Jee 2014 Booklet2 Hwt Solutions Dynamics of a Particle

Vidyamandir Classes

Solutions to Home Work Test/Physics Dynamics of a Particle 1.(C)

HWT - 1

Action - reaction pair, equal and opposite.

2.(ABC) Act on two different objects has equal magnitudes have opposite directions. 3.(C)

F

Force by ceiling on chain

T

F = T + W2

5.(A)

W2

and

T = W1



F = W1 + W2

W1

T

10 kg 6.(A)

T  F

mL  x L

a

T

T

10g

T

T

mx F T  a L

L

F L  x

Total mass, m

F

L x

L

x

a

rope 7.(B)

T W2

W1

Both scales will read 10 kg.

Going up

a T

L x

Coming down

F T

x

kv mg kv

v mg acceleration a1

v

acceleration a2 

Clearly, a2  a1

t2  t1

T 8.(CD)

T = Mg if acceleration = 0 M 9.(D)

lift

Mg

If the scooter accelerates, then f  0 Total force on man by seat, F 

N 2  f 2  F  500 N Δt1 and Δt3

man

F = 500 N for Δt2 10.(D)



f 500N

System cannot remain in equilibrium.

f A

B

Direction of motion

N N = 500 N 500a f  g = 500 N

F

smooth

N

F N

B A F mg

mg

VMC/Dynamics of a Particle

25

HWT-Solutions/Physics

Vidyamandir Classes

Dynamics of a Particle 1.(A)

Friction on B is upward, to balance weight.

3.(D)

The angle of which the block just begins to slide is

HWT - 2

m

  tan 1   





  does not depend on the mass of the block. 4.(AB)

Friction is towards east if the car accelerates. In rolling without acceleration, there is no horizontal force on the car. no friction. 

5.(D)

Time period are equal



z 2  ω1 ω2

ω1  ω2

6.(C) m  g cos 

mg sin 

mg sin 

m  g cos 

coming down

going up a   g  sin    cos  

s

1 2 at 2

a  g  sin    cos  



t  1/ a

1 tcomming down 2 g  sin    cos    4 g  sin    cos  

tgoing up  

7.(C)

  0.6 tan 



At top

water



9.(A)

2

r

mg N

mg  N 

Normal force from bucket N

 is decreasing as the car climbs the bridge. mg cos   N 

m r

10.(B)

f1 



m 2 r N is increasing

m r

mg 

m 2 r

v



m

mg

2

r

2

f 2  ma 

F



2

N  mg cos  



m 2 r

 2 f12  f 22  m a 2    r 

   

2

f2

car f1

m 2 r

VMC/Dynamics of a Particle

26

HWT-Solutions/Physics

Vidyamandir Classes

Dynamics of a Particle 1.(BD)

Net force, F net 

HWT - 3

mv 2 (radially) r

F net  F  F others

As the particle moves in the circle,

F net changes direction As F is a constant, F others must change direction as well as magnitude. 2.(C)

T = m1g T/2 = m1g/2 m2, m3 accelerate 

2m2 m3 T  g 2 m2  m3



2m2 m3 g mg  1 m2  m3 2

T m1 T/2 m2

4 1 1   m1 m2 m3 3.(B)

mg  T  ma

T

T  mg  ma

man a  g 1  n 



T   mg

4.(B)

m3

a

mg

M  m 2 Mmg T  M m 

2 Mmg ~ 2mg M

T m M

5.(C)

Te a

Chain

m2

m1

6.(B)

Total force on pulley, F  F 

4m1m2 g m

4m1m2 g m1  m2

Fmax  mg

N



 mg F  sin   cos  Fmin 

m1 g  T  m1a

m1g

F cos    N F sin   N  mg

a

2

2m1m2 g m1  m2

m1  m2  m

T  m2 g  m2 a

m

m1  m2  m

T 

T

 mg 1  2

VMC/Dynamics of a Particle

F

f  N

 f mg

27

HWT-Solutions/Physics

Vidyamandir Classes 7.(A)

