Law s of Mo tio n

Cha pter

c=fJ
-- -- -- -- -- --- --- --- - - -- -- ----
To p ic-1: 1st. llnd & lllrd Laws of Motion
[=:J
l·(N)
F(N/

,.® MCQ, with One Correct Answer

A particle of m ass mis moving in the xy-p lane s uch that
0 75

(D)
()5 - - - - ---

(C)
its velocity at a point (x, y) is given as ~ = a ( yx + 2xy) ,
where a is a non-zero constant. What is the force F acting
on the particle? I \1h . 211231 0 Fig (c) IO l(s) 0 Fig (d/ 2n

(a)
2
F=2ma (xx + yy) (b) F= ma 2 (yx +2xy) The impulse is highest in figure.
2 2 (a) Fig (A) (b) Fig (D) (c) Fig (C) (d) Fig (B)
(c) F=2ma (yx+xy ) (d) F = ma (xx +2yy)
The figure represent s the momentu m time (p-t) curve for a
A bu llet IO g leaves the barrel of gun ·with a velocityo f600 pa rticle movi ng a long an axis under the influence of the
mis. Jfthe barrel ofgun is 50 cm long and mass ofgun is 3 force. Identify the region s on the graph where the
kg, then value of impulse supplied to the g un wi ll be: mag nitude of th e force is maximu m a nd minimum
j\1.1111 \pri1 11, .Z02lfl 11 respectively ? lf(l:J - t 2) < t 1. p I\ Jain .Jan. 30. 2023 (l)f
(a) 12Ns (b) 6Ns (c) 36Ns (d) 3Ns
@
l Three force F 1 = ION, F = 8 N, F 3
= 6 N are actin g on a (a) c and a
particle of mass 5 kg. T~e forces F2
a nd F3 are applied
ai ns at rest. If the force (b) band c
perpend icular so that partic le rem
F is removed , then the accelera tion of the particle is: (c) c and b
1
f\l ,i, \prill2. W2 ~(11I 1
2 2 (d) aandb 11 '2 '3
(a) 2 m s- 2 (b) 0.5 m s 1 (c) 4.8 m s- (d) 7 ms-
An average force of 125 N is applied on a machin e g un '>. A machine gun ofmass IO kg fires 20 g bullets at the rate of
1
firing bullets each of mass IO g at the speed of250 mi s to 180 bullets per minute with a speed of I 00 m s- each . The
recoi l velocity of the g un is : f '\1ain .Ja n . JfJ, 2023 (11 )1
keep it in p osition. The number ofbullets fired per second
by themach inegun is : /\lam \p11 11, 202~(1), (a) 0.02 mis (b) 2.5 mis (c) 1.5 mis (d) 0.6 m/s
(a) 5 (b) 50 (c) 100 (d) 25 I fl. Force acts for 20 son a body of mass 20 kg, starting from
A body of mass 500 g moves along x-ax is such that it's res t, after which the force ceases a nd then body describes
velocity varies with displacem ent x accordin g to the relation 50 min the next IO s. T he value of force will be ;
I \foin .Jan. 29. 2023 (IJH
v = J ofx mis the force acting on the body is: (d) ION
(a) 40N (b) SN (c) 20N
/\lai11 \vn l ll 2023 (11)/ T he air escapes from the
11. A balloon has mass of IO gin air.
(a) ~' l66 N (b) 25N (c) 125 N (d) SN velocity 4.5 crn/s. If the
ba lloon at a uniform rate with
At any rns tant the velocity of a pa rticle of mass 500 g is balloon s hrinks in 5 s complete ly. Then, the average force
' 2')
( 2 t i +3 t J ms - I . I f the force act ing on th e particle at actin g on that balloon wi ll be (i n dyne).
l' fai11 Jul) 28, 2022 (f)I
t - Is is {f + xJ} N . Then the va lue ofx wi ll be; (a) 3 (b) 9 (c) 12 (d) 18
/\11111.\pri l H .W l H I J/ I!. A ball of mass 0. 15 kg hits the wall with its init ial speed of
(a) 3 (bJ 4 (c) 6 (d) 2 12 ms I a nd bounces back without c hanging its initial speed
Figure<, (a) , (b) , (c) and (d) show variation force with
of Ifthc force applied by the wall o n the ooll
during the contact
tJ1ne. f \1 1111 I ti,. I 1111 • I lfJf is I 00 N. Calcul ate the time duration of the contact of ball
/-(NJ wi rh th e wall. j \tuin J uh 26, .:!022 t H>I
(11) 0.018 s (b) 0.036s (c) 0.009s (d) 0.072 s
I l ;\ block of metal weighin g 2 kg is resting on a fric tiooless
plane (as s hown in figure). It is s truck by a jct re1leasing
(A)
wa ter al u rate of I kgs I nnd at a speed of IO ms- • Then,
2
th e initial accelera tion of the block, in m s- , will be :
h g /b/ 20 • 1/~J I \J i1in Jum.' !.9 .Z022 (Ill
Laws of Motion A31

r--_,.a=?
(c) J/n[1+Hv,) (d) Jzrg1an- [J¥v,)
1

20. A ooll is thrown vertically up (taken as+ z-axis) from the

(a) 3 (b) 6 (c) 5 (d) 4 ground The correct momentwn-beight (p-h) diagram is:
14. A block of mass 2 kg moving on a horizontal surface with !Main 9 April 2019 1)

