Ch 3 Kinematics

Chapter # 3

Rest & Motion : Kinematics

[1]

Objective - I 1.

Sol.

At motor car is going due north at a speed of 50 km/h. It makes a 90o left turn without changing the speed. The change in the velocity of the car is about (A) 50 km/h towards west (B*) 70 km/h towards south-west (C) 70 km/h towards north-west (D) zero ,d eksVj dkj mÙkj dh vksj 50 fdeh-/?kaVk dh pky ls tk jgh gSA ;g viuh pky vifjofrZr j[krs gq, cka;h vksj 90° dks.k ls ?kwe

tkrh gSA dkj ds osx esa ifjorZu gS & (A) 50 fdeh0/?kaVk] if'kpe dh vksj (C) 70 fdeh0/?kaVk] mRrj&if'pe dh vksj B

(B*) 70 fdeh0/?kaVk nf{k.k&if'pe dh vksj (D) 'kwU;

Change in velocity of the car = Vf – Vi

=

502  502

towards south-west = 50 2 km/h = 70 km/h towards south-west 2.

Fig. shows the displacement time graph of a particle moving on the X-axis.

(A) the particle is continuously going in positive x direction (B) the particle is at rest (C) the velocity increases up to a time to, and then becomes constant. (D*) the particle moves at a constant velocity up to a time to, and then stops. X-v{k ds vuqf n'k xfr'khy d.k dk foLFkkiu≤ ys[kkfp=k] fp=k es aiznf'kZr gS &

Sol.

(A) d.k fujUrj /kukRed x fn'kk esa xfr'khy gSA (B) d.k fojkekoLFkk esa gSA (C) le; to rd osx c<+rk gS rRi'pkr~ fu;r gks tkrk gSA (D*) d.k le; to rd fu;r osx ls xfr'khy jgrk gS] rRi'pkr~ fojkekoLFkk esa vk tkrk gSA

D Slope of "x-t" graph gives velocity Here (t = 0 to t = t0) slope is constant so we can say that velocity is constant. Than aftrer to, slope is zero. So velocity is zero. So velocity is zero.  velocity zero means particle stops.

manishkumarphysics.in

Chapter # 3 3.

Sol.

4.


[2]

A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let x A and x B be the magnitude of displacements in the first 10 seconds and the next 10 seconds. (A) x A < x B (B) x A = x B (C) x A > x B (D*) the information is insufficient to decide the relation of x A with x B. le; t = 0 ij d.k dk osx iwoZ dh vksj u gSA bldk Roj.k fu;r ,oa if'pe dh vksj gSA ekukfd izFke 10 lsd.M rFkk mlls vxys 10 lsd.M esa foLFkkiu ds ifjek.k Øe'k% xA rFkk xB gS & (A) x A < x B (B) x A = x B (C) x A > x B (D*) x A rFkk x B esa laca/k fuf'pr djus ds fy, lwpuk vi;kZI r gSA

D W E   a= constant u V= u + at Some time after velocity is zero O = u – at t = u/a then after particle is return back for this information, we not decided that particle is turn back before 10 sec or after 10 sec. A person travelling on a straight line moves with a uniform velocity 1 for some time and with uniform velocity 2 for the next equal time. The average velocity  is given by

2 1 1 1 1 1 1   2   (B)  1  2 (C)  (D)   2 2  2 2 2 ,d O;fDr ,d ljy js[kk ds vuqfn'k dqN le; ds fy, ,dleku osx 1 ls xfr djrk gS] rRi'pkr~ mrus gh le; ds fy, ,dleku osx 2 ls xfr djrk gSA vkSlr osx  dk eku gS & (A*)  

(A*)   Sol.

1   2 2

(B)  1  2

(C)

2 1 1    2 2

(D)

1 1 1    2 2

A S1 = u1t S2 = u2t Total displacement = S  S1 + S2 S = u1t + u2t = (u1 + u2) t Total time = t + t = 2t Total displacement u 1  u 2 t u 1  u 2 average velocity = Total time taken = = 2t 2

5.

A person travelling on a straight line moves with a uniform velocity 1 for a distance x and with a uniform velocity 2 for the next equal distance. The average velocity is given by

1   2 2 1 1 1 1 1 (B)  1  2 (C*)      (D)      2 2 2 2 2 ,d O;fDr ,d ljy js[kk ds vuqfn'k x nwjh rd ,d leku osx 1 ls xfr djrk gS rFkk mlls vkxs leku nwjh fu;r osx 2 ls r; djrk gSA vkSlr osx dk eku gS & (A)  

1   2 (B)  1  2 2 Total displacement = 2x Total time taken = t1 + t2 (A)  

Sol.

