kinemat-hints and solution

8/18/2019 Kinemat-Hints and Solution

1/15

Solution to Subjective Assignments

LEVEL – I

1. Comparing the general expression 0v = (v0 cos θ0) î + (v0 sin θ0) ĵ with thegiven equation v0 = 20 î + 10 ĵ , we obtain

v0 cos θ0 = 20 and v0 sin θ0 = 10

!he x"component o# velocit$ v a#ter a time interval t = 1 sec, v x = v0 cos θ0 = 20m%sec because the hori&ontal component o# velocit$ o# a pro'ectile remains

constant and the corresponding $"component o# velocit$ v is given as v$ = (v0 sin

θ0) gt putting v0 sin θ0 = 10 m%sec, g = 10 m%sec2 and t = 1 sec, we obtainv$ = 10 10 * 1 = 0

⇒ !he required velocit$ = v = vx î + v$ ĵ = 20 î

2. !he velocit$ o# the bod$ at an$ time t #rom the instant o# pro'ection is given as

= 0 gt

e#erring to the derived #ormulae, time average velocit$ is given as

∫ = 0t

0dt+

t

1+ -utting = (0 gt) we obtain

∫

−=−⊥= 0

0

t

0

000

t 2

tg+dt)gt+(+

.here t0 = total time o# ascent /ince = 0 at t = t0 ⇒ t0 =g

+0

-uttingg

+t 00 = , we obtain

2

++ 0=

3. !he velocit$ o# the bomb 'ust a#ter its release is equal to the velocit$ o# the

aeroplane !he displacement r can be directl$ written b$ re#erring the derived

ormula as r = $ 1g$

v2 20 +

$ putting

v0 = 100 m%sec, g = 10, $ = 1000 m, we obtain

r = 132 m (approximatel$)

4nd φ = tan"1

0v2

g$ = tan"1

×100

2

100010

⇒ φ = tan"1 00

Bhatnagar IIT-JEE/PMT Academy. C-5, ama Par!,"ttamnagar, #e$ %e&hi


2/15

4. 5et the initial velocit$ has the

magnitude v0, inclined at an angle

θ0 with hori&ontal !he time ta6enb$ the #ootball to cover 70 m is

equal to 7 seconds

⇒ t =00 cosv

70

θ

⇒ v0 cos θ0 = 70%t = 70%7 = 10 m%s

!he 6inematical equation #or

vertical motion is

$ = (v0 sin θ0)t"2gt

2

1

50 mt ' 5 (

)50, - *.5+

θ0

v0

y

t ' 0

y ' *.5 m)0,0+

⇒ (v0sin θ0) (t) =2gt

2

1" $

⇒ (v0sin θ0) (7) =2)7)(10(

2

1 " 17

⇒ v0sin θ0 = 287

7123= m%s

'9)sinv(i9)cosv(v 00000 θ+θ=

) '928i910(v0 += m%s

⇒ v0 = 2::8 m%s ; θ0 = tan−1 10

28= 67.960.

5. !he distance travelled b$ the roc6et in the #irst 1 minute in which resultant

acceleration is verticall$ upwards and 10 m%s2 will be

h1 = 0 × :0 +2

1 × 10 × :02 = 1


3/15

6. 5et the$ meet a#ter time t #rom the instant o# release at point - as shown in the

#igure /ince /1 is downwards,

/1 = /g + /i = @gt2 + 1t (a)

; /2 is upwards ⇒ /2 = /i " /g

⇒ /2 = 2t " @ gt2 (b)

(a) + (b) ⇒ /1 + /2 = (1 + 2)t

⇒++

h t

+=

(/1 + /2 = h ; 1 = 2 = )

⇒2V

) t =

t = t

p

t = 0

v1 = v

t = 0

v2 = v

1s

2s

7. &!'5et the total distance covered = h m, and the totaltime ta6en = ! seconds

4ccording to the question,

the path covered during !th second = h%2

⇒ Aistance covered during (!" 1) seconds = h%2⇒ (1% 2)g(!"1)2 = h%2,where h =(1%2) g!2

⇒ (1%2)g (! 1)2 = (1% 8)g!2

⇒ (!"1)%! = ± 1%√2⇒ ± ! = √2! " √2⇒ !(√2 1) = √2

t '0

A

t 'T-*

t 'TC

B

⇒ ! =12

2

±s

/ince ! B 1, we accept the other value

⇒ ! =12

2

−= √2(√2+1)= &2* 2' s.

