LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019 3 LAWS OF MOTION & FRICTION Physics Secrets by Shiv R. Goel (meant only for WAVES classroom students) SOLUTION EXERCISE-1 (GRADE- I) 1. B 2. AB sin = 5 4 and cos = 5 3 3 4 5 A T = 1000 cos i ˆ + 1000 sin ) j ˆ ( = N j ˆ 800 i ˆ 600 A B T T = j ˆ 800 i ˆ 600 3. D Action rection pair act on different bodies and are of some type. N N mg mg Earth Normal force exerted by table on book is the reaction force of normal force exerted by the book on the table. Normal force exerted by the ground on the table is the reaction force of normal force exerted by the table on the ground. 4. C F drag is opposite to the direction of motion 'ma' is not a force, acceleration is the effect of net force on a body. gravitational force acts downwards. As the helicopter moves with constant velocity there force acceleration is zero forces must balance each other Thus F rotor acts such that is balnces the cesultant of F drag & F gravitational 5. C String will break easily if the tension in the string is more. 2cos= W w T T T T = cos 2 w as increases, cos decreases, T increases drawing FBD for the lower half 6. D 60ー T B 100 T A T B cos 60ー = 100 T B = 200 N T B sin 60ー = T A T A = 100 3 N 7. B 8. A 2T sin = 10
15
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LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
3LAWS OF MOTION & FRICTION
Physics Secrets by Shiv R. Goel(meant only for WAVES classroom students)
SOLUTIONEXERCISE-1 (GRADE- I)
1. B2. AB
sin = 54
and cos = 53
3
45
AT
= 1000 cos i + 1000 sin )j( =
Nj800i600
AB TT
= j800i600
3. DAction rection pair act on different bodies and areof some type.
N
N
mg
mg
Earth
Normal force exerted by table on book is thereaction force of normal force exerted by the bookon the table.Normal force exerted by the ground on the tableis the reaction force of normal force exerted by
the table on the ground.4. C
Fdrag is opposite to the direction of motion'ma' is not a force, acceleration is the effect of netforce on a body.gravitational force acts downwards.As the helicopter moves with constant velocitythere force acceleration is zero forces must balanceeach otherThus Frotor acts such that is balnces the cesultantof Fdrag & Fgravitational
5. CString will break easily if the tension in the stringis more.2cos = W
w
T
T T
T =cos2
w
as increases, cos decreases, T increasesdrawing FBD for the lower half
6. D
60° TB
100
TA
TB cos 60° = 100 TB = 200 NTB sin 60° = TA
TA = 100 3 N7. B8. A
2T sin = 10
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
0.1T T
25m
10m
tan = 251.0
~ sin
2T × 251.0
= 10
T = 1250 N ]
9. D10. C11. B12. B13. D14. A15. A16. C
a = g sin
1001g = 0.1 m/s2
17. Tx = 2T
; TTy = 23T
Tx= 23T
; TTy = 2T
Wx = 0; Wy = –W; WX = 2W
; WWy =
23W
; Nx = 2N
;
Ny = 23N
; NX = 0 ; Ny = N]
18. B
(B).50N
100N
70N
25N
T
70N
Short and thick
a = 5100–120
95 – T = 2.5 × 4= 4 m/s2
T = 85 N19. A
20. P|| = 5P3
,W|| = 5W4
,
T|| = T, P = 5P4
,W = 5W3
, TT = 0
21. CCheck all options.For example, see option(C).Fx = 2(–1)2 – 2 = 0Fy = (–1) – (–1) = 0Fz = 5(–1)2 – 2(2)2 – 3(–1) = 0
22. DM P
m
P = (m + M)aM T T = Ma
23. C
a =21
21mmmm
g
5 = 2m2m
× 10
5m + 10 = 10m – 2030 = 5 mm = 6 kg
24. BF F = ma
F
F cos = ma N +
F sin = mg25. B26. B27. B28. C29. C30. C31. C32. C33. D34. B35. A36. C37. C,D38. C39. [(A) — Q, (B) — P, (C) — R, (D) — S]
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
When block will slip over the wedge, wedge willmove in right side. Thus acceleration of blockwill make more than angle from the horizontal.Normal force will be normal to the surface of thewedge & block. Since wedge move in right side,
40. A
f = 0.9
F
in limiting condition,f = × F cos
F sin = f = Fcos tan = = 0.9 = tan–1 (0.9)
41. B
37º
N
= 1/3100 g
100kgT
T
25 g
T
a
N + T sin37º = 100gN = 100g – T sin 37ºT – 25 g = 25a
T = 25 g + 25 aFor block to start moving
T cos 37º = × (100 g – T sin 37º)
(g + a ) =2s/m
3ga
53
31
5425
)g100(31
42. D43. B
(B)32
3
K'
for a particular meterial of spring
1K K= constant
32K
= K ×
K' = 2K3
44. B
3 kg 2 kg 10N2ms–2
T T 2 m/s2
for 2 kg mass10 – T = 2 × 2
T = 6 Nfor 3 kg mass
T = 3 × a a = 2 m/s2
45. D
.(D)
N N
mg sin30º mg cos30º = 0.5
30º
N = mg cos30º f = mg cos30º
contact force = 22 Nf =
23102
231025.0
2
= Nt55
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
46. A
j4iF
F
1 kg
= 0.3
Drawing FBD of block,
i + 4 jN
1 g = 10 Nt
f 0.3 N
In vertical direction,N + 4 = 10N = 6f 0.3 × 6
As the horizontal component of F is 1 Nt, (< 1.8Nt) friction is static which is also equal to 1 Ntin opposite direction.
