EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ

EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ

INTEXT EXERCISE: 1 14.

1. As the bus slows down your body continues to move with

the same speed due to inertia and therefore you feel thrown Because of equillibrium T = 16 N
forward. There is no actual force acting on you.
Reading in each spring balance is equal
2. 1HZWRQ¶V¿UVWODZ to T = 16 N
3. At the highest point, the speed is zero. The force due to
gravity is downwards towards the earth’s center. Hence the
acceleration due to gravity is downwards too.
4. According to Newton’s third law both forces must be equal.
5. Reaction force is a pusing force on you due to G.
15.
6. 2T sin 37c = 100 N

16. If the woman is not able to lift the crate, the

forces must be balanced P + C = W
7. When the net force becomes zero, the velocity becomes
constant.
8. Because of inertia the dog stays at rest whereas the truck
moves forward (back of truck moves towards the dog) 17. System moves with constant velocity
9. Since the top block is moving with constant speed. The net and therefore net force must be zero.
force on it must be zero.
P = T + 25N
10. Luggage due to its inertia tries to move with the same speed,
whereas the train slows down.

18. Free body diagram of the knot.

Using the conditions of equilibrium
T1 = T2 cos 45c
1000N = T2 sin 45c

Due to this the separation between the front of the train

compartment and the luggage decreases. 19. By symmetry R A = RB

INTEXT EXERCISE: 2 2R A cos 60c = 10N

11.

20.

T2 = 50 N

T = 6g
4 T = 2g = 19.6N
T = T1 + 4g 1 INTEXT EXERCISE: 3
6g - 4g 21. Apparent weight = m ^ g - ah when acceleration is
12. a = 10 = 1.96 m/s2. downward.
mB g
13. Reading in the scale is equal to the tension (force) acting at 22. a =
m A + mB
its ends.
23. Fv = 6it –8 tj + 10 kt
" " " "
F = ma F =m a
Clearly tension on both ends is equal to W . Hence the 6 2 + 8 2 + 10 2 = m1 (1)
reading is W. m = 10 2 kg

48
Physics
24. T =
2m1 m2
& 10 =
2 # 1 # m2 # 10 By virtual work theorem, Tv2 + 2Tv1 = 0
m1 + m g 1 + m2 & m2 = 1 kg
& Ta2 + 2Ta1 = 0
25. Impulse is area under F-t graph & a2 + 2a1 = 0
26. Acceleration is equal to the slope of v-t graph. at t = 2s So option (a) is correct
15
a =+ 3 =+ 5m/s 2 32. Tv - 2Tv' = 0
force = ma & v = 2v'
= 0.25N 33. - 2Ta + Ta' = 0
t = 4s a=0 force = 0 & a' = 2a
15 34. - 3Ta1 + Ta2 = 0
t = 6s a =- 3
& a2 = 3a1
=- 5m/s 2 35. dx dy
force = ma x 2 + d 2 = y 2 & x dt = y dt
= 0.25N RppRsite to mRtiRn
-
& xv A = y (20) & v A = 25ms 1
27. y
d
37°
x
a1 + a2 2
36. a = 2 = 3.5m/s
m3 g - T1 = m3 a
37. By virtual work
T1 - ^m1 g sin θ + T h = m2 a - T # X2 - 7TX1 = 0
T - m1 g = m1 a
a2 + 7a1 = 0
Adding the equations we get,
m g - m g sin θ - m g
a = 3 m +2 m + m 1
1 2 3

Use m1 = 50g, m2 = 100 g, m3 = 500 g

28.
38. By virtual work
X A + XB + XC + XD = 0
a A + aB + aC + aD = 0

10 sin α - T = 1 ]ag
T - 10 sin β = 1 ]ag
10 ^sin α - sin βh Assuming downward as + ve
a= 2 - 1 - 7 - 2 + aD = 0
29. aD = 10 m/s 2 downwards

39. By virtual work,

- 2TX A - TXB - 2TXC = 0
2X A + XB + 2XC = 0
F - 150 = 15 ]1 g f = 165N 2a A + aB + 2aC = 0
30. 5 - N = 0.5 ]2g
N = 4 Newtons

40.

INTEXT EXERCISE: 4
31. ///////////////////////

m2 ­ a2 Equating components along the string

VA cos i = u
a1 ­ m1 VA = u sec i

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 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
INTEXT EXERCISE: 5 EXERCISE  1
41. We don’t include the pseudo force in an inertial reference
frame. (Moving with constant velocity) 1. 1HZWRQ¶V¿UVWODZ Fnet = 0 & v = constant
42. Since acceleration of frame, aframe = 0 m/s2
2. 1HZWRQ¶V¿UVWODZGHVFULEHVLQHUWLD
Hence value of pseudo force = |aframe × mass of object| = 0 N
43. R = m ^ g + a h = m ^ g + 4g h = 5 mg 3.
44. F pseudo =- ma frame
dv F-T =bM l bF l
L x M
=- m dt
4. F - W = c W m 3g
=- m ]2t g it g
45. x = t3 + t2 + t 5. Opposite force causes retardation.
dx 6. Force exerted by string is always along the string and of pull
v = dt = 3t2 + 2t + 1 type. When there is a contact between a point and a surface
dv the normal reaction is perpendicular to the surface and of
a = dt = 6t + 2 push type.
ma
F pseudo =- ma frame 7. Force on m1 = Force on m2 & a1 = 2 2
m1
=- ]6t + 2g 2it v t

