D C Pandey Solution

For JEE MainIntroductory Exercise 5.1

1. w1 = weight of cylinder

w2 = weight of plank

N 1 = normal reaction between cylinder and plank

N 2 = normal reaction on cylinder from ground

N 3 = normal reaction on plank from ground

f1 = force of friction on cylinder from ground

f2 = force of friction on plank from ground

2. N = normal reactions

w = weights

3.

4. Force Fx Fy

F1

→ 4 30cos °= 2 3 N

4 30 2sin ° = N

F2

→ − °4 60cos

= − 2 N

4 60 2 3sin ° = N

F3

→ 0 −6 N

F4

→ 4 N 0

5. Taking moment of forces about hinge force A (so that moment of hinge force becomes zero).

Clockwise torque of W = anti-clockwise torque of T

∴ wl

T l2

30

= °( sin )

or T w= Ans.

6. Net force in vertical direction = 0

∴ T wcos 30° =

or Tw= 2

3Ans.

7.

∑ =Fx 0

∴ F − ° − ° =3 60 10 60 0cos sin

∴ F = 10.16 N Ans.

∑ =Fy 0

∴ R + ° − ° =3 60 10 60 0sin cos

∴ R = 2.4 N Ans.

8.

Chap ter 5 Laws of Motion � 1

w1

N2

N1N3

N1

w2

f1 f2

w1N2

N1

N3 w2

N4N3

N w

T

w

T

N

30°

a

a

a 30°

60°

60°60°

R

3 N

10 N

x y

60°

F

SOLUTIONS

T

T

a A

mg

a

B

For A T macos 45° =

or T ma= 2 ...(i)

For B mg T ma− ° =cos 45

∴ mg ma ma− =

or ag=2

Ans.

Substituting in Eq. (i), we get

Tmg=

2Ans.

Introductory Exercise 5.2

1. (a) a = Net pushing force

Total mass

= −+ +

120 50

1 4 2

= 10 2m/s Ans.

(b)

120 1 10− = × =R a

∴ R = 110 N Ans.

(c) F manet = = ( ) ( )2 10

= 20 N Ans.

2. Since surface is smooth, acceleration of both is

g sin sin ,θ = ° =10 30 5 2m/s down the plane

The component of mg down the plane ( sin )= mg θprovides this acceleration. So, normal reaction willbe zero.

3.

Nmg=4

is given

mgmg

ma− =4

∴ ag= 3

4 Ans.

4. Constant velocity means net acceleration = 0.

Therefore net force should be zero. Only two forces T and mg are acting on the bob. So they should beequal and opposite.

Asked angle θ = °30 Ans.

T mg= = =( ) ( )1 10 10 N Ans.

5. T mg mamg

sin sinθ − ° = =302

∴ T mgsin θ = ...(i)

T mgcos cosθ = °30

or Tmg

cos θ = 3

2 ...(ii)

Solving Eqs. (i) and (ii), we get

θ =

−tan 1 2

3Ans.

and T mg= =7

25 7 N Ans.

6. a = Net pulling force

Total mass

=+ +

8

1 2 1

= 2 2m/s

TB 8 1 21− = =T m aB ( ) ( )

∴ TB = 6 N

TA T m aA = =2 2 2( ) ( )

∴ = 4 N Ans.


2 � Mechanics - I

1 kg120 N

R

a

mg

aN

mgq

T

30°

mg

T

30°

q

a

30°

1. m g g gA = + ⋅ °( ) ( ) ( ) sin2 2 30 as a = 0

∴ mA = 3 kg Ans.

2. T g4 4= and T g1 1= ( ) as a = 0

∴ T

T4

1

4= Ans.


Total mass

= −+

2 1

2 1

g g( )

= =g

3

10

32m/s

After 1 s,

v at= = 10

3m/s

At this moment string slacks ( ` )T = 0

String is again tight when,

s s1 2=

∴ 10

3

1

2

1

22 2t gt gt− = ( )g = 10 2m/s

On solving we get, t = 1s

.0

3Ans.


Total mass

= + −+

F 10 20

1 2( )F = 20 N

= 10

32m/s Ans.


1. 2 kg 2 2T a= ( ) ...(i)

1 kg 10 1 2− =T a( ) ...(ii)

On solving these two equations, we get

T = 10

3N

and 2 220

32a T= = =m/s acceleration of 1 kg

block

2.

