Laws of Motion (01!10!09)

Name : Roll No. : Topic : Mohammed Asif

Ph : 9391326657, 64606657

Multiple choice questions (only one is correct) 1. A body of specific gravity 6, weighs 0.9kg when placed in one pan (say pan A) and 1.6kg when placed on the other pan (pan B) of a false balance. The beam is horizontal when both the pans are empty. Now if the body is suspended form pan A and fully immersed in water, it will weigh a) 0.5kg b) 0.6kg c) 0.75kg d) 0.8kg 2.

The pulley has mass M >m. String is massless. The above system is released from rest from the position shown. Then

a) Body (1) will slowly come down, till equilibrium is attained at level AA’ shown. b) The system will perform oscillations with equilibrium position at level AA’ and amplitude 0.5m c) The response of the system depends on whether pulley-string interface has friction or not. d) The system will continue to be in the same initial position. 3.

In the position shown above Let T1 be the tension in the left part of the string, T2 = Tension in the right part, N = Normal reaction on block (2) by the resting surface. Then

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4.

a) T1 = T2 = 0 and N = mg

b) T1 =mg, T2 = 0, N =mg

c) T1 = T2 = mg and N = 0

d) T1 = T2 =

A body of mass m was slowly hauled up the hill and down the hill onto the other side by a force → F which at each point was directed along a tangent to the trajectory. The work performed by this force, if the coefficient of friction is µ1 uphill and µ 2 dowbhill, is

b) mg ( 2h + µ11 + µ 2 2 ) d) mg ( µ11 − µ 2 2 )

a) mg ( µ11 + µ 2 2 ) c) mg ( h + µ11 + µ 2 2 ) 5.

mg mg and N= 2 2

The bob of a pendulum is taken to position A and given an initial velocity u in the direction shown, in the following cases.

If the minimum value of u so that the pendulum reaches position OB in case (i) and position OC u2 is ( g = 10 ms −2 ) in case (ii) are u1 in case (i) & u2 in case (ii), then u1 a)

(3 + 2 )

b)

(2 + 3 )

c)

(2 + 3 )

d)

(3

2 −2

)

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6.

A conveyor belt carrying powdery material is at an angle 370 to the horizontal and moves at a constant speed of 1ms-1 as shown. Through a small hole in the belt, the powdery material drops down at a constant rate of 1kg per second. What is the force to be applied on the belt along the direction of its motion so as to maintain its constant speed of 1ms-1?

a) -1 N 7.

d) None of the above

b) 10 cm

c) 10.5 cm

d) 11 cm

A pendulum consists of a mass m attached to the end of a light string 0.5m long. It can oscillate in the vertical plane. If it is let go in the horizontal position and has an inelastic collision with the floor, e = 0.5, the rebound velocity is (ms-1) (the angle turned is π / 6 ) (g = 10ms-2)

a) 9.

c) +1 N

Small body A on a hemispherical body B which is on a wedge C which is on a smooth horizontal surface. System is released from rest from the position shown when α = 37 0 ,  B = 10 cm,  = 5cm . 0 When α = 53 , C is 4.5cm  B at that instant is (neglect friction)

a) 9 cm 8.

b) Zero

5 2

b)

5 4

c)

35 4

d)

45 4

The track is in the vertical plane. The track is rough with friction coefficient µ . A particle at A is allowed to slide down and goes upto B and returns. Heights of A and B are 0.2m and 0.1m respectively, as shown. The maximum possible value of µ is

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a) 10.

1 4

b)

1 7

c)

1 6

d)

1 5

A particle is projected form a point on a horizontal floor. After it has three collisions with the  10 12 

floor, it is found that the ratio of maximum height to minimum peaks reached by it is  36  . 2  The coefficient of restitution is a) .5 b) 0.64 c) 0.8 d) 0.9

11.

