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Mohammed
Asif
Name :
Roll No. :
Topic :
Ph : 9391326657, 64606657
Multiple choice questions(only one is correct)
1. The vertical displacement of cylinder A in metres is given by4
2t
y = where t is in seconds. B
will hit the ground at t =
a) s2
1b) s
2
1c) 1 s d) s2
2. A small ring C is made to move along the rotating rod AB between r = r0 +d and r = r0 d, and
its equation given by r = r0 + d T
t2sin
, where t is the time counted form the instant the ringpasses the position r = r0 and T is the period of oscillation. Simultaneously the rod rotates about
the vertical axis through end A with a constant angular velocity 0.
The value of r for which the radial (r-direction) acceleration is zero is
a)
+
=20
21
1
0
T
rr
b)
=20
21
1
0
T
rr
c)
+=
2
021
0
T
rr d)
=
2
021
0
Trr
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3. A particle (mass 1/ 2 kg) initially (at t = 0) rest is acted upon by four forces and for the time
intervals indicated.
^^^
1kjiF ++=
N during (t = 0 to t = 1s)
^^^
2232 kjiF +=
N for (0, 2)
^^^
332kjiF
+=N for (1, 2)
^^^
432 kjiF +=
N for (2, 3)
The kinetic energy of the particle at t = 3s is
a) 9J b) 18 J c) 27 J d) 36 J
4. A projectile of mass kg21 is projected at t = 0 form origin O on horizontal ground (treated as
xy plane with vertical upwards as +z axis) with an initial impulse
sNkji^^^
5.75.76 ++. At t = 1s, the
projectile receives a furter impulse of^
6 i Ns. The range of the projectile is
a) 48m b) 75m c) 96m d) 108m
5. A projectile projected at t =0 with an initial velocity 20ms-1 at an angle 370 to horizontal collides
elastically with a fixed inclined plane at t = 2s and returns to the initial point. The angle of the
inclined plane is
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a)2
1sin 1 b)
2
1cos 1 c)
2
1tan 1 d) 2tan 1
6. A train is moving with constant acceleration along a straight track. The front of the train has a
velocity 20ms-1 when it moves past a signal post, a passenger at the middle compartment has a
velocity of 320 ms-1 when he moves past the same signal post. When the rear of the train
moves past the same signal post, it will have a velocity of (ms-1)
a) 420 b) 520 c) 620 d) 820
7. The retardation a of a particle moving along a straight line is given by a = kS where S is
displacement, k is constant. S vs time, t graph is
a) b) c) d)
8. Two ships A and B are 10km apart with ship A due north of B and sailing west at 40kmph. Ship
B is sailing north at 20kmph. They will be closest to each other in (minutes)
a) 6 b) 9 c) 12 d) 18
9. Three vectors
CBA ,, satisfy the relation 0.0. ==
CAandBA . Then
a) 0. =
CB b) 0=
+
CBA c) 0=
CBA d) 0. =
CBA
10. A boat which has a speed of 10km/hr in still water crosses a river of width 0.5km along the
shortest possible path in 6 minutes. A man whose swimming speed is 9km/hr in still water
crosses the river along the shortest possible path in
a) min39 b) min36 c) min65 d) 8 min
11. A projectiles initial velocity is u and angle of projection is . When it makes an angle with
vertical, its speed is
a) sincosu b) ecu coscos c) coscosu d) seccosu
12. For angles of projections ( ) ( ) + 4545 and , the ranges of the projectiles are R1 and R2 andthe maximum heights reached are H1 and H2 respectively, then H2 H1 is equal to
a) 2tan2
21 RR +b) 2tan
421 RR +
c)2tan2
21 RR +d)
2tan4
21 RR +
13. Two particles A and B which are 100m. apart at time (t = 0) are moving in the same directionon a straight line. The velocity and deceleration of A are 20m/s and 2m/s 2 and velocity and
acceleration of B are 2.5m/s and 0.75m/s2 at (t = 0). Find the time at which B crosses A and also
the vrel verses t graph where vrel is the relative velocity of A with respect to B.
