Kinema Tics

Chapter # 3

Rest & Motion : Kinematics

SOLVED EXAMPLES 1. Sol.

2.

Sol.

An old person moves on a semi–circular track of radius 40.0 m during a morning walk. If he starts at one end of the track and reaches at the other end, find the distance covered and the displacement of the person. The distance covered by the person equals the length of the track . It is equal to  R =  × 40.0 m = 126 m. The displacement is equal to the diameter of the semi-circular track joining the two ends. It is 2 R = 2 × 40.0 m = 80.0 m. The direaction of this displacement is from the initial point to the final point. The distance travelled by a particle in time t is given by s = (2.5 m/s2)t2. Find (a) The average speed of the particle during the time 0 to 5.0 s and (b) The instantaneous speed at t = 5.0 s. (a)The distance travelled during time 0 to 5.0 s is s = (2.5 m/s2) ( 5.0 s)2 = 62.5 m The average speed during this time is vav = (b)

62.5 m = 12.5 m/s 5s

s = (2.5 m/s2)t2

ds = (2.5 m/s2) (2t) = (5.0 m/s2) t. dt At t = 5.0 s the speed is

or

v =

ds = (5.0 m/s2) (5.0 s ) = 25 m/s dt

3.

Figure shows the speed versus time graph for a particle . Find the distance travelled by the particle during the time t = 0 to t = 3s

Sol.

The distance travelled by the particle in the time 0 to 3 s is equal to the area shaded in the figure. This is a

1 1 (base) (height)= × (3s) (6 m/s) 2 2 = 9m. Thus , the particle covered a diatance of 9 m during the time 0 to 3s. right angled triangle with height = 6 m/s and the base = 3s. The area is

4. Sol.

A table clock has its minute hand 4.0 cm long. Find the average velocity of the tip of the minute hand (a) between 6.00 a.m. to 6.30 a.m. and (b) between 6.00 a.m. to 6.30 p.m. At 6.00 a.m. the tip of the minute hand is at 12 marks and at 6.30 a.m. or 6.30 p.m. it is 180º away.Thus , the straight line distance between the initial and final positions of a tip is equal to the diameter of the clock. Displacement = 2R = 2 × 4.0 cm = 8.0 cm. The dispalcement is from the 12 mark to the 6 mark on the clock panel. This is also the direction of the average velocity in both cases. (a) The time taken from 6.00 a.m. to 6.30 a.m. is 30 minutes = 1800 s. The average velocity is

8.0 cm Displacement = = 4.4 × 10–3 cm/s 1800 s time (b) The time taken from 6.00 a.m. to 6.30 p.m. is 12 hours and 30 minutes = 45000 s. The average velocity is vav =

vav =

8.0 cm Displacement = 45000 s = 1.8 × 10–4 cm/s time manishkumarphysics.in

Page # 1

Chapter # 3 5.

Sol.


A particle starts with an initial velocity 2.5 m/s along the positive x direction and it accelerates uniformly at the rate 0.50 m/s2 .(a) Find the distance travelled by it in the first two seconds. (b) How much time does it take to reach the velocity 7.5 m/s ? (c) How much distance will it cover in reaching the velocity 7.5 m/s ? (a) We have , x = ut +

1 2 at 2

1 (0.50 m/s2)(2s)2 2 = 5.0 m + 1.0 m = 6.0 m Since the particle does not turn back it is also the distance travelled. = (2.5 m/s)(2s) +

(b) We have , v = u + at or, 7.5 m/s = 2.5 m/s + (0.50 m/s2)t or,

t =

7.5 m / s  2.5 m / s 0.50 m / s 2

= 10s

(c) We have, or, v2 = u2 + 2ax or, (7.5 m/s)2 = (2.5 m/s)2 + 2(0.50 m/s2)x

(7.5 m / s) 2  (2.5 m / s) 2 or,

6. Sol.

x

=

2  0.50 m / s 2

= 50 m.

A particle having initial velocity u moves with a constant acceleration a for a time t.(a) Find the displacement of the particle in the last 1 second.(b) Evaluate it for u = 5 m/s , a= 2m/s2 and t = 10 s. (a) The position at time t is

1 2 at 2 The position at time (t –1s) is s = ut +

s’ = u(t –1s) +

1 a(t–1s)2 2

1 2 1 at – at(1s) + a(1s2) 2 2 Thus , the displacement in the last 1 s is st = s – s’ = ut – u(1s) +

= u(1s) + at(1s) –

1 a(1s)2 2

a (2t – 1s) (1s) .................(i) 2 (b) Putting the given values in (i) or, st = u(1s) +

 m 1 st =  5  (1s) +  s 2

 m  2  s2

  (2 × 10s –1s)(1 s) 

 m = 5m + 1 2  (19 s) (1s)  s  = 5m + 19m = 24m.

7.

A ball is thrown up at a speed of 4.0 m/s. Find the maximum height reached by the ball. Take g= 10 m/s2.

Sol.

Let us take vertically upward direction as the positive Y–axis . We have u = 4.0 m/s and a = –10 m/s2.At the highest point the velocity becomes zero. Using the formula v2 = u2 + 2ay, 0 = (4.0 m/s)2 + 2(–10 m/s2)y manishkumarphysics.in

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Chapter # 3


16 m 2 / s 2 or,

y =

20 m / s 2

= 0.80 m.

8.

A particle moves in the X–Y plane with a constant acceleration of 1.5 m/s2 in the diretion making an angle of 37º with the X–axis . At t = 0 the particle is at the origin and its velocity is 8.0 m/s along the X–axis . Find the velocity and the position of the particles at t = 4.0 s.

Sol.

ax = (1.5 m/s2)(cos37º)

and

4 = 1.2 m/s2 5 ay = (1.5 m/s2)(sin37º)

= (1.5 m/s2) ×

3 = 0.90 m/s2 5 The initial velocity has components ux = 8.0 m/s and uy = 0 At t = 0 , x = 0 and y = 0 The x-component of the velocity at time t = 4.0 s is given by ux = ux + axt = 8.0 m/s + (1.2 m/s2) (4.0 s) = 8.0 m/s + 4.8 m/s = 12.8 m/s The y-component of the velocity at time t = 4.0 s is given by uy = uy + ayt = 0 + (0.90 m/s2) (4.0 s) = 3.6 m/s The velocity of the particle at t = 4.0 s is

= (1.5 m/s2) ×

v=

u2 x  u 2 y =

(12.8 m / s) 2  (3.6 m / s) 2 = 13.3 m/s

The velocity makes an angle  with the X–axis where tan  =

uy ux

9 3 .6 m / s = 12.8 m / s = 32

The x-coordinate at t = 4.0 s is x = uxt +

1 2 at 2 x

1 (1.2 m/s2)(4.0 s)2 2 = 32 m + 9.6 m = 41.6 m The y-coordinate at t = 4.0 is = (8.0 m/s)(4.0 s) +

1 2 1 ayt = (0.90 m/s2)(4.0 s)2 = 7.2 m 2 2 = Thus , the particle is at (41.6m , 7.2 m) at 4.0 s.

y = uyt +

9.

A ball is thrown from a field with a speed of 12.0 m/s at an angle of 45º with the horizontal. At what distance will hit the field again ? Take g = 10.0 m/s2

Sol.

The horizontal range =

u 2 sin 2 g (12 m / s) 2 sin(2  45 º )

=

10 m / s 2 manishkumarphysics.in

Page # 3

Chapter # 3


144 m 2 / s 2 =

10.0 m / s 2

= 14.4 m.

Thus , the ball hits the field at 14.4 m from the point of projection. 10. Sol.

A swimmer can swim in still water at a rate 4.0 km/h .If he swims in a river flowing at 3.0 km/h and keeps his direction (with respect to water) perpendicular to the current, find his velocity with respect to the ground.  The velocity of the swimmer with respect to water is v S,R = 4.0 km/h in the direction perpendicular to the  river. The velocity of river with respect to the ground v R,G = 3 km/h along the length of the river. The velocity  of the swimmer with respect to the ground is v S,G where    v S,G = v S,R + v R,G Figure shows the velocities. It is clear that ,

 v S ,G =

(4.0 km / h) 2  (3.0 km / h) 2 = 5.0 km/ h

The angle  made with the direction of flow is tan  = 11. Sol.

4 4.0 km / h = 3 3.0 km / h

A man is walking on a level road at a speed of 3.0 km/h. Raindrops fall vertically with a speed of 4.0 km/ h. Find the velocity of the raindrop with respect to the man. We have to find the velocity of raindrops with respect to the man. The velocity of the rain as well as the velocity of the man are given with respect to the street. We have    v rain,man = v rain,street – v man,street Figure shows the velocities.

Figure 3.13

It is clear from the figure that  v rain,man =

(4.0 km / h) 2  (3.0 km / h) 2 = 5.0 km/ h

QUESTIONS

FOR

SHORT

ANSWER

1.

Galileo was punished by the Church for teaching that the sun is stationary and the earth moves around it. His opponents held the view that the earth is stationary and the sun moves around it. If the absoulte motion has no meaning are the two viewpoints not equally correct or equally wrong?

2.

When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal. Comment on this statement. manishkumarphysics.in

Page # 4

Chapter # 3


3.