T Ma= ...(i)

Mg T Ma− = ...(ii)

On solving these two equations. we get

ag=2

and TMg=2

Ans.


Total mass

= −+

=30 20

3 22 2m/s

2 kg T a− = × =20 2 4

∴ T = 24 N

Now, Mg T= 2

∴ MT

g= 2

= 48

10

= 4.8 kg Ans.

4. m1

T m a a a= = =1 2 0 3 2( ) ( . ) ( ) 0.6 ...(i)

m2 F T m a− =2 2

∴ 2T a= −0.4 0.2 ...(ii)


a = 2

72m/s Ans.


1 kg u = 0

g

103

m3

s1

2 kg

gs2

2 kg 2 T

a

mg = 10 N

2a

T

1 kg

MT

a mg

T

M

M

2 kg

3 kg

2 T2 T

P2

T

P1

T

a

a

P ® Stationary2

and 212

35T = N Ans.



Total mass

aMg

m m M=

+ +1 2

...(i)

m1 N m gcos30 1° = ...(ii)

N m asin 30 1° = ...(iii)

z

From Eqs. (ii) and (iii), we get

a gg= ° =tan 303

Substituting this value in Eq. (i) we get,

M = 6.83 kg Ans.

2. (a) With respect to box (Non-inertial)

X X u t a tx x= + +021

2

= + −X t t0210 5 Ans.

V u a t tx x x= + = −10 5 Ans.

(b) Vx = 0 at 2s = t0 (say)

∴ To return to the original position,

time taken = =2 40t s Ans.

3. (a) In car's frame (non inertial)

ax = − 5 2m/s (due to pseudo force)

ux = 0

az = 0

uz = 10 m/s

Now apply, v u at= + and s s ut at= + +021

2 in X

and Z-directions.

(b) In ground frame (inertial)

a u ax x z= = =0 0 0, ,

and uz = 10 m/s

4. Relative to car (non-inertial)

a1 is due to pseudo force

ax = − + = −( )5 3 8 2m/s

Block will stop when

v u a t tx x x= = + = −0 10 8

or at t = 1.25 s

So, far t ≤ 1.25 s

X X u t a t X t tx x= + + = + −02

021

210 4

V u a t tx x x= + = −10 8

After this u gs > 5 2m/s as us > 0.5

Therefore, now the block remains stationary withrespect to car

5.

N f mgcos sin37 37° + ° = ...(i)

N f masin cos37 37° − ° = ...(ii)


f = 3.6 m

= 9

25mg Ans.

Subjective Problems

4 � Mechanics - I

30°

N

m1

a

30°

u = 102a = 5 m/s

(due to pseudo force)

ms

x

2a = 5 m/s1

2a = u g = 3 m/s2 kax

u = 10xms

mg

37°

qf

37°a

37°N

T

2 a

F2Tm1 m2

a

30°m2

m1

a

Ma

1. See the answers.

No friction will act between sphere and groundbecause horizontal component of normal reactionfrom rod (on sphere) will be balanced by thehorizontal normal reaction from the wall.

2. See the solutions

3. See the solutions

4. (a) At P

At P, F TT

2 452

= ° =cos ...(i)

w TT= ° =cos 452

...(ii)

At Q, FT

12

= ...(iii)

From these three equations, we can see that

F F wT

1 22

60

230 2= = = = = N Ans.

5. N A cos 30 500° = N ...(i)

N NA Bsin 30°= ...(ii)


N A = 1000

3N

and N B = 500

3 N Ans.

6. T w1 45cos ° =

and T1 45 30sin ° = N

∴ w = 30 N Ans.

7. N F= = 40 N

f mg= = 20 N

Net moment about C = 0

∴ Anti-clockwise moment of f = clockwisemoment of N

∴ ( ) ( )2020

240

= ⋅ x

or x = 5 cm Ans.

8. w = 100 N

∑ (forces in horizontal direction) = 0

∴ T f N+ ° = °cos sin30 30 ...(i)

∑(Forces in vertical direction) = 0

∴ N fcos sin30 30 100° + ° = ...(ii)

∑ (moment of all forces about C) = 0

∴ ( ) ( ) ( )T R f R=or T f= ...(iii)

On solving these three equations, we get

T f= = 26.8 N

and N = 100 N Ans.


