos20=.940

a 40 J

b. 42.2 J

c. 42.6 J

d. 45 J

Question 15.Consider the following two statements.

STATEMENT 1 Linear momentum of a system of particles is zero.

STATEMENT 2 Kinetic energy of system of particles is zero.

(A) A does not imply B and B does not imply A.

(B) A implies B but B does not imply A

(C) A does not imply B but b implies A(D) A implies B and B implies A.

Question 16.The Position vector of the center of mass of uniform semi circular ring of radius R and

Mass M whose center coincided with the origin

a.r=(2R/)j

b.r=(R/)jc.r=(4R/)jd. none of the these

Question 17.Chosse the correct option

a.if Workdone by the conservative force is positive then Potential energy decreases

b. Rate of change of momentum of many particles system is proportional to net external force on the

system

c.The workdone by the conservative force in closed loop is zero

d. None of the above


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Question 18.Two electrons(e) are the at the point (-a,0) and (a,0).Their mass is m.They are released

from rest.The acceleration of the center of mass of the system when the electron at 4a distance apart

a. e2/64m0a

2

b. zero

c. e2/16m0a

2

d. none of the above

Question 19.The potential energy of a certain particle is given by

U=20x2+35z

3.Find the vector force on it

a. -40xi-105z2k

b. 40xi-105z2k

c.-10xi-105z2k

d 40xi+105z2k

Matrix Match type

Question 20.

Column I

a. Frictional force

b. Gravitational force

c. Electrical force

d Viscous force

Column II

P. Workdone by the force in closed loop is zero

Q. Workdone by the force in closed loop is not zero

Question 21. A mass is whirled in a circular path with constant angular velocity and its angular

momentum is L.If the string is now halve keeping the angular velocity same then angular momentum

is

a. L

b. L/4

c. L/2

d. 2L

Question 22.A mass is moving with constant velocity along a line parallel to xaxis away from origin.its

angular momentum with respect to origin isa. is zero

b. remains constant

c. goes on increasing

d. goes on decreasing

Question 23.A cylinder rolls up the incline plane reaches some height and then roll down without

slipping throughout this section.The direction of the frictional force acting on the cylinder area. Up the

incline while ascending and down the incline while descending

b.Up the incline while ascending and desending

c. down the incline while ascending and up the incline while descendingd.down the incline while ascending and desending


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Question 24.A uniform sold sphere rolls on the horizontal surface at 20 m/s.it then rolls up the incline

of 30.If friction losses are negligible what will be the value of h where sphere stops on the incline

a. 28.6 m

b 30 mc. 28 m

d. none of these

Question 25. A cylinder of Mass M and radius R rolls down a incline plane of inclination .Find thelinear accleration of the cylinder

a. (2/3)gsinb.(2/3)gcosc gsind none of these

Question 26 An ice skater spins with arms outstretch at 1.9 rev/s.Her moment of inertia at this time is

1.33 kgm2.She pulls her arms to increase her rate of spin.Her moment of inertia after she pulls her

arm is .48kgm2.What is her new rate of spinning

a. 5.26 rev/s

b. 5.2 rev/s

c 4.7 rev/s

d. none of thes

Question 27. Moment of inertia of a uniform rod of lenght L and mass M about an axis passing

through L/4 from one end and perpendicular to its lenght

a. 7ML2/36

b.7ML2/48

c. 11ML2/48

d.ML2/12

Question 28. A wheel starts from rest and spins with a constant angular acceleration. As time goes

on the

acceleration vector for a point on the rim:

a. increases in magnitude but retains the same angle with the tangent to the rim

b.increases in magnitude and becomes more nearly radial

c. increases in magnitude and becomes more nearly tangent to the rim

d. decreases in magnitude and becomes more nearly radial


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Question 29.

Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A. The ratio of

the

magnitude of the &radical acceleration of a point on the rim of B to the magnitude of the radial

acceleration ofa point on the rim of A is:

a. 4

b . 2

c 1/2

d 1/4

Question 30. For a wheel spinning with constant angular acceleration on an axis through its center,

the ratio of

the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

a 2

b 1/2

c 4

d 1/4

Question 31. A wheel initially has an angular velocity of 18 rad/s. It has a constant angular

acceleration of 2.0 rad/s2 and is

slowing at first. What time elapses before its angular velocity is18 rad/s in the direction opposite to its

initial

angular velocity?

a 3 sec

b 6 sec

c 18 sec

d none of these

Question 32. One solid sphere X and another hollow sphere Y are of same mass and same outerradii. Their moment of inertia about their diameters are respectively Ix and Iy such that

(A) Ix= Iy

(B) Ix > Iy

(C) Ix < Iy

(D) Ix/Iy=Dx/Dy

Where Dx and Dy are their densities.