2 1 1 (C*)      2 2

x x = v v 1 2


1 1 1 (D)      2 2

Chapter # 3


[3]

Total displacement 2x average velocity = Total time taken = x x  v1 v 2

v=

6.

Sol.

7.

2 1 1  v1 v 2



1 1 2 = v v v 1 2

A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is (A) a upward (B) (g-a) upward (C) (g-a) downward (D*) g downward Åij dh vksj a Roj.k ls xfr'khy fy¶V ls ,d iRFkj NksM+k tkrk gSA eqDr djus ds i'pkr~ iRFkj dk Roj.k gS & (A) a Åij dh vksj (B) (g-a) Åij dh vksj (C) (g-a) uhps dh vksj (D*) g uhps dh vksj

D Stone is released from an elevator The acceleration of the stone after the release is aSe = -g = g downward

A person standing neat the edge of the top of a building throws two balls A and B. The ball A is thrown vertically downward with the same speed. The ball A hits the ground with a speed A and the ball B hits the ground with a speed B. We have : ,d Hkou ds Åij fdukjs ij [kM+k O;fDr nks xsans A ,oa B Qsadrk gSA xsan A m/okZ/kj Åij dh vksj rFkk xsan B m/okZ/kj uhps dh vksj leku pky ls Qsadh x;h gSA xsan A tehu ls A pky ls Vdjkrh gS rFkk xsan B tehu ls B pky ls Vdjkrh gSA ge dg

ldrs gS fd &

Sol.

8.

(A) A > B (B) A < B (*C) A = B (D) the relation between A and B depends on height of the building above the ground. vA rFkk vB esa lac a/k Hkou dh tehu ls Å¡pkbZ ij fuHkZj djrk gSA

C After come back at level 1 velocity of the ball 'A' is same and equal to ball 'B'. v2 = u2 + 2as Ball A+ Ball B travels the same height so we can say that uB = uA In a projectile motion the velocity ç{ksi.k xfr esa osx (A) is always perpendicular to the acceleration

ges'kk Roj.k ds yEcor~ gksrk gSA

(B) is never perpendicular to the acceleration

dHkh Hkh Roj.k ds yEcor~ ugha gks ldrkA

(C*) is perpendicular to the acceleration for one instant only

,d {k.k ds fy, Roj.k ds yEcor~ gksrk gSA

(D) is perpendicular to the acceleration for two instants. Sol.

nks ckj Roj.k ds yEcor~ gksrk gSA C

In a projectile motion at maximum height, y component of the velocity is zero. Only instant velocity is perpendicular to the acceleration.


Chapter # 3 9.

Sol.


[4]

Two bullets are fired simultaneously, horizontally and with different speed from the same place. Which bullet will hit the ground first? (A) the faster one (B) the slower one (C*) both will reach simultaneously (D) depend on the masses.

{kSfrt fn'kk esa ,d gh LFkku ls ,d lkFk nks xksfy;ka fHkUu&fHkUu pkyksa ls nkxh x;h gSA dkSulh xksyh tehu ij igys igqapsxh & (A) rst xfrokyh (B) /kheh xfr okyh (C*) nksuksa ,d lkFk gh igqapsxh (D) muds nzO;ekuksa ij fuHkZj djsxkA C

Both bullet will hit the ground simultaneously because downward acceleration of the both bullet is same & equal to g. t fired at same place. S = ut + 1/2 at2 Sy = uy + 1/2 ay + t2 Sy = O + 1/2 gt2 t= 10.

2S y g

.

The range of a projectile fired at an angle of 15o is 50 m. If it is fired with the same speed at an angle of 45o, its range will be (A) 25 m (B) 37 m (C) 50 m (D*) 100 m 15o dks.k ij iz{ksfir iz{ksI; dh ijkl 50 eh- gSA ;fn bldks leku pky ls 45o dks.k ij iz{ksfir fd;k tk;s] rks bldh ijkl gksxh

& Sol.

(A) 25 eh-

D

(B) 37 eh-

(C) 50 eh-

(D*) 100 eh-

u 2 sin 2 R= g u 2 sin 30º 50 = g

.......... (1)

u2 = 100 .......... (2) g Given  fired with the same speed atom angle of 45º R' =

u 2 sin 2  45º u 2 = sin90º g g

{ sin90º = 1}

u2 R' = = 10 m g 11.