+h ~) +h
speed of 4 ms- 1 enters a rough surface ranging from
x =0.5 m to x = 1.5 rn . The retarding force in this range of
rough surface is related to distance by F = -J.....: where
1
k = l 2 Nm- • The speed of the block as it just crosses the (a)
rough surface will be: l:vlain J une 28. 2022 (IT)J
(a) Zero (b) 1.5 rns- 1 (c) 2.0 ms- 1 (d) 2.5 ms- 1
I 5. The initial mass of a rocket is 1000 kg. Calculate at what
rate the fuel should be burnt so that the rocket is given an
acceleration of 20 ms- 2 . The gases come out at a relative
speed of 500 ms- 1 with respect to the rocket :
[Use g = IO m/s2f j\ fain \ug. 26. 2011 (DI
(c) +h (d) -4th
2 I . A particle of tna5.5 mis moving in a straight line with mcmmtum
(a) 6.0 x 102 kg s- 1 (b) 500 kg s- 1 p. Starting at time t =0, a furce F = A1 acts in the same directi~
on the moving particle during time interval T so that its
(c) I0kg s- 1 (d) 60kg s- 1
momentum changes from p to 3p. Herek is a constant The
t 6. A particle of mass M originally at rest is subjected to a
valueofTis: !Main 11Jan.2019 Hl
force whose direction is constant but magnitude vari es
with time according to the relation

22 .
(a) 2t ~) 2Jf (c) !! (d) If
A particle of mass m is acted upon by a force F given by

Where F0 and Tare L'Onstants. The force acts only for the the empirical law F = ~ v(t). If this law is to be tested
time inter. 'lll 2T. Tho vdocity v ofthe (Xlltlcle after time 2T is: t
1\ 1.iin Jul~ 2i. 202 I (11)1 experimentally by observing the motion starting from rest,
(a) 2F0TtM (b) F0T 2M the best way is to plot: !Main O nli ne April 10, 20161
(c) 4F 0T/3M (d) F0Tl3M I
(a) log v(t) against - (b) v(t) against t2
1.,. A spaceship in spac~ sw\!eps stationary interplanetary dust. t
dM (r) 1 l
As a result. its mass increases at a rate --;;;- = bv (r) . (c) log v(t) against 2 (d) log v(t) against t
t
where,. (r) is its instantaneous velocity. The instantaneous 23 . A body of mass 5 kg under the action of constant force
acceleration of the satellite is : I'.\ lain Sep.05. 2020 ([T)l
3 F=F,;i+ fyJ has velocity at t = 0 s as v =(6i-2])m!s
bv
(a) -b/ (t) (b)
M(t) and at t = IOs as v = +6 Jm / s . The force F is:
3 !Main Online April ll, 20141
2bv3 bv
(c) M(r) (d) - 2M(r) (a) (-3i+4J)N (b) ( -ii+1J) N
18. A small ball of mass m is thrown upward with velocity u
from the grOW1d The ooll experiences a resistive fcrce mhl- (c) (3i-4]}N (d) ( ii--11) N
where v is its speed. The maximum height attained by the
ball is: l'.\lain Sep. 04, 2020 (I[)J 24. A particle moves in the X-Y plane llllder the influence of a
force such that its linear momentum ts
t _1 ku 2 l ( k1/l
(a) -tan -
2k g
(b) -lnll+-J
k 2g
p(r) = A [i cos(kz)- }sin(kt)], wbereA andkareamsaaots.
The angle between the fcrce and the tmmentwn is (2007)
l ( 2
(a) Cf (b) JOl (c) 45° (d) ~
(c) _!_tan-1 ku
k 2g
2
(d) -lnll+-J
2k
ku
g
\

® 3 NumcricAns..r
19. A ball is thrnwn upward with an initial velocityv0 from the 25. A man of60 kg is running oo the road and suddenly jumps
surface of the earth. The motion of the ball is affected by a into a stationary troUy car of mass 120 kg. Then the troUy
drag force equal to myv2 (wherem ismassoftheool~ vis its car stans moving with velocity 2 ms- 1. The velocity of
instantaneous velocity and y is a constant). Time taken by the running man was _ _ _ _ _ _ ms- 1• When
the ball to rise to its zenith is : (~tain 10 .\ pril 20191) he jumps into the car. fl\laio J une28. 2022 (1)1
 - - -·--- ----- - -

A32
Physlc:a
26. A batsm an hits back a ball of mass 0.4 kg straig (b) cann ot be an inert ial fram e beca use the
ht in the eanh is
direc tion of the bowl er witho ut chan ging its revol ving roun d the sun.
initia l speed
of 15 ms- 1• The impu l se impa rted to
the bal I is (c) is an inert ial fram e beca use New ton 's laws
_ _ _ _ _ __ Ns. j'.\1ai n June 26. 20 22 (11)1 arc
appli cable in this fram e .
27. A boy push es a box of ma ss 2 kg
with a force (d) cann ot be an inert ial fra~1e beca use the
earth is
F= (20i + IO})No n a frictionless surfa ce. lfthe box was rotat ing abou t its own axis.
initia lly at rest , then _ _ _ _ mis displ acem
x-axi safte r 10 s.
ent along the -v-'
=:,,;,
9 Asser tion and Reason Type Questicns
j\ 1ain Feb. 16.20 21 (l)f
® 4 Fill in th e B la nks
JI . Sta tl' m l' n t- t : It is e.asie r to pull a heav y objec
push it on a lewl grou nd and
t than to