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a)
sec
5.5
342535 +
b)
sec
5.5
342535 +
c) 20 sec d) 18 sec
14. At t = 0 particle 1 starts form rest at A, moves along circle of radius
1metres witeh constant
tangential acceleration, reaches B at t = 1s. Distance AB =
2meters. Particle 2 starts at C at
t = 0, moves along straight line with constant acceleration, reaches A at t = 1s AC = metres.Magnitude of acceleration of 1 relative to 2, at t = 1 is (ms-2) (treat 102 = )a) 364 b) 464 c) 564 d) 664
15. Figure shows displacement Vs time for a particle. Which of the following statements is true?
a) It moves along a straight line for t < 0
b) It moves along curved path for t > 0
c) It moves with constant acceleration for t > 0
d) It has zero velocity for t < 0
Assertion & Reasoning Type
Each question consists of two statements: one Assertion (A) and the other is Reason (R). You
have 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 Ac) A is true but R is false d) A is false but R is true
16. Assertion(A): In projectile motion with a given velocity(u) it is not possible that the body hit a
given point on the ground in more than one direction
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Reason(R): In projectile motion the velocity and acceleration vector will never be in the same
Direction
17. Assertion(A): When two bodies are moving in tow concentric circles with same angular velocity then the body A I stationary with respect to B
Reason(R): The distance between them is constant.
18. Assertion(A): In motion of a body the graph between distance and time, the graph should nothave negative slop with time axis.
Reason(R): In motion of a body the distance connot decreases.
19. Assertion(A): A very small block of mass m attached to the end of a light metal rod of negligible
mass which is free to rotate about the other end in a vertical circular motion. Its
velocity at the lowest point is just sufficient to complete the circle, then at the
instant block is at the highest point tension in rod is zero.
Reason(R): In vertical circular motion of a point mass attached to a string at the topmost point
0,, min
2
==+ Tvfarsor
mvmgT
grv =
20. Assertion(A): The path of a projectile as seen form another projectile is a parabola.
Reason(R): Both are moving with same acceleration so_
1
_
2
__
0 uuuanda relrel ==
For question nos. 21 to 25
Two columns are given in each question. Match the elements of Column-I with Column-II
21. If during the motion of a particle, x is used for distance, u is for speed,_
s is for displacement
and_
v is for velocity (i and a are used to show instantaneous value and average value) then
match following:
Column-I Column-II
i) Motion along a circle with constant tangential acceleration p)_
sx =
ii) Motion along a straight line with retardation given by__
vka = q)__
ii vu =
iii) Motion along a straight line with constant acceleration r) avav vu_
=
iv) Projectile Motion s)dt
vd
dt
vdi
i
_
_
=
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t)dt
vd
dt
vdi
i
_
_
=
22. Figure shows the potion (x) of a particle as function of time (t) under a variable force F, then
match the following:
Column-I Column-II
i) Velocity is positive and maximum at p) A
ii) Velocity is negative and magnitude decreasing at q) B
iii) Acceleration is positive and maximum at r) C
iv) Acceleration is positive and its magnitude is increasing at s) D
v) Velocity is positive and acceleration is negative at t) E
23. For particles are moving in XY plane such that at t = 0, (Here_
ABV . Velocity of w.r.t B, vA:
actual velocity of A w.r.t each)Column-I Column-II
i) ( ) ( ) ( )jiVjiVjiV BDCAAB ==+=2
5,25,24
___
then, magnitude of_
DCV is p)
2212 +
ii) If
=
2
jivD then
_
Av q) 17
iii) In time st2
1= due to some external forces velocity of each body is oriented along X-axis
and___
,, ACCDD VVV are in positive X direction with same magnitude as above, then find out
DBV_
such that__
, AB vandvare same along X axis. r) 2212 +
iv) The magnitude of average acceleration of A along X-axis is s) 10
t) 7
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24. Find out the trajectory of a particle if the condition of motion is
Column-I Column-II
i) Its X and Y co-ordinates as a function of time are
( ) 2/124 CttY += CtX += 2 p) Exponential
ii) ==dt
vdbut
dt
vdv
dt
d__
_
0,0 constant q) Hyperbola
iii) Tangential and normal accelerations are variable butthere resultant is always constant r) Straight line
iv) Velocity along X-axis Vx = constant but Vy = kY (where k is a constant and Y is coordinate
along Y axis) with an intial value of 0= tatVV XYO s) Circle
v) Velocity is positive and acceleration is negative at t) parabola
25. The angle of inclination of an inclined plane is 450
Column-I Column-II
i) A particle is thrown with an angle with an inclined plane
such that it strikes the inclined plane normally then is p)
22
1tan 1
ii) In the previous part if it strikes the inclined plane horizontally
then ( 045+ is q)2
45 0
iii) A particle is just thrown horizontally form the top of the inclined
plane and its range is ng
u
.