A car travels at a speed of 60 km/h due north and the other a speed of 60 km/h due east .Are the velocities equal ? If no, which one is greater ? If you find any of the questions irrelevent, explain.

4.

A ball is thrown vertically upward with a speed of 20 m/s. Draw a graph showing the velocity of the ball as a function of time as it goes up and then comes back.

5.

The velocity of a particle is towards west at an instant . Its accleration is not towards west, not towards east, not towards east , not towards north and not towards south. Give an example of this type of motion.

6.

At which point on its path a projectile has the smallest speed?

7.

Two particles A and B start from rest and moves for equal time on a stright line. The particle A has an acclereation a for the first half of the total time and 2a for the second half. The particle B has an acceleration 2a for the first half and a for the second half. Which particle has covered larger distance?

8.

If a particle is accelerating , it is either speeding up or speeding down. Do you agree with this statement?

9.

A flood packet is dropped from a plane going at an altitude of 100m. What is the path of the packet as seen from the plane? What is the path as seen from the ground? If someone asks “what is the actual path”, what will you answer?

10.

Give examples where (a) the velocity of a particle is zero but its accleration. But it’s acceleration is not zero (c) the velocity is opposite in direction to the acceleration.

11.

Figure shows the x coordianate of a particle as a function of time . Find the signs of vx and ax at t = t1, t =t2 and t = t3.

12.

A player hits a baseball at some angle. The ball goes high up in space. The player runs and the catches the ball before it hits the ground. Which of the two(the player or the ball) has greter displacement ?

13.

The increase in the speed of a car is proportional to the additional petrol put into the engine. Is it possible to accelerate a car without putting more petrol or less petrol into the engin?

14.

Rain is falling vertically. A man running on the road keeps his umbrella tilted but a man standing on the street keep his umbrella vertical to protect himself from the rain but both of them keep the umbrella vertical to avoid sun-rays. Explain.

Objective - I 1.

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 ( B*) 70 fdeh0 / ? k a V k nf{k.k&if'pe ( C) 70 f d e h 0 / ? k a V k ] mRrj&if'pe dh vksj ( D) ' k w U ; 2.

dh

vksj

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. manishkumarphysics.in

Page # 5

Chapter # 3 X-v{k


ds vuqfn'k xfr'khy d.k dk foLFkkiu≤ ys[kkfp=k] fp=k es aiznf'kZr gS &

(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

3.

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;kZIr gSA

4.

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*)  

5.

1   2 2

(B)  1  2

(C)

2 1 1    2 2

(D)

1 1 1    2 2

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)  

(A)  

1   2 2

(B)  1  2

2 1 1 (C*)      2 2

1 1 1 (D)      2 2

6.

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

7.

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 &

(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 laca/k Hkou dh tehu ls Å¡pkbZ ij fuHkZj djrk gSA manishkumarphysics.in

Page # 6

Chapter # 3 8.


In a projectile motion the velocity

ç{ksi.k

xfr

esa

osx

H C V _ C h -

3_Obj.I_8

(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.

nks ckj Roj.k ds yEcor~ gksrk gSA 9.

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 10.

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 & (A) 25 eh11.

(B) 37 eh-

(C) 50 eh-

(D*) 100 eh-

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 (D*) RA rFkk RB ds 12.

(B) RA = RB

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 (C) 30o mÙkj ls if'pe dh vksj 13.

(C) RA > RB

e/; laca/k lqfuf'pr djus ds fy;s lwpuk vi;kZIr gSA

(B) 30o mÙkj ls iwo Z (D) 60o iwo Z ls mÙkj

dh vksj dh vksj

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 &

manishkumarphysics.in

Page # 7

Chapter # 3


(A) 2u cos

(B*) u/cos

(C) 2u/cos

(D) ucos

OBJECTIVE - II 1.

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 2.

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

(B) the acceleration of the particle is a (D*) at t =2s particle is at the origin. [HCV_Ch_3-obj-II_2]

dh vksj xfr'khy ,d d.k ds fy;s

(A) d.k (C) d.k

dk dk

x = ut(t–2s) + a(t–2s) 2 vkjfEHkd osx u gSA Roj.k 2a gSA

(B) d.k dk Roj.k a gSA (D*) le; t = 0 ls- ij d.k ewy fcUnq ij [HCV_Ch_3-obj-II_2]

gSA

Sol.

at t = 2 second x = 0 i.e. particle is at origin.

3.

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)

It is possible to have a situation in which

fdlh fLFkfr esa ;g lEHko gS fd

 d 0 dt

 d d   0 but  0. dt dt

 ijUrq d   0 . dt

(C) 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 (D) 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.

(A), (B), (C)

4.

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 5.

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 manishkumarphysics.in

Page # 8

Chapter # 3


(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.

;fn fdlh le; vUrjky esa osx 'kwU; gks bl le; vUrjky esa Roj.k 'kwU; gksxkA Sol.

(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

6.

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

7.

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.

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 8.

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 eqMr+ kA (C) The particle has zero displacement

d.k dk foLFkkiu 'kwU; gSA

Sol.

gy

(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 (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
Page # 9

Chapter # 3 10 sec •

gS A


_.kkRed

o

_.kkRed

{ks=kQy

leku

ugha

gSA

vr%

foLFkkiu

'kwU;

ugha

v-t xz k Q dk t = 0 ls t = 10 sec rd o t = 10 ls 20 sec. rd {ks = kQy leiu gS A vr% vkSlr pky leku gSA vr% v kS l r pky lek u gS A •

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.

iznf'kZr lEiw.kZ le;karj esa vkSlr osx _.kkRed gSA 10.

Sol.

The accelerations of a particle as seen from two frames S1 and S2 have equal magnitude 4 m/s 2. (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. [HCV_Ch_3-obj-II_10] 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 dksbZ 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 .

WORKED OUT EXAMPLES 1. Sol.

A man walks at a speed of 6 km/h for 1 km and 8 km/h for the next 1 km. What is his average speed for the walk of 2 km ? Distance travelled is 2 km. Time taken =

1km 1km + 6 km / hr 8 km / hr

Average speed =

 1 1 7 =   h = h. 6 8 24

2 km  24 48 = km/h 7 hr 7

 7 km/h. manishkumarphysics.in

Page # 10

Chapter # 3 2.

Sol.


The I.Sc. lecture theatre of a college is 40 ft wide and has a door at a corner. A teacher eneters at 12.00 noon through the door and makes 10 rounds along the 40 ft wall back and forth during the period and finally leaves the classroom at 12.50 p.m. through the same door. Compute his average speed and average velocity. Total distance travelled in 50 minutes = 800 ft. 800 ft/min = 16 ft/min 50 At 12.00 noon he is at the door and at 12.50 pm he is again at the same door. The displacement during the 50 min interval is zero. Average velocity = zero.

Average speed =

3.

The position of a particle moving on X-axis is given by x = At3 + Bt2 + Ct + D The numerical values of A , B , C, D are 1, 4, –2 and 5 respectively and SI units are used. Find (a) the dimensions of A, B, C and D, (b) the velocity of the particle at t = 4s, (c) the acceleration of the particle at t = 4s , (d) the average velocity during the interval t = 0 to t = 4s, (e) the average acceleration during the interval t = 0 to t = 4s.

Sol.

(a) Dimensions of x, AT3 , Bt2, Ct and D must be identical and in this case each is length. Thus, [At3] = L, or, [A] = LT–3 2 [Bt ] = L or, [B] = LT–2 [Ct] = L or, [C] = LT–1 and [D] = L. (b) or,

(c)

x = At3 + Bt2 + Ct + D dx = 3At2 + 2Bt + C dt Thus , at t = 4s , the velocity = 3(1 m/s3) (16 s2) + 2(4 m/s2) (4s) + (–2 m/s) = (48 + 32 – 2) m/s = 78 m/s v = 3At2 + 2Bt + C

v=

dv = 6At + 2B. dt At t = 4s u = 6(1 m/s3)(4s) + 2(4 m/s2) = 32 m/s2 (d) x = At3 + Bt2 + Ct + D Position at t = 0 is x = D = 5 m. Position at t = 4s is (1 m/s3)(64 s3) + (4 m/s2)(16 m/s2)(16 s2) – (2 m/s)(4s) + 5m = (64 + 64 –8 + 5)m = 125 m. Thus , the displacement during 0 to 4s is 125m – 5m = 120m

or, v =

Average velocity =

120 m = 30 m/s 4s

(e) v = 3At2 + 2Bt + C Velocity at t = 0 is C = –2 m/s. Velocity at t = 4s is = 78 m/s.

v 2  v1 Average acclearation = t  t = 20 m/s2. 2 1 4.

From the velocity-time graph of a particle is given in figure, describe the motion of the particle qualitatively in the interval 0 to 4s. Find (a) the distance travelled during first two seconds, (b) during the time 2s to 4s, (c) during the time 0 to 4s, (d) displacement during 0 to 4s, (e) (acceleration at t =


1 sec). 2

Page # 11

Chapter # 3

Sol.