30°

NA

30°

w = 50 g = 500 N

NB

w

T1

45°

T = 30 N2

f

mg = 20 N

CF

Nx

w = 100 newton

T

30°

C

30°f

30°N

F1

T45°

PF2

Q

w

9. H TT= ° =cos 602

...(i)

V T+ ° =sin 60 40 ...(ii)

∑ =M0 0

∴ 402

60l

T l

= °( sin )

or T = 40

3N Ans.

H = 20

3N

and V = 20 N

∴ Net hinge force = + =H V2 2 40

3N Ans.

10. (a) a = Net pushing force

Total mass

= −+ +

=100 40

6 4 103 2m/s

(b) Net force on any block = ma

(c) N a− = =40 10 30

∴ N = 70 N Ans.

11. (a) T a T T a T a1 2 1 210 20 60 30= − = − =, ,

On solving, we get T T1 210 30= =N, N Ans.

(b) T a T a2 220 60 30= − =,

On solving, we get

T2 24= N Ans.

12. (a) T g a1 2 2− =

∴ T g a1 2= +( )

= +2 ( )9.8 0.2

= 20 N Ans.

(b) T g a2 5 5− =

∴ T g a2 5= +( )

= +5 ( )9.8 0.2

= 50 N Ans.

13. (a) a = Net pulling force

Total mass

= − ×200 16

16

9.8

= 2.7 m/s2

(b) 200 50 51− − =T a

∴ T1 200 5 50= − × −2.7

= 136.5 N Ans.

(c) T g a2 9 9− =

∴ T g a2 9= +( )

= +9 ( )9.8 2.7

= 112.5 N Ans.

14. Force diagram on both sides is always similar.Therefore motion of both sides is always similar.

6 � Mechanics - I

T1

a

a

10 kg

T1 T2

a

T2 60 N

20 kg

30 kg

T2

a

60 N20 kg

a

30 kgT2

2 g1.9 kg

0.1 kg

T1

20.2 m/s

5 g

1.9 kg

0.2 kg

T2

20.2 m/s

2.9 kg

T1

50 N

5 Kg

200 Na

9g

7 kg

2 kg

T2

a

40 NN

a

10 kg

mg = 40 N

H

V

O

T

60°

For example, if Monkey accelerates upwards, then T g> 20 . But same T is on RHS also.

Therefore, bananas also accelerate upwards.

15. T T T TA B= =3 ,

In such type of problems

aT

∝ 1

∴ | | | |a aB A= 3

or a aB A= − 3 , as aA and aB are inopposite directions.

16. X X XA B C+ +2

= constant on double differentiating with respect totime, we get,

a a aA B C+ + =2 0 Ans.

17. X XA B= sin θ ...(i)

Double differentiating with respect to time, we get

a aA B= sin θ Ans.

18. a a ar3 6= + = ...(i)

a a ar2 4= − = ...(ii)


a = 1 2m/s


Total mass

= + − −+ + +

=( )3 4 2 1

3 4 2 14 2g

m/s Ans.

4 kg 40 41− =T a ∴ T1 24= N Ans.

3 kg T T a1 230 3+ − = ∴ T2 42= N Ans.

1 kg T a3 10 1− = ( ) ( ) ∴ T3 14= N Ans.

20. P2 2 21 2T T= ∴ T T1 2= ...(i)

1 kg T a2 1= ( ) ( ) ...(ii)

2 kg T aa

r1 20 22

− = −

...(iii)

3 kg 30 32

1− = +

T aa

r ...(iv)

On solving these equations, we get

T T1 2

120

11= = N

a a12120

11= = m/s

a aa

r22

2

50

11= − = − m/s


q

A

XA

XB

q

1

ar

a

a

ar2 3

2 kg

3 kg

2T2

P2

P1

a

1 kg

T1

T1

ar

a r

a2

a2

T2

T2

T2

Fix line

A

B

TXC

XB

XA

C

Monkey

T

20 g

Bananas

T

20 g

2 T

A

2 T

B

T

T

or a2250

11= m/s (downwards)

a aa

r32

2

70

11= + = m/s (downwards)

21. T Mg Ma− ° =sin 30 ...(i)

2 2 22

Mg T Ma− =

...(ii)

solving these equations, we get

ag=3

Ans.