The potential energy function along the positive x axis is given by U ( x ) = −ax +

b , a, b are x

constants. If it is known that the system has only one stable equilibrium configuration, the possible values of a and b are a) a = 1, b = 2 b) a = 1, b = -2 c) a = -1, b = 2 d) a = -1, b = -2 12.

The acceleration of the 1kg block immediately after the string is cut is (g = 10ms-2).

a) 4ms-2 13.

b) 4.1ms-2

c) 16ms-2

d) 40ms-2

A circular pan with its side wall inclined inward at 530 as shown is rotating about its central vertical axis with a constant angular velocity. A ball placed at the edge rotates along with it as shown. If the ball exerts a force of 22.5 N on the side wall and 23.5N on the bottom surface, the mass of the ball is

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a) 1kg 14.

c) 3kg

4kg

Acceleration of 10kg block, when system released from rest, is

a) 0.5ms-2 15.

b) 2kg

b) 1 ms-2

c) 1.25 ms-2

d) 1.66ms-2

Case(A)

Case (B)

The reaction force between the 5kg and 6kg block in case A and case B will be a) equal and non zero b) unequal with case A being more c) unequal with case being more d) equal and zero For Question No. 16 to 20: Each question consider of two statements: one is Assertion (A) and the other is Reason(R). You are to examine these two statements and select the answer using the code given below. a) Both A and R individually correct, but R is the correct explanation of A b) Both A and R individually correct, but R is the not correct explanation of A c) A is true but R is false d) A is false but R is true 16. Assertion (A): If a block is released form rest, when reaches the bottom point of the wedge, its speed is same irrespective whether the wedge is fixed or the wedge is fee to move. Reason(R): Mechanical energy is conserved in both cases. 17.

Assertion (A): A body is at rest on floor. You lift it vertically up and bring it to rest at a point h above the ground. The work done by you is zero. Reason(R): Any non-zero work done by a force on a body results in change in kinetic energy of the body.

18.

Assertion (A): The negative of the work done by the conservative internal forces on s system equals to change in its potential energy. Reason(R): Work energy theorem. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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19.

Assertion (A): In a tug of war that team wins which applies more tension force on string then their opponents. Reason(R): The winning team must be having stronger players.

20.

Assertion (A): If a block starts moving at t = 0 with an acceleration ‘a’ then its kinetic energy at time t with respect to two non-inertial frames which have same acceleration in a direction is not always same. Reason(R): The work energy theorem ∆K −∆W is not for non-inertial frame.

Match the following Questions 21 to 25 21. Two columns are given in each question. Match the elements of Column-I with Column-II. Column-I Column-II i) Friction force p) Contact force ii) Normal reaction q) Electromagnetic force iii) Tension in a string r) Gravitational force iv) Force between two charges of mass m s) Nuclear force a) (i-p,q), (ii-p,q), (iii-q,s), (iv-q,r) b) (i-p,q), (ii-p,q), (iii-q), (iv-q,r) c) (i-p,q), (ii-p), (iii-q,s), (iv-q) d) (i-q), (ii-p), (iii-q,s), (iv-p,q,r) 22.

In Column-I there are some motions of a body and Column-II contains the list of concepts that can be used for the analysis of these motions Column-I Column-II i) A body moving in a vertical circle p) Conservation of energy ii) A body moving in a horizontal circle q) Conservation of momentum iii) A body dropped form a height on a block attached to the top of a vertical spring r) Centripetal force iv) Rocket propulsion s) Centrifugal force a) (i-p,q,r), (ii-q,r,s), (iii-p,q,s), (iv-p) b) (i-p,s), (ii-p,r), (iii-q), (iv-p,q,r) c) (i-p,r), (ii-q,s), (iii-p), (iv-p,q) d) (i-p,r,s), (ii-r,s), (iii-p,q), (iv-q)

23.