At t = 0, the particle is at rest, say at the origin. After that the velocity is positive , so that the particle moves in the positive x direction. Its speed increses till 1 second when it starts decreasing. The particle continues to move further in positive x direction. At t= 2s , its velocity is reduced to zero, it has moved throught a maximum positive x distance . Then it changes its direction, velocity being negative , but increasing in magnitude. At t = 3s velocity is maximum in the negative x direction and then the magnitude starts decreasing. It comes to rest at t= 4s. (a) Distance during 0 to 2s = Area of OAB

1 × 2s × 10 m/s = 10 m. 2 (b) Distance during 2 to 4s = Area of BCD = 10m. The particle has moved in negative x direction during this period. (c) The diatance travelled during 0 to 4s = 10m + 10m = 20m. (d) displacement during 0 to 4s = 10m + (–10m) = 0 (e) at t =1/2s acceleration = slope of line OA = 10 m/s2 (f ) at t = 2s acceleration = slope od line ABC = –10 m/s2. =

5.

Sol.

A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/ s and one second later the speed becomes 150 m/s. Find (a) the acceleration and (b) the distance treavelled during the (t +1)th second. HCV_Ch-3_WOE_5 ,d d.k fojke ls 'kq: gksdj fu;r Roj.k ls xfr djrk gSA le; t lSd.M ij] bldh pky 100 m/s rFkk ,d lsd.M ckn pky 150 m/s gks tkrh gS] crkb;s (a) Roj.k vkSj (b) (t +1)th lsd.M esa r; dh xbZ nwjh (a) Velocity at time t is 100 m/s = a.(t second)......................(1) and velocity at time (t +1) second is 150 m/s = a.( t + 1) second...............(2) Subtracting (1) from (2) a = 50 m/s2.........(3) (b) Consider the interval t second to (t + 1) second, time elapsed = 1s intial velocity = 100 m/s final velocity = 150 m/s. Thus , (150 m/s)2 = (100 m/s2) + 2(50 m/s2) x or, x = 125 m.

6.

Sol.

A boy streches a stone against the rubber tape of a catapult or ‘gulel’ (a device used to detach mangoes from the tree by boys in Indian villages) through a distance of 24 cm before leaving it. The tape returns to its normal position acclerating the stone over the stretched length. The stone leaves the gulel with a speed 2.2 m/s. Assuming that the acceleration is constant while the stone was being pushed by the tape, find its magnitude. Consider the accelerated 24 cm motion of the stone. Initial velocity = 0 Final velocity = 2.2 m/s Distance travelled = 24 cm = 0.24 m Using v2 = u2 + 2ax

4.84 m 2 / s 2 a= = 10.1 m/s2. 2  0.24 m 7.

A police inspector in a jeep is chasing a pickpocket on a straight road . The jeep is going at its maximum speed v(assumed uniform). The pickpocket rides on the motercycle of a waiting friend , when the jeep is at manishkumarphysics.in

Page # 12

Chapter # 3


a distance d away, and the motercycle starts with a constant acceleration a . Show that the pickpocket will Sol.

be caught if v  2ad . Suppose the pickpocket is caught at a time t after the motercycle starts. The distance travelled by the motercycle during this interval is

1 2 at .............................................(i) 2 During this interval the jeep travels a distance s + d = vt.............................................(ii) By (i) and (ii), s =

1 2 at – vt + d = 0 2

v  v 2  2ad a The pickpocket will be caught if t is real positive This will be possible if or,

t=

v2  2ad or, v 

2ad .

8.

A car is moving at a constant speed of 40 km/h along a straight road which hends towards a large verticle wall and makes a sharp 90º turn by the side of wall. A fly flying at a constant speed of 100 km/h, starts from the wall towards the car at an instant when the car is 20 km away. flies until it reaches the glasspane of the car and returns to the wall at the same speed. it continues to fly between the car and the wall till the car makes the 90º turn. (a) What is the total distance the fly has travelled during this period ? (b) How many trips has it made between the car and the wall ?

Sol.

(a) The time taken by the car to cover 20 km before the turn is

20 km 1 = h.The fly moves at a constant 40 km / h 2

speed of 100 km/h during this time . Hence the total distance covered by it is

km 1 × h = 50 km. h 2 (b) Suppose the car is at distance x away (at A) when the fly is at the wall (at O) . The fly hits the glass plane at B , taking a time t. Then AB = (40 km/h)t, and OB = (100 km/h)t, Thus x = AB + OB = (140 km/h)t 100

x t = 140 km / h , or

5 x. 7 The fly returns to the wall and during this period the car moves the distance BC. The time taken by the fly in this return path is or

OB =

 5x / 7  x   = 140 km /h 100 km / h   40 x 140

Thus ,

BC

=

=

or,

OC

= OB – BC =

2 7 3 x 7


Page # 13

Chapter # 3


If at the begining of the ground trip (wall to the car and back) the car is at distance x away, it is

3 x away 7

when the next trip again starts . Distance of the car at the begining of the 1st trip = 20 km. Distance of the car at the begining of the 2nd trip =

3 × 20 km. 7 2

3 Distance of the car at the begining of the 3rd trip =   × 20 km. 7 3

3 Distance of the car at the begining of the 4th trip =   × 20 km. 7 3 Distance of the car at the begining of the nth trip =   7

n 1

× 20 km.

Trips will go on till the car reaches the turn that is the distance reduces to zero. This will be the case when n becomes infinity. Hence the fly makes an infinite number of trips before the car takes the turn. 9.

A ball is dropped from a height of 19.6 m above the ground rebounds from the ground and raises itself up to the same height . Take the starting point as the origin and vertically downward as the positive X-axis. Draw approximate plots of x versus t , v versus t and a versus t. Neglect the small interval during which the ball was in contact with the ground. (g = 9.8 m/s2) HCV_Ch-3_WOE_9 xsan dks tehu ls 19.6 ehVj ÅapkbZ ls fxjk;k tkrk gS vkSj ;g mNydj mlh Å¡pkbZ rd okil tkrh gSA izkjfEHkd fcUnq dks dsUnz ekus rFkk uhps okyh fn'kk dks /kukRed ekus rks x rFkk t, v rFkk t, a rFkk t ds e/; oØ [khafp,A ml le; dks ux.;

ekfu, ftl nkSjku xsan tehu ds lEidZ esa gSA Sol.

Since the accelertaion of the ball during the contact is different from ‘g’ , we have to treat the downward motion and the upwards motion separately for the downward motion :

1 2 at = (4.9 m/s2)t2. 2 The ball reaches the ground when x = 19.6m. This gives t = 2s. After that it mives up, x decreases and at t = 4s, x becomes zero, the ball reaching the initial point. We have at t =0 x=0 x t = 1s x = 4.9m t = 2s x = 19.6m 20 m t = 3s x = 4.9m t = 4s x=0 10 m Velocity : During the first two seconds, v = u + at = 9.8 (m/s2)t 2 at t = 0 v=0 4 t (seconds) 3 1 at t = 1s, v = 9.8 m/s at t = 2s, v = 19.6 m/s a = g = 9.8 m/s2 , x = ut +

During the next two seconds the ball goes upwards , velocity is negative, magnitude decreasing and at t = 4s , v = 0 . Thus at t = 2s, v = –19.6 m/s at t = 3s v = –9.8 m/s at t =4s v = 0.


Page # 14

Chapter # 3


At t = 2s there is an abrupt change in velocity from 19.6 m/s to –19.6 m/s. Infact this change in velocity takes places over a small interval during which the ball remains in contact with the ground. Acceleration : The acceleration is constant 9.8 m/s2 throughout the motion (except at t = 2s) 10. Sol.

2

9.8 m/s

t (second) 1

2

3

4

–10 m/s2

A stone is dropped from a ballon going up with a uniform velocity of 5.0 m/s. If the ballon was 50 m high when the stone was dropped, find its height when the stone hits the ground .Take g = 10 m/s2. At t = 0, the stone was going up with a velocity of 5.0 m/s. After that it moved as a freely falling particle with downward acceleration g. Take vertically upwards as the positive X-axis. If it reaches the ground at time t, x = –50m u = 5 m/s, a = –10 m/s2.

1 2 at 2

We have

x = ut +

or,

–50m = (5 m/s).t +

1 × (–10 m/s2)t2 2

1 41 s 2 or, t = –2.7s or 3.7s Negative t has no significance in this problem. The stone reaches the ground at t = 3.7s. During this time the ballon has moved uniformly up. The distance covered by it is 5 m/s × 3.7s = 18.5 m Hence , the height of the ballon when the stone reaches the ground is 50m + 18.5m = 68.5 or,

11.

t

=

A football is kicked with a velocity of 20 m/s at an angle of 45º with the horizontal. (a) Find the time taken by the ball to strike the ground, (b) Find the maximum height it reaches, (c) How far away from the kick does it hit the ground ? Take g = 10 m/s2. ,d QqVckWy dks osx 20 m/s ls {kSfrt ls 45º ds dks.k ij fdd (iSj ls) ekjh tkrh gSA (a) QqVckWy dk tehu ls Vdjkus eas yxs le; dh x.kuk dhft,A (b) vf/kdre Å¡pkbZ tgk¡ rd ;g QqVckWy igq¡prh gSA (c) tgk¡ ls geus bls fdd ekjh gS] mlls fdruh nwjh ij ;g ckWy Vdjk,xh \ g = 10 m/s2 ysaA HCV_Ch-3_WOE_11

Sol.