28. The magn itude of the force (in nC\\1ons) act mg ...,t.ltl 'mcn t- Z : The magn itude of fricti onal force
on a body depends
varies with tim e t (in micro secon ds) as <sho,,1 on th e natur e of the two surfa ces in conta ct.
1 111 the fig 12008 1
AB, BC and CD are straig ht line segm cnb. The (a) State ment - I ,s True . State ment-:! is True
magm tudc ; Statement-
of the total impu l se of the force on the lxxiy fro 2 1s a corre ct exp la natio n for S tatem ent- l
m 1 4 P"
tot = 16µsi s .. ....... ... ... N~. 111n ~ - 1 \f ,1rh, I (b) State men t- I is True . Stntc.· mcnt -2 is True
; Statement-
~oo ~ - -- 2 1:- NOT a corm.·t expla natio n for State ment -I
1
z
r.r10
(c) Sta teme nt - I is True . Statc ment -2 is False
(d) State men t - I is Fa lse. State ment -2
~ 400 is True
0 ,~ ,,.11u tll:n t- 1 A cloth cove rs a table . Some dishe
"- 1110 -+"-"- -;.:; s· are kept
on 11. The cloth can tx: pulle d out witho ut dislo
' -f - - + ' - - f ..l· t -+- t - -t ~ l dging the
0 ;! J (, ~ Ill 1;
d1,hc s from the table .
14 l'-
Tunr 1u " ~ "'' ·•ll ,mnt -~ : For every actio n there is an equal
and
(Ji,~ True / F alse oppt,,1tc reacti on.
12007 1
29 . A rm:kc t m oves fon, ard h~ pu,h m g th (a) State ment- I 1s True . Statc ment
-2 is True ; Statement-
e ,urrounJ1nf! J 1r
bal'. kwar ds . ::! "' a com.'Ct exp la na tion for State ment - I
11t11<11
~~ ,, MCQswith 0 1e or Ma-c th nn OJeCbrr (h) Sta teme nt- I is T rue. State ment -2 is True
; Statement-
ca An~ er
:! is NOT a corre ct expla natio n for State ment
J O. A rcfor ence frame attac hed tn thc earth -I
I 19Sh - :! , t.ir 1..., 1 (c) State ment - I is True , State ment -2 is False
(a) is an inerti al fram e h:~ dcfin111 on
(U) Statement - I is False , State ment -2
is True .

c5
E -Jl_ __ _Top
_ ic-2
__ : Mot
__ ion
_of
_Con
__ nec
__ted _
Bod
__ies. __
Pul ley
_&_Equ
__ ilibr
_ium of__
For ces
__ _ __
~ MCQ; ·with OleC'ocrect AnS\" "7 A block of ✓
3. 3kg is attac hed to
T
1. The position v,xto r ofa partic le relate d to time a strin g whos e othe r end is
tis given by attac hed to the wall . An 30°

r = (10ti + l 5t 1J+ 7k) m unkn own force F is appli ed so

that the strin g make s an angl e
of30° with the wall. The tensi on
The direc tion of net force expe rienc ed by the
parti cle is: Tis: (Give n g = 10 ms-2)
[:\lai nAp ri1 15.20 2J (f)I
!Mai n J an. 30, 2023 (11)1 .fJ~g
(a) Posit ive y-axi s (b) Posit ive x-axi s (a) 20N (b) 25N (c) ION (d) 15N
(c) Posit ive z-axi s (d) In x-y plane 4. Give n below are two state ment s :
2. As show n in figur e, a 70 kg garde n rolle r is State ment -I : An eleva tcr can go up er down
push ed with with unif ~
a force of F = 200N at an angle of30 ° with horiz
speed when its weig ht is balana:d with the tension
ontal . The
of its cable-
nonn al reaction on the roller is State ment -H : Fcrce exert ed by the floor ofan eleva
IMain J an. 31 , 2023 (1)1 taoo the
foct ofa persoo standing on it is mere than his'her weigh
(Giv en g = IO m s- 2 ) t wbCO
the elew tcr goes down with incr~ ing speed.
(a) 800✓2 N In the light ofthe above statements, choose the CC1Tect
ansWCf
frcm theop ticns given below : !Mai n Jan. 24, 2023 (Ill
(b) 600N (a) Both state ment I and state ment 11 are false
(c) ~N (b) State ment I is true but State ment ll is false
(c) Both State ment I and State ment ll are true
(d) 200✓3 N (d) State ment I is false but State ment Il is true
 Laws of Motion A33
As per given figure. a weightless pulley Pis attached on a m1 101.:g
double inclined frictionless surface. The tension in the
string (massless) will be (if g = IO m/s 2)

(a)

(b)
(4 ✓3 + l)N
4( ✓3 + l)N
IMain J a n. 2-t 2023 (1)1
~1
(a) 3360N (b) 3380N (c) 3120N (d) 3240N
11 . For a free body diagram shown in the figure, the four forces
(c) 4(✓3-l)N
are app lied in the 'x' and 'y' directions. ~at ad~i~ional
(d) (4 ✓3 - l)N force must be applied and at what angle with pos1t1ve x-
axis so that the net acceleration of body is zero?
6. Two lx_xiies of!11asses_m 1= 5 kg and m~= 3 kg are connected [:\ lain J uly 25, 202 2 (H )I
by a hg~t s~nng going over a smooth light pulley on a
~mooth mchned plane as shown in the figure. The system y 7N
(a) fiN .45°
ts at rest . Th e force exerted by the inclined plane on the
body of mass m I will be : [Take g = IO rns-2] (b) fiN. 135°
6N 5N
l\ foin Jul~ 29, 2022 (IOI
2 X

(c) fj N. 30°

~ -, (d) 2N,45° 8N

(a) 30N (b) 40N (c) SON (d) 60N 12. A block of mass M placed inside a box descends vertically
7. A block · A' tak es 2 s to slide dO\\Tl a frictionless incline of with acceleration 'a' . TI1e block exerts a force equal toone-
30° and len gth ' /'. kept insi de a lift going up with unifonn fourth of its weight on the floor of the box. The value of'a'
velocity •v·. Ifthc incline is changed to 45°. the time taken
will be : 11\lain .l une 29, 2022 (11 )1
by th e block, to s lid e down the inclin e. wi ll be
3
approx imatel y: ['111i n Jul~ 27. 2022 (11)1 (a) K (b) K (c) g (d) g
(a) 2.66 s (b) 0.83s (c) 1.68s (d) 0.70s 4 2 4
A uniform metal chain of mass m and length •L' passes I J. In the arrangement shown in fig-
over a massless and frictionless pulley. It is released from ure a 1, 32, a 3 and a 4 are the ac-
rest with a part of its length · f' is hanging on one si de and celerations of masses m 1, m 2,
rest of its length 'L - (' is hanging on the other side of the m 3 and m 4 respectively. Which
l of the following relation is true
pull y. At a certain point of tim e. when i = - . the
X for thi_s arrangement?
acceleration of the chain is f. The val ue ofx is
!Main J une 26. 2022 (IJ)I