47. B
. 37º
N
= 1/3100 g
100kgT
T
25 g
T
a
N + T sin37º = 100gN = 100g – T sin 37ºT – 25 g = 25a
C) = 53°Fsin = 80NN = 60 fC 6N,f = 5N downwards (static friction acts )
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
EXERCISE-1 (GRADE- II)1. A
o45 o45
mgm
1T1T
1TM
2T
Mg
2T1 cos 45o = mg 1mgT
2
T2 sin = Mg + T1cos45o
T2 sin = Mg +mg2
…..(1)
T2 cos = T1 sin 45o
T2 cos =mg2
….(2)
(1) (2)2Mtan 1m
2. (D) For A : T = mgFor B: T + mg cos = mg sin mg mg cos mgsin
+ cos = sin Squaring both sides2 + 2 cos2 + 2 2 cos = 1 – cos2 2 2 2 2( 1) cos 2 cos 1 0
2 2 2 2
2
2 4 4( 1)( 1)cos
2( 1)
=2
2
11
=
2
2
11
(ignoring –ive sign)
2
2
1cos1
3.B 3T = 50g + 25g T = 25g = 250 N
T TT
50 g25g
A,CBefore burning BC, the Free Body Diagramsare shownT
2 = T
1 + m
2g —(1)
kx = T2 = m
1g ——(2)
where x is extension in the spring.Just after buring, T
1 will become
zero , but T2 will remain same
T2 – m
2g = m
2a a =
1 2
2
(m m )gm
8.A,C 2T cos = F FT
2cos
FT
TIf increases, cos decreases and hence tensionincreases.
Now T > F F F
2cos
1 > 2 cos cos < 1/2 > /3
9.A,C Tension will be equal to apparent weight of 5 kgblock, which is 5(g + a) = 5 [10 + 2] = 60 N
10.B 2TvA = 3TvB A B3v v2
A 03v v2
A/B A Bv v v 00
3vv
2 = 0v
211.B T1 = 2T2
2T 2T
1 1T m g
B
2 3
12 3
2m m gm g 2
m m
1 2 3
4 1 1m m m
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 201913. (c)
2 2F f N , f varies from 0 to N, where N = Mg
So F varies from N to 2 2( N) N
2 2N F ( N) N
2Mg F Mg 1
15.C Given 1 2 3 4ˆF F F F 100 3i
..(1)
2 3 4ˆF F F 100 1 i
..(2)
1 3 4ˆF F F 100 24 j
..(3)
(1)×2 – (2) – (3) we get
1 2ˆ ˆF F 700i 2400j
1 2F F ˆ ˆa 7i 24jm
2 2 2a 7 (24) 25 m / s 17.(B) Let f is the resistive force.
fT f
T
mg
A B
for A: T = f , for B: T + f = mgfrom above T = mg/2
19.B 70 + 5g – 100 = 5a
a
100 N
5 g
70 N
2.5 ga
T
70 N
a = 4 m/s2
2.5 g + 70 – T = 2.5 a T = 85 N
20.(B) Let tension in the spring is T, thenT
a
T3 kg 2 kg
22 m/ s
10 N
10 – T = 2 × 2 T = 6 NT = 3a a = T/3 = 6/3= 2 m/s2
21.(A) 21S at2
2 1t
a
22
1
at2t a
1 gsin g cos4 gsin
3 tan4
22.(a) (i) A (ii) A (b) (i) B (ii) AIn part (a), both blocks move togetherIn part (b), blocks move separately.