=- 16it N 8. m
dv
dt = kt & # dv = mk # tdt
0 0
46. F pseudo =- ma frame
& v = m b t2 l
k 2
& F pseudo = 2 4 + 9 = 2 13 N
47. N = m ^ g + a h = 0.5 ]10 + 2g = 6 N
F2
48. 9. &
m F4
ma Where F4 = F3 + F1
a
q mg

According to question sin i = x ]1 in xg

1
Fnet = F2 + F4 Ÿ
1
So tan i = Fnet a
x2 - 1
N
q 10.
s 10 2
co a = 2 = 5 m/s
Vy = a y t = 15 m/s
a
m q ma Vx = 5 + 5.3 = 20 m/s
q
sin 11. Point A is mass less so net force on it
m
g q mg cos q + ma sin q
most be zero
mg
& F–T sin θ = 0
To keep the block stationary relative to the inclined plane [Equilibrium of A in horizontal
mg sin i = ma cos i direction]
F
a = g tan i & T = sin θ
g
&a= 12.
x -1
2

49.

T – mg = 0 [ Equilibrium of block]
Acceleration relative to wedge is ^g + ah sin i T = 10
Reading of spring balance is same as tension across spring
50. If frame is inertial then resultant force must be zero. balance.
If frame is non-inertial then the resultant of all forces and the 13. F1 + F2 + F3 + F4 + F5 = 0 . If F5 = F (given force)
pseudo force must be zero.
Then a =-c F1 + F2 + F3 + F4 m =- F
m m

50
Physics
14. 2T cos θ = mg
15. 10 - T1 = 2a 23.
T1 - T2 = 3a T2 = 5a
16. Gas molecules will tend for move towards the back of
compartment lowering the pressure in front. Reading of spring balance is same as tension across the
17. balance.
& T = 10 g = 98N
T = 2 a (Newton's II law for 2 kg block)
& a = 49 m/s 2
2mg sin 30º - mg sin θ 24. Weight of man in stationary lift is mg
Acceleration =
2m + m
g (2 sin 30º - sin θ )
Given a = g/6. so = g/6
3
1
& sin T = 2
& T = 30c mg–N = ma
18. F1 t1 = F2 t2 [Newton’s II law for man]
19.
& N = m (g - a)
Weight of man in moving lift is equal to N.
mg 3 g
& m ( g - a) = 2 & a= 3

25.

N = Mgcos θ $ force exerted by plane on the block.

26. N = m ^ g + a h
10 - T2 = 1 a (Newton's II law for A)
27. N
40sin30°=20N

T2 + 30 - T1 = 3a (Newton's II law for B)

5kg 40cos30°=20— 3N
FBD 5kg
T1 - 30 = 3a (Newton's II law for C) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

g
& a= 7 50N

6g Net vertical force acting on the body is equal to zero.

& T2 = 7
28. mamin = mg - Tmax
20. 75 mg g
= mg - 100 mg = 4 & amin = 4
2g g 10 20 -2
29. For BC = 0, a =
2 + 5 + 1 = 4 = 4 = 8 ms
F = m1 4 [Newton’s II law for m1 ]
(2 + 1) g 3g 30 -2
F = m2 6 [Newton’s II law for m2 ] For BC = 2m, a =
2 + 5 + 1 = 8 = 8 ms
F = (m1 + m2) a [Newton’s II law for ^m1 + m2h ] 30. Just after release T = 0 due to non–impulsive nature of
& F = b 4 + 6 l & 1 = b 4 + 6 la spring. So acceleration of both blocks will be g .
F F 1 1
31. Case (i) : F1 = 2T1
& a = 2.4 m/s 2
21. v 2 = v 2 + 2as F1
0 2 = 1 2 + 2m 4mg - 2mg 2mg g F1
x a1 = = 6m = 3
6m
x= - m 2 2
v = v + 2as
2F g
` T1 - 2mg = 2m # 3 T
' T
0 2 = 3 2 + 2mF x & F ' = 9F 8mg 16mg
& T1 = 3 ` F1 = 3
22.
Case (ii)
F2 = 2T2 F2
4mg - 2mg g
a2 = 6m = 3
g T
Mg sin θ - T = Ma [Newton’s II law for block 1] ` 4mg - T2 = 4m # 3 T

T = Ma [Newton’s II law for block 2] 8mg 16mg

Mg sin θ T2 = 3 ` F2 = 2T2 = 3
2 T = Mg sin θ T= 2

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 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
32. a1 =
2mg - mg
= g; a2 =
2mg - mg
= 3
g 39. T2 = ^m1 + m2 h a
m 3m T m
mg + mg - mg g T1 = m1 a ` T1 = m +1m
2 1 2
a3 = 2m = 2 ; a1 > a3 > a2
40. 1 .....(1)
h = 2 gt12
33. Coin

F – N = Ma [Newtons law for block of mass M]