2
then ( )n1
cot
r)
( )4tan 1
iv)If 21 and are two angles with the inclined for which range is same then s) D
Write the final answer to each question in this section in the column provided.
26. The car A has a forward speed of 36km/hr and is acceleration at 1ms-2. Determine the velocity
of the car relative to observer B at time t = 15s who rides in a non-rotating chair on the giant
wheel. The angular velocity of the giant wheel is constant at1
10
srad
. The radius of the
giant wheel is .50
m
The diagram shows the position at t = 0.
(If your answer in vector form is xi +yi then value of (x. y) is
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27. A stone projected form top of a cliff at t =0 with velocity 14ms-1 at an angle 530 above horizontal
collides at t =t1s with another stone projected at t = 2s form the same point with same velocity
but at different angle. Determine t1. (g = 10ms-2)
28. A man traveling in train moving with a speed of 80km/hr fires at an object moving away form
the train at right angle to it with a speed of 60km/hr. The line connecting the object and the
man makes an angle of 450
to the train at the time of shooting. If the velocity of bullet is700km/hr, find the angle at which he should aim in order to hit the object.
29. A gun is fired form a moving platform which is oscillating whose oscillation is given by a is the
amplitude of the oscillation. If maximum and minimum range of the projectile are R1 and R2then find elevation of the gun from horizontal in terms of given quantities assume that the
amplitude of platform is negligible with respect to these rang and solve this for values R1=100m,
R2=800m, sec/7,3
1radma == and g = 9.8m/s2 {in the oscillation as given above the body
move to and for about there equilibrium position and its velocity is maximum at equilibrium
position equal to a}
30. A balloon starts rising form the ground at time (t = 0). The ascending rate is constant and equal
to 20m/sec die to wind the balloon gathers the horizontal velocity component v x =^
iyK where
K is a constant and is the height of ascend. An aeroplane is moving with velocity 80m/sec in the
same vertical plane. At time (t = 0) the horizontal distance of plane and balloon is 3200 m as
shown in the figure and a bomb is dropped form the plane and it collide with balloon some
where in space, K = 4sec-1, g = 10m/sec2 then find out value of (x+y) where x, and y are the
coordinates of point of collision. (take x-axis as horizontal and y-axis as vertical and origin at
the starting point of balloon)
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KEY1) b 2) a 3) c 4) b 5) d
6) b 7) c 8) a 9) c 10) c
11) b 12) b 13) c 14) a 15) d
16) d 17) d 18) a 19) d 20) d
21) (i-q);
(ii-p,q,r,s);
(iii-p,q,r,s,t);
(iv-q).
22) (i-t);
(ii-r);
(iii-s);
(iv-r); (v-p)
23) (i-p);
(ii-r);
(iii-q);
(iv-t).
24) (i-q);
(ii-s);
(iii-t);
(iv-p).