When two bodies collide they come in contact at t = t1 and loses contact at t = t2. This (t2-t1) is a very small time interval. Consider this small time interval and match the following. i) Elastic collision p) Kinetic energy decreases and potential energy increases and then potential energy decreases and kinetic energy increases ii) Inelastic collision q) Kinetic energy + potential energy is conserved iii) Perfectly inelastic collision r) Momentum is conserved iv)Oblique elastic collision s) Kinetic energy decreases and potential energy increases and then situation remains same a) (i-p,q,r), (ii-q,r,s), (iii-p,q,s), (iv-p) b) (i-p,s), (ii-p,r), (iii-q), (iv-p,q,r) c) (i-p,r), (ii-q,s), (iii-p), (iv-p,q) d) (i-p,r,s), (ii-r,s), (iii-p,q), (iv-q)

24.

For each of the following four cases of Column-I match the range of force F in column B so that the block m is not slipping on the surface with which it has contact. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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For all cases m = 2kg; µ = 0.2 Take g = 10m/s2 Column-I

Column-II

p) (48N, 80N)

500   500 N, N  11  19 

q)

380   220 N, N 11  19 

r) 

380  220  N, N  17  23 

a) (i-p), (ii-q), (iii-q,r), (iv-p,q,r,s) c) (i-q), (ii-p,s), (iii-q,s), (iv-r,s) 25.

s) b) (i-p), (ii-q), (iii-r,s), (iv-q,r) d) (i-q,r), (ii-p,q), (iii-r), (iv-p,q,r)

A body is moving in a vertical circle of radius R, considering motion from top point of circle bottom most point of circle, match the following. Column-I Column-II i) Tangential accelertaion p) Always increases ii) Centripetal acceleration q) Always decreases iii) Angular velocity r) First increases then decreases iv) Potential energy s) Variable depending on angle made with vertical by string t) Variable independent of the made angle with vertical by string a) (i-r,s), (ii-q,r), (iii-p,t), (iv-p,q) b) (i-q,s), (ii-r,s), (iii-p,s), (iv-q,r) c) (i-p,s), (ii-q,r), (iii-q,s), (iv-p,r) d) (i-r,s), (ii-q,s), (iii-p,s), (iv-p,s)

Write the final answer to each question in this section in the column provided.

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26.

The block ABCDE of mass 5m has BC part spherical, of radius 1m, is on frictionless horizontal surface. BC is quarter circle. A small mass m is released at B and slides down. How far away form D does it hit the floor? (g = 10ms-2)

27.

The kinetic energy of a particle of mass 1kg moving along a circle of radius 1m depends on time t as K = t4. Fid the force acting on the particle as function of t.

28.

Spring is already in a compressed potion with initial compression = 5cm. The 10kg block is allowed to fall. Determine the maximum compression of the spring before the block rebounds. (g = 10ms-2).

29.

A long plank of mass M = 8kg, length  = 1m rests on a horizontal surface. Coefficient of friction µ1 = 0.2 . A small block, mass m = 2kg rests on the right extreme and of M, on the rough top surface. At t = 0, a force F= 25.5N is applied on M towards the right. (see figure). The block m does not slide on M. At t = 3s, force F is increased to 31N. The block m falls off M’s surface at t =7s. Determine the coefficient of friction between m and M.

30.

A particle is released on the smooth inside wall of a cylindrical tank at A with a velocity u which makes an angle α with the horizontal tangent. When the particle reaches a point B, a distance h below A, determine the angle β(as an inverse function of cosine) made by its velocity with the horizontal tangent at B.

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Passage-I (31 to 33): Two blocks of masses 10kg and 5kg are placed on a rough horizontal floor as shoen in figure. The strings and pulley are light and pulley is frectionless. The coefficient of friction between 10kg block and surface is 0.3 while that between 5kg block and surface is 0.2. A time varying horizontal force. P=5t Newton (t is in sec) in applied on 5kg block as known. [Take g=10ms-2]

31.

32.