At rest
N – N’ = ma [Newtons law for block of mass m]
u
N’ = M’a [Newtons law for block of mass M’]
u
F
& N ' = M ' M+m+M ' h

F
N = (m + M ') M + m + M '
Relative to lift = g - 0 = g
34. 2
srel = urel t2 + 1/2 arel t2
1 .....(2)
srel = h ` h = 2 gt22

1 2=1 2
2 gt1 2 gt2 & 6t1 = t2@
41.
T2 – 8g = 8a T1 + T2 = mg
[Newton’s II law for 8 kg block] If upper spring is cut
& T2 = 8 × 2.2 + 8 × 9.8
= 96 N
T1 – 12 g – T2 = 12 a
mg - T2 = m # 6 .....(i)
[Newton’s II law for 12 kg block]
& T1 = 12 × 2.2 + 12 × 9.8 + 96
T1 = 240 N If lower spring is cut :

35. mg– 3 mg = ma [Newton’s II law for man]

4
g mg - T1 = ma .....(ii)
& a= 4

36.
adding (i) and (ii)
2mg–T1 + T2 = m (a + 6)
F – k x = m1 a1 [Newton’s II law for M1] 2mg–mg = m (a + 6)
kx = m2 a2 [Newton’s II law for M2] mg = m (a + 6)
By adding both equations. g = a+6
2
a = 4m/s .
F - m1 a1
F = m1 a1 + m2 a2 & a2 = m2 42.
37. F = m1 a1 + m2 a2 [Newton’s law for system]
2
200 = 10×12 + 20×a a = 4m/s .
38.

The length of string AB is constant.

speed A and B along the string are same u sin θ = V

V
u = sin θ

30 – T2 = 3 a [Newton’s II law for 3 kg block] 43. X1 - XP + X2 - XP = Constant

T2 – T1 = 6 a [Newton’s II law for 6 kg block]
Differentiating
T1 – 10 = 1 a [Newton’s II law for 1 kg block]
a1 + a2 = 2aP
By adding three equations
30 – 10 = 10 a & a = 2 m/s 2 .

52
Physics
44. 5.
N2
sin ]180 - 37cg
mg
sin ]180 - 37cg
=

By string constant 2a1 + a2 = 0

Also 4g - 2T = 4a1
6. Let L1 and L2 be the portions (of length) of rope on left and
]1 g g - T = 1 ^a2 h right surface of wedge as shown
45. 2TX A - TXB = 0 ` Magnitude of acceleration of rope
2VA = VB M
[L1 sin α - L2 sin β] g
a= L M =0
^` L1 sin α = L2 sin βh
7. From graph F = 2t 2
dv = 2 2
dt a = 5 t
V = 2 t3
15
8. F sin ]kxg = mg
EXERCISE  2 At the instant of lowing contact.
]kxg
1. 2T sin θ = W a = F cosm
W
T = 2 cosec θ Using adx = vdv
F cos ]kxg
x
v2 = 2 # m dx
0

2. = 2mF sinkkx

v = 2F sin kx
km
2h
T1 cos 45º = T2 cos 45º 9. First case, t1 = g , where h is the initial height
& T1 = T2 ; (T1 + T2) sin 45º = mg
RI WKH FRLQ IURP WKH ÀRRU RI the elevator. Second case,
2 T1 = mg 2h
t2 = g+a
mg
T1 =
2 ` t2 < t1
T1 10. For t < 0 . System is in equilibrium and hence
T sin θ = Mg +
2
mg F1 = F2 = F
T sin T = Mg + 2 .......... (i) 11. Reading of spring balance is less than m
T mg
T cos T = 1 = 2 .......... (ii) if a (.) and reading of spring balance is
2
greater than m if a is upwards
dividing (i) and (ii) 12.
M + m/2 2M
tan θ = m/2 = 1 + m
3. T = mg
2T cos T = Mg
A + B + C + D + E = 300i............. (1)
2mg cos T = Mg
M B + C + D + E = –100i.................. (2)
cos T = 2m 1 1
A + C + D + E = 2400j.................. (3)
M 1 2m equation (1) - equation(3) give
4. T = 2m2 m3 B = 300i–2400j................ (4)
2 m2 + m3 g
equatiRn ^1 h - equatiRn ^2 h give
m1 g 2m2 m3 A = 400i............................ ^5 h
2 = m2 + m3 g
Adding equatiοn ^4 h and ^5 h
4m m
m1 = m +2 m3 & A + B = 700i–2400j
2 3
1 1 = 4 A+B
a] A + Bg = 100
m2 + m3 m1
= 7i–24j