Two projectiles A and B are projected with angle of projection 15o for the projectile A and 45o for the projection B. If RA and RB be the horizontal range for the two projectiles, then (A) RA < RB (B) RA = RB (C) RA > RB (D*) the information is insufficient to decide the relation of RA with RB. iz{ksI; A, 15o dks.k ij rFkk iz{ksI; B, 45o dks.k ij iz{ksfir fd;s x;s gSA ;fn RA rFkk RB bu nksuksa dh {ksfrt ijkl gS] rks & (A) RA < RB (B) RA = RB (C) RA > RB (D*) RA rFkk RB ds e/; lac a/k lqfuf'pr djus ds fy;s lwpuk vi;kZI r gSA


Chapter # 3 Sol.


[5]

D

u 2 sin 2 R= g angle's are given but not given the information about 'u'. 12.

Sol.

A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction. (A*) due north (B) 30o east of north (C) 30o north of west (D) 60o east of north ,d unh if'pe ls iwoZ dh vksj 5 eh-@fefuV dh pky ls cg jgh gSA unh ds nf{k.kh fdukjs ij fLFkr ,d O;fDr tks fLFkj ikuh esa 10 ehVj@fefuV dh pky ls rSj ldrk gS] U;wure le; esa unh ikj djuk pkgrk gSA mldks fuEu fn'kk esa rSjuk pkfg;s & (A*) mÙkj dh vksj (B) 30o mÙkj ls iwoZ dh vksj o (C) 30 mÙkj ls if'pe dh vksj (D) 60o iwoZ ls mÙkj dh vksj

A

d t= V br cos 

tmin  (cos ) max mean =  = 0º d t= V br

So he should swim in a north direction. 13.

In the arrangement shown in fig. the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed fp=k esa iznf'kZr O;oLFkk esa] ,d vrU; Mksjh ds fljs P rFkk Q ,d leku pky ls uhps dh vksj xfr'khy gSA f?kjuh A rFkk B dh fLFkfr fu;r gSA nzO;eku M dh] Åij dh vksj pky gS &

(A) 2u cos Sol.

(B*) u/cos

(C) 2u/cos

(D) ucos

B

Along the string velocity of block M is same the velocity of the each & every particle of the string. So 2vcos = 2u v = u/cos always take the component of the resultant.


Chapter # 3


[6]

OBJECTIVE - II 1.

Sol.

Consider the motion of the trip of the minute hand of a clock. In one hour (A*) the displacement is zero (B) the distance covered is zero (C) the average speed is zero (D*) the average velocity is zero

?kM+h dh feuV okyh lqbZ dh uksd dh xfr ij fopkj dfj;sA ,d ?kaVs esa & (A*) foLFkkiu 'kwU; gSA (B) r; dh xbZ nwjh 'kwU; gSA (C) vkSlr pky 'kwU; gSA (D*) vkSlr osx 'kwU; gSA AD Minute hand complete one rotation in 1 hour. So Displacement is zero. Distance = 2r 0 Distance average speed = Total time taken  0 Displacement average velocity = Total time taken = 0

2.

Sol.

3.

A particle moves along the X-axis as x = ut(t–2s) + a(t–2s) 2 (A) the initial velocity of the particle is u (C) the acceleration of the particle is 2a X-v{k dh vksj xfr'khy ,d d.k ds fy;s x = ut(t–2s) + a(t–2s) 2 (A) d.k dk vkjfEHkd osx u gSA (C) d.k dk Roj.k 2a gSA at t = 2 second x = 0 i.e. particle is at origin.

(B) the acceleration of the particle is a (D*) at t =2s particle is at the origin.

(B) d.k dk Roj.k a gSA (D*) le; t = 0 ls- ij d.k ewy fcUnq ij gSA

Pick the correct statements: lgh dFku dk p;u djks & (A) Average speed of a particle in a given time is never less than the magnitude of the average velocity.

fdlh le; vUrjky esa d.k dh vkSlr pky] vkSlr osx ds ifjek.k ls de ugha gksrh gSA

(B)

 d d   0 but It is possible to have a situation in which  0. dt dt

fdlh fLFkfr esa ;g lEHko gS fd (C)

(D)

 d 0 dt

 ijUrq d   0 . dt

The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.

fdlh le; vUrjky esa ,d d.k dk vkSlr osx 'kwU; gS rks ;g lEHko gS fd rkR{kf.kd osx le; vUrjky esa dHkh 'kwU; ugha gksA

The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite acceleration are not allowed)

,d ljy js[kk esa xfr dj jgs d.k dk fdlh le; vUrjky esa osx 'kwU; gSA rks ;g lEHko gS fd rkR{kf.kd osx bl le; vUrjky esa 'kwU; u gksA (vuUr Roj.k dh vuqefr ugha gSA)

Sol.