T0

T0 cos   T T0 sin   mg

T  mg / tan



B

A





T P 8.(AB)

mg

 i   50 cm / s

 f  i 

 area under the curve  10

mass

0.07    0.50  m/s 0.07     0 .5 m / s

aa  9.(A)

F(N)

At rest T1  10 g  T2  6 g

0.0008 0.01 t

0.004

 f  i 1    100m / s 2 time 0.01

10g

T1

Just after breaking, T1  10 g  T2  0 4a = 6g

a

4kg

4kg

T2

4g

a = 3g/2 = 15 m/s2

6kg 10.(B)

m

m

F1= 2mg

m 2m

m

2mg  mg a2  3m = 2/3

2mg  mg a3  2m = g/2

mg 2mg  mg a1  m a1  a3  a2

Dynamics of a Particle

1.(C)

kv  kv    a1    g  a2    g   m M    |a1 |  | a2 | 

h

1 a



h1  h2

T

2.(C)

HWT - 4

M

m

mg

mg

kv

kv T

2mg T = mg

2mg

VMC/Dynamics of a Particle

mg

28

HWT-Solutions/Physics

Vidyamandir Classes 5.(A)

f = 10N

f  mg sin

10  20 N sin 30 m  2kg mg 



mg f   mg cos 

6.(A)

rough

mg sin 

mg sin 

smooth a2  g  sin    cos  

a1  g sin t1  t2 / 2

a

1 t

7.(C)



2

a1  4a2





3 tan  4

N 2  mg N1  f

N1  2   f  2   N 2 1.5 2 f  2 f  1.5mg

f 

8.(C)



1 3

f min 

9.(C)

Fmin 

mg / 2/

3

C

4

N2 f

3mg 8

f   N2

N1

A

Torque about C = 0

0   3/ 8



3

B mg

 mg 1  2

F

 12.5 kgF

3

25kg

30N

N Net force on block by incline



f

mg

VMC/Dynamics of a Particle

29

HWT-Solutions/Physics

Vidyamandir Classes

Dynamics of a Particle t2

1.(A)

f1

R

N

N

A

f1 mg

2.(C)

Refer to HWT-4 Solution No 1

3.(B)

T3  3mg

T3

2  m  2m  g m  2m

T2  2T1 

4.(B)

F A

mg

T1 

HWT - 5



8mg 3

T2

4mg 3

T1



right side goes down.

m 2m

Refer to HWT - 3 Solution No.10

6.(ABCD) Suppose the car and plant have the same acceleration, a in opposite directions T + f = m1a N f  T  m2 a



T  0 if m1  m2

C

a

T

10kg

R m2

N

T

P

f

m1g

If m1  m2  T  0 7.(A)

a

m1

3m

f m2g

20kg F=200N  = 0.1

N

12m/s2

N  10 g

T   N  12  10 

T

T  120  10  130 N

N

10kg mg

N a

130N

N   20 g

200  130    20 g   20a

a  2.5 m / s 2

200 N

20g 8.(B)

N  ma mg  F f  s N mg  s  ma 

a g

a

M

m

f

 m

N

mg

s

VMC/Dynamics of a Particle

a

30

HWT-Solutions/Physics

Vidyamandir Classes 9.(A)

N

N  f  mg  0

f



| N  F |  mg (Net force by the plane)

 10.(B)

100gm at rest  T  100 gm   g

P1

=1N

mg

2T m1

P2 T 100gm

200gm

 200  200  10  T  a 1000 1000

T

a  5 m / s2 Acceleration of pulley P1 = 5/2 m/s2

a (200g)

2T

5 2T  m1 g  m1 2

5/2 m/s2

m1  2 / 10  5 / 2 



m1g

m1  160 g

Dynamics of a Particle 1.(B)