27.(I) A (II) D (III) BFriction on 2 kg and 4 kg will be limiting, but on 6 kg
friction will be less than limiting.T + 8 + 4 = 20 T = 8 N
4 kg 2 kg20 NT
2f 8 N 1
f 4 N
EXERCISE#2 (ONE OR MORE THAN ONE )
1.(BD) (fs)max = 0.25 × 10 = 2.5 NFmax = 2NSo block does not move
a = 03. (BCD) Minimum value of force F to keep block at rest
F = mg (F = N) 0.5 F = 0.1 × 10 F = 20 N
That means if F 20 N friction will be equal tomg (10 N)
FN
mg
N
F = NAs block can not move
perpendicular to the wall
5. (BCD) 30 = µsmg30 = 0.3 m (10)m = 10 kgSo total mass must be greater than 10 kgSo m 4 kg
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
EXERCISE#4 (SUBJECTIVE)
1 considering equilibrium of part PQ of the ropetension at point P should be equal to the weigth ofpart PQ.
x L
T
xgLm
LmxgT ]
Q.2 Applying newton's II law on whole system
F – 40 – 21
× 10 = (4 + 21
)a
54 – 40 – 5 = 4.5a9 = 4.5aa = 2 m/s2
Applying Newton's II law on lower half
T – 21
(45) = 25.4
× a
B
T
T = 245
+ 25.4
× 2 = 245
+ 4.5
= 22.5 + 4.5 = 27 N ]Q.3 Applying Newton's II law on the two blocks :
8 Fx
8kg
T
N
Fx – T = 8a
2
T
2kg
T – 2g = 20
Adding 1 & 2Fx – 2a = 10a
for a > 0,Fx > 20(i) & (ii)Fx – T – 4T + 8g = 0
T = 5g8Fx
for string toblock , T 0Fx – 8 g
[Ans. (a) Fx > 20 N (b) Fx – 80 N]
Q.4 [Ans. T = 7Mg8
]
Sol. As pulleys are massless
T1 = 2F ....(1)T2 = 2T1 = 4F ....(2)T = 2T2 = 8F ....(3)For equilibrium of mT2 + T1 + T = Mg4F + 2F + F = Mg7 F = Mg
F = 7Mg
putting in (3)
T = 7Mg8
]
Q.5
F
200N
2F
F
(a) Considering the whole system net upward force= 2Fthe maximum acceleration of man would be whenthe upper rope has maximum tension
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
Maximum upward force, 2F = 300 Ndownward force w = 200 NApplying Newton's II law300 – 200 = 20 amax
amax = 20200–300
= 5 m/s2
(b) (i) when painter is at rest2 F = wF = 100 N
(ii) When painter moves with an acceleration,a = 2m/s2
2F – w = 20 × 2
F = 2240
= 120 N
Q.6 [2 sec]
m1 m2
F = 30t N
4kg 1kg
3F
3F
3F
3F2
m1
40
N 3F2
t103F
t203F2
for m1 to lose contact with the floor N = 0
3F2
= 40 = 20 t
t = 2 sec ]
Q.7 [Ans : 6 N]
3kg2kgP = 10 N A B
due to constraint both blocks will move with sameacceleration
T
netsys M
Fa a = 510
= 2 m/s2
AN
a
for block AP – N = ma × a10 – N = 2×2N = 6 N
Q.8 [3 : 1]
[Sol.5kg 2kg 1kg
AB
CF = 16N
N2N1N
The three blocks will move with same acceleration,
asys = 816
= 2 m/s2
5kg16N N1
16 – N1 = 5 × aN1 = 6 N
1kgN2 N2 = 1 × a
3NN
2
1
= 2 N ]
Q.9 [Ans 5N, 16/31 kg ]
[Sol.37°
1kg3 m/s2
A
Ba
37ºa
AN2
53º
Applying Newton's II law to A in verticalMa g – N1 cos37º = MA = × a
.......(i)by constraintsaA cos 37º = 3 sin37º
.......(ii)
37ºN 1
3m/s2
M gB
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
applying Newton's II lawN1 cos53º = mB × 3N1 × 1 × 3 ..........(iii)from equation (i) & (ii) cos can find mA and a
Q.10 [Ans. (a) 31
mm
2
1 (b) a = 3/4 m/s2 ]
F = m1 × 3F = m2 × 1
(a) dividing the equation
31
mm
2
1
(b) If the masses are combinedF = (m1 + m2)a=m1 × 3substituing the value of m2
m1 a + 3m1 a = 3 m1
a =43 m/s2 ]
Q.11 [Ans : m'm]
[Sol. m m'
force on both the masses is same i.e. kxkx = makx = m'a'
'mm
'aa ]
Q.12 [Ans : 0.5 m]
[Sol.