25) (i-s);
(ii-r);
(iii-p);
(iv-q).
26) 63 27) 8s 28) 53
0
29) 45
0
30) 4200
Solutions
1. 22
2 42
1
42
1 === msaat
ta BAA
sttm2
1''.4.
2
11 2 ==
2. It is clear that the ring C, at any instant has two types motions a circular motion with angular
velocity 0 and radius say r and SHM with equilibrium position r0, amplitude d, and angular
frequency T
2=
. Acceleration will be zero if centripetal acceleration is equal and opposite toSHM acceleration
(SHM acceleration directed towards equilibrium position and hence will be opposite to
direction of centripetal acceleration.
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( )
+
=
=
2
0
0
2
0
2
0
21
1.
2
T
rxxT
xr
+
==2
0
00
21
1
Trxrr
3.
^^^^
333. kjitF ++=
( )^^^^
33303 kjita tp +++==( )
( )J
m
ptatKE 27
2
12
27
23
22
=
===.
4.
^^^
111 2 kjiu ++=
Z = 0 at statgtt 302
115 2 ==
At t =3s,
x = 12 x 1 x (12 + 12) x 2 = 60m
y = 15 x 3 = 45m
mRange 75456022=+=
5. Collision takes place while descending time of flight = sg
u4.2
sin2=
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Returns to initial point collides normallyAt t = 2s,
Velocity = 58816 22 =+ and
82012sin==
gtu
2
1tan 1=
( )2tan2
1tan
22
11 =
==
6. Let be acceleration and lbe length of train. Then using V2 = u2 + 2as
( ) ( ) 80012..220320 2
2
=+=
aa
( ) 20002220 222 =+= VaV
7. Skdt
dvSka == .
SdSkvdvSkdt
Sd
ds
dv== .
At S =0, v = say2
2uCu =
222 uSkv +=
2222 Skudt
SdSkuv ==
dt
Skuk
xddt
ksu
Sd=
=
2
222
Cu
kS
kt += 1sin1
At t = 0, S = 0 0=C
u
kS
kt 1sin
1 =
tkk
uStk
u
kSsinsin ==
Sinusoidal.
8. Velocity vector diagram
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Displacement diagram
BA is the closest approach
Time520
sin10'
/
==
BAV
AA
min61.0520
1.
5
110 ===
hr
9. Given
CAandBA rr0// =
CBACBtoA
10. Velocity vector diagram
kmphVr
75=
Now
kmphvm 6=
min65min606
5.0
6
5.0===
kmph
mTime
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11. ( ) cossincos90cos uvuv ==
ecu
uv coscos
sin
cos ==
12. Clearly( )g
uRR
==
452sin2
21
( )12
2cos 212
+
==RR
g
u
( ) ( )[ ] += 45sin45sin2
222
12g
uHH
( )( )[ ] +++= 45sin45sin45sin45sin2
2
g
u
sin45cos2.cos45sin2.2
2
g
u=
2sin.2.
cossin2
12.
2
12.
2
22
g
u
g
u ==
2cos
2sin.
2
1.2cos
2
2
=
g
u
2tan.4
2tan2
1.
2
2121 RRRR +=+=
13. If you solve it by formula
OBOA vvrelUS ==_
0
__
OBOA vvrelU =_
0
_
___
0 OBOA aarela ==Then your answer is wrong like (a) or (b).
In this case before B crosses A the particle A stops at time
sec102
20==t
So we can use this formula only upto 10th sec the distance traveled by A before it stops is 100m
So for B the time to travel 200m is
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200 =2
.75.0.2
15.2 tt+
So t = 20sec
Now till t = 10sec the 75.2_
=rela and after this it will be only2
_
/75.0 smarel = .
