INSIGHT PHYSICS CLASSES

8/14/2019 INSIGHT PHYSICS CLASSES

1/19

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

Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 1

Ph: 040 64606657, 9391326657. www.asifiitphysics.vriti.com
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2/19

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




3/19

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.




4/19

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




5/19

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

_

_

=




6/19

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




7/19

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




8/19

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)




9/19

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.




10/19

( )

+

=

=

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=




11/19

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




12/19

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




13/19

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




14/19

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)




15/19

( )( )

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+

=




16/19

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




17/19

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




18/19

^^

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 ==




19/19

( ) mxx 2.6715

1424 11 =

=

sx

t 8

5

314

2.67

53cos14 01

1 =

==

28.



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