l'fain Jul~ 28. 2022 (I [)I

I i' ~
, ! I ,' L- l
(a) 6 (b) 2 (c) 1.5 (d) 4
9. A monkey of mass 50 kg climbs on a rope which can
withstand the tensioo (T) of350 N. Ifmonkeyinitiallyclimbs
down with an acceleration of 4 m/s2 and then climbs up with
an acceleration of5m/s 2• (d) 2a 1 + 2a2 + 3a 3 + a4 = 0
Choose the corret option (g = IO m/s 2). 14. The boxes of masses 2 kg and 8 kg are connected by a
[~fai n Jul~ 26, 2022 (1)1 massless string passing over smooth pulleys. Calculate
(a) T = 700 N while climbing upward the time taken by box of mass 8 kg to strike the ground
(b) T = 350 N while goint downward starting from rest. (use g = 10 rn/s 2)
(c) Rope will break while climbing upward 1'1ain Aug. '1 7. 2021 (1 1)1
(d) Rope will break while going downward (a) 0.34 s
10. Three masses M = 100 kg. m1= 10 kg and m2 = 20 k~ are
arranged in a system as shown in tigu_re. All th~ surfaces
(b) 0.2 s
are frictionless and strings are rnextens1ble and
weightless . The pulleys are also weightless and
frictionless . A force Fis applied on the system so that the (c) 0.25 s
2
mass m, moves upward with an acceleration of 2 ms- .
The value off is : (d) 0.4 s
(Takeg= IOms- 2) [l\htin.Ju1~26,2022 (l)I
A34
15. A mass of 1o kg is suspen~ed by _a rope _of length 4 mh, 20. The pulleys and string s shown in the figure are SltlOoCh
from the ceiling . A force F 1s applie d honzo nta 11 y at t e
and of neglig ible mass. For the system to remain •
mid-point of the rop~ s~ch that th~ top half of the rope
makes an angle of 45 with the vertical. Then F equals: equilibrium , the angle 0 should be 12001 \j
(Take g = 1O ms- 2 and the rope to be massless)

16.
(a) IOON
!Main 9 J an. 20 19 11, 7 .Jan. 2020 Il l
(b) 90N (c) 70N
Two blocks of mass Ml= 20 kg and M2.
= 12 kg are connected by a metal rod ot
mass 8 kg. The system is pulled vertically
up by apply ing a force of 480 N as
shown. The tension at the mid-p oint of
(d) 75 N
480
N
(a) C1'

(b)

(c)
3(1'

45°
~~
-l,
17 I
the rod is: !Main OnUne ApriJ 22, 20131 I
(d) (fJJ
00
(a) 144N
(b) 96N ~ 3 Numeric Answer
(c) 240N M2 21. Four forces are acting at a
(d) 192 N point P in equili br iu m as 1,
17. Two particles of mass m each are tied at the ends of a light shown in figure . The ratio of
string oflength 2a. The whole system is kept on a frictionless forcer F 1 toF 2 is I : xwhe re
horizontal surface with the string held tight so that each x =-
mass is at a distance 'a' from the centre P (as shown in the --
j\laia Ju]~ 25. 2022 (I)] F:
figure) . 22 . A system of 10 balls each of mass 2 kg are connected
Now, tlie mid-point of the string is pulled vertically upwards via
massless and stretchable string. The system is allowed to
with a small but constant force F. As a result, the particles slip over the edge of a smooth table as shO'wn in figwc.
move towards each other on the surface. The magnitude
Tensi on on the string betwe en the 7th and 8th ball
of accele ration , when the separa tion between them
1s _ _ _ _ _ _ N when 6 th ball just leaves the table.
becomes 2x, is 12007 1
10
0 ball [~t ajn J un e 26, 2022 ([01
F a F
(a)
2m ✓a2 - x2
F X
(b) m m
2m ✓a2 - x2 p
IE )IE )I ..
I ball
F X a a
(c) 23. A block of mass 200 g is kept
2m a station ary on a smooth in-
clined plane by apply ing a
F ✓a 2 -x 2
2~
mjnim um horizontal force
(d) 2m x F -~
F = J; N as shown in fig-
18. The string between blocks of
ure . The value of x =
mass m and 2m is massless and
inexte nsible . The system is
su spend ed by a massl ess
spring as shown . If the string
is cut find the magnitudes of
24.
l~tain J une 25, 2022 (Il)}
/2.w
A car is moving on a plane inclined at 300 to the bcrizootal
with an acceleration of 10 ms- 2 parall el to the plane
accelerations of mass 2m and
m (immediately after cutting)
upward. A bob is suspended by a string from the roof of
the car. The angle in degrees which the string makes with
(2006 - 3'\1,- l l the vertical i s _ _ ()lain A~.31, 2021 (1)t
g (Take g = 10 ms- 2)
(a) g , g (b) g, 2 25. Consider a frame that is made up of two
fl'
g thin massless rods AB and AC as shown
t
g g _a'_ ~
(c) 2 'g (d) 2'2 in the figure. A vertic al force P of "
19. A string ofnegligible mass going over a clamped pulley of
mass m supports a block of mass Mas shown in the figure
.
magnitude 100N isapp liedat pointA of
the frame. t
,
n
The force on the pulley by the clamp is given by 1200 1SJ
Suppose the force is Presolved parallel It
to the arms AB and AC of the frame. The
(a) J'i. Mg magnitude of the resolved component ~
(b) .fi. mg . aloog tbearm ACisx N. J.!5' \ _ it
The value ofx, to the nearest integer. is _ _ . 7~ c
(c) ,f(M + m)2 + m2 g [Given : sin(35 °) =0.573,
_"
l...
cos(3 5°) = 0.8 19
(d) ✓(M+m)2 +M2 g sin( 1100) = 0.939,
cos(l 10°)= --0.342] I\<fain Murch 1~ 202 l (Ill
 Laws of Motion A35
option(s) is/are correct? [g is the acceleration due to
~ ) ' True /F alse
gravity. Neglect friction] I Adv. 20191
2M
~ti. The pulley arrangements ofFigs. (a) and (b) are identical. u,
The mass of the rope is negligible. In (a) the mass mis lifted
up by a_tta~hing a mass 2m to th e other end of the rope. In
(b) . m ts ltfted up by pulling the other end of the rope with
a constant downward force F= 2 mg. The acceleration of m
is th e sam e in both cases --.--~11984 - 2 Marks!