&b' !he height o# #all = h = (1%2) g!2 = (1% 2)(


4/15

θ=θ−=∆2

sin22r ) cos 1(2r r 222

⇒2 sinr 2 r

θ=∆

ow the magnitude o# average velocit$

= v =t

r

∆∆

t

2 sinr 2

v

θ

=

(i)

we 6now that

r

v ;

t=ωω=θ (when w =

constant)

⇒v

r t θ= (ii)

r ∆

v

Ft=t

t=0

-θ v

1r

2r

G

-utting t #rom (ii) in (i) we obtain

θ

θ

=θ

θ= 2

sinv2

v

r

)2%(sin r 2v ⇒

-

.2'& sin2

v

v θ=

10. .e 6now that velocit$ o# rain wrt man v rm is given as mr rm v v v −=

5et21 mm

v ; v are the initial ; #inal velocities o# the man, then

2mr 2rm1mr 1rmv"vv ;v"vv ==

⇒ 2m2rm1m1rmr vvv v v +=+=

e#erring the vector diagram we obtain

21

21 rmmr

vvv +=1m

v

1rmv

θ2rm

v

2mv

21rm

2r vvv += (a)

-utting C = ∆v = nvwe obtain

θ= cot nvv1rm (b)

Hsing (a) ; (b) we #ind22

r ) cot v(vv θη+=-cot/1v v 22" +=

θα

1rmv 2rm

v

1mv

2mv

A

C 4

v nv


!"t#I#$%&S' #(#4


5/15

LEVEL – II

1. 5et a#ter a time t the particles attain the ground level !he particles acquire equal

vertical velocit$ a#ter time t, that is equal to gt

!he hori&ontal components o# the velocities remain uncharged !here#ore the

total velocities o# the particle a#ter time t are given as

'9 gti9vv 11 +−=′ ; '9 gti9vv 22 +=′

/ince the$ move perpendicularl$ 'ust be#ore touching the ground, there#ore

0v v 21 =

⇒ 0 ) '9 gt i9(v ) '9 gti9v( 21 =++−

⇒ g2t2 = v1v2 ⇒ g

vvt

21=

!he vertical displacement o# each particle =2

1h = gt2 where h = height o# the

pole

;g

vvt

21=

⇒g2

vv 21=

=

2

21

g

vv g

2

1 h

+1 +2 =0 =0

=t =t +2 +1

2v ′ gt

1v′

gt

2. 4t the point o# collision, x1 = x2 and $1 = $2

⇒ vo cosθ1t1 = vo cosθ2t2 or t2 =

θ

θ

2

1

cos

cos

t1

.here θ1 and θ2 are the angles o# pro'ection and t1, t2 are the time o# #light

/imilarl$ #or equal vertical displacement ($1 = $2)

vosinθ1t1 − 2

1g 21t = vosinθ1t1 −

2

1g 212t

vosinθ1t1 − vosinθ2

( )222112

1 tt2

gt

cos

cos−=

θθ

vot1( )

θ

θ−θ=

θθ−θ

2

2

12

22

21

2

21

cos

coscost

2

g

cos

sin

⇒ t1 =( )

12

22

221o

coscos

cossinv2

θ−θ

θθ−θ

t2 = t1 =2

1

cos

cos

θθ

∴ !ime interval ∆t = t1 − t2 =( )

21

21o

coscosg

sinv2

θ+θθ−θ

/ubstituting the given values,


!"t#I#$%&S' #(#+


6/15

∆t = 10 s

3. orward Iourne$

ector C represents wind velocit$

4 represents velocit$ plane relative to

ground 4C represents velocit$ o# plane in still air

800cos i9 + (800 sinθ + 70) '9 = 4

tan87° = θ+θ

cos800

70sin800

4+

0

1 E

S

2

3

A

⇒ tan θ +<

secθ = 1

⇒ tan2θ − 2tanθ + 1 =:8

1tan2 +θ

⇒ :3 tan2θ − 12< tanθ + :3 = 0

⇒ tanθ =12:

7180

12:

70


7/15

∴ s = s1 + s2 = u1t + u2t = (u1 + u2)g

uu 21

/ubstituting the values, the distance between the particles =


8/15

⇒ i9)20v(v x −=ωµ + v$ '9

/ince the wind appears to blow #rom north east, the x ; $ components o# mv ω

must be equal

⇒ (vωm)$ = (vωm)x and both are negativehence vx " 20 = 10 " 20 = "10 m%s

⇒ wind speed = 10√2 m%s and the wind is blowing #rom north west

9 xdx

+d+a ==

⇒ dxx+d+ = ⇒ ∫ = dxx+d+

⇒ 2%320

2

x3

2

2

++=

− ⇒3%2

20

2

8

)++(3x

−=

-utting = 2 0 we obtain , x =

3%8

0

3%2

2

02

+3+

8

C

=

10. 5et us consider that velocit$ o# train with respect to ground

=v! and velocit$ o# car in the #irst part o# 'ourne$

= ve and velocit$ o# car in the last part o# 'ourne$

=v′e

4lso distance o# the point where the ob'ect #ell and where the car turns bac6

=x6m

?enceM ve − v! =:

16m%min (1)

v′e + v! = :

1

6m%min (2)

1v

x

e

= (3)

ev

N

′ = 2 (8)

/olving these (1), (2), (3) and (8)

ve = 1338 6m%hr, v′e = :: 6m%hr v! = 338 6m%hr x = 1338 6m


!"t#I#$%&S' #(#,


9/15

Solutions to 5bjective Assignments

LEVEL – I

1. !he stone is sub'ected to earthPs gravitational #ield a#ter losing contact with theelevator !here#ore its acceleration will be equal to g = < m%s 2 pointed verticall$

downwards

2. 5et this pro'ectiles be pro'ected with velocities

'9)sinv(i9)cosv( '9$i9vv1111$x1 11

θ+θ=+=

and '9)sinv(i9)cosv( '9$i9vv 2222$x2 22 θ+θ=+=

!he velocities o# the pro'ectiles a#ter a time t are given as

'9)gtsinv(i9)cosv(v11111

−θ+θ=′

'9)gtsinv(i9)cosv(v22222

−θ+θ=′

'9)sinvsinv(i9)cosvcosv(vv2211221121 θ−θ+θ−θ=′−′

/ince the relative velocit$ 21 vv ′−′ is a constant and so does not var$ with time,the locus o# one wrt the other is a straight line

3. v = a1t1 = a2t2

∴ t1 = (a2%a1) t2 = (8%2)t2 = 2t2t1 + t2 = 3 or 2t2 + t2 = 3t2

t2 = 1 sec and t1∴ v = 2 × 2 = 8 m%s

4. N =ct2 ; $ = bt2

⇒ ct2dt

dx= ; bt2

dt

d$=

/ince v =22

dt

d$

dt

dx

dt

ds

+

=

⇒ v = 22 )bt2()ct2( +

⇒ v = 2t 22 cb +

+. 5et G- = r 4ngular speed about the origin = ω =tp0

t0p v where,r

v = !he

component o# velocit$ o# - wrt G perpendicular to G-

⇒r

sinv

θ=ω where r = b cosec θ

⇒b

sinv

2θ=ω

⇒ (C) is correct


!"t#I#$%&S' #(#9


10/15

!he angular speed o# an$ point - wrt another point F = ωpq = E(!he componento# relative velocit$ between them is perpendicular to pq) % (!he distance o#

separation pq)D

!here#ore $ou can neither writeb

v

sinop

v nor

cosecb

v

op

v=

θθ=

!he last choice (A) is dimensionall$ wrong

6. 5et the particle be pro'ected verticall$ up"ward with velocit$ u #urther, let t bethe time when pro'ectile be at height h !hen

h = ut "2

1gt2 or

2

1gt2 ut + h = 0

∴ t2 " 0g

h2t

g

u2=+ (1)

#rom equation (1)

sum o# roots, t1 + t2 = " g

u2

a

b= (2)

product o# roots, t1t2 = gh2

a

c = (3)