Force =

HWT - 6

Change in momentum 2mv  Δt time

2  10  1013  5    10 N 0.01

2.(A)

We can write a A  2a B  aC  0  a A  aC  aB   2  a  f   2 1 a  f    down or  f  a  up  2  2 

a

f A

C

B 3.(D)

T1  8 N

T1  tan 30 T2

W = 40N

T2  8 3 N 

f  40    8 3   3 / 5  0.35

30

T1

N

friction = T2  8 3 N

T2

W = 8N

T2 f 40

VMC/Dynamics of a Particle

31

HWT-Solutions/Physics

Vidyamandir Classes 4.(D)

 R  Mg  sin 37 

f  Ma

mg sin 37  f cos 37  R sin 37  ma

Put f   R R 

m

mg  ma cos 37

M

5mg 5 f  4 3 f  R

R

37

15mg 15 mg R f  3  5 3  5

 R  Mg  

3  f M 5  3 4f m  mg  R    5 5



3 15mg  3Mg  5 Mg   75  mg 3  5  mg  12mg   60  mg

B

N

25



M m

f

f a

a

20

M

15m 2  15Mm  25 m 2  20 Mm  0 If M = m 15  15  25  20   0 2

 7  3

mg

3

a

6.(B)

x1  2 x3  x2  

 0.4 g

7

3

x1

Differentiate w.r.t. time  v1  2v3  v2  0 v  v2 v3  1 2 a1  a2 a3  2

a3   1 m / s

C x3

v3  0    2 m / s

2

B

v3  5   2  1 5   3m / s

7.(B)

Mass = slope of F  a graph =

8.(C)

5 sin 37  M sin 53

9.(A)

3 4M  5 5

14  4  2.5 kg 40

M = 3.75 kg

  0 .5

PN mg   N 

x2

A

At t  o1 v1  0 a1   2 m / s 2 v2  4 m / s a 2  0 

5

37

mg

R

5.(D)

7 3

37

P P

N

mg  20 N 

 N  10  0.5  5N 1g  5  1a

N

mg 10N

N

a

a  5m / s 2

1g

VMC/Dynamics of a Particle

32

HWT-Solutions/Physics

Vidyamandir Classes 10.(B)

f  F  s N F  0.5  60 g

 s  0 .5

a

N

 s  0 .4

60kg

F

 300 N

N

300

60a  300  0.4  600 

k N

F

a  1m / s 2

60g

Dynamics of a Particle 1.(C)

20 g  T  20  a 

HWT - 7

T a

a  7m / s 2 T  60 N

monkey

A 0.2

2 2.(A)

60g

20g

25

0.5

8 25  0.5  2  8  g

 3.(C)



B cannot move

Acceleration =

5   3  1

5   3  1

No. friction between A and B

g

= g/9 m/s2

T

5 kg

3 kg

T

T

1g  T   g / 9

T 

g/9

8g N 9

1kg

1g 4.(B)



Friction = 8 N towards left.

F1 F2

 5.(B)

F2 = 2 N

F1 = 10 N

Friction = 2N towards right

Acceleration = slope = 2 m / s 2



a   k g   2

6.(B)

F2

 k  0 .2

r  2m1 a   t m / s2 t2 At t  25   2 ar  at



2

= 2d m / s 0

v2  2 r

at ar

  |m / s 3

VMC/Dynamics of a Particle

33

HWT-Solutions/Physics

Vidyamandir Classes 7.(D)

a

Tmax for no slipping = s  4  5 g

4

a

36  4m / s 2 4 5

36   0 .4

= 36N

5kg

T 4

Mg  36  4m M  6kg

Mg M N

T

8.(A)

T

1kg   0 .8

37 mg

37 B 9.(A)

g/3 9g B  9g 

9g 3

B = 12g

B

g/4 mg B  mg 

mg 4

m

P

10.(A)