53
 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
13. 19. (a) – Pulling force on bricks = 2F
(b) – Pulling force on bricks = F
(c) – Pulling force on bricks = F
(d) – Pulling force on pulley = F/2
Resultant acceleration 20. Maximum tension in string Tmax sin 30° = 40
2 40
g 1 g & Tmax = 1 = 80N
m +e go +
3 3
c
2
= . . g cos 150º 2
2 2 2 2 2
For monkey Tmax –mg = ma & a = 6m/s –2
g
. sin 150º 21. Acceleration of system
2 1
= b 44 + 11 l g = 53 # 10 = 6 ms -2
tan a = = -
3g g 3
+ . cos 150 o
2 2 Relative acceleration of blocks = 12ms –2
a = 30c
Now 2 + 4 = 1 (12) t 2 & t = 1 sec
i.e. Resulting acceleration is in vertical direction 2
22. In (A) T = kx1 = 2g
14. Acceleration of two mass system is a = F leftward
2m In (B) T = kx2 = 3g–3× = 12 g
g
N 5 5
g 4
F In (C) T = kx3 = 2g–2× = g
60°
3 3
30° x1 5x2 3x3
2 = 12 = 4
FBD of block A 23. F1 - T = m1 a & T - F2 = m2 a
mF
N cos 60°–F = ma = 2m solving N = 3F F -F
& a = m1 + m2
1 2
15. Before cutting the spring
` T = F1 - m1 c m1 + m2 m = 1m2 + m2 1
F -F m F +m F
1 2 1 2

24. 25 - 20 = 2a
2
& a = 2 . 5m / s
25. a1 + a2 = 2a .....(1) T - Mg = Ma1 .....(2)

T2 = mg T - mg = ma2 .....(3)
a1 - a2 = 2a .....(4)
After
2mg - mg = 2 ma a1 & a2 are accelerations of M & m w.r.t. earth.
a = g/2 26. 1 (3 + a) = 20
2
T3 = mg/2 & ai = gm/s

a f = a 3 kg = 3
mg mg 2-1 9 2g
T2 - T3 = mg - 2 = 2 ` Δa =
3
16. T = M # a .....(i) 27. a = 0.3 g = g
2 0.9 3
T .....(ii) g 2g
20 - 2 = 2 # a V = 0+ 3 2 =
3
1 Ma
20 - 2 # 2 = 2a V
` the string tight again after f = g = 2 sec
3
T
& 2 = 1 # g & T = 20N
28.
2
a = 5 m/s
& M = 8 kg
17. Acceleration
Net force 3 # 250 - (100) g sin θ
= Total mass = 100 2F + N - Mg = Ma
750 - 260 -2 2F - N - mg = ma
= 100 = 4.9 ms
18. Let acceleration of masses 29. By setting string length constant
w.r.t. pulley be a L = 3l1 + 2l2
Mg–T–Ma0 = Ma & 3v0 = 2v A
T + ma0 –mg = ma 3
v A = 2 v0
& (M–m) g– (M–m) a0 = (m + M) a v A = v A - vB
& a = (g–a0) v
= 20
But a0 2 g so a 1 0 and T 1 0
towards right
& Tension in string will be zero

54
Physics
V1 + V2 34.
30. a VP =
2
Pulley P1 Vp

V1
uo mu By virtual work
V2
- 2TX2 + TX1 = 0 2a2 = a1
un
2X2 = X1
P1

2c M2 m = M1
x nv1 nv2 v2 p nu
F - 2T T
P2 P3 nv
vn
vn T
M
F - 2T = 3
0 + v1 3F 12 a2 = F - 2T = 2 m/s 2
T = 7 = 70 N M2 7
u= 2 ..... (i)
Pulley P2 35.
v +v
v = 1 2 2 & 2v = v1 + v2 ..... (ii) If pulley C is massless the tension in the
–v + u strings must be zero.
Pulley P3 v = 22 ..... (iii) Therefore acceleration of A & B must be
Eliminate v1 & v2 to get g
& 2u + u - 2u = 2v & 3u = 4v
3
v= 4u
36.
31. Solving problem in the frame of pulley 2a1 = a2

3.25 cos T–1 sin T = 3 cos 30 + 1 sin 30

3 1 Mg - 2T T - Mg sin 30c
3.25 cos T– sin T = 2 + 2 2e o =e o
2M M
3.25 cos T– sin T = 2 Mg
13 cos T–4 sin T = 8 2 mg - 2T = T -
2
13 1– sin 2 T = 8 + 4 sin T 5Mg
T =
169 - 169 sin T = 64 + 16 sin 2T + 64 sin T
2 6
5g g g
T - Mg sin 30c = - = ms 2
185 sin T + 64 sin T–105 = 0
2
a2 = M 6 2 3
3 37.
& sin T = 5
3
& tan T = 4
32.

by virtual work, - TX1 - 5TX2 = 0

a1 + 5a2 = 0
By symmetry we can conclude that block will move only in
vertical direction. Length of string AB remains constant b mg T l + 5 d 0 m n= 0
- m g - 5T
M
Velocity of point A and B along the string is same. 0

u
V cos T = u & V = cos T
6g - T c M + m = 0 m
1 25
33. l1 + l2 + l3 + l4 = C 0

"2
"3 "1 6Mm0 g
d,1 + d,2 + d,3 + d,4 = 0 "4 T = m + 25 M
v 0
dt dt dt dt T 6m0 g
2 m/s a1 = g - M = g - 25M + m0

= d 25M + m 0 n g
–v–v + 0 + v + 2 = 0 & v = 2m/s
25M - 5m
0

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 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
38. F.B.D. of mass m is w.r.t. trolley
T sin (α–θ) + mg sin θ–FP = 0
[Equilibrium of mass in x direction w.r.t. trolley]
& T sin (α–θ) + mg sin θ–mg sin θ = 0
& T sin (α–θ) = 0
since T cant be zero , sin (α–θ) must be zero
Equating compounds along the length of rod
45. FBD of block w.r.t. wedge
u cos i = v sin i
VB = u cot i N
30°
39. Acceleration of bolt with respect to car mg
mg 30°