The motion of block starts at t = t0, then t0 is a) 14s b) 8s c) 9s

t0 is in between 2 c) 12.5N and 17.5N d) 12.5N certain value

The friction force between 10kg block and surface at t = a) zero and 10N

33.

d) 12s

b) 10N and 35N

The acceleration of 5kg block at t = 2t0 is a) 12 ms-2

b)

14 ms 5

−2

c)

14 ms 9

−2

d) 2 ms-2

Passage-II (34 to 36): A block of mass 4kg is pressed against a rough wall by two perpendicular horizontal force F 1 and F2 as coefficient of static friction between the block and floor is 0.6 and that of kinetic friction is 0.5. [Take g=10ms-2]

34.

For F1 = 300N and F2 = 100N, find the direction and magnitude of friction force acting on the block. a) 180N, vertically upwards b) 40N, vertically upwards Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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2 5  −1  2  d) 91.6N, making an angle of tan   with the horizontal in upward direction. 5 

−1 c) 107.7N, making an angle of tan   with the horizontal in upward direction.

35.

For F1=150N and F2=100N, find the direction and magnitude of friction force action on block. 2 5  −1  2  b) 75N, making an angle of tan   with the horizontal in upwards direction 5  −1  2  c) 170.7N, making an angle of tan   with the horizontal in upwards direction 5  −1 a) 90N, making an angle of tan   with the horizontal in upwards direction

d) Zero 36.

For data of Question No.35, find the magnitude of acceleration of block. a) Zero b) 22.5 ms-2 c) 26.925 ms-2 d) 8.175 ms-2

37.

System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass 2m and m just after the string is cut will be

a)

g upwards, g downwards 2

c) g upwards, 2g downwards

b) g upwards,

g downwards 2

c) 2g upwards, g downwards

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1) 6) 11) 16) 21) 26)

c b c d b

31) 36)

a d

24 m 5

2) 7) 12) 17) 22) 27)

d c d d d

32) 37)

c a

β 2 +t 6 N

3) 8) 13) 18) 23) 28)

c c a c c 5 2

33)

c

(

)

17 −1 cm

4) 9) 14) 19) 24) 29)

a b d d a

µ2 = 0.1

5) 10) 15) 20) 25) 30)

d b c c d

34)

c

35)

b

  u cos α cos −1   2 gh  1+ 2 u 

Solutions 1.

i)

m11 = m2 2

……….(1)

ii)

m11 + M2 = m2 2 + 0.92 ⇒ M1 = 0.92

…………(2)

iii)

m11 +1.61 = m2 2 + M2 ⇒ 1.61 = M2

………….(3

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     

iv)

m11 + 1 = m2 2 + x2 ⇒ 1 = x2 ⇒ x =

1 2

From (2) M1 = 0.92  0.9 3 ⇒ 1.21 = 0.92 ⇒ 1 = = = 0.75 2 1.2 4 x = 0.75 kg

2.

(d) Self explanatory [acc = 0, u = 0

∴at rest]

3.

(c) Self explanatory [acc = 0, u = 0

∴at rest]

4.

Uphill: F = mg sin θ + µ1 mg cos θ F.ds = mg ds sin θ + µ1 mg ds cos θ

∫ F .ds

= mg

∫ dy

+ µ1 mg

= mg .h + µ1 mg 1

∫ dx ………(1)

Downhill:

− F = mg sin θ − µ2 mg cos θ F .ds = Fds ⇒ ∫ F .ds = −mgh + µ2 mg 2

∴ W = (1) + ( 2) = mg ( µ11 + µ 2 2 ) 5.

……….(2)

Case (i)

muC2 TA ≥ 0 ⇒ = mg cos 45 ⇒ u12 > gR R Case (ii)

…………..(1)

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TC ≥ 0 ⇒

mvC2 ≥ mg ⇒ vC2 > gR R

1 1 1   mu 2 = mv C2 + mgR 1 −  2 2 2  ⇒ vC2 = u 2 − 2 gR + 2 gR ⇒ u 2 − 2 gR + 2 gR > gR

(

)

⇒ u 22 = 3 − 2 gR

⇒

6. 7.