Qiven that3

1

t

t

2

1 = or t2 = 3t1 (8)

#rom equation (2) and (8), we get

8t1 = g2

uort

g

u21 = (7)

#rom equation (3) and (8) we get

32

1t =

g

h2

g2

u3or

g

h22

=

or3h8

g2u2 = (:)

4lso, maximum height, ? =g2

u2

()

#rom eq (:) and () we obtain,

? =8

?3orh

3

h8=

7. $ geometr$ o# the #igure the length o# the string is = 8x 4 + x

⇒dt

dx

dt

dx8

dt

d 4 +=

/ince the length o# the string is constant,dt

d = 0

-utting

4

4 vdt

dx ;v

dt

dx==

8

v v 4 = numericall$

4 = 4v"v = v 4 + v =

8

v + v = 127v = 127v


!"t#I#$%&S' #(#10


11/15

/ince the displacement o# is #our times that o# 4, their speeds can not be equal

,. !ime ta6en =mv

d

= 2w

2mw vv

d

− = 22


12/15

1+. !he height at which stone was released = 80melocit$ o# balloon in the upward direction at the time o# release o# stone

= 127 ×


13/15

1. = 2dx

tdt

S(i)

31 t$

2 3=

2d$ t

dt 2= S(ii)

t = 1, vx = 1, v$ =*

19 9v i '2

= +r

2

2

d x2t

dt= S(iii)

2

2

d $t

dt= S(iv)

at t = 1 s ax = 2 and a$ = 1

ˆ ˆa i j∴ = +r

2. Hsing constrained equation, / *c0(v vθ =

Gn di##erentiation / *c0(a aθ =

3. 2 2($ h) x h− + + = l

2 2

d$ x dx0

dt dtx h+ =

+

2 2

d$ x dx

dt dtx h= −

+

Ady vdt 5

= −

, ' - c m

h

'

1

c

m

y

4

3J u J v

7= S(i)

2 2

42 2 2

d $ hv

3dt (x h )2

=+

4 3

1:a v

(7)=

4

1:a v

127= S(ii)


!"t#I#$%&S' #(#13


14/15

4. i#

2 2 2 2u cos u sin

g 2g

θ θ>

*tan /−θ <

i#

2 2 2 2u cos u sin

g 2g

θ θ

<1tan 2−θ >

+. inal velocit$ o# each ball

v )/gh+=!his is independent o# path

urther,1 2h

tsin g

= ÷θ

or *

t(in

µθ

#or same h

1

2

t sin

t sin

β∴ =α

7. 4s velocit$ o# the particle increases #rom&ero and #inall$ it decreases to &ero,hence acceleration can not remainpositive #or all the time o# motion>inimum magnitude o# acceleration canbe #ound out b$ the velocit$" time curveAisplacement = 4rea bounded b$ the graph and time axis

>agnitude o# minimum acceleration = 8 m%s2∴ (4) ; (C)

COMPREHENSIONS

1$

dv2t

dt=

∴ 2$d$

v tdt

= =

and $ =t

2 Ari#t = 0 × time o# crossing

3 0x v t= S(i)3t

$3

= S(ii)

#rom (i) and (ii)


!"t#I#$%&S' #(#14


15/15

3

3

0

x$

3v=

8 . 0 2 - 0 2

* 0 m / (

5 m / (

α

v ,

v y5 - m / (

v

!ime a#ter which velocit$ vector becomes perpendicular to initial velocit$ vector is

u 10 2t

gsin 10 sin :0 3= = =

θ seconds

5et v$ be the vertical component o# velocit$ at that instant thenv$ = u + at

v$ =10 2

7 33

×−

v '

v ' 5,

α

v ',

5

-

* 0

-

$

7v

3= −

2

2 7v 73

∴ = + ÷

10

v3

= m%s2v

gcos,

α =

2

vgcos

=α

= 203 3

= m

7 4cceleration o# particle is equal to acceleration due to gravit$

: at = g sin α73

1010

3

= × = 7 m%s2

%A( (E 5LL5I8

1 jo o 99u :0cos30 :0sin30→

= i +

( )$ 9a g '= −a 0=

x


!"t#I#$%&S' #(#1+

kinemat-hints and solution

Documents