At the point of crossing the velocity of B = 17.5m/s and 0_
=Av so 5.17_
=relvSo ans is graph (c)
14. Particle A:
Tangential acceleration aT = 2ms-2
[ ] 11 2,1,1,0
===== mxvandtSu tat
Centripetal acceleration (at t = 1)
4
2
== Rv
an
Particle A: 2=a[ ]1,,0 === tSu
^^^^^
2/1
_
26224 jiijia += +=
( ) 2222/1_
36426=+= msa
15. Obvious.
16. It is possible to hit a point with 2 different angle and velocity and acceleration are not parallel.
Show assertion is incorrect but reason is true. So ans is (d)
17. In this case Vrel = w(r2 r1)
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( )( )
wrr
rrww =
=12
12'
So motion of A with respect to B is circular
(d) is correct answer
18. Distance can not be decreased with increases in time. So slope could not be negatives
so ans. Is (a)
19. The rod may have negative (compressive) tension so minimum velocity at topmost point is zero.
And T = -mg.
20. Obvious Ans is (d)
21. In case of motion of a body with some retardation its velocity decreases so
dt
vd
_
is negative whiledt
vd_
is positive
22. The velocity time graph and the points are as shown the figure. So ans. are clear.
23. (i) ( )1.................____
ACBADBDC vvvv ++=
J2
23
12
7 +=
17_
=DCv
(ii) DBDABA vvvv
____
++=
jijS
iS
ji
2
1
2
1
2222++=
ji2
14
2
2+
=
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10_
=Av
(iii) At st2
1=
iaviviv ACDCD 1,17,1___
===
0___
== ABBA vvv
So form equation (1)
7_
=DBv
(iv) ( ) iiivtA 670
_
=++=
( )22122/1
26+=
+=
iiax
24. (i) CCxxCCxCtty +++=+= 222242
++= CxCxy2
12222
Which is a hyperbola(ii) Obvious only in circular motion
(iii) Net acceleration is constant so parabola. Example projectile motion.
(iv) kYvy =
kYdt
dY =
== kteYYkdtdtdY
0
kk
v
k
vY x
y == 00 and
xv
kx
x
eYYvxt 0; ==
25. (i)045,
cos
sin2 ==
g
uTf
At last point velocity along inclined plane is 0
tan2cot
cos
sin2sincos0 =
=
g
ugu
2
1tan =
( )2cot21tan 11 =
=
ii) In this case the condition is( )2tantan4 +=
( ) 01 454tan = iii) In this case range is
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22
secsin2
g
u
02
452.2
12 == forg
u
22222
== ng
u
nn
1
tancot
11
=
iv)2
45 0
26. ( )^
1 5 ita taA
==
( )
^^^
21 51 01 5 ijita tVA =+==
1550
10
== msR
Velocity of B at an angle to horizontal
( )
++
=
^^
c o5s i n5 iiVA
(tan genital)
Putting 5.1370
0 +=+= t radian0000 5330727037 ==+=
8.0sin6.0cos =+= and
15= tAt
Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 17
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^^
34 jiVA ++
( )
= BABA VVtatV 15/
^^^^^
32342 5 jijii =So (x, y) = -63
27. Let point of projection be (0, 0).
Let collision point be (x1, y1)
3
453tan,
5
353cos 00 ==
Using equation of trajectory,
53cos14
..
2
153tan
22
2
10
11
xgxy =
9
25..
14
5
3
4.
2
1
211
xxy = (1)
And for the second stone
2
2
1
211 cos.
14
5tan
xxy = ..(2)
Equating (1) & (2)
=
+
tan
3
4
cos
1
3
5
cos
1
3
5.
14
51
2
12xx ..(3)
Time of flight till collision, t1 is given by
0
11
53cos14
xt = and for the second stone,
( )cos14
2 12x
t =
2cos
1
3
5
14
1
=
x
.(4)
Putting (4) in (3) 6.08.0sin = or
-0.8 does not satisfy (5)
Hence ( ) 01 376.0sin ==
Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 18
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( ) mxx 2.6715
1424 11 =
=
sx
t 8
5
314
2.67
53cos14 01
1 =
==
28.
Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 19
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