(a) - oft h e spnng,

At an extension ofxo · t h e magnttu
· de
4
of acceleration of th e block connected to the spring
111 F = 2 mg . 3g
m 2m ISIQ
(a) (b) 4Mg
(b) Xo =k
Mc-Qi with One or More than One Correct Answer
X
One end of a hori zontal uni form beam of weight Wand (c) When spring achieves an extension of for the
_Q_
2
length L is hinged on a verti cal wall at point O and its other first time, the speed of the block connected to the
end is supported by a light inex tensibl e rope. The other
end of th e rope is fi xed at point Q. at a height L above the spring is 3g
. /M
~si
hin ge at point 0 . A block of weight a W is attach ed at the
point P of th e beam, as shovm in the fi gure (not to scale).
(d) a2 - a 1 =a 1 -a3
29. In the arrangement shown in the Fig, the ends P and Q of
The rope can susta in a max imum tension of ( 2 ✓2) W. an unstretchable string move downwards with uniform
Wh ich of th e foll owing statement(s) is(are) correct ? speed U. Pulleys A and Bare fixed. 11982 -3 Mark s I
!Ad, . 2021 1
Mass M moves upwards with a speed
(a) 2Ucos 0

(b) U/ cos 0

w (c) 2U / cos 0

L (d) U cos 0

® 10 Subjective Problems
(a) The vertical component of reaction force at O does
30. Two blocks of mass 2.9 kg and 1.9 kg are
not depend on a
suspended from a rigid support S by two s
(b) The horizontal component of reaction force at O is
inextensible wires each of length l meter,
equal to Wfor a = 0.5
see fig. The upper wire has negligible mass
(c) The tension "in the rope is 2Wfor a = 0.5
and the lower wire has a uniform mass of0.2
(d) The rope breaks if a > 1.5
kg/m. The whole system ofblocks wires and 2.9 kg
28 A block of mass 2M is attached to a massless spring with
spring-constant k. This block is connected to two other support have an upward acceleration of0.2
blocks of masses M and 2M using two massless pulleys m/s 2. Acceleration due to gravity is 9.8 m/
and strings. The acceleration of the blocks are a, _, a2 and s2. lf 989 - 6MarksJ
a as shown in the figure. The system is relea5cd from rest
3
with the spring in its unstretched state. The maximum · (i) Find the tension at the mid-point of the lower wire.
extension of the spring is x 0 • Which of the following (ii) Find the tension at the mid-point of the upper wire.
 . Laws of Motion
Chap ter . ,

7. (b) We have, Impulse = Area under F-t curve

~
§J
Topic-1: 1st. llnd & lllrd Laws of Motion

---------------- - lfi
for fig (D), area is highest. So, impulse is maximum .
d-
1. (a) ·:v=a(y;+2xy)
dv a (dy A dtydxA )
8. (c)
I~ p =
dt
⇒ ___!:=Slope of curve
dt
A~, _shown in figure maximum slope represent (c) and
:. a=dt= dtx+2 aummum slope represent (b).
9. (cl) Given, mass of machine gun, M = l Okg
=a( vy; +2v x y) = a( 2xa; +2ayy) =2a 2 [ x; +yy] mass of bullet, m = 20 g = 20 x I o-3 kg
2. (b) By momentum conservation, velocity o~bullet. V= I~ ms- 1• let V be the recoil velocity
m 1u 1 + m2 u2 = m v + m v of gun. using conservation of momentum
1 1 2 2
0 = 3(- v) + 0.01 (600 - v) nmv= MV
2m/s
V::::
Impulse on gun = 3 x 2 = 6 Ns ⇒ 20 x l0- 3 x ~ x l00 = 10V
3. (a) Given, 60
·: n = 180 bullets per minute
Mass of the particle, m = 5 kg ⇒ V= 0.6 m/s
As the particle is at rest, So resultant of F and F t = 20 sec
10. (b)
should be opposite to ~
2 3
CJ-:+' I

Fnct "." ✓F/ + F/ = ✓6 2 + 82 = 10 Assume surface to be frictionless

50m

Fnct 10 Then . 50 = V x 10 ⇒ V = 5 mis

Acceleration, a = - = - = 2m / s2 As v = u + at ⇒ V = 0 + a x 20
m 5
4. (b) Given, mass of each bullets, m = 1Og ⇒ 5 = a x 20 ⇒ a = _!_ m/s 2
4
Speed of each bullets, v = 25 0 m/s
ForceF = nm v I
So. F = ma = 20 x - = 5N
Here n = number ofbullets fired per second 4
F 125 dm 10
:. n = - = - - - - - = 50 11. (b) Force, F = - v = - x 4.5 = 9dyne
mv IO x 10- 3 x 250 dt 5
5. (b) Given, 12. (b) lnitial momentum Pi =0.15x l2(i)