22. T a= 1 ( ) ...(i)

10 1− =T a( ) ...(ii)


T = 5 N Ans.

23. 2 50 5T a− = ...(i)

40 4 2− =T a( ) ...(ii)


a = 10

72m/s or

g

7Ans.

and 220

72a = m/s or

2

7

gAns.

24. Figure (a)

N mg= = 40 N

µ sN = 24 N

Since, F Ns< µblock will remain stationary and

f F= = 20 N

Figure (b)

N mg= = 20 N

µSN = 12 N and µKN = 8 N

Since, F NS> µ , block will slide and kinetic friction ( )= 8 N will act.

aF f

m`= − = − =20 8

26 2m /s

Figure (c)

N + =20 60 ⇒ N = 40 N

µSN = 8 N

and µKN = 4 N

Since, F NS> µ , block will slide

Kinetic friction ( )= 4 N will act.

aF f

m= − = − =20 4

6

8

32m/s Ans.

25. af

mg= = =µ 3 2m/s

(a) Relative motion will stop when velocity ofblock also becomes 6 m/s by the aboveacceleration.

v at=

∴ tv

a= = =6

32 s Ans.

(b) S at= = =1

2

1

23 2 62 2( ) ( ) m Ans.

26. 2 kg block has relative motion towards right.Therefore maximum friction will acts on it towardsleft.

f N= = =µ ( ( ) ( )0.4) N1 10 4

a124

14= = m/s

a224

22= − = − m/s

8 � Mechanics - I

T

a 4 kg 2 a5 kg

40 N50 N

2 T

N

mg = 60 N

20 N

20 N = F

a

f = mmg

1 kg

a

T1 kg

T

a

10 N

M a

mg sin 30° 2M

2T

a2

2 mg

T

(a) Relative motion between them will stop when,

v v1 2=∴ u a t u a t1 1 2 2+ = + ...(i)

∴ 2 4 8 2+ = −t t ...(ii)

or t = 1 s Ans.

(b) Substituting value of ' 't in Eq. (ii), either onRHS or on LHS, common velocity = 6 m/sAns.

(c) s u t a t1 1 121

2= +

= + =( ) ( ) ( ) ( )2 11

24 1 42 m Ans.

s u t a t2 2 221

2= +

= + − =( ) ( ) ( ) ( )8 11

22 1 72 m Ans.

27. Relative motion of 2 kg block is towards right.Therefore maximum friction on this will act towards left

f N= = =µ ( ) ( ) ( )0.6 N2 10 12

a2212

26= − = − m/s

a1212

112= = m/s

(a) Relative motion between them will stop when,

v v1 2=or u a t u a t1 1 2 2+ = +or − + = −18 12 3 6t t ...(i)

(b) Substituting value of ' 't in equation

(i) either on RHS or on LHS we have, commonvelocity = − 4 m/s Ans.

(c) s u t a t1 1 121

2= +

= −

+

( ) ( )187

6

1

212

7

6

2

= −12.83 m Ans.

s u t a t2 2 221

2= +

=

+ −

( ) ( )37

6

1

26

7

6

2

= − 0.58 m Ans.

28. N = 20 N = applied force

µS N= = 16 N

and µKN = 12 N

Driving force F mg= (downwards)

= 20 N

Since, F N> µS , block will slide downwards andkinetic friction of 12 N will act in upward direction

∴ aF N

m= − µK (downwards)

= −20 12

2

= 4 2m/s (downwards) Ans.

29. f N mgmax = =µ µ= × ×0.8 2 10

= 16 N

Block will not move till driving force F t= 2becomes 16 N. This force becomes 16 N in 8 s

For t ≤ 8 s

a = 0

For t ≥ 8 s

aF f

m

t= − = −max 2 16

2

a t= − 8

i.e., a t- graph is a straight line of slope +1 andintercept −8

corresponding a t- graph is shown below

30. N mg= = 60 N

µS N = 36 N

µK N = 24 N

Driving force F t= 4

Block will move when,

F NS= µ


3 m/s2 kg12 N

a2

1 kg18 m/s12 N

a1

– +

45°

2a (m/s )

t (s)8

8 m/s

2 m/s1 kg

a1

2 kg4 N

– +

a2

4 N

or 4 36t = ⇒ t = 9 s

For t ≤ 9 s

a = 0

For t ≥ 9 s

At 9 s block will start moving. Therefore kinetic friction will act

∴ aF N

m

tK= − = −µ 4 24

6

= −

2

34t

∴ a t- graph is a straight line

At t a= =9 2 2s m/s,

The corresponding a t- graph is as shown in figure.