………….(2)

u ( 2) = 3 − 2 = 3 2 − 2 ⇒ u2 = = (1) 1 / 2 u u1 2 2 2 1

Freaction =

(3

2−2

)

dm .vrel . But vrel = 0 . dt

No external force horizontally,

∴ xCM does not shift

→ = 0.3( sin 53 − sin 37 ) = 0.06 m

Taking

xA/ B x B / C = unknown xC = given = −0.005 m

∴m A ( x A / B + x B / C + xC ) + mB ( x B / C + xC ) + mC ( xC ) = 0 ⇒ 1( 0.06 + x B / C − 0005 ) + 2( x B / C − 0.005 ) + 3( − 0.005 ) = 0 ⇒ x B / C = 0.01 ⇒ x B = x B / C + xC

= 0.01 − 0.005 = 0.005 m. ⇒  B = 0.10 + 0.05 = 0.105 m = 10 .5 cm

8.

Before collision: Using loss of P.E = Gain of KE

Components u y = − 5. ux = −

3 2

5 2

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v y = −e.u y = +

5 35 ∴v = v y2 + v x2 = ms −1 2 4

vx = u x = −

9.

15 ms −1 4

Work done by friction, wf. We know wf has to be minimum

= µmg ( 0.4 +0.3) = 0.7, µ mg .

But wf = loss of P.E = 0.1mg ∴1mg ≥ 0.7 µ mg ⇒ µ <

1 7

10.

^

^

u = u s θ i j+ nu c θ oi s ^

^

u1 = e s uθ i j+ nu c θ oi s ^

^

^

^

u2 = e2us θ i j+nu c θ oi s u3 = e3us θ i j+n u c θ oi s H =

e 2 sin 2 θ e 6 u 2 sin 2 θ H3 = = e6 H 2g 2g

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H 1 1012 2 36  2 6  ∴ = 6 = 36 ⇒ e 6 = 12 =  2  H3 e 2 10  10  6 2 64 ⇒e = 2 = = 0.64 100 10

11.

6

dU ( x ) b = −a + ( − ) 2 dx x dU − ax 2 − b = 0 ⇒ equillibrium ⇒ =0 dx x2 ⇒ x ≠ 0 and x 2 = −

b a

⇒ b and a are of opposite signs. ⇒ options (b) or (c) possible For stable equation. d 2U should be +ve dx 2 b b ⇒ ( − )( − 2 ). 3 = 3 > 0 ⇒ ( for x > 0 ) b > 0 x x 12.

T’ =70N, T = T’ -30N=40N On cutting thread, A1kg =

40 = 40 ms −1 1

All other’s acc = 0

13.

N1 = N 2 sin 37 + mg

N 2 cos 37 = mω R 2

……..(1) ……...(2)

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∴N1 = 23 .5 N 2 = 22 .5

⇒ (1) ⇒ 23 .5 = 22 .5 × 0.6mg ⇒ mg = 23 .5 −13 .5 = 10 N ⇒ m = 1kg

14.

f static = 20 N 20 ∴A = =2 10 25 − f 25 − 20 ∴a = = 5 5

a < A ⇒ not possible

Hence no relative motion ⇒A =a =

15.

25 = 1.66 ms −2 10 + 5

Case(A)

1 − R1 = 40 a ⇒ R1 = 1 − 40 a

Case (B)

1 − R2 = 15 a ⇒ R2 = 1 −15 a ∴R2 > R1

16.

R is clearly true, but A is false. Initial energy is P.E. If wedge is free to move, it will have KE ⇒ KE of block will be less than that if wedgeis fixed.

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17.

You did positive work and gave positive energy. Gravity did equal negative work and made net ∆KE zero.

18.

Self explanatory.

19.