(a) Take the origin at the point where the ball is kicked , vertically upward as the Y-axis and the horizontal in the plane of motion as the X-axis . The intial velocity has the components ux = (20 m/s)cos45º = 10 2 m/s and uy = (20 m/s)sin45º = 10 2 m/s When the ball reaches the ground , y = 0 Using

y = uyt –

1 2 gt , 2

0 = (10 2 m/s)t –

1 × (10 m/s2) × t2 2

or, t = 2 2 s = 2.8 s. Thus , it takes 2.8s for the football to fall on the ground. (b) At the highest point vy = 0. Using the equation vy2 = uy2 – 2gy



O = 10 2m / s

2

– 2 × (10m/s)2 H

H = 10 m Thus the maximum height reaches is 10 m. (c) The horizontal distance travelled before falling to the ground is x = uxt = (10 2 m/s)(2 2 m/s) = 40 m. 12.

Sol.

A helicopter on flood relief mission, flying horizontally with a speed u at an altitude H, has to drop a food packet for a victim standing on the ground. At what distance from the victim should be the packet be dropped ? The victim stands in the verticle plane of the helicopter’s motion. The velocity of the food packet at the time of release is u and is horizontal. The vertical velocity at the time of release is zero. manishkumarphysics.in Page # 15

Chapter # 3


Verticle motion : If t be the time taken by the packet to reach the victim , we have for verticle motion, H=

1 2 gt 2

or,

t =

2H g .....................(i)

Horizintal motion : If D be the horizontal distance travelled by the packet, we have D = ut. Putting t from (i) , D=u

2H g

The distance between the victim and the packet at the time of release is

D H 2

13.

2

2u2H  H2 g

=

A particle is projected horizontally with a speed u from the top of a plane inclined at an angle  with the horizontal. How far from the point of projection will the particle strike the plane? ,d d.k dks {kSfrt :i ls u pky ls  dks.k ds urry ds Åijh fcUnq ls ç{ksfir fd;k tkrk gSA ;g d.k ç{ksi.k fcUnq

ls fdruh nwjh ij ry ls Vdjk,xk \ Sol.

HCV_Ch-3_WOE_13

Take X,Y-axes as shown in figure. Suppose that the particle strikes the plane at a point P with coordinates (x,y) . Consider the motion between A and P. Motion in x direction : Initial velocity = u Acceleration = 0 x = ut......................(i) Motion in y direction : Initial velocity = u Acceleration = g y

=

1 2 gt .......................(ii) 2

Eliminating t from (i) and (ii)

Also

gx 2 Thus ,

gu

2

1 x2 g 2 u2

y

=

y

= xtan 

= xtan  giving x = 0 or,,

2u 2 tan  g

Clearly the point P corresponds to x =

then y = xtan  =

2u 2 tan  g

2u 2 tan 2  g

The distance AP = l =

x2  y2

=

2u2 2 tan  1 tan  g

=

2u2 tan  sec  g

Alternatively : Take the axes as shown in figure. Consider the motion between A and P. manishkumarphysics.in

Page # 16

Chapter # 3


Motion along the X-axis : Initial velocity = ucos  Acceleration = gsin  Displacement = AP

1 (gsin  )t2............................(i) 2 motion along the y-axis initla velocity = – u sin Acceleration = g cos  Displacement = 0 Thus , AP = (ucos  )t +

Thus ,

0 = –utsin  +

or,

t=0,

1 2 gt cos  2

2u sin  g cos 

2u sin  Clearly , the point P corresponds to t = g cos  Putting this value of t in (i)

 2u sin   g sin  AP = (ucos  )  g cos   + 2   =

14.

Sol.

15. Sol.

2u2 sin  2u 2 sin  tan 2  + g g

 2u sin      g cos   =

2

2 u2 2 u2 sinsec2 = tansec g g

A projectile is fired with a speed u at an angle  with the horizontal, Find its speed when its direction of motion makes an angle  with the horizontal. {kSfrt ls  dks.k cukrs gq, ,d oLrq dks u pky ls ç{ksfir fd;k tkrk gSA bldh pky crkb,] tc bldh xfr dh fn'kk {kSfrt ls  dks.k cukrh gSA HCV_Ch-3_WOE_14 Let the speed be v when it makes an angle  with the horizontal. As the horizontal component of velocity remains constant, vcos  = ucos  or, v = ucos  sec  A bullet is fired horizontally aiming at an object which starts falling at the instant the bullet is fired. Show that the bullet will hit the object. The situation is shown in figure. The object starts falling from the point B. Draw a vertical line BC through B. Suppose the bullet reaches the line BC at a point D and it takes a time t in doing so.

Consider the vertical motion of the bullet. The initial vertical velocity = 0. The distance travelled vertically = BD

1 2 1 2 gt . In time t the object also travels a distance = gt = BD. Hence at time t, the object will also be 2 2 at the same point D. Thus, the bullet hits the object at point D. =

16.

Sol.

A man can swim in still water at a speed of 3 km/h. He wants to cross a river that flows at 2 km/h and reach the point directly opposite to his starting point. (a) In which direction should he try to swim (that is , find the angle his body makes with the river flow) ? (b) How much time will he take to cross the river if the river is 500 m wide ? (a) The situation is shown in figure. The X-axis is chosen along the river flow and the origin at the starting position of the man. The directionmanishkumarphysics.in of the velocity of man with respect to ground is along the Y-axis (perpendicular Page # 17

Chapter # 3


to the river). We have to find the direction of velocity of the man with respect to water.  Let v R,G = velocity of the river with respect to the ground = 2 km/h along the X-axis

 v m,r = velocity of the man with respect to the river = 3 km/h making an angle  with the Y-axis.  and um,g = velocity of the man with respect to the ground along the Y-axis. We have    um,g = v m,r + v r,g .....(1) Taking components along the X-axis 0 = –(3 km/h)sin  + 2 km/h 2 3 (b) Taking components in equation (i) along the Y-axis  um,g = (3 km/h)cos  + 0  um,g = 5 km/h

or, sin  =

or,

Time =

Displaceme nt in y direction Velocity in y direction 0.5 km

= 17.

Sol.

5 km / h

=

5 h. 10

A man can swim at a speed of 3 km/h in still water to cross a 500 m wide river flowing at 2 km/h. He keeps himself always at an angle of 120º with the river flow while swimming. (a) Find the time he takes to cross the river. (b) At what point on the opposite bank will he arrive? The situation is shown in figure

 Here v r,g = velocity of the river with respect to the ground.  v m,r = velocity of the man with respect to the river  v m,g = velocity of the man with respect to the ground. (a) We have    v m,g = v m,r + v r,g .....(i) Hence , the velocity with respect to the ground is along AC. Taking y-components in equation (i),

 v m,g sin  = 3 km/h cos30º + 2 km/h cos90º = 3 3 km/h. 2 Time taken to cross the river =

Displaceme nt along theY  axis velocity along theY  axis manishkumarphysics.in

Page # 18

Chapter # 3

Rest & Motion : Kinematics 1/ 2 km

=

1

3 3 / 2 km / h

=

3 3

h

(b) Taking x-components in equation (i),

 1 v m,g cos  = –3 km/h sin30º + 2 km/h = km/h 2 Displacement along the X-axis as the man crosses the river = (velocity along the X-axis) (time)  1  1  1km  h  =  ×  =  km 2 h 6 3   3 3  18.

A man standing on a road has to hold his umbrella at 30º with the vertical to keep the rain away. He throws the umberella and starts running at 10 km/h.He finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to (a) The road (b) the moving man,

Sol.

When the man is at rest with respect to the ground , the rain comes to him at an angle 30º with the vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation when the man runs is shown,in the figure.

 Here v r,g = velocity of the rain with respect to the ground  v m,g = velocity of the man with respect to the ground  and v r,m = velocity of the rain with respect to the man.    We have v r,g = v r,m + v m,g ........(i) Taking horizontal components , equation (i) gives   v r,g sin30º = v m,g = 10 km/h or,

 v r,g

=

10 km / h sin 30 º

= 20 km/h

Taking vertical components equation (i) gives   v r,g cos30º = v r,m or,

 3 v r,m = (20 km/h) 2 = 10 3 km/h

19.

Sol.

A man running on a horizontal road at 8 km/h finds the rain falling vertically. He increses his speed to 12 km/ h and finds that the drops make angle 30º with the vertical. Find the speed and direction of the rain with respect to the road.    v rain,road = v rain,man + v man,road We have The two situations given in the problem may be respresented by the following figure.


Page # 19

Chapter # 3


 v rain,road is same in magnitude and direction in both the figures. Taking horizontal components in equation (i) for figure,  v rain,road sin  = 8 km/h ..............(ii)  Now consider figure (3-W14b). Draw a line OA  v rain,man as shown. Taking components in equation (i) along the line OA.  v rain,road sin(30º +  ) = 12 km/h cos30º ..............(iii) From (i) and (ii) sin( 30 º   ) sin 

or,

sin 30 º cos   cos 30 º sin  3 3 = sin  4

1 3 cot  + 2 2

=

3 3 4

or,

cot 

=

3 2

From (ii),

Sol.

12  3 8 2

or,

or,

20.