500 Finalmomentu~ Pr =0.15x 12(-i)

mass of body, m = = 0.5kg
1000 l.1Pl= 3.6 kgm/s or, 3.6 = F~t
3.6 = 100 ~t :. ~t = 0.036 sec
velocity, V = I o✓x ⇒ v2 = I OOx
dp dm
dv vdv 13. (c) F=-= V-=I Ox\= lON
2v- = 100 ⇒ a= - =50rn/s 2 dt dt
dx dx
:. Force,F =ma=0 .5x50= 25N F IO
a=-= -= 5 mfs2
6. (a) Mass of particle, m 2
m = 500 g = 0.5 kg
velocity of a particle, vdv kx
I.S ''
-
V
A
= 2ti + )t j
2A 14. (c) a= -
dx
; adx = vdx; J--dx=
m
J vdv
05 4
k ., 6
- dv ,. ,. 2 2 v- -1
4 ⇒ -~r21==.:...-;
a=-=2 i+6tj ⇒ --[1.s 2
-o.s 2 ]= v - -
dt 2m 2 2x2 -
⇒ -3x4=v2-16
- A A

at t = I , a = 2i + 6j
Force acting on the particle, ⇒ v2 = 4 ⇒ V = 2 m/s
15. (cl) Thrust force on rocket is given by
F=ma= o.s(2I + 6]) =I+ 3]
F=I+ x] Hencex=3
Fthrust = ( Vrel . ~7)
Laws of Motion A211

1 Let time t required to rise to its zeni th (v = O) so,

(v,·e/·~ - mg) =ma 0 d
J-=-; J t
= dt [for Hrmx. v = O]
~ 3
500 ( :;') - I 0 x IO = I 0 x 20
3
Vo g+ yv 0

~ dm = ( 60 kg/s)
dt
:. t =
I
Jig tan
_/
lJivJg l) 0

(c) From the Newton's second law of motion,

20. (d) v2 = u2 - 2gh
f = ma
✓u 2 -2gh
⇒ •=>•=~[,f;t)']
or v=
2
Momentum, P = mv = m✓u -2gh

⇒ •=!>~[,f;t)']
u2
At h = 0, P= mu and at h=-,P=O
g . . .
upward direction is positive and downward d1rect1on 1s

⇒ !dv=~ '[[1f; 1 negative.

21. (b) From Newton's second law
)}
dp
-=F=kt
dt
V= Fo[t+-l-(T-t)3]2T
⇒

⇒ v=
M 3T 2

~ {[ 2r + ~ (r-2r) J-[o+ ~, ]}
32 3
0
3
Integrating both sides we get,

J;· dp= fa'

kT 2 /p
ktdt⇒[p1:· = k[ ~ I
⇒ 2p= 2 ⇒ T=2Vk
⇒ V= ~ [4;] R dv R
(b) From the Newton's second law, 22. (a) From F = -2 v(t) =} m- = 2v(t)
t dt t

Jdt = J
F = dp = d(m v) = v(dm) ... (i)
dt dt dt
dv Rdt
Integrating both sides mt2
dM(t) 2
We have given - - = bv (t) ... (ii) R 1
' dt In v = - - :. ln v oc -
Thrust on the satellite, mt t

-v( :7)
23. (a) From question,
2 3 Mass of body, m = 5 kg
F= = -v(bv ) = -bv [Using (i) and (ii)]
Velocity at t = 0,
3 -bv3 u=(6t-2})mls
⇒ F=M(t)a=-bv ⇒ a=-­ Velocity at t = I Os,
M(t)
(d) v=+ 6} mis
H ! (g+kv
2
) = a (acceleration) Force, F=?
. V- U 6 j - (6f - 2j) - 3f + 4 j
u Acceleration, a = - - = ----'----'------'--'- = m, s:?
1 10 5
- 2
Force, F = ma
F =mkv -mg (._.mg and mkv2 act opposite to
each other) -- 5 x -(- 3-i+ -4 }) = (- 3~ J )N
1+ 4~.
5
- f 2
G=-= -(kv +g] 24 . (d) Oiven : momentum p (t) = A[icos(kt) -) sin (kr) ]
m
dv - · dp • •
~ V· - = - [kv 2 + g] (-.-a = V :;,) And, force, F = - = Ak [- isin( kt ) - j cos (kt)]
dh dt ·
0 h

fz-= Jdh ⇒ -I 1n[kv

v-dv Ju =- h Hcre, F.P = O Out F.i~= l'fJCos0
~ 2
+g .'. CllS O - 0 =-~ t) "" 90".
u kv + g o 2k ,,
I lencc, angle between the forc-e momcJllum, 0 -" 90°

~ ..!_
2k
1n[ ku2 +
--
g
g]- - h 2~. (6) Ry law of conservation of li1K-;1r momentum t; = p
➔ ➔

1
⇒ 60 x V-= ( 120 t- 60) " 2 ⇒ 601 '~ J60 ⇒ I'-= 6 m/s
(a) Net acceleration )
dv 26. (12) lmpuh.l' - A I'
dt '=' a == - (g+yv2) ►
- r, - ~ . .: 1111' ( 1111' ) - ~ 1111 ' = 2 X 0.4 X 15 -: 12 Ns
 A212
N

. F . A
2. (c) l
27. (~00) a =-= 10i + 5j
fl/

1>1sp l:11.:cmc111 of the box along x-axis,

I ~ I
r "" - a , t ~ = - x IO x I 00 = 500 m
2 2 mg \
28. (0.005) Arc11 under the F - t graph gives the impulse P sin 30°
imported lo the body.
The magnituJe of total impulse of force on the body from From FBD ofroller
t ~ 4 ,,s to t <=- 16 µs N=mg+ F sin30°
., area (BCDFEB)
=- n.rca of BCFEB + area CDFC
= 700 + 200 X 2I =800N
f'
(inN) 3. (a) T 8-
800
0
600