Acceleration of block w.r.t wedge

mg 2 - mg ` 2 j
3 1
= c 2 mg
3 -1
& t= m
Bolt have acceleration perpendicular to inclined plane with
respect to car so it will hit the surface of car at point Q as Now from S = ut + 1 at 2 . 1 = 1 c 3 - 1 m gt 2
shown so distance from point P = 3m 2 2 2
4
40. The free body diagram of cylinder is as shown.
^ 3 - 1h g
& t= = 0.74 s
Since net acceleration of cylinder is horizontal,
N AB cos 30° = mg .....(i) EXERCISE  3
or N AB = 2 mg
3 1. N = mg + F sin 30°
and NBC –N AB sin 30° = ma 1
= 700 + 200 # 2 = 800 newton
or NBC = ma + N AB sin 30°...... (ii)
N
Hence N AB remains constant and
NBC increases with increase in a.
41. T sin θ = m ^ g sin α + a0h

T cos θ = mg cos α 30°

g sin α + a0
& tan θ = c g cos α
m F
g sin α + a
mg
θ = tan –1 c g cos α 0 m
2. i = 30°
42. Slope of vrel –t curve is Constant.
3g
& arel = Const. a = a1 – a2 ! 0 cos i = T

Inference is that at least one reference frame is accelerating 3 3g

& 2 = T
both can’t be non - accelerating simultaneously.
43. (Force diagram in the frame of the & T = 20 N
car) Applying Newton’s law
T
perpendicular to string
mg sin T = ma cos T
30°
a
tan θ = g F
Applying Newton’s law along string q
2 2
& T–m g + a = ma
T = m g + a 2 + ma
2 3g
3.
44. 2M
Y
1M
q q
q F1 2 1
x 2
1
2 F1 2
F2 F2

56
Physics
i = 45° 8. It is given that v = 10 x m s -1
Taking components along x & y dv
)RUFHLVGH¿QHGE\ F = m ... (i)
1 2-1 1 dt
F1 = 2 - = = Multiplying and dividing equation (i) by dx
2 2 2
dv dx
1 2+1 3 F = m dt dx
F2 = 2 + = =
2 2 2 dx dv dv
& F = m dt dx = m dx ... (ii)
F1: F2 = 1: 3 Finding the values of the terms in equation (ii)
x=3 dv 10 5
4. 50 = V # 10 dx = 2 x = x
V = 5 m/s dv 5
v dx = a = 10 x # = 50 m s -2
V = 0 + a # 20 x
1
5 = a # 20 Hence, the force is F = 2 # 50 = 25 N
1 9. Given, the position vector of the particle is
a = 4 m/s2
r = 10t it + 15t2 tj + 7kt
1
F = ma = 20 # 4 = 5 N The velocity of the particle can be calculated as follows:
20 kg t = 20 sec dr
F v = dt
V
= dt _10t it + 15t2 tj + 7kt i
d
u=0
50 m
a T = 10it + 30t tj
dv
And, the acceleration of the particle is given by a = dt
5. 4g sin 60° - T = 4a ... (1)
= dt ^10i + 30t tj h = 30tj
d
4 g sin 60° Hence, the force on the particle can be written as
T a
F=ma
= 30 mjt
10. 2 N Sin T = mg
4 g sin 30° mg
N= cosec T= mg R
T - g sin 30° = a ... (2) 2 2 l2
Solving (1) and (2) we get, R2 -
4
20 3 - T = 4T - 20
T = 4 ^ 3 + 1hN
R
N N
q l/2 l/2
6. Centripetal force will be provided by the spring force. Let the When l =R,
mg

elongation in the spring be x, then we can write

N = mg R mg
& kx = m~2 ]r + xg 2 3 2
=
3R
& 7.5x = 2.5 ]0.2 + xg 4
0.5
& x = 5 = 0.1 11. R Fx = 0 ŸT2 = 160 N
R Fy = 0 ŸT1 = 120 N
Therefore, required tension will be T = kx = 0.75 N.
Spring force = 120 N
7. The equation for velocity is
120 N
Spring constant, k = = 3000 N/m
v = _ 2t it + 3t2 tj i x 4 cm
Thus, the acceleration is y T2
d _ 2t it + 3t2 tj i
= ^2i + 6t tj h
dv 37o
a = dt = T1
dt
120
At, t = 1 s, a = 2it + 6tj 160 37o
200
2it + 6tj t t
The force is written as F = ma = 2 = i + 3j
Comparing it with the given value of force, x = 3

57
 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
12. F.B.D of bead w.r.t truck. N sin θ = mA
N a N + mA sin θ = mg cos θ
ar = g sin T + A cos T
37o m (12) mg sin T cos T
Solving A =
M + m sin 2 T
mg sin T cos T
2

mg sin 37° ar = g sin T +

M + m sin T
2

mg
48 30 18 Mg sin T + mg sin T
Ÿa = 12 cos 37°10 sin 37°  - = m/s2 =
M + m sin 2 T
5 5 5
Now,
1 18 36 ] M + mgg sin T
# # t2 = & t=2s =
2 5 5 M + m sin 2 T
13. (T + 450) - 1000 = 100a .....(1) A m cos θ
ar = ] M + mg
& T - ]450 + 250g = 25a .....(2)
tan β = y = ar sin T = sin T
a
14. ax ar cos T - A A
N1 90q
N2 cos T - a
r

150q 120q = sin T

cos T - b M + m l cos T
m
mg
= b MM+ml θ
tan
N1 2g
& N1 c m
2 = 1
sin 120c = sin 90c 3 3
g & N1 = 5
5.