=



= cot–1

3 2

 v rain,road = 8 km / h = 4 7 km/h. sin 

Three particles A , B and C are situated at the vertices of an equilateral triangle ABC of side d at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB , B along BC and C along CA. At what time will the particles meet each other ? The motion of the particles is roughly sketched in figure. By symmetry they will meet at the

centroid O of the triangle. At any instant the particles will from an equilateral triangle ABC with the same centroid O. Concentrate on the motion of any one. particle say A. At any instant its velocity makes angle 30º with AO. The components of this velocity along AO is vcos30º. This component is the rate of decrease of the distance AO intially,


Page # 20

Chapter # 3


AO =

2 2  d d   3 2

2

d =

3

Therefore , the time taken for AO to become zero =

Time =

d 3 = v cos 30 º

3v  3

=

2d 3v

Displaceme nt in y direction Velocity in y direction 0.5 km

=

2d

5 km / h

=

5 h. 10

EXERCISES 1.

A man has to go 50 m due north, 40m due east and 20m due south to reach a field. (a) What distance he has to walk to reach the field ? (b) What is his displacement from his house to the field? ,d O;fDr ,d eSnku rd igqp a us ds fy, 50 eh- mRrj dh vksj] 40 eh- iwoZ dh vksj rFkk 20 eh- nf{k.k dh vksj pyrk

gSA

(a) eSnku (b) mlds

rd igqapus ds fy, mlds }kjk pyh xbZ nwjh fdruh gS \ ?kj ls eSnku rd foLFkkiu fdruk gS \

[Ans : (a) 110 m (b) 50 m,tan–1

3 north to east ] 4

2.

A particle starts from the origin , goes along the X-axis to the point (20m , 0) and then returns along the same line to the point (–20m,0) . Find the distance and displacement of the particle during the trip. ,d d.k X-v{k ds vuqfn'k ewy fcUnq ls pyuk izkjEHk djds fcUnq (20eh-, 0) rd igaqprk gS] blds i'pkr~ iqu% mlh js[kk ds vuqfn'k fcUnq (–20eh-,0) rd ykSVrk gSA lEiw.kZ xfr esa d.k ds }kjk r; dh xbZ nwjh rFkk foLFkkiu Kkr dhft;sA [Ans : 60 m, 20 m in the negative direction]

3.

It is 260 km from Patna to Ranchi by air and 320 km by road.An aeroplane takes 30 minute to go from Patana to Ranchi wheras a delux bus takes 8 hour. (a)Find the average speed of the plane .(b) Find the average speed of the bus.(c) Find the average velocity of the plane.(d) Find the average velocity of the bus. iVuk ls jkaph dh nwjh ok;q ekxZ ls 260 fdeh- rFkk lM+d ekxZ ls 320 fdeh- gSA iVuk ls jkaph tkus esa ok;q;ku dks 30 fefuV rFkk MhyDl cl dks 8 ?kaVs yxrs gSA (a) ok;q;ku dh vkSlr pky Kkr dhft;sA (b) cl dh vkSlr pky Kkr dhft;sA (c) ok;q;ku dk vkSlr osx Kkr dhft;sA (d) cl dk vkSlr osx Kkr dhft;sA [Ans :(a) 520 km/h, (b) 40 km/h, (c) 520 km/h Patna to Ranchi,(d) 32.5 km/h Patna to Ranchi]

4.

When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the readind is 12416 km.(a) What is the average speed of the car during this period ? (b) What is the average velocity? [Ans :(a) 32 km/h (b) zero ] ,d O;fDr dkj ls Hkze.k gsrq ?kj ls fudyrk gS] ehVj dk ikB~;kad 12352 fdeh- gSA tc og nks ?kaVs i'pkr~ ?kj ykSVrk gS rks ikB~;kad 12416 fdeh gSA (a) bl le;karj esa dkj dh vkSlr pky fdruh gS ? (b) vkSlr osx fdruk gS ? [Ans :(a) 32 km/h (b) zero ]

5.

An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average accleration? ,d /kkod viuh vf/kdre pky 18.0 km/h çkIr djus esa 2.0 s le; ysrk gS rks d.k ds vkSlr Roj.k dk ifjek.k gksxk \ Ans. 2.5 m/s2 Vmax = 18 km/h = 5 m/s using v = u + at dk mi;ksx djus ij 5 = 0 + a(2)  a = 2.5 m/s2 [Ans :2.5 m/s2]

Sol.

6.

The speed of a car as a function of time is shown in figure. Find the distance travelled by the car in 8 seconds and its acceleration. le; ds Qyu ds :i esa fdlh dkj dh pky fp=k esa iznf'kZr dh xbZ gSA dkj ds }kjk 8 lsd.M eas r; dh nwjh rFkk

bldk Roj.k Kkr dhft,sA


Page # 21

Chapter # 3


as a function of time is shown in figure. Find the distance travelled by the car in 8 seconds and its acceleration.

[Ans : 80 m, 2.5 m/s2] 7.

The acceleration of a cart started at t = 0, varies with time as shown in figure . Find the distance travelled in 30 seconds and draw the displacement-time graph. t = 0 ij 'kq: gksus okyh ,d xkM+h ds Roj.k dk le; ds lkFk ifjorZu fp=k esa fn[kk;k x;k gSA 30 lSd.M esa r; nwjh crkb;s

rFkk foLFkkiu≤ vkjs[k [khafp,A Acceleration (in ft/s

2

)

HCV_Ch-3_Ex._7

5.0 0

10

20

30

Time (in second)

-5.0

[Ans : 1000 ft.]

8.

Figure shows the graph of velocity versus time for a particle going along the X-axis. Find (a) the acceleration, (b) the distance travelled in 0 to 10 s and (c) the displacement in 0 to 10 s. X-v{k ds vuqfn'k xfr'khy d.k dk osx≤ vkjs[k fp=k esa iznf'kZr fd;k x;k gSA Kkr dhft;s % (a) Roj.k (b) 0 ls 10 lsd.M esa r; dh xbZ nwjh rFkk (c) 0 ls 10 lsd.M esa foLFkkiu

Ans. (a) 0.6 m/s2 (b) 50 meter 9.

(C) 50 meter

Figure shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find the average velocity during 0 to 10s., (b) instantaneous velocity at 2,5,8 and 12s. fp=k esa x-v{k ds vuqfn'k xfr'khy d.k ds le; ds Qyu ds :i esa X funs'Z kkad dk vkjs[k O;Dr fd;k x;k gSA Kkr dhft;s % (a) 0 ls 10 lsd.M ds e/; vkSlr osxA (b) 2,5,8 ,oa 12 lsd.M ij rkR{kf.kd osxA manishkumarphysics.in

Page # 22

Chapter # 3

10.


[Ans : (a) 10 m/s , (b) 20 m/s, zero, 20 m/s, –20 m/s] From the velocity-time plot shown in figure, find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period. fn, x, osx≤ xzkQ dh lgk;rk ls d.k }kjk çFke 40 lSd.M esa r; dh xbZ nwjh Kkr djks rFkk bl vUrjky esa vkSlr

osx Hkh Kkr djksA

v 5 m/s

0

30 10

20

40

t(s)

–5 m/s

Sol.

Ans. 100m, zero Magnitude of area of v-t graph gives distance travelled 1   Distance =  [20 x 5] x 2 = 100 m 2 

 Displacement =

gy

v-t xzkQ

1 1 [20 x 5] + [20 x (–5)] = 0 2 2

 Area of v  t graph gives displacement. Displacement   = = 0 Ans. Time

ds {ks=kQy dk ifjek.k r; dh xbZ nwjh çnku djrk gSA

1   nwjh =  [20 x 5] x 2 = 100 m 2   rFkk v  t xzkQ dk {ks=kQy foLFkkiu çnku

 foLFkkiu = 11.

and

djrk gSA

1 1 [20 x 5] + [20 x (–5)] = 0 2 2

 foLFkkiu  = = 0 Ans.

le;

Figure (3-E6) shows x-t graph of a particle. Find the time t such that average velocity of the particle during the period 0 to t is zero. [Ans : 12s] ,d d.k dk x-t vkjs[k fp=k esa iznf'kZr fd;k x;k gS A le; t dk og eku Kkr dfj;s ftlds fy;s 0 ls t ds le;karj esa d.k

dk vkSlr osx 'kwU; gksA


Page # 23

Chapter # 3


position B of the particle such that the average velocity between the positions A and B has the same direction as the instataneous velocity at B. [Ans :x = 5m, y = 3m] ,d d.k fcUnq A ls pyuk izkjEHk djrk gS rFkk iznf'kZr oØ ds vuqfn'k xfr djrk gSA d.k dh og fLFkfr B Kkkr dhft;s ¼ yxHkx ½ rkfd fLFkfr;ksa A rFkk B ds e/; d.k ds vkSlr osx dh fn'kk rFkk B ij d.k ds rkR{kf.kd osx dh fn'kk ,d leku

gksA

13.

An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s2 for 5.0s. Find the distance travelled during the period of acceleration. [Ans :35m] 4.0 m/s osx ls xfr'khy ,d oLrq dks 5 ls- ds fy;s 1.2 m/s2 dh nj ls Rofjr fd;k tkrk gSA Rofjr fd;s x;s le;karj

eas r; dh xbZ nwjh Kkr dhft;sA 14.