400

200
D
0 2 4 6 8 10 12 14 16 T
(in ps)
9.

= ..!..(200+ 8()()) X 2 X 10-6 + ..!.. X IQ X 800 X J0- 6 From the free body diagram shown above,
2 2
= 0.00 J + 0.004 = 0.005 Ns Reaction cos e = ✓3g ·: e = 30° Jj = ✓3g
29. T 2 T
False: The forward motion ofrocket
⇒ T=20N
is due to the exhaust gases, thrown
backward not due to surrounding air 4. (b) From FBD oflift,
push ing backwards. T 1 = T 2 + mg, m = mass oflift
Rocket
Here exhaust gases thrown ⇒ T1 -T2 =mg
backwards is action and rocket
moving forward is reaction. ⇒ Tnct = mg . a
Exhaust gases
This phenomenon takes place in the
Action So, {I) is true a=O ia
absence of air as well. 10.
From FBD of person,
30. (b, d) Earth is an accelerated frame and hence, cannot be
N+ma=mg mg g
an inertial frame.
Earth is revolving round the sun and is rotating about its N=mg-ma ⇒ N<mg
own axi s. So, (II) is false.
31. (b) It is easier to pull a heavy object than to push it on a
level ground. This is because the nonnal reaction in the case
5. (b) For 4 kg block
of pulling is less as compared by pushing. (f= µ N). Therefore 4g sin 30° - T = 4a ... (i)
the frictional force is small in case of pulling. For l kg block
T- lgsin30°= la
The magn.i tude of frictional force depends on the nature of
4(T - g sin 30°) = 4a ...(ii)
the tw o s urfaces in contact. But is not the currect
From (i) ~om (ii), we get
explanation of statement- I . 4(T - g sm 30°) = 4g sin 60° - T
32. (b) Cloth can be pulled out without dislodging the dishes
from the table because of inertia. 5T = 20✓3+20
Law of inertia is the Newton's first law of motion. T = (4 ✓3 +4)N
6. (b) For equilibrium condition, m~g 111 1~ :-tnll
For every action there is an equal and opposite reaction. :=

This is Newton's third Jaw of motion. . m2 3 ~

~ Topic-2: Motion of Connected Bodies, Pulley sm8 =;-, =5 \\

~
(\\ , .,. .,, II('

~ & Equilibrium of Forces \\\ \ i ~ ,,~ ' " ''' -" ..,

4
_ . A
2
A dr A

j + 7k ; v = - = 10i +30tj
_ A A
cos O=-
5
...::::--1 () ,11 ,t.1

1. (a) r = 10ti+l5t Normnl force (N) on m1 -' 5g rn:- u

dt
- d~ = 5 x IO x ~ = 40 N
·: F= ma
A ( - · )

a=dt = 30j 5 . ~ pl:>11'·

7, (c) Acceleration of hl cx: k on smooth uidinc;
So net force is along + y direction. a =- g sin H
~
-~- Using, s == ut + 2at
1 2 y component=!
Angle with x-axis: tan 0 = x component l
Ma
A213

2 So, 9 = tan- 1( 1) = 45°

s == _!_ g sin 30°(2)
2 12. (c) For observer in box
When the incline is changed to 45° N
N-"- Ma = Mg
⇒ N= M(g -a)
I . 45o t 2
s == -gsm
2 Ma
As distance tra velled is same ⇒ ~== M(g -a )
4
(½)(4) == ~ t ⇒ 2
t = .J2✓2 ::: 1.68 ⇒ a =-
3g
4
(d) Acceleration in s uch system is given as
8. ➔
( m 2 - m 1)
a = .2..--,,_______;_ g
l.ll..LI.J.J.W1..1JJJ..Jw u. <•) I T•a =o
( m 2+ m1 )

⇒ Jk. = 0,( L - fl ) - lP) g ⇒ L = _!: = L

2 1, L 4 x m,
So. x =- 4 m2
9. (c) Given 1hat ma s" of monkey. m = 50 kg :!T
Accclcrnt1on d ue to gravity. g -= IO m!si
Tension CD ~ 350 N
Given monkey clim b~ downward. accclernt1on of monkey.
u - (i4 111 /Nl
When monke y c ltmbs upward. a<XClcntuon o f monkey.
n =- 5 111/ ~2
(For 11p wnrd) 14 . (d) From free body d iagram 2T
T
T mg mn
⇒ T 111 g t mn - 50{10 + 5) "' 750N 80 - 2T = 8a ...(i)
Rope will hrcnk whi le climbmg upward T - 20 = 4a __ (ii)
(For dow nward) Multiply equat ion (ii) by
a
T - m(g n) - 50(10 - 4) - ) 00N 2 3nd adding with equation
Rope wi ll not brc..,k whi le chmbmg downward (i) WC get
IO. (11) L l'I :, 1 he the acceleration of I 00 kg block 40 10 i80 20
(8 + 8) a = 40 ⇒ a=-=-mls
Fl3D 11f I00 kg block w.r.L ground □T 16 4
F- T N 1 . l 2 2S
IOO a 1 ••• (i) ➔ +- -+ a, Usmg S = - at 2 ⇒ t =-
2 a
F N,
FBD of 20 block w.r.t. 100 kg O.2x2x4 2
=>
10
=t ⇒ t = 0.4 sec
T- 20 g = 20(2) ⇒ T = 40 + 200 T

~ T - 240 ... (ii)

j 2mls 2 15. (a) The free body diagram is
By l.ami's theorem
NI = 20a I .. . (iii)
mg F
FBD of 10 kg block w.r.t. l 00 kg 20g ⇒ --=-
1350 135
~ 2mfs2
⇒ F=mg= lOON 10kg