Question 33.Which of the following is noninertial frame of refrence

a. A train which speeding Upb. A train with constant speed


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c. A train which speeding down

d A train at rest

Question 34.what is of these is true for Projectile motion

a. Velcoity is perpendicular to acceleration at the highest point

b. Horizontal components of velocity remains constant throughout the pathc. Range of the projectile is given by Horizontal component of velocity X Time of flight

d. None of the above

Question 35. A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a

force P is applied at the free end of the rope, the force exerted by the rope on the block is

a)Pm/m+M

b)P

c)PM/m+M

d)Pm/M-m

Question 36.A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring

and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration

of 5 m/s2, the reading of the spring balance will be

(A) 24 N

(B) 74 N

(C) 15 N

(D) 49 N

Question 37. A particle moves in a straight line according to

x=t3-4t2+3t

Find the acceleration of the particle at displacement equal to zero

a.(-8,-2,10)

b. (-1,-2,10)

c. (8,2,10)

d. (1,2,10)

Detailed solutions

Solution 1.

Let H be the height

then

First Half

H/2=(1/2)gt12

----(1)

or

(1/2)gt12=H/2

Also v=gt1

Second Half

H/2=vt2+(1/2)gt22or


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(H/2)=gt1t2+(1/2)gt22

or

(1/2)gt22

=(H/2)-gt1t2 ---(2)

From 1 and 2

(1/2)gt22

=(1/2)gt12

-gt1t2t2

2+2t1t2 -t1

2=0

or t2=[-2t1+(4t1t2 +4t1t2 )]/2or t2=[-2t1+2t18]/2t2=.44t1

so t1 > t2

Hence a is correct

Solution 2

Let v1 and v2 be the final velocities of the mass

Since the linear momentum is conserved in the collision

Momentum before =Momentum after

1*12+2*-24=1*v1+2*v2

Which becomes

-36=v1+2v2 ----1

Now

e=(v2 -v1)/(u1 -u2)

or 2/3= (v2 -v1)/[12-(-24)]

or

v2 -v1=24 ----2

Solving 1 and 2

v2=-4

v1=-28

Hence a is correct

Solution 3.

Total momentum=(18 kgm/sec)i

velocity of Center of mass=(3 m/s)i.

Mass of one object=4 kg

Velocity of this object=(1.5 m/s)i

let m be the mass of other object

And v be the velocity

Now we know total momentum =Total massX velocity of center of mass

(18 kgm/sec)i=(m+4)(3 m/s)i.


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or m=2 kg

Now Vcm=(m1v1+m2v2)/(m1+m2)

or

3i=(4*1.5i + 2v)/6or 18i=6i+2v

v=6i m/sec

hence a is correct

Solution 4 Since no external force is present,Momentum is conserved in the collision

Since the collison is in elastic ,KE is not conserved

Solution 5.

Net momentum before explosion zero

Since momentum is conserved in explosion

Net momentum after collosion is zero

Momentum of first part after explosion=2i

Momentum of second part after explosion=3j

So momentum of third party after explosion=-(2i+3j) as net momentum is zero

Now Net change is momentum of this part =-(2i+3j)

Now we know that

Average force X time =Net change in momentum

Average force=-(2i+3j) 103

hence a is correct

Solution 6.

Intial velocity of bullet=u

Intial velocity of block=0

So net momentum before collison=mu

Let v be the velocity after collisionThen Net momentum after collision=(M+m)v

Now linear momentum is conserved in this collision

so

mu=(M+m)v

or v=mu/(M+m)

So kinetic energy after collision

=(1/2)m2u

2/2(M+m)

Hence a is correct


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Solution 7.

Deceleration due to friction=g

Intial velocity=P/M

Now v2=u

2-2as

as v=0

P2/M

2=2gsor =P2/2gsM2

Hence a is correct

Solution 8. For a conservative force

Workdone between two points is independent of the path

Workdone in a closed loop is zero

And -W=Uf-Ui

So for positive work potential energy decreases

So a and b are correct

Solution 9.Frequency =2(l/g)

So it is independent of the mass

Solution 10

Total Energy at height h1

=(1/2)mv12+2gh1

Total energy at height h2

=(1/2)mv22+2gh2

Since we know that total energy remains constant during a free fall

Total Energy at height h1=Total energy at height h2

or (1/2)mv12+2gh1=(1/2)mv2

2+2gh2

or v12-v2

2=2g(h2-h1)

Solution 11As shown in fgure,the height attained by the bob when the string subtends an angle is

h=l-lcosor h=l(1-cos)So potential energy at this point is given by

=mgh=mgl(1-cos)

When the bob passes through equilibrium position,this potential energy is converted into kinectic

energy

if v be the velocity of the bob the KE=(1/2)mv2

Now (1/2)mv2=mgl(1-cos)

or v=radic:2gl(1-cos)


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Solution 12. A rocket works on the principle of linear momentum.

Solution 13 Considering velocities to the right as positive

The intial momentum of the system is

=[(W+w)/g]v

Let v be the increment in velocity then

Final momentun of the car is

(W/g)(v+v)

While that of man is

(w/g)(v+v-v)

Since no external forces act on the system,the law of conservation of momentum gives then

[(W+w)/g]v=(W/g)(v+v)+(w/g)(v+v-vor v=wu/(W+w)

Solution 14

The incline is shown in figure

If the package travels the entire length s of the incline ,the frictional force will perform work -Nswhere is coefficient of friction and N is normal reaction.Let h be the height of the incline plane then the gravtational potential energy of the package will

increase by mgh.