15. FBD of (sphere + block) system

N’
37o
o N’
N 37
aB = a A sin θ
N 6. a S
i - 2 (a + b)Uj
N 3 7. aC = a A cos θ .....(1)
From FBD of system “sphere + block”, =
60 4 2mg - T = 2m.ac .....(2)
N = 45 N
T cos θ = m . a A .....(3)
EXERCISE 4 2mg = cos T ` cosc T j + 2m.ac
m a

& 2mg = a + 2m k ac
1. a AC = a A - aC m
cos θ
2

` a A = a AC + aC . Also a = b + c
2g cos 2 T
2. & ac =
1 + 2 cos 2 T
8.

T ' = mg
T = m1 a1x
T ' sin D + mg sin D = T = m'g
T cos θ = m2 a2x
m' = ] M + mg sin α
3. L = 1 g sin θt 2 9. T = 10 (10 + 1 - 0.5 cos 60c)
2
T = 10 (10.75)
x = 1 gt 2
2 T = 107.5N
L = sin θ 10. T = 5N
x 10g sin 30c - 5 = 10a
4.
& ablock = 4.5 m/s ]"g . So, apulley = 2.25 m/s2 ]!g
2

11. on 5 kg Fnet = 0 & a = 0

Spring force doesn’t change instantaneously.

58
Physics
12. From geometry rA = 10m N3 = Mg + N1 cos θ .....(1)
N1 cos θ = mg .....(2)
N1 sin θ = N2 .....(3)
N1 sin θ = F .....(4)
VA cos α = VB cos β From (1), (2), (3), (4) we get required condition
13. v cos T = u 17.
v = u sec T amG = a 2 + a 2 - 2a 2 cos a
dv = dT = 2a sin α/2
dt u sec T tan T dt fff.. (i)
b
tan T = y
dT b dy 18. 2V1 = V2 + VP
sec 2 T dt = – 2 dt
y
- 2a1 = a2 + aP
=+ b2 cos 2 T cos
u
T
y
2
= 1b b 2 cos Tu
y
= b T tan 2 T......... (ii)
u cos 19. N1 = mg cos 37c + ma sin 37c

2 N2 = mg cos 37c - ma sin 37c

dv u
& dt = b tan3 θ from (i) and (ii)
N3 = m ^ g + ah cos 37c
N4 = m ^ g - ah cos 37c
14.

Similarly for acceleration

b1 = g sin 37c - a cos 37c
b2 = g sin 37c + a cos 37c
b3 = ^ g + ah sin 37c
dx =- 20 sin T dT
v=
dt dt b4 = ^ g - ah sin 37c
dy d T 4 20.
u=
dt =- 16 sin T dt & u = 5 v = 0.8 v
15. As cylinder will remains in contact with wedge A
Vx = 2u

As it also remain in contact with wedge B Force on pulley = T 2

u sin 30° = Vy cos 30°–Vx sin 30°
sin 30c u sin 30c 21.
Vy = Vx cos 30c + cos 30c
Vy = Vx tan 30c + u tan 30c
Vy = 3u tan 30c = 3 u

V = V x2 + V y2 = 7 u For going downward slowly a = 0

Method - II mg = T + N
In the frame of A N + Mg = T
] M + m g g = 2T
3u sin 30º = Vy cos 30º
T = 100N
& Vy = 3u tan 30º = 3 u
Therefore (d) is wrong and (c) is correct
and Vx = 2u & V = V x2 + V y2 = 7 u
For maximum acceleration
16. N1 T + N - ^mg h = ma
N2 T - ^ N + Mg h = Ma
N3 N1 2T - ] M + m g g = ] M + m g a
mg
300 - 200 = 20 a
From FBD a = 5 m/s 2

59
 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
22. Both men have different masses and same tension force 28. F.B.D. of block B w.r.t. wedge
acting on them, therefore they can’t have same acceleration.
(c) is incorrect.
23. aB = 4a A T = 10aB .....(1)
40g - 4T = 40a A .....(2)
From (1) & (2)
40g = 4 (10 # 4a A) + 40a A
& 400 = 200a A
& a A = 2m/s 2
for block A
TB = 80N
N cos 45º = 1.7a .....(i)
24.
for block B
0.6g sin 45º + 0.6a cos 45º = 0.6b .....(ii)
N + 0.6 a cos 45º = 0.6 g cos 45º ....(iii)
by solving (i), (ii) & (iii)
3g 23g
a = 20 and b =
20 2
T = m1 g
Now vertical componentof acceleration of
when thread is burnt, tension in spring remains same = m1 g 23g
(m1 –m2) B = b cos 45º = 40
m1 g–m2 g = m2 a m2 g = a = upwards and horizontal component of acceleration of
17g
B = b sin 45º–a = 40
29.