4B  16kg 3g

Mass of sand = 7kg

P F

a

a

m M

mg F

mg

P   F  mg   ma

 PM  F   M  m



F  Mg  Ma

Dynamics of a Particle

HWT - 8

3kg

1.(B)

2kg P = 10N a

A

a

B

10  2m / s 2 23

F  3 a   6N

VMC/Dynamics of a Particle

F

3kg

34

HWT-Solutions/Physics

Vidyamandir Classes 2.(B)

T 

2m 2 sin 30 mg g 2  2m  2

T 3   3T  T      2   2   

T  3.(A)

T

m

30

2

T

T

30

T

M

3mg 2

Fmin  mg  sin 30   cos 30   0 Fmax  mg  sin 30   cos 30   115 N

F

  3/ 4

m = 10kg

N

N F

F

T

30

30

f

30 mg 4.(D)

mg

N cos 37  mg  f sin 37

v2/r

mv 2 N sin 37  f cos 37 = r

N

37



1 3

f = MN = N / 3 4 3N N   mg 5 5 3

Put

mv 2 3mg 4 3mg   r 4 3 3 4 3

v2  5.(B)

F 1 / m

6.(C)

Radial force =

 9  4 3  gr



37

 

4 3

mg

v  84m / s

mv 2 4  22  16 N = r 1

Tangential force = 3  4  12 N

Total force =

122  162 = 20 N

M 10.(C)

2T m1 has no acceleration

T

T = 1g = 10 N 1g T

m2

m1

2 g  10  2a

a

a  5 m / s2 am 

2g

50  5 / 2 m / s2 2

5/2M 2T

VMC/Dynamics of a Particle

towards right

5 2T  M 2 4T M   8kg 5

35

HWT-Solutions/Physics

Vidyamandir Classes

Dynamics of a Particle 2mV 2  1  20  = 2000 N Δt 1 / 50

1.(C)

Force =

2.(D)

amax of car =  g s

HWT - 9

1 2 at 2



t

1 a



1 

t

3mg

3.(C)

T

a

Before cutting T   3mg T  mg After cutting T   3mg

2m

2mg T

T 0

a = g/2

m a

a  g

mg

F

5.(B)

T

T

a0 A

m

x   cos 1   a

a

a 

m

x a0



2T sin   F

T 

F  2 sin 

F 2 1

x2 a2

ma0  T cos

a0 

Fa 2

2m a  x

2

.

6.(A)

t1  t2

9.(A)

Refer to HWT – 2 Solution - 1

7.(D)

x a



a0 

Fx 2m a 2  x 2

Refer HWT – 1 Solution – 9

10.(ACD) N 2   M  m  g (A) N1 sin   M | a1 | (C) ma 2   Ma1

8.(D)

m M

Refer HWT – 1 Solution - 10

N1

N1



Action – reaction pair N1  N 2 ) (D)

a1

N2

mg mg

VMC/Dynamics of a Particle

36

HWT-Solutions/Physics

Vidyamandir Classes

Dynamics of a Particle

HWT - 10

Δv 20m / s   400 m / s 2 Δt 0.055

1.(C)

aav 

3.(A)

If B  0 Net force = mg  B  mg

4.(C)

F  mg  0.3 N

5.(C)

a1 slope 1  a2 slope 2

B

mg



6.(C)

20 / 5 1 12 / 3



F1 ma  1 1  5/ 7 F2 m2 a2

F(N) v 50 250 10

20

20

30 t(s) 10

–50

20 t

–250 7.(B)

T  2a 3g  T  3a T  12 N

8.(C)

P P O 2  1 m m

9.(A)

a  k g

= 1.2  10  12m / s 2 v  u  at

a

v  12m / s

u  12m / s V  0 t  1s

10.(C)

m1  m2

m2 m1 T 

2m1m2

 m1  m2 

g

VMC/Dynamics of a Particle

37

HWT-Solutions/Physics