~ l 0 a, -240 = 10(2)
16. (d)
480
a= 20+ 12+8 = 12 m/s2
::::) 31 = 26 m/s 2 20 kg
480 - T=24a
F= 240- 20(26) = 100 x 26
4 kg
~ F=3 360N 480 -T=24x 12
11.
(a) Le1
~ • addtion force required be =F 480-T=288 T
F+5 i 6~ • •
- 1 + 7j-8j=0
T= 192N
~ F::::i +]. 1r 1=.J12 +1 2 = ✓2
 A214
17. (b)
------------- ----------
From figure, acceleration
of mass m is due to the force T
F 21 (3) Along horizontal
F I + J cos 45° = 2 sin 45°
cose 2 I I
Fj= ✓2- ✓2= ✓2
Tcos0 = ma
Along vertical
Tcos0 f 2 = I sin 45 + 2 sin 45
⇒ a=-- ... (i) ma+-- x ~ x --+m
m
3 Fj I
also.F=2Tsin0 ⇒ T= - .
F F2 = 3 sin 45 = ✓2 so, F2 =3. so, X == 3
2 SID 0
Putting this value of T in eqn. (i) 22. (36) We have, acceleration of system as
F ) cos0 6mg Jg
a== ( 2sin0 -;;;- a=--=-
IOm 5
F F X
=--- taking 8, 9. IO together
2mtan 0 2m ✓0 2 -x2
18. (c) Before the string is cut the tensi on T has to hold 3g 3x2x3xl0
T = 3ma = Jm x - = = 36N
both the masses 2m and m therefore. 5 5
T = Jmg
23. (12) Let draw FBD of block clearly for equilibrium
When the string is cut, the mass m is a freely ful ling body and
its acceleration = acceleration due to gravity = g. F cos 60° = mg sin 60°
For mass 2m. j ust after the string is
T rema ins 3mg because of the extension
3mg C
UL T rr ⇒ -=
F
tan60°
F cos 60°
;if

.
o f stnng. m 2m mg
F
⇒ ✓x = ✓3
g
3mg - 2mg = 2m x a ⇒ =a mg 2mg
19. (cl) At equilibrium T = Mg
2 0.2 x l0
F.B.D. of pulley ⇒ J; =2✓3 ⇒ x = 12
T
24. (30)

.Ilg
F 1 = (m + M)g
The resultant force on pulley
3o_o_+_m_a
From figure, tan (30° + 0) = _m..:::g_s_in_
F= JFj 2 +T 2 = [J (m + M ) 2
+M 2 ]g _ mg cos 30°
As pulley is on rest. So force applied by clamp should be l
equal to 'F and opposite to it. tan0+ -
20. ⇒ tan(3oo+0)= 5+10 = 1+2 ⇒ .[3 [3
(c) The tension in both strings will be same due to 5 ✓3
symmetry. fj l _ _!__ t118
.[3
⇒✓3 tan8+1=3- ✓3tan8
From figure,
T = mg ⇒ 2✓3tan8=2 ⇒ tan8= ~ 0=30°
➔
2S. (82) P makes angle of 35° with AC
A
So, component along AC = I 00 cos 35 = 81 .9 N :::: 8~ 1"
mg 26.
mg False : From FBD, shown in case (a) for mass m
For equilibrium T - mg == ma ... (i)
For mass 2m
✓2 mg= Tcos 0+Tcos0 = 2Tcos 0
Case (a)
✓2 mg= 2(mg)cos0 2mg - T :::; 2ma ... (ii) r. r
cos0= Ji ⇒
I
e = 45°
From (i) and (ii) at
(l == g/3
+J
... (iii) mg
2 m!!
Laws of Motion A215
case (b) T - mg = ma'
Let 'x ' be the extension of the spring at a certain instant
2mg - mg = ma'
::::> 2T - KJc = 2Ma I N
[ ·.· T = 2mg] 2Mg - T = 2Ma 3 ~a
:. a' = g T T=2mg Mg - T = Ma 2 Kx 2~
... (iv)
Hence, from eq (iii) & (iv) a < a' On solving we get,
mg F=2mg 4g 3kx - 3K ( 8mg) 2Mg ... (i)
(a, b, d) a, = 7-14M = 14M x - 3K

r T
Comparing it with a = - ci(x - x0 )

L
(1)2= - -
3k
w=
f3k
'JT4M
1 w
L
and T = - - + - -
14M
4Mg 2kx
... (ii)
7 7
For a 1 = 0 (Maximum extension of spring) we have from (i)

Since OQ = OP 4g 3kx
:. L P = L Q = 45° 7-14M = O
At eq uilibrium. about point 'O'
3kx 8Mg
T 4g=- x= - -
R,_+ r;: = W + aW .. . (i)
2M 3k
. ✓2
16Mg
T ••
X
0
=2x= -3k-
and Rx = ✓2 ... (ii)
Torque about point 'O' is zero For x= x: =¼(163~g) = 4~g
So, w½+aWL = 1l :. T = ✓2 ( : +aw) . .. (iii) .
Fromeqn. (t)a = - - - - x - - = -
1 7 14M 3x 7
4g 3k 4Mg 2g

.
Rx= ✓2
T
= (W +aW ) At x = ·1 particle is at mean position and its velocity= Aw
2
TI1ereforc for a = 0.5
= x 0 ✓ 3k = 8Mg ✓ 3k
w
R,. =- +aW =-+ 0.5W
w 2 14M 3k 14M
" 2 2 29. (b) Here from figure, AN= x (= constant as pulley A and
or Rr = W
i.e., the horizontal component of reaction force at, 0. Rx = Bare fixed), NO= z. Then velocity of mass m = dz . Also
Wfora = 0.5 dt '
d f.
Now torque about point P let DA = f. then - =U
L W dt
TJ,L = W -
2
⇒ R.
,I
= -2 FromMNO
The verti ca l component of reaction force at O does not x2 + z2 = f.2
depend on a
... (i) u
Differentiating Q
As per question, rope can sustain a maximum tension of equation (i) w.r.t to t
2./i W dz df
O+ 2z - = Udt- ⇒ zv = eU
2✓2w = ✓2(: +aw) ⇒ 2 =½+a dt M

.. Q =:: -
3 ⇒ VM = -; LJ = z~ f = C~ 0
(-.- COS 0 = 1)
2 30. I = Mass of unit length of wire - 0.2 kg/m.
(d) According to a,
➔
co ns t rai nt relation Im
from figure,

,- ~
a - a2 + a3
2
~ a2 + a = 2a
~ 3 I m 1 = I \l kg
a)+ a = a +
~ •- J I 31 m. = ~.9kg
-... a, - a - i"J 111 1g ·• Ail! •
......, 0 . 3 - a2 - a I
Pllon (d) is correct ~Al