A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s2 to his scooter. How far will it travel before stopping? [Ans : 12m] 43.2 fdeh @ ?kaVk osx ls xfr'khy O;fDr czd s yxkdj mlds LdwVj esa 6.0 m/s2 voeUnu mRiUu djrk gSA :dus ls iwoZ

og fdruh nwjh r; djsxk\ 15.

Sol.

A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train and (b) The maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed. ,d Vªsu fojkekoLFkk ls pyuk izkjEHk djds vk/ks fefuV rd 2.0 m/s2 Roj.k ls pyrh gSA blds i'pkr~ czd s yxkus ls Vªus ,d fefuV esa fojkekoLFkk esa vk tkrh gSA (a) Vªsu ds }kjk r; dh x;h dqy nwjh] (b) Vªus ds }kjk izkIr fd;k x;k vf/kdre osx rFkk (c) vf/kdre dh vk/kh pky ij Vªus dh fLFkfr¼;k¡½A HCV_Ch-3_Ex.I_15 [Ans :(a) 2.7 km , (b) 60 m/s, (c) 225 m and 2.25 km] (b) Plotting the velocity time graph according to the given conditions. tan  = 2 maximum velocity achieved would be 60 m/sec. v 30 so v = 60º

tan  =

60 =1 60 Suppose at time t1 and at t2 train achieves half the maximum velocity.

Now

tan  =

30  tan  = t 1

t1 =

30 15 second 2

30 30 tan  = 90  t 90 – t2 = ; t2 = 60 second Ans. 1 2  After 15 sec and 60 sec the train achieves half the maximum velocity. 16.

A bullet travelling with a velocity of 16 m/s pentrates a treetrunk and comes to rest in 0.4m. Find the time taken during the retardation. 16 eh-@ls- osx ls xfr'khy ,d xksyh isM+ ds rus esa 0.4 eh- ?kqldj :d tkrh gSA voeanu esa yxk le; Kkr dhft;sA Ans : 0.05 s

17.

A bullet going with a speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before comming to rest. Find the deceleration. 350 eh-@ls- pky ls xfr'khy xksyh dadjhV dh nhokj esa 5.0 lseh- ?kqldj :d tkrh gSA voeanu Kkr dhft;sA manishkumarphysics.in

Page # 24

Chapter # 3

Rest & Motion : Kinematics 5

2

[Ans : 12.2 × 10 m / s ] 18.

A particle starting from rest moves with constant acceleration . If it takes 5.0 s to reach the speed 18.0 km/ h find (a) the average velocity during this period,and (b) the distance travelled by the particle during this period. [HCV_Ch-3_Ex._18] fLFkj voLFkk esa fLFkr ,d d.k fu;r Roj.k ls pyuk izkjEHk djrk gSA ;fn bldks 18.0 fdeh-@?k.Vk pky izkIr djus e s a 5.0 l s d . M y x r s g S ] r k s K k r d h f t ; s A (a) b l l e ; k U r j e s a v k S l r o s x r F k k (b) b l l e ; k U r j e s a d . k } k j k r ; d h x b Z n w j h A [Ans :(a) 2.5 m/s , (b) 12.5 m]

19.

A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km/h and the brakes cause a decleration of 6.0 m/s2.find the distance travelled by the car after he sees the need to put the brakes on. ,d Mªkboj vko';drk iM+us ds 0.20 lsd.M i'pkr~ czsd yxk ikrk gS ;g Mªkboj dk ^^izfrfØ;k le;** dgykrk gS A fn og 54 fdeh-@?k.Vk dh pky ls dkj pyk jgk gS rFkk czsd ls mRiUu voeanu 6.0 eh-@ls-2 mlds }kjk czsd yxkus

dh vko';drk vuqHko gksus ds i'pkr~ r; dh x;h nwjh Kkr dhft;sA Ans :22 m 20.

Complete the following table :

fuEu lkj.kh dks iwjk dhft;sA Driver X Reaction time 0.20s Speed = 54 km/h Braking distance a = ………………….. A( decleration Total stopping on hard braking = distance 6.0 m/s2) b =………………….. Speed = 54 km/h Braking distance e = ………………….. B( deceleration Total stopping on hard braking = distance 7.5 m/s2) f =…………………..

Car Model

Driver Y Reaction time 0.30s Speed = 72 km/h Braking distance c = ………………….. Total stopping distance d =………………….. Speed = 72 km/h Braking distance g = ………………….. Total stopping distance h =…………………..

MªkbZ oj X i zfr fØ; kl e; 0.20 l s d. M

MªkbZ oj Y i zfr fØ; kl e; 0.30 l s d. M = 54 fdeh-@ ?k.Vkdhnw j h pky = 72 fdeh-@?k.Vkczsd yxkus A (i w j hr kdr l sczsd pky a = ………….. dhnw j h c = ………. y xkusi j czsd yxkusdh ayxhdq y nw jh 2 : duses : dusesayxhdq y nw jh voea nu = 6.0 eh-@l s-) b =……….. d =……….. dkj dkekW My

= 54 fdeh-@ ?k.Vkdhnw j h pky = 72 fdeh-@?k.Vkczsd yxkus l sczsd pky e = ………….. dhnw j h g = ………. y xkusi j czsd yxkusdh ayxhdq y nw jh 2 : duses : dusesayxhdq y nw jh voea nu = 6.0 eh-@l s-) f =……….. h =……….. B (i w j hr kdr

[Ans :(a) 19m

(b) 22m

(c) 33m

(d) 39m

(e) 15m

(f) 18m

(g) 27m

(h) 33m]

]

21.

A police jeep is chasing a culprit going on a moterbike. The motor bike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike? [HCV_Ch-3_Ex_21]

,d iqfyl thi] eksVjlkbZfdy ij lokj ,d pksj dk ihNk dj jgh gSA eksVj lkbZfdy ,d eksM+ ij 72 km/h dh pky ls fudyrh gSA thi bldk ihNk 90 km/h, ls dj jgh gS vkSj thi eksM+ ij eksVjlkbZfdy ds igq¡pus ds10 lSd.M ckn igq¡prh igq¡prh gSA ;g ekfu, fd os fu;r pky ls pyrs gSA eksM+ ds fdrus nwjh ckn thi eksVjlkbZfdy dks idM+ manishkumarphysics.in

Page # 25

Chapter # 3


ysxhA ? [Ans : 1.0 km] 22.

A car travelling at 60 km/h overtakes another car treavelling at 42 km/h. Assuming each car to be 5.0 m long, find the time taken during the overtake and the total road distance used for the overtake. 60 fdeh-@?kaVk dh pky ls xfr'khy ,d dkj] 42 fdeh-@?k.Vk dh pky ls xfr'khy ,d vU; dkj ls vkxs fudyrh gSA ;g ekurs gq, fd izR;sd dkj 5.0 e h - y E c h g S ] v k x z f u d y u s e s a y x k

le ;

Kkr

df j; s

rF k k

ikjd

dj u s

esa

lM+d

ij

pyh

xbZ

dqy

nwjh

Kkr

df j ; s A

Ans : 2s, 38m] 23.

A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g = 10 m/s2. ,d xsan 50 eh-@ls- dh pky ls Å/okZ/kj Åij dh vksj iz{ksfir dh xbZ gSA Kkr dfj;sA (a) vf/kdre Å¡pkbZ] (b) vf/kdre Å¡pkbZ rd igqapus esa yxk le;] (c) vf/kdre Å¡pkbZ dh vk/kh Å¡pkbZ ij pkyA (g = 10 eh-@ls-2 ysaI) Ans :(a) 125m , (b) 5s, (c) 25 2 m/s]

24.

A ball is dropped from a ballon going up at a speed of 7 km/s. If the ballon was at a height 60 m at the time of dropping the ball , how long will the ball take in reaching the ground? Åij dh vksj 7 eh-@ls- dh pky ls xfr'khy xqCckjs ls ,d xsan NksM+h tkrh gSA ftl le; xsan NksM+h x;h gSA xqCckjs dh Å¡pkbZ 60 eh- gS] tehu rd igq¡pus esa xsan dks fdruk le; yxsxk \ Ans : 4.3s]

25.

A stone is thrown vertically upward with a speed of 28 m/s.(a) Find the maximum height reached by the stone.(b) Find its velocity one second before it reaches the maximum height. (c) Does the answer of part (b) changes if the intial speed is more than 28 m/s such as 40 m/s or 80 m/s? ,d iRFkj dks 28 eh-@ls- pky ls m/okZ/kj Åij dh vksj QSadk x;k gSA (a) iRFkj }kjk r; dh xbZ vf/kdre Å¡pkbZA (b) vf/kdre Å¡pkbZ izkIr djus esa ,d lsd.M iwoZ osx (c) ;fn vkfjfEHkd osx 28 eh-@ls- ls vf/kd tSls 40 eh-@ls;k 80 eh-@ls- gks rks D;k Hkkx gks rks D;k Hkkx (b) dk mÙkj ifjofrZr gks tk;sx \ Ans :(a) 40m, (b) 9.8 m/s, (c) No]

26.