q
q
Tsin q T 2q

2q

for m1 a=0 mg mg sin2 q

T sin T= mg sin 2T
25. F
F = 2 T cos T T = 2 cos T sin 2i
T = mg
sin i
= 2mg cos T
T - cos T . T -
= 48 N
On increasing θ , cos θ decreases and hence T increases. 30.
26. T = mgeff = weff
= 5 ]10 + 2g T1
= 60 N (1)
= 6 kg f m T2 T2 (2)
27. There is no horizontal force on block A, therefore it does m
not move in x-direction, whereas there is net downward Mass 2 kg is at rest, so T2 = 20 N
force (mg–N) is acting on it, making its acceleration along So for (2), T2  mg = ma2
negative y-direction. 20  10 = 1 × a2
a2 = 10 m/s2
Block B moves downward as well as in negative x-direction.
Downward acceleration of A and B will be equal due to i.e. pulley connected to block and monkey (2) is moving up
constrain, thus w.r.t. B, A moves in positive x-direction. with 5 m/s2
T1 = 2 T2 = 40 N
So for (1),
T1  mg = ma1
Due to the component of normal exerted by C on B, it moves 40  10 = 1 × a1
in negative x-direction. a1 = 30 m/s2

60
Physics
End of rope is coming down with 5 m/s2. So, acceleration of 7 3
(1) w.r.t. his rope is 30 + 5 = 35 m/s2 upwards. Acceleration ma # 4 = mg # 4
of monkey (2) with respect to his rope is zero because he is g 3
just holding the string. a= 7
mg
31. a = Acceleration of ball w.r.t block = g sin 60 + a cos 60
2M + m g 3 g 3
mg = 2 + 14
v = 2e
2
o# h
2M + m 8g 3 4g 3
0.25 # 10 = 14 = 7
= 2= G # 16.2 # 10 -2
L = 2 #c
2 + 0.25 1 4g 3 m # 2
t
5 # 16.2 # 10 -2 7
v2 = 7L
0.25 t= &n=7
v = 0.6 m/s 2g 3
35.
32. Let plank moves distance x upwards. Length of string,
passing through hands of B, loosened by 3x.
Length of rope passing through “B” = 3x

°
T1

30
Length of rope passing through “A” = 4x

=
V1

q
as A itself moves distance x upwards.
60°
`x=4
33. a T1
8 T V2 m
T
T1 sin ]ig V1 = T1 V2
a
1 V1 sin ]ig = V2
& a1 sin ]ig + V1 cos i dt = a2
di

10 - T = 1a ...(1) & a1 sin ]30g = a2 ]a V1 = 0g

T = 8a ... (2) & a1 = 2a2
10 = 9a mg - T = ma2 ...(i)
10 ...(ii)
a= 9 T cos 60° = ma1
velocity when 1 kg reaches at ground. a1 = 2a2 ...(iii)
10 On solving equation (i), (ii) and (iii) we get
v2 = 0 + 2 # 9 # 3
a2 = 2m/s2
20
v= 3 36. N cos 37c = mg A B
2#3 3 & F = N sin 37c N
t= 10 = 5 F
F 3
distance between 1 kg and 8 kg & mg = Tan37c & F = 4 mg N
37°

mg
x = vt = 2
34. 37. N = F = 3 mg = 5 mg
sin 37c 4 3
a 5k
4
N a
Ng = mg + N cos 37c
ma 4
Ng = mg + F cot 37c = mg + F 3
60° N
38. y = x tan 37c
aB = a A 3 (-)
4
mg 39-40
block
ball Insect will move with acceleration a1 ( = acceleration of m2 )
Let block starts moving with acceleration a towards relative to ground

a1 = d m1 + m2 n g
right. From F.B.D of block m -m mg
use m1 = 2
1 2
g+a
N sin 60 = ma
With respect to block 41. FBD of Block in ground frame :
N + ma sin 60 = mg cos 60 applying N.L. 150 + 450–10M = 5M
ma 600
sin 60 + ma sin 60 = mg cos 60 & 15M = 600 & M = 15
ma ]1 + sin2 60g = mg cos 60 sin 60 & M = 40 Kg

61
 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
42. If lift is stopped & equilibrium is reached then Before string between pulley & C is cut
T = 450 N 450 + N = 400 TAB = 3mg, TCD = 4mg
N & N =- 50 After string between pulley & C is cut
So block will lose the contact with TAB = 0
weighing machine thus reading of
6mg 3
Mg = 400 M weighing machine will be zero. a = 4m = 2 g
T
TCD + 2mg = 2m b 2 g l
T = 40 g 3

So reading of spring balance will be 40 TCD = 4 mg

40 g Kg.
43.
T = 450 N 950 - 400
N = 400 N
a= 40
a
40 Kg 450 45 2
& a = 40 = 4 m/s Ans
Previous Year (JEE Main)
Mg = 400 N