Tension in rope remains same through out but it is the frictional force provided by the ground that helps a team to win.

20.

Assertion is right because only the charge in kinetic energy in both the frames are same (including work of pseudo force) but kf is dependent on the ki as kf – ki = ∆k and ki depends on the velocities of these reference frames at time t =0 reason is wrong so assertion is right and reason is wrong

21.

(b) self explanatory

22.

(d) self explanatory

23.

(c) self explanatory

24.

(a) 1st case

20 cos 37 = f ≤ uN (for no motion) 16 ≤ 0.2( 20 + F + 20 sin 37 )

F ≥ 48 N

F ∈( 48 , ∞)

………….(1)

nd

II case

mg − F sin 37 = f ≤ µN

(to prevent downward slipping)

3F ≤ 0.2( F cos 37 ) 5  4 3 F +  ≥ 20  25 5  500 F≥ N 19

20 −

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F sin 37 − mg = f ≤ µN

(to prevent upward slipping)

3F  4 − 20 ≤ 0.2 F  5  5 3 4  F −  ≤ 20  5 25  500 500   500 F≤ N so F ∈  N, N 11 11  19 

……..(2)

IIIrd case

mg sin 37 − F sin 37 = f ≤ µN

(for no downward slipping)

 4  3 12 − F   ≤ 0.2 F   + 16  5  5  3 16  4 F +  ≥ 12 − 5  25 5  44 25 220 F≥ × = N 5 19 19

− mg sin 37 + F sin 37 = f ≤ µN

(for no upward slipping)

 4  3F − 12 ≤ 0.2 F   + 16  5  5  76 25 380  3 4  76 F −  ≤ ⇒F ≤ × = N 5 11 11  5 25  5 380   220 N,  So, F ∈  ……………(3) 11   19

IVth Case

f = mg sin 37 − F cos 37 ≤ µN

(for no downward sliding with respect to wedge)

4F 3  12 − ≤ 0.216 + F  5 5  16 220 4 3  F + ⇒F ≥ N  ≥ 12 − 5 23  5 25  F cos 37 − mg sin 37 = f ≤ µN (for no upward sliding with respect to wedge) 4 3F   F − 12 ≤ 0.216 +  5 5   Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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380 17 380   220 F ∈ N, N 17  19  F≤

……….(4)

25.

(d) Self explanatory

26.

When m leaves C, let its velocity (horizontal) be u. Then block 5m will have velocity

∴Energy equation: ⇒u =

u to left. 5

5 gR 50 = ms −1 3 3

∴ t = time to reach ground = 2h = 2 ×1 = 1 s g

∴ Distance of hitting point from D = =

27.

u 6u 6 ×t + u ×t = t= 5 5 5

50 1 . = 3 5

10

5

24 m 5

1 1 mv 2 = v 2 = t 4 ………..(1) 2 2 mv 2 1.v 2 ∴ Fn = = = v 2 = 2t 4 R 1 dv = 2 2t v = 2 t2 v 2 = 2t 4 dt dv dv Ft = m = 1. =2 2t dt dt K=

∴ Resultant F = Ft 2 + Fn2 = 2 t 2 + t 6 28.

Initial energy: 1 2 1 2 kx = × 4000 × ( 0.05 ) = 5 J 2 2 P.E mass = mgh =100 ×0.2 = 20 J ∴ Total Einitial = 25 J P.E spring =

Final energy P.E spring =

1 1 2 k ( x + x 2 ) = × 4000 ×( 0.05 + x ') 2 2

(

)

1 ×4000 × 0.0025 + x ' 2 +0.1x' 5 + 2000 x' 2 +200 x ' 2 P.E mass = −mgx ' = −100 x '

∴ 2000 x'2 +100 x − 20 = 0 ⇒ x' =

5 200

(

)

17 −1 m =

5 2

(

)

17 − 1 cm

Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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29.