A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped. ,d Åaps Hkou ds Åij cSBk gqvk O;fä 1 lsd.M ds fuf'pr le;kUrjky ij xsansa fxjk jgk gSA tc NBh xsan fxjk;h

tk jgh gks rks rhljh] pkSFkh rFkk ikapoh xsan dh fLFkfr;k¡ Kkr dfj;sA Ans :44.1 m, 19.6 m and 4.9m below the top 27.

A healthy youngman standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height. (1.8 m)? ,d rkdroj LoLFk O;fä 11.8 eh- Åaps Hkou ls 7 eh- nwjh ij [kM+k gS] og ns[krk gS fd Åijh eafty ls ,d cPpk fQlydj fxj x;k gSA cPps dks viuh ckgksa esa (ÅapkbZ 1.8 eh-) >syus ds fy;s mls fdl pky (,dleku ekusa) ls Hkkxuk

pkfg;sA

Ans : 4.9 m/s 28.

An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform ? 12.1 eh- Å¡ps csj ds isM+ ij ,d fpfM+;k cSBh gqbZ gSA bl isM+ ds uhps ls 6 fdeh- izfr ?k.Vk dh pky ls ,u-lh-lhijsM xqtj jgh gSA fdlh fof'k"V {k.k ij fpfM+;k ,d csj fxjkrh gSA dkSulk dSMsV (ml {k.k ij dSMsV ds isM+ ls nwjh crkb;s) onhZ ij ;g csj fxjskxk ? Ans : 2.62m

29.

A ball is dropped from a height .If it takes 0.200s to cross the last 6.00m before hitting the ground, find the height from which it was dropped.Take g = 10 m/s2. [HCV_Chp-3_Ex-29] fdlh Å¡pkbZ ls ,d xsan fxjk;h x;h gSA ;g tehu ls Vdjkus ls iwoZ bldks vafre 6 eh- r; djus esa 0.200 lsd.M le; yxrk gS] Kkr dfj;s fd xsan dks fdl Å¡kbZ ls fxjk;k x;k Fkk \ g = 10 m/s2. Ans : 48m

30.

A ball is droped from height of 5 m onto a sandy floor and pentrates the sand up to 10 cm before comming to rest. Find the retardation of the ball in sand assuming it to be uniform. ,d xsan dks 5 eh- Å¡pkbZ ls jsrhys Q'kZ ij fxjk;k x;k gS] ;g fojkekoLFkk esa vkus ls iwoZ 10 lseh- xgjkbZ rd /kal manishkumarphysics.in

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tkrh gSA jsr esa xsan dk voeanu (,delku ekurs gq,) Kkr dfj;sA Ans :490 m/s2] 31.

An elevator is descending with uniform acceleration. To measure the acceleration , a person in the elevator drops a coin at the moment the elavator starts. The coin is 6 ft above the floor of the elevator at time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

,d fy¶V ,d leku Roj.k ls uhps dh vksj vk jgh gSA tc fy¶V pyuk izkjEHk djrh gSA fy¶V esa fLFkr ,d O;fä bldk Roj.k ekius ds fy;s ,d f lD dk fxjkrk gS A t c f lD ds dks f xjk ;k x;k gS ml {k . k ij flDdk f y ¶V ds Q'kZ ls 6 Åij gSA O;fDr izsf{kr djrk gS fd flDdk 1 lsd.M esa Q'kZ ls Vdjkrk gSA bl vkadM+ksa ds vk/kkj ij fy¶V dk Roj.k Kkr dfj;sA Ans : 20 ft/s2] 32.

A ball is thrown horizontally from a point 100m above the ground with a speed of 20 m/s. Find(a) the time it takes to reach the ground, (b) the horizontal distance it travells before reaching the ground, (c) the velocity (direction and manitude) with which it strikes the ground. (g = 10 m/s2) tehu ls 100 eh- Å¡pkbZ ij fLFkr ,d fcUnq ls ,d xsan 20 eh-@ls- dh pky ls {kSfrt fn'kk esa Qsadh tkrh gSA Kkr dfj;sA (a) bldks tehu rd igqapus esa yxk le; (b) tehu rd igqapus ls iwoZ blds }kjk r; {kSfrt nwjh (c) tehu ls Vdjkrs le; bldk osx (ifjek.k ,oa fn'kk) (g = 10 m/s2) Ans : (a) 4.5s, (b) 90m

33.

(c) 49m/s, (d)  tan–1 5 = 66º with horizontal

A ball is thrown at a speed of 40 m/s at an angle of 60º with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take g = 10 m/s2. ,d xsan {ksfrt ls 60° dks.k ij 40 eh-@ls- pky ls Qsadh xbZ gSA Kkr dfj;sA (a) blds }kjk r; dh x;h vf/kdre Å¡pkbZ (b) xsan dh ijklA )g = 10 m/s2) Ans :(a) 60m , (b) 80 3 m

34.

In a soccer practice session the football is kept at the centre of the field 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of 64 ft/s at an angle of 45º to the horizontal. Will the ball reach the goal-post ? QqVcky ds vH;kl ds fy;s] xksy ds 10 QhV Å¡ps [kEHkksa ls 40 xt nwj eSnku ds dsUnz ij QqVcky j[kh gqbZ gSA xksy djus dh ps"Vk djus ds fy, fdd yxkus ls QqVcky {ksfrt ls 45° dks.k ij 64 [email protected] dh pky ls xfr'khy

gksrh gSA D;k QqVcky xksy esa igqapsxh \ Ans : Yes 35.

A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated at a distance of 2.0 m from the goli of the second player. This second player has to project his goli by keeping the thumb of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wine. If the height from which the horizontally, with that speed it be projected so that it directly hits the stationary goli without on the ground earlier?

Hkkjr ds xk¡oksa esa dapksa dk [ksy izpfyr gS] ;g dkap dh NksVh&NksVh xksfy;ksa ls [ksyk tkrk gS] ftUgsa daps dgrs gSA ,d f[kykM+h dh xksyh] nwljs f[kykM+h dh xksyh ls 2.0 eh- nwj gSA nwljs f[kykM+h dks viuh xksyh cka;s gkFk dh rtZuh rFkk e/;ek Å¡xyh ds chp Qalkdj viuh xksyh ds LFkku ij nka;s gkFk dk vaxwBk j[kdj] nka;s gkFk dh rtZuh Åaxyh ls xksyh dks iz{ksfir djuk iM+rk gSA ;fn iz{ksfir dh xbZ xksyh] izFke f[kykM+h dh xksyh ij Vdjk tkrh gS rks nwljk f[kykM+h thr tkrk gSA ;fn xksyh dks 19.6 lseh dh Å¡pkbZ ls {kSfrt fn'kk esa iz{ksfir fd;k tkrk gSA rks xksyh dk og iz{ksi.k osx Kkr dfj;s] ftlls xksyh fLFkj xksyh ls Vdjkus ls igys gh tehu ls u Vdjk,A Ans ; 10 m/s 36.

Figure shows a 11.7 ft wide ditch with the approach roads at an angle of 15º with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch? fp=kkuqlkj 11.7 QhV pkSM+s ukys dh lEidZ lM+dsa {kSfrt ls 15º dks.k ij gSA lM+d ij xfr'khy eksVj lkbfdy dh

U;wure pky fdruh gks fd og ukys dks lqjf{kr :i ls ikj dj ys \ ;gk¡ ;g ekfu;s fd eksVj lkbfdy dh yEckbZ 5 QhV gS rFkk ;g lM+d dks rc NksM+rh gS tc bldk vxyk Hkkx lEidZ lM+d ls ckgj fudy tkrk gSA 11.7 ft 15º

Assume that the length of the bike is 5 ft, and it leaves the road when the front part runs out of the approach manishkumarphysics.in

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road. 37.

[Ans : 32 ft/s]

A person standing on the top of a cliff 171 ft. high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall? ,d O;fDr 171 QhV Å¡ph ehukj ds Åij [kM+k gS] ;g ehukj ls {kSfrt fn'kk esa 228 QqV nwjh ij tehu ij [kM+s vius fe=k dh vksj ,d iSdsV Qsadrk gSA ;fn og iSdsV dks Bhd vius fe=k dh fn'kk esa 15.0 QhV@ls- dh pky ls Qsadrk

gS

rks

iSdsV

fdruh

nwjh

iwoZ

gh

fxj

tk;sxk

\

Ans :192 ft.] 38.

A ball is projected from a point on the floor with a speed of 15 m/s at angle of 60º with the horizontal. Will it hit a vertical wall 5m away from the point of the projection and perpendicular to the plane of projection without hitting the floor ? Will the answer differ if the wall is 22m away? Q'kZ ij fLFkr ,d fcUnq ls ,d xsan 15 eh@ls- dh pky ls {ksfrt ls 60º dks.k ij iz{ksfir dh tkrh gSA D;k ;g xsan Q'kZ Q'kZ ij Vdjk, fcuk iz{ksi.k fcUnq ls 5 eh- nwj fLFkr ,oa iz{ksi.k ry ds yEcor~ fLFkr ,d m/okZ/kj nhokj ij VDdj ekjsxh \ ;fn nhokj 22 eh- nwj fLFkr gks rks mÙkj esa dksbZ ifjorZu gksxk] D;k \ Ans :Yes, Yes]

39.

Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle  with the horizontal.

iz{ksI; dh mu nks fLFkfr;ksa ds e/; tc ;g vf/kdre Å¡pkbZ dh vk/kh Å¡pkbZ ikj djrk gS] vkSlr osx Kkr dfj;sA bldks {ksfrt ls  dks.k ij u pky ls iz{ksfir fd;k x;k gSA Ans :ucos  ,horizontal in the plane of projection 40.

A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode verticallly below the plane is the statement true if the plane files with uniform speed but not horizontally?

{ksfrt fn'kk esa ,d leku pky ls xfr'khy ok;q;ku ls ,d ce fxjk;k tkrk gSA O;ä dfj;sa fd ce] ok;q;ku ds Bhd m/okZ/kj uhps foLQksfVr gksxkA D;k ;g dFku rc Hkh lR; gksxk tc ok;q;ku ,dlekuosx ls xfr djs fdUrq ksfrt ry esa ugha \ 41.

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s2 and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car. ,d jsy dkj esa [kM+k gqvk yM+dk ,d xsan Bhd m/okZ/kj Åij dh vksj Qsadrk gSA dkj {ksfrt iVjh ij 1 eh-@ls-2 Roj.k ls xfr dj jgh gSA rFkk m/okZ/kj fn'kk esa iz{ksi.k osx 9.8 eh-@ls- gSA dkj esa xsan yM+ds ls fdruh nwj fxjsxh \ Ans :2m

42.

A staircase contains three steps each 10 cm high and 20 cm wide (Figure ) What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane? lh<+h esa 10 lseh- Å¡ph ,oa 20 lseh- pkSM+h rhu lhf<+;k¡ gSA lcls Åijh lery yq<+d jgh xsan dk U;wure {ksfrt osx

fdruk gks fd ;g lh/kh lcls uhps lery ij Vdjk;s \ Ans : 2m /s]

43.

A person is standing on atruck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8m. Find the speed and the angle of projection (a) as seen from the truck , (b) as seen from the road. lery {ksfrt lM+d ij 14.7 eh-@ls- dh ,d leku pky ls xfr'khy Vªd esa ,d O;fä [kM+k gqvk gSA ;g O;fä ,d xsan bl izdkj Qsadrk gS fd tc Vªd 58.8 eh- py pqdk gksrk gS rks xsan iqu% Vªd esa ykSV vkrh gSA Kkr dfj;sa iz{ksi.k dks.k rFkk pky (a) Vªd ls ns[kus djus ij (b) lM+d ls ns[kus ijA Ans : (a) 19.6 m/s upward (b) 24.5 m/s at 53º with horizontal]

44.

The benches of a gallery in a cricket stadium are 1 m wide and 1m high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at 35 m/s at angle of 53º with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit ? fØdsV LVsfM;e dh xSyjs h esa 1 eh- pkSM+h rFkk 1 eh- Å¡ph cSaps gSA ,d cYysckt tehu ls 1 eh- Å¡pkbZ ls xsna dks ekjdj manishkumarphysics.in

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yEck NDdk yxkrk gSA xsan {ksfrt ls 53º dks.k ij 35 eh-@ls- pky ls xfr izkjEHk djrh gSA igyh cSap cYysckt ls 110 eh- nwj fLFkr gS rFkk cSapsa xfr ds ry ds yEcor~ gSA fdl cSap ij xsan fxjsxh \ Ans : Sixth] 45.

A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/ s. find the minimum and maximum angles of projection for sucessful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level. , d O ; f D r u n h d s f d u k j s i j c S B k g S A o p g 1.0 e h - y E c h u k o d h l h / k e s a u k o d s d s U n z l s 5.5 e h - n w j c S B k g S A o g u k o , d l s o Q S a d u k p k g r k g S A o g l s o d k s d s o y 10 e h - @ l s - d h p k y l s Q S a d l d r k g S ] l Q y r k i w o Z d Q S a d u s

ds fy;s iz{ksi.k dks.k dk U;wure rFkk vf/kdre yhft;s fd iz{ksi.k fcUnq rFkk uko dk fdukjk g S A

eku Kkr dfj;sA ;g eku ,d gh {ksfrt lery esa

[Ans :Minimum angle 15º, Maximum angle 75º but there is an interval of 53º between 15º and 75ºm, which is not allowed for sucessful shot.] 46.

A river 400 m wide is flowing at a rate of 2.0 m/s.A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. (a) Find the time taken by the boat to reach the opposite bank.(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank. 400 eh- pkSM+h unh 2.0 eh-@ls- dh pky ls izokfgr gks jgh gSA unh ds cgko ls yEcor~ fn'kk esa,d uko ty ds lkis{k 10 eh-@ls- osx ls tk jgh gSA (a) uko dks foijhr fdukjs rd igq¡pus esa yxk le; Kkr dfj;sA (b) uko lkeus okys

fdukjs ij vkjEHk fcUnq ds lh/ks okys fcUnq ls fdruh nwj igqapsxh \ Ans :(a) 40s, (b) 80 m 47.

A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. (a) If he heads in a direction making an angle  with the flow, find the time he takes to cross the river.. (b) Find the shortest possible time to cross the river. 5 fdeh-@?k.Vk dh pky ls izo kfgr gks jgh 500 eh- pkSM+h unh dks ,d rSjkd ikj djuk pkgrk gSA mldh ikuh ds lkis{k pky 3 fdeh@?k.Vk gSA (a) ;fn og ikuh ds izo kg dh fn'kk ls  dks.k ij rSjrk gS rks unh ikj djus esa yxk le; Kkr dfj;sA (b) unh ikj djus esa yxus okyk U;wu re le; Kkr dfj;sA [Ans : (a) 10 minute , (b) 10 minute ] θ sin

48.

Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to his point.Find the minimum distance that he has to walk.

fiNys iz'u ij fopkj dfj;sA O;fä nwljs fdukjs ij vius izkjEHk fcUnq ds Bhd lkeus okys fcUnq ij igqapuk pkgrk gS ;fn og nwljs fdukjs ij fdlh vU; fcUnq ij igqaprk gS rks Bhd lkeus okys fcUnq rd iSny pyuk iM+sxkA iSny pyus ds fy;s vko';d U;wure nwjh Kkr dfj;sA 2 km] 3 An aeroplane has to go from a point A to another point B,500 km away due 30º east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B. ,d ok;q;ku dks fcUnq A ls 500 fdeh- nwj ,oa iwoZ ls mÙkj dh vksj 30º dks.k ij fLFkr fcUnq B rd tkuk gSA ok;q mÙkj dh vksj 20 eh-@ls- dh pky ls izokfgr gks jgh gSA ok;q;ku dh gok esa pky 150 eh-@ls- gSA (a) fcUnq B rd igqpaus ds fy;s pkyd dks ok;q;ku fdl fn'kk esa mM+kuk gksxk \ (b) ok;q;ku dks A ls B rd igqpaus esa yxk le; Kkr dfj;sA Ans :(a) sin–1(1/15) east of the line AB, (b) 50 min]

Ans : 49.

50.

Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound t1 time after he sees the event. A and B interchange their position and the experiment is repeated. This time B hears the drum t2 time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.


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,d eSnku esa x nwjh ij nks fe=k A o B [kM+s gS rFkk ok;q A ls B dh vksj izokfgr gks jgh gSA A ,d Mªe dks ctkrk gS rFkk B mldks ctkrs gq, ns[kus ds t1 i'pkr~ mldh /ofu lqurk gSA A o B miuh fLFkfr;k¡ ijLij ifjofrZr dj ysrs gS rFkk iz;ksx dks nksgjkrs gSA vc B ?kVuk dks ns[kus ds t2 le; i'pkr~ /ofu lqurk gSA fLFkj ok;q esa /ofu ds osx v rFkk ok;q ds osx u dh x.kuk dfj;sA nksuksa fe=kksa ds e/; izdk'k dh xfr esa yxs le; dks ux.; eku yhft;sA x1 1 Ans : 2  t  t 2  1 51.

 x1 1  ,    2  t1 t 2

  ] 

Suppose A and B in the previous problem change their positions in such that the line joining them becomes perpendicular to the direction of wind while maintaining the separation x. What will be the time lag B finds between seeing and hearing the drum beating by A ? ekukfd fiNys iz'u esa of.kZr A rFkk B vius fLFkfr;ks¡ bl izdkj cny ysrs gS fd mudks feykus okys js[kk ok;q izokg ds yEcor~ gks tkrh gS rFkk buds e/; nwjh x gh jgrh gSA A ds }kjk ctk;s x;s Bksy esa B ns[kus rFkk lquus esa fdruh

le; i'pkr izsf{kr djsxk \ x Ans : 52.

v 2  u2

Six particles situated at the corners of a regular hexagon of side ‘a’ move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other. a Hkqtk ds le"kV~Hkqt ds dksuksa ij fLFkr N% d.k ,d leku pky v ls xfr'khy gSA izR;sd d.k viuh fn'kk vius ls vkxz okys

dksus ij fLFkr d.k dh vksj j[krk gSA izR;sd d.k dks vkil esa feyus esa yxk le; Kkr dfj;sA Ans : 2a / v


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Kinema Tics

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