44. 1. The speed of a ball just before collision with ground

2t
F is u = 2 # gH = 2 # 10 # 9.8 = 14 m/sec
]Downwardsg
45°
A B The speed of ball just after collision is
v = 2gh = 2 # 10 # 5 = 10 m/sec
F.B.D. of block “A” ^Upwards h

N
N1
So, a = Dv = 10 + 14 = 120 m/s2
Dt 0.2
f
F cos 45° t
2. F = 10i + 5jt
m = 100g = 0.1 kg
mg F sin 45°

a = m / 100it + 50tj S = u t + 2 a t 2 = 2 a t 2 ]as u = 0g

N = mg + F sin 45c = mg + t F 1 1

fmax = nN = 2 ^100it + 50tj h 2 2 = 200it + 100tj = ait + bjt

1

= ^mg + t h
1 a
2
a = 200, b = 100 ` b =2
N' is zero, till F cos 45c # fmax 3. N

t # ^mg + t h
1
2
t # mg
t # 20 sec . Mg
During T # 20 sec, N' = 0 and acceleration both blocks
When lift is at rest, N = mg
remains to be zero. t.20 sec, System starts moving forward
with an acceleration ‘a’ & 60 # 10 = 600 N
When lift moves with downward acceleration: In frame of
F cos 45c - fmax
where a = lift, pseudo force will be in upward direction.
^ m A + mB h
N'+Ma
t - ^m A g + t h
1
2
=
5
t - mA g
= Mg
10
at t = 40 sec, a = 2 m/ sec 2 N' = M (g - a) = 60 ]10 - 1.8g & N' = 492 N
at t - 60 sec, a = 4 m/ sec 2
Dp
4. Fav =
45. Before spring 2 is cut TAB = 3 mg Dt
After spring 2 is cut, TAB = 3mg 0.12 # 25
= 0.1 = 30 N

5. Net force = 8it + 4tj + 4kt

a = m = 2it + tj + kt
F
6. Statement I is correct, because lift is moving with zero
acceleration. Statement II is incorrect as force exerted will be
less than the weight.

62
Physics
7. Drawing the FBD of the point where F is applied 12. At equilibrium
T
30° mg
F = T sin i = # sin i
cos i
F 10 # 10 = 100N
= cos 45° # sin 45°
13. F = ma = F0 e -bt
v t

T' = 3 kg
# dv = Fm # e0 -bt
dt
0 0

T ' = 3g & T cos 30° = 3 g & T = 2g = 20 N F -bt t

v = m0 ; e E
-b 0
v = mb ^1 - e -bt h
8. V F0

12 t1 14. l A = 2l , lB = b 3l l
5 5
12 Kl = K A l A = KB lB

t2 Kl = K A b 25l l
5K & 5K
KA = 2 KB = 3
t1 8 t2
8
15. Vertical component of acceleration of A
a1 = ^ g sin Th . sin T = g sin 60c. sin 60c = g. 34
6t = 4t
2
1
2
2 That for B
9. Given that a particle is projected with velocity v0 along the
X-axis. A damping force is acting on the particle which is a2 = g sin 30c. sin 30c = g 14
proportional to the square of the distance from the origin i.e., ` ^a AB h= =
3g g g 2

F =- ax2 . 4 - 4 = 2 = 4.9 m/s

According to Newton’s Second law, ma =- ax2
0 x Previous Year (JEE Advanced)
&a=
- ax 2 vdv a 2
m & dx =- m x & # vdv = # - ma x dx 2

1. v = a ^ yxt + 2xyt h
v0 0
1

& b v2 l =- m b x3 l & 2 =- m 3 & x = b 3mv 0 l

2 0 3 x 2
2 3 3
a -v a x 0
v 0 0 2a v x = ay vy = 2ax
10. dvx dy
dt = a dt = 2a x
2

45° ` F = ma = 2ma2 ^ xxt + yyt h

dvy
dt = 2avx = 2a y
2

F 2.
100 N

100 N

10 kg
T
= 100 , along the vertical direction
2
T ma cos T = mg cos ]90 - Tg
= F , along the horizontal direction.
2 & a = tan θ & a = dy
g g dx
& F = 100N
11. Here, v = K _ yi^ + xj^ i
^ & d ^kx 2h = a & a
x = 2gk
dx g
3. F = 2T sin θ
dx ^ dy ^ _ ^ ^i
dt i + dt j = K yi + xj
dx dy dy dy/dt Kx
,
dt = Ky and dt = Kx dx = dx/dt = Ky
ydy = xdx
Integrating both sides # ydy = # xdx or y2 = x2 + constant

T cos θ F cos T F x
a= m a = 2m sin T = 2m
a2 - x2

63
 EĞǁƚŽŶ͛Ɛ>ĂǁƐŽĨDŽƟŽŶ
4. After string is cut, FBD of m 5. 2mg cos θ = 2 mg
mg 1 =
a= m =g. cos θ = cos 45c & θ = 45c
2
FBD of 2m (when string is cut tension in the spring takes
¿QLWH WLPH WR EHFRPH ]HUR +RZ HYHU WHQVLRQ LQ WKH VWULQJ
immediately become zero.)
3mg
3mg - 2mg g
2m
a= 2m = 2 -

2mg

64