ABC

(a)

Distance Average speed = Total time taken Displacement average velocity = Total time taken



Displacement  Distance Average speed  average velocity


Chapter # 3

(b)

Let

SO

(c)

(d)

4.

Sol.

(a) (b) (c) (d) 5.


[7]

 v  cos tˆi  sin tˆj  dv =  sin tˆi  cos tˆj dt  dv 0 dt

 v = cos2 t  sin 2 t = 1  dv =0 dt Let the path of the particle is Displacement = 0 Average velocity = 0 But instantaneous velocity is never zero in the interval. In the straight line motion Displacement = 0 Average velocity = 0 But in this motion at one point instantaneous velocity is zero. An object may have (A) varying speed without having varying velocity (B*) varying velocity without having varying speed (C) nonzero acceleration without having varying velocity (D*) nonzero acceleration without having varying speed.

,d oLrq ds fy;s lEHko gS & (A) ifjorZu'khy osx ds fcuk ifjorZu'khy pky (B*) ifjorZu'khy pky ds fcuk ifjorZu'khy osx (C) ifjorZu'khy osx ds fcuk v'kwU; Roj.k (D*) ifjorZu'khy pky ds fcuk] v'kwU; Roj.k BD Speed increase, that causes magnitude of velocity increases. Magnitude of velocityis constant but direction is continuous changes. Acceleration = rate of change of velocity May be possible that acceleration change the direction of the particle but not change the magnitude of the velocity. Mark the correct statements for a particle going on a straight line:

,d ljy js[kk esa xfr dj jgs d.k ds fy, lgh dFku dk p;u dhft, &

(A) If the velocity and acceleration have opposite sign, the object is slowing down.

;fn osx o Roj.k dk fpUg foifjr gks rks oLrq eafnr gks jgh gSA

(B) If the position and velocity have opposite sign, the particle is moving towards the origin.

;fn fLFkfr o osx dk fpUg foijhr gks rks d.k ewy fcUnq dh vksj xfr dj jgk gSA

(C) If the velocity is zero at an instant, the acceleration should also be zero at that instant.

;fn fdlh {k.k osx 'kwU; gks rks ml {k.k Roj.k Hkh 'kwU; gksuk pkfg,A

(D) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval. Sol.

;fn fdlh le; vUrjky esa osx 'kwU; gks bl le; vUrjky esa Roj.k 'kwU; gksxkA (A), (B), (D) During retardation, acceleration opposes velocity. Velocity implies the direction of motion of a body.

eanu ds nkSjku] Roj.k osx dk çfrjks/k djrk gSA osx oLrq dh xfr dh fn'kk n'kkZrh gSA


Chapter # 3 6.

Sol.

(a) (b) (c) (d) 7.


[8]

The velocity of a particle is zero at t = 0 (A) The acceleration at t = 0 must be zero (B*) The acceleration at t = 0 may be zero. (C*) If the acceleration is zero from t = 0 to t = 10 s, the speed is also zero in this interval. (D*) If the speed is zero from t = 0 to t = 10 s the acceleration is also in the interval. t = 0 ij d.k dk osx 'kwU; gS & (A) t = 0 ij Roj.k fuf'pr :i ls 'kwU; gksxkA (B*) t = 0 ij Roj.k 'kwU; gks ldrk gSA (C*) t = 0 ls t = 10 ls- ds e/; Roj.k 'kwU; gS] bl le;kUrj esa pky Hkh 'kwU; gksxhA (D*) ;fn t = 0 ls t = 10 ls- ds e/; pky 'kwU; gS rks bl le;kUrj esa Roj.k Hkh 'kwU; gksxkA

BCD Free falling body t=0  v=0 & a=g Body at rest From t = 0 to t = 10s, the speed is also zero in this interval. Because (v = u + at) If body is rest in the interval of t = 0 to t = 10s.

Mark the correct statements: (A*) The magnitude of the velocity of a particle is equal to its speed. (B) The magnitude of average velocity in an interval is equal to its average speed in that interval. (C) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero (D) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

(a)

lgh dFkuksa dk p;u dhft;s & (A*) fdlh d.k ds osx dk ifjek.k mldh pky ds cjkcj gSA (B) fdlh le;karjky esa vkSlr osx dk ifjek.k] ml le;karj esa vkSlr pky ds cjkcj gSA (C) ,d ,slh ifjfLFkfr lEHko gS fd fdlh d.k dh pky lnSo 'kwU; gks fdUrq vkSlr pky 'kwU; ugha gksA (D) ,d ,slh ifjfLFkfr lEHko gS fd fdlh d.k dh pky dHkh Hkh 'kwU; ugha gks] fdUrq fdlh le;karj esa vkSlr pky 'kwU; gksA A  Speed = v

(b)

Average velocity =

Sol.