From t = 0 to t = 3sec

f2 < f2 max so both blocks are moving together and then F – f1 max = (m + M) a ………(1) f1 max =0.2 x 10 x 10 = 20N So 25.5 – 20 = 10a ⇒ a = 0.55 m / s −1 ……….(2) Displacement S1 =

1 × 0.55 × 32 2

=2.475m ……….(3) Velocity = 0.55 x 3 = 1.65m/sec ……….(4) From t = 3 to t = 7 sec f2 =f2 max = 20 µ2 31 − 20 − 20 µ2 = 8 am

20 µ2 = 2am 1 1 = ( aM − am ) 4 2 2 1 ⇒ aM − am = 8

After solving these equation. aM =

30.

9 , am = 1, µ2 = 0.1 8

Initial velocity u can be split into tangential component u cos α and vertical component u sin α ∴ vz at B can be attained by, v z2 = u 2 sin 2 α + 2 gh mu 2 cos 2 α Normal reaction N = r

∴ tangential component at B = u cos α ∴ velocity at B = v z2 + u 2 cos 2 α = u 2 + 2 gh ∴∠ βis given by

cos −1

u cos α 2 gh 1+ 2 u

31 to 33. From constraint theory we can relate the acceleration 10kg and 5kg blocks. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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Limitng friction force between 5kg blocks and surface is, f L = 0.2 × 5 ×10 = 10 N Limiting friction force between 10kg block and surface is, f L2 = 0.3 ×10 ×10 = 30 N 1

For 5kg block, P-2T - f1 = 0 [in equilibrium, i.e., when blocks are not moving] For no motion of blocks, f1 > f L and f 2 ≤ f L P − 2T ≤ f L1 and T ≤ f L2 So, 1

⇒

2

P ≤ f L1 + 2 f L2

So, for motion to take place, P ≥ f L1 + 2 f L2 5t 0 = f L + 2 f L = 10 + 2 × 30 ⇒ 1

⇒ At t =

2

t 0 =14 s

t0 = 7 s , the equation for 5kg and 10kg blocks are 2 35 − 2T − f1 = 0 and T − f 2 = 0 35 = f1 + 2 f 2

⇒

And we know at t = 7s both the blocks are at rest so f1 ≤ f L and f 2 ≤ f L 1

2

Solving above equation we get, 0 ≤ f1 ≤ 10 N ; 12 .5 N ≤ f 2 ≤ 17 .5 N 12 .5 ≤ T ≤ 17 .5 N

And At t = 2t0 = 28s, equation for blocks are 140 – 2T – 10 = 5a and T – 30 = 10 x 2a

⇒ 34.

a=

14 ms −2 9

The forces acting on the block are F1, F2, mg, normal contact force and frictional force. Here fractional force won’t act along vertical direction as the component of resultant force along the surface acting on body is not along vertical direction and direction of the friction force is either opposite to the motion of block (direction of acceleration of block) if it is moving or not moving.

So, f L = µN1 = 0.6 × 300 = 180 N Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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−1  2  Resultant of 4g and F2 is 107.7N making an angle of tan   with the horizontal

5 

As force applied along the surface is > f L , so the block doesn’t move and friction is static in nature. f =107 .7 N

2 5 

−1 making an angle of tan   with the horizontal in upward direction.

35.

For F1 = 150 N , f L = 0.6 ×150 = 90 N As component of resultant force along the surface is 107.7N and is greater than f L , so kinetic friction comes into existence, i.e., friction force acquires the value f = µk N1 = 0.5 ×150 = 75 N . Its direction is opposite to component of resultant force along the surface.

36.

Acceleration of block =

37.

Initially under equilibrium of mass m: T = mg Now, the string is cut. Therefore, T = mg force is decreased on mass m upwards and downwards on mass 2m.

∴

am =

107 .7 − 75 = 8.175 ms −2 4

mg =g m

(downwards) And

a2 m =

mg g = 2m 2

(upwards)

Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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23