= 30º
K = 40 N/m
m1
m2
m2
2 kg
3 kg
m1
T
20
30
T
in equilibrium conditionfor m1 ,m1 g sin30 + kx = T10 + 40x = T .........(i)for m2m2g = T Þ T = 30 ..........(ii)from (i) & (ii) x = 0.5 m
Q.13 [Ans: x1 : x2 : x3 : 15 : 18 : 10]
[Sol.
20 20
Tkx = T1
a1
for constant extension a of the blocks is same;kx1 – 20 = 2a120 – kx = 2a1
a1 = 0, kx1 = 20
x1 = k20
m
(b)
kxa2
2 kx2
3020
30 – kx2 = 3a2kx2 – 20 = 2a2
solving,a = 2 m/s2
x2 = k24
m
(c)
kx3
a3
kx3
10 20
kx3 – 10 = 1a320 – kx3 = 2a3
Solving
a3 = 310
m/s2
x3 = k340
]
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
Q.14 (a) 30° (b)23
]
[Sol.
g sin
(a) S t r i n g w o u l d m a k e a n g l e with thevertical if a = g sin.
a = g sin = 630
= 5 m/s2
sin = 21
= 30º
(b)
0.1a(pseudo force)
T
0.1g
for equilibriumT sin = 0.1a cosT = 0.1 × 15 × 3
=23
N ]
Q.15 [Ans. (M + m1 + m2)
g
mm
1
2 ]
[Sol. m2
m1
MF
for alll blocks to move togetherF = (M + m1 + m2) a ......(i)
Tm1
T = m1 a .......(ii)Sol. Note that the force exerted by the string accelerates
m1
m2
T
N
m g2
T = m2 g ........(iii)
Solving (ii), (iiii)
a =1
2m
gm
Substituting a in (i)
F = (m1 + m2 + M)1
2m
gm]
Q.16 [Ans 55]
[Sol. 37°
5 m/s2
N25N
50N
37º
Considering motion of the block w.r..t theinclined plane
Pseudo force on block FP = 25 NApplying Newton's law on the block in directionperpendicular to the inclined surface,
N = 25 sin 37º + 50 cos 37ºN = 15 + 40N = 55 Newton
By Newton's Third law,this force exerted by inclined plane on the blockis equal to the force exerted by the block on theinclined plane ]
Q.17 [Ans : 0.56]
[Sol.mg sin37º
37°mgcos37º
NN = mgcos37º
a1 = mº37cosmgº–37sinmg
if friction was absenta2 = g sin37º
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
The pi g starts from rest in both theeases &distance travelled is same,
211ta
21
= 222ta
21
also t = 2t2
º37cos4º37sin3
]
Q.18 [Ans : (a) 10.8 N ; (b) 0.33 m/s2 ; ]
[Sol.= 37º
a
N
3.5gk =0.25 15 N
3.5kgfk
Resolving the froces in vertical and horizontal andapplying Newton's II law(a) N = 3.5 g + 15 sin 37º = 35 × 9.8 + 9= 43.3fk = N= 0.25 × 43.3= 10.8 N15 cos 37º – fk = 3.5 a
(b) a = 5.38.10º–37cos15
= 0.33 m/s2 ]
Q.19[Ans : 1/3]
[Sol. a = g/3
37º
N
mg cos37ºmg sin37ºmg
N
Applying Newton's Law in direction perpendicularand parallel to the plane,
N = mg cos 37º
a = mº37cosmgº–37sinmg
3g
= g53
– g54
= 31
]
Q.20 [Ans : 2000 N]
[Sol.A
90 kgB 10kgF
= 0.5
acceleration of the blocks, a = 100F
FBD of B,
a
N
f
10 gApplying Newton's II Law in horizontal for blockB,
N = 10 × 100F
For limiting condition f = Nf = 10 g
N 10g
0.5 × 10F 10 g
F 2000 N. ]
Q.21 [Ans1 kg]
[Sol. m1kg37°
=0.5A T
Tf
1g cos37º1g sin37º
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
for block A to slide up with constant velocity,a = 0
mg = T.......(i)T + f = 1g sin 37º.......(ii)N = 1g cos 37º.......(iii)f N.......(iv)
Considering acceleration of blocks in the directionof applied force F,for block B,T – 40 = 4a.......(i)
N
50
50T
fA
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
for block A,Applying Newton's II Law
50 – T – f = 5a.........(ii)Adding (i) & (ii)
f = 10 – 9a..........(ii)i f a 0,friction f is kineticfk = N= 0.8 × 50= 40 NThis gives.
a = – 310
But then direction friction of friction we assumedis wrong a = 0friction Ps static and just sufficient to balance forcesfrom equation (ii)
f = 10 N
Q.25 [Ans.