Now let assume v speed be given to the package so as to reach the top

Then kinectic energy at the intial point=(1/2)mv2

Now applying work energy thoerm

K.Ef-K.Ei=Workdone by the gravitational force + workdone done by the frictional force

Now since K.Ef=0

Also Workdone by the gravitational force=-(change in gravitational potential energy)=-mgh

Therefore

-(1/2)mv2=-mgh-Ns

or (1/2)mv2=mgh+Ns

Now s=3

N=mgcosh=ssin

Substituting all the values

(1/2)mv2=42.2J

Solution 15

Net momentum=m1v1+m2v2

Net Kinectic Energy=(1/2)m1v12+(1/2)m2v2

2

Let v1=v ,v2=-v and m1=m2=mThen Net momentum=0 but Net Kinectic Energy is not equal to zero


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Now lets v1= v2=0

Then Net Kinectic Energy=0 and Net momentum=0

Hence (c) is correct

Solution 16.

Consider a differential element of lenght dl of the ring whose radius vector makes an angle with thex-axis .If the angle subtended by the length dl is d at the center then,dl=Rd

Let be the mass per unit lengthThen mass of this element is dm=Rd

Xcm=(1/m)(Rcos)(Rd) integrating from 0 and =0Ycm=(1/m)(Rsin)(Rd) integrating from 0 and or Ycm=2R/Hence A is correct.

Solution 17-

The workdone by a conservative force is equal to the negative of the potential energy.When the

wokdone is positive ,the potential energy decreases.

The rate of change of total momentum of a many -particle system is proportional to the net force

external to the system ;the internal forces between particles cannot change the momentum of the

system.The workdone by the conservative system is zero in closed loop

Hence a,b,c are correct

Solution 18

Since there are no external force on the system,center of mass remains at rest so acceleration is zero

Solution 19.

U=20x2+35z

3

F=-(/x)i--(/y)j--(/z)kor

F=-40xi-105z2k

Solution 20

Frictional force is non conservative force

Gravitational force is conservative forceElectrical force is conservative force

Viscous force is non conservative force

And for conservative force, Workdone by the force in closed loop is zero

And for non conservative force,Workdone by the force in closed loop is not zero

Solution 21

Angular momentum for this is defined as

=mr2

First case


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L=mr2

Second case

Lf=m(r/2)2

So Lf=L/4

Solution 22

L=(mv)Xr

or

L=mvrsinNow rsin=perpendicular distance from x axis which is constantSo Angular momentum is constant

Solution 23:

Imagine the cylinder to be moving on a frictionless surface.In both the cases the acceleration of the

CM of the cylinder is gsin.This is also the acceleration of the point of contactof the cylinder and the inclined plane..Also no torque (about the center of the cylinder) is acting on the

cylinder since we assumed the surface to be a frictionless and the forces

acting on the cylinder is mg and N which passes through the center of the cylinder.Therefore the net

movement of the point of contact in both the cases is in downward direction

Therefore frictional force will act in upward direction in both the cases

Solution 24

Let h be the height

The rotational and translational KE of the ball at the bottom will be changed to Gravitational energy

when the sphere stops .

We therefore writes

(1/2)Mv2+(1/2)I2=Mgh

For a solid sphere I=(2/5)Mr2and also =v/r

So (1/2)Mv2+ (1/2)(2/5)Mr

2(v/r)

2=Mgh

or

v2

/2 +v2

/5=gh

or h=28.6 m

Solution 25

Net force on the cylinder

Fnet=mgsin -for ma=mgsin -fWhere f is the frictional force

Now =fXR=I

Now in case of pure rolling we know thata=R=> =a/R


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So f=Ia/R2

From 1 and 2

a=mgsin /[m+(I/R2)]

Now I=mR2/2

So a=(2/3)gsin

Solution 26

Law of conservation of angular momentum

I11=I22or

1.33(1.9)=.482or

2=5.26 rev/s

Solution 27

Using parallel axis theorem

I=Icm+Mx2

where x is the distance of the axis of the rotation from the CM of the rod

So x=L/2-L/4=L/4 Also Icm=ML2

/12

So I=ML2 /12+ML2 /16=7ML2 /48

Solution 28

Tangential acceleration=radius* angular acceleration

Since angular acceleration is constant ...Tangential acceleration is constant

Radial acceleration=r* (angular velocity)2

Since angular velocity increase with time...Radial acceleration increase with time

So resulttant acceleration increase with time and becomes more radial as time passes

Solution 29.

Radial acceleration=r* ()2

For wheel A

Radial acceleration of A =r* ()2

For wheel B

Radial acceleration of B=r* (2)2=4r* ()2

So Radial acceleration of B/Radial acceleration of A=4:1


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Solution 30

At rim

v=r

At point between the center and rim

v=(r/2)

Ratio =2

Solution 31

0=18=-18

anugular acceleration()=-2

Now

=0+tor t=18

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Solved Examples of System