31. N mg= °cos 60

=

=( ) ( )6 101

230 N

µSN = 18 N

µKN = 12 N

Driving force F mg= °sin 60

=

( ) ( )6 10

3

2

= 52 N

(a) Force needed to keep the block stationary is

F F NS1 = − µ (upwards)

= −52 18

= 34 N (upwards) Ans.

(b) If the block moves downwards with constantvelocity ( , )a F= =0 0net , then kinetic frictionwill act in upward direction.

∴ Force needed is,

F F NK2 = − µ (upwards)

= −52 12

= 40 N (upwards) Ans.

(c) Kinetic friction will act in downward direction

F ma3 52 12 6 4− − = = ( ) ( )

∴ F3 88= N (upwards) Ans.

Objective QuestionsSingle correct option

1. amg F

mg

F

m= − = −

m mA B> ∴ a aA B> and A will reach earlier


Total mass

= + −+ +

=( )2 2 2

2 2 2 3

g g

FBD of C

mg T mamg− = =3

∴ T mg= = =2

3

2

320( ) 13.3 N Ans.

3. T gA = =10 100 N

T TB Acos 30° =

∴ TB = 200

3N

T TB Csin 30° =

∴ TC = 100

3N

10 � Mechanics - I

F

F 1 m Ns

52 N

F 2m NK

52 N

F 3

mN = 12 N

K

a

a

mg

FF = air resistance

T

mg

aC

2a (m/s )

t (s)9

2

4. amg T

mg

T

m= − = −

a gT

mg

mg

m

gmin

max= − = − =

2

33

Ans.


Total mass

= × − ×+

=10 10 5 10

10 5

10

32m/s

5 kg T a− × = × =5 10 550

3

∴ T = 200

3N

This is also the reading of spring balance.

6. Since 2 30mg mgsin ° =∴ System is stationary

FBD of m T mg=

7. tS

a a= ∝2 1

∴ t

t

a

a1

2

2

1

=

2

1

1

1=

−=

−g

g gK K

sin

sin cos

θθ µ θ µ

as, sin cosθ θ= °at 45

On solving the above equation, we get

µK = 3

4

8. F mg mg1 = +sin cosθ µ θF mg mg2 = −sin cosθ µ θ

Given that F F1 22=∴( sin cos ) ( sinmg mg mgθ µ θ θ+ = 2

− µ θmg cos )

On solving, we get tan θ µ= 3 Ans.

9. Only two forces are acting, mg and net contact force

(resultant of friction and normal reaction) from the

inclined plane. Since the body is at rest. Thereforethese two forces should be equal and oppsite.

∴ Net contact force = mg (upwards)

10. During acceleration,

v a tF

mt= =1 1

11

∴ Fmv

t1

1

= ...(i)

during retardation,

0 2 22

2= − = −v a t vF

mt

∴ Fmv

t2

2

= ...(ii)

If t t1 2= then F F1 2= t t1 2< then F F1 2> and

t t1 2> then F F1 2<11. Angle of repose

θ µ= =

= °− −tan ( ) tan1 1 1

330

So, particle may be placed maximum upto 30°, asshown in figure below

h R R R= − ° = −

cos 30 1

3

2Ans.

12. Net pulling force F g g1 15 5= −F g1 10 100= = N

Net stopping force F2 5 10= × ×0.2

= 10 N

Since F F1 2> , therefore the system will move(anticlockwise) with an acceleration,

aF F= −

+ +1 2

15 5 5 = =90

2523.6 m/s

15 kg block

15 10 15 152× − = × = ×T a 3.6

∴ T2 96= N

5 kg block

T a1 5 10 5 5− × = × = × 3.6

∴ T1 68= N

∴ T

T1

2

68

96

17

24= =


30°

R 30°

h

a

mg

T

D C Pandey Solution

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