Total Displacement Total time taken

Distance Average speed = Total time taken

(c) (d)

|Average velocity|  |Average speed|  |Displacement|  |Distance| Speed of a particle is always zero means particle at rest. Distance travelled by the particle is zero. So average speed = 0 Speed of a particle is never zero means, particle travelled some distance. So average speed =

8.

Distane 0 time

The velocity-time plot for a particle moving on a straight line is shown in fig.

,d ljy js[kk esa xfr'khy d.k dk osx≤ xkzQ n'kkZrk gS (A) The particle has constant acceleration

d.k dk Roj.k fu;r gSA

(B) The particle has never turned around.

d.k dHkh Hkh ugha eqM+rkA

(C) The particle has zero displacement

d.k dk foLFkkiu 'kwU; gSA

(D) The average speed in the interval 0 to 10s is the same as the average speed in the interval 10s to 20s. 0 ls 10s esa vkSlr pky esa rFkk 10s ls 20s esa vkSlr pky dk eku leku gSA


Chapter # 3 Sol.

gy


[9]

(A), (D) Slope of given v-t graph (i.e. acceleration) is constant. • From 0 to 10 seconds, velocity is in positive and then negative. That means the particle trans around at t = 10 sec. • The positive and negative areas are not equal, So displacement is not zero. • Area of v-t graph from t = 0 to t = 10 sec is same as that from t = 10 to 20 sec. Hence average speed is same. (A), (D) fn, x, v-t xzkQ dk
vr% vkSlr pky leku gSA 9.

Fig. shows the position of a particle moving on X-axis as funcation of time. fp=k esa X-v{k ds vuqfn'k xfr'khy d.k dh fLFkfr] le; ds Qyu ds :i esa O;Dr dh xbZ gS &

(A*) The particle has come to rest 6 times d.k 6 ckj fojkekoLFkk esa vkrk gSA (B) The maximum speed is at t = 6 s t = 6 ls- ij vf/kdre pky gSA (C) The velocity remains positive for t = 0 to t = 6 s t = 0 ls t = 6 lsd.M ds e/; osx /kukRed jgrk gSA (D) The average velocity for the total period shows is negative. Sol.

(a) (b) (c)

(d)

iznf'kZr lEiw.kZ le;karj esa vkSlr osx _.kkRed gSA A

Slope of x-t graph gives the velocity. Here 6 times slope is zero. So we can say that the particle has come to rest 6 times. Just after t = 6 sec slope of x-t graph is zero. Then –ve. So we easily can say that speed is not maximum at t = 6s. Slope of x-t graph some time +ve & some time –ve  or zero. So we can say that in interval of t = 0 to t = 6s some time velocity is +ve, –ve or zero. Average velocity =

Total Displacement xf  xi x = = = +ve Total time taken t t


Chapter # 3 10.

Sol.


[10] 2

The accelerations of a particle as seen from two frames S1 and S2 have equal magnitude 4 m/s . (A) The frames must be at the rest with respect to each other. (B) The frames may be moving with respect to each other but neither should be accelerated with respect to the other. (C) The acceleration of S2 with respect to S1 may either be zero or 8 m/s 2. (D*) The acceleration of S2 with respect to S1 may be anything between zero and 8 m/s 2. nks funsZ'k ra=kksa S1 rFkk S2 ls fdlh d.k ds Roj.k dk izsf{kr eku ,d leku 4 eh-/ls2 gS & (A) funsZ'k ra=k fuf'pr :i ls ,d nwljs ds lkis{k fLFkj gSA (B) funsz'k ra=k ,d nwljs ds lkis{k xfr'khy gks ldrs gS] fdUrq dksb Z Hkh nwljs ds lkis{k Rofjr ugha gks ldrk gSA (C) S1 ds lkis{k S2 dk Roj.k 'kwU; vFkok 8 eh-/ls2 gks ldrk gSA (D*) S1 ds lkis{k S2 dk Roj.k 'kwU; rFkk 8 eh /ls-2 ds e/; dqN Hkh gks ldrk gSA    a s 2s1 = a s 2p + a ps1  a s 2s1 = 4 2  4 2  2.4.4. cos  Hence a s2s1 may lie between 0 & 8 depending upon value of .


Ch 3 Kinematics

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