2
112g
, 22mg
]
Sol.
A AB Bf
30° mgsin mgcos
N T
a
Tmgsin
mgcos
N mgcos
for block Amgsin – T = ma ...(1)for block Bmgsin + T –Bmgcos = Ma ...(2)solving (1) & (2)
g –32
·23
g = 2a a =
2g
2
11
T = mgsin – ma = 2mg
–
2
112
mg = 22
mg
Q.26 [Ans. (a) 4g
, (b) 8m5
]
Sol.m1
m2
TT
m1 + m2 = m
Tmax = 32mg15
let m1 > m2m1 g – T = m1 aT – m2 g = m2 a
T =21
21
mmgmm2
...(i)
a =
21
21
mmgm–m
...(ii)
Tmax =
32gm–m15 21
for rimiting condition,
gmmmm2
21
21
= g32
)m–m(15 21
212
21 mm4)mm(1615
(m1 – m2)2 = (m1 + m2)
2 – 4 m1m2
(m1 – m2)2 =
16)mm( 2
21
(m1 – m2)2 =
4)mm( 2
21
substituting the value of (m1 – M1) in equation (ii)
a = 4g
Let maximum value of greater mass be M
Tmax = gm
)M–m(M2 = g
3215
gm
)M–m(M2 = 32
mg15
M = 22m
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
Q.27 [Ans: a1 = 31g15
m/s2, a2 = 49g15
m/s2, ]
[Sol. Applying Newton's second law in vertical,
N – mg = ma sin37º
N – mg = m( 53
a)
N = m(g + 53
a1)
Also, f = ma cos 37º
N = m( 54
amax)
amax = a1
103
m
1a
53g = m
1a54
a1 = 15g/31 m/s2
Now, for maximum acceleration,
N = m
2a
53g
N = m
2a54
On solving, a2 = 15g/49 m/s2 ]
Q.28 [Ans : 0.54]Sol. m = 4.1 kg
F = 40 Nby figure, vf= 5m/s; v = 0.5m/s
a = tv–v if
= 15.0–5
= 4.5 m/s2
ma = F – mg4.1 × 4.5 = 40 – × 4.1 × 9.8 = 0.54
EXERCISE # 5 PREVIOUS YEAR1. (C)
Tcos Tcos
Tsin Tsin
2 mgmg mg
A
T T
CBTT
For equilibrium in vertical directon for body Bwe have
2 mg = 2T cos
2 mg = 2 (mg) cos[ T = mg (at equilibrium)]
cos = 21
= 45º
Q.2 (D) At equilibrium T = Mg
mg TT
Mg
T
F.B.D. of pulleyT=Mg
F F = (m+M)g1
The resultant force on pulley isF = 22 TF
F = [ 22 M)Mm( ]g
3. (A) The forces acting on the block are shown. Sincethe block is not moving forward for the maximum
LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019
force F applied, thereforeF cos 60º = f = N ....(i)[Horizontal Direction]For maximum force F, the frictrional force is thelimiting friction =N]and F sin 60º + mg = N ....(ii)From (i) and (ii)
60º
Fcos60º
FFsin60º
mgf
N
F cos60º = [F sin60º + mg] F =
=
23
321–
21
10332
1
= N20
415
4. (C) JUst before the string is cut by equilibrium of massm, T' = mg ....(i)By equilibrium of mass 2m, T = 2mg + T
....(ii)From (i) and (ii), T = 2mg + mg = 3mg
5. (B)
8. (A) µtan
mg
P
N
force for which f = 0 P0 = mg sin Case I P = mg (sin – µ cos ) P < P0
mg
PfN
P + f = mg sin mg (sin – µcos) + f = mg sin f = µ mgcos
Case II P = mg (sin + µ cos) > P0
mg
PN
f
P = f + mg sinmg (sin + µ cos) = f + mg sin f = µ mg cos ]