Kinemat-Hints and Solution

Solution to Subjective Assignments LEVEL – I 1.

 Comparing the general expression v 0 = (v0 cos 0) ˆ i + (v0 sin 0) ˆj with the given equation v0 = 20 ˆ i + 10 ˆj , we obtain v0 cos 0 = 20 and v0 sin 0 = 10.

The x-component of velocity v after a time interval t = 1 sec, v x = v0 cos 0 = 20 m/sec because the horizontal component of velocity of a projectile remains constant and the corresponding y-component of velocity v is given as vy = (v0 sin 0) – gt; putting v0 sin 0 = 10 m/sec, g = 10 m/sec2 and t = 1 sec, we obtain vy = 10 – 10 × 1 = 0  The required velocity = v = vx ˆ i + vy ˆj = 20 ˆ i 2.

The velocity V of the body at any time t from the instant of projection is given as V = V0 – gt. Referring to the derived formulae, time average velocity is given as 1 t0 V   V dt; Putting V = (V0 – gt) we obtain t 0 t0 t   V    ( V0  gt ) dt   V0  g 0  t0 0 2  V0 Where t0 = total time of ascent. Since V = 0 at t = t0  t0 = . g V0 V Putting t 0  , we obtain V  0 . g 2

3.

The velocity of the bomb just after its release is equal to the velocity of the aeroplane. The displacement r can be directly written by referring the derived Formula as r = y

2v 02 1 gy

By putting v0 = 100 m/sec, g = 10, y = 1000 m, we obtain r = 1732 m (approximately) And



 = tan-1 

  = tan 0.707



 gy v 0  = tan-1 2  







 10  1000 100   2 

-1

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Part-I-KM(S) -PH-2

4.

Let the initial velocity has the magnitude v0, inclined at an angle 0 with horizontal. The time taken by the football to cover 50 m is equal to 5 seconds. 50  t= v 0 cos 0  v0 cos 0 = 50/t = 50/5 = 10 m/s. The kinematical equation for vertical motion is 1 – y = (v0 sin 0)t- gt 2 2 1  (v0sin 0) (t) = gt 2 - y 2 1  (v0sin 0) (5) = (10)(5)2 - 1.5 2 123.5  24.7 m/s.  v0sin 0 = 5  v 0  ( v 0 cos 0 )ˆi  ( v 0 sin 0 )ˆj  v 0  (10 ˆi  24.7 ˆj ) m/s.  v0 = 26.64 m/s & 0 = tan1

5.

y v0

t=0

0

(0,0) y = 1.5 m

x (50, - 1.5) 50 m

t=5s

24.7 = 67.960. 10

The distance travelled by the rocket in the first 1 minute in which resultant acceleration is vertically upwards and 10 m/s2 will be 1 h1 = 0  60 +  10  602 = 18000 m 2 and velocity acquired by it will be v = 0 + 10  60 = 600 m/s Now after 1 minute the rocket moves vertically up with initial velocity of 600 m/s and acceleration due to gravity opposes its motion. So it will go to a height h2 till its velocity becomes zero.  0 = (600)2 – 2  10  h2  h2 = 18000 m. (a) Maximum height reached by the rocket from the ground H = h1 + h2 = 18 + 18 = 36 km. (b) Let t be the time taken after 1 min to reach the maximum height  0 = 600 – gt  t = 60 sec. i.e. After finishing fuel the rocket goes up for 60 s i.e. 1 minute more.

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM(S) -PH-3

6.

Let they meet after time t from the instant of release at point P as shown in the figure. Since S1 is downwards, t=0 S1 = Sg + Si = ½gt2 + V1t (a) & S2 is upwards  S2 = Si - Sg  S2 = V2t - ½ gt2

7.

t=t p  s2 p

v2 = v t=0

h 2V

(a) Let the total distance covered = h m, and the total time taken = T seconds. According to the question, the path covered during Tth second = h/2.  Distance covered during (T- 1) seconds = h/2  (1/ 2)g(T-1)2 = h/2, where h =(1/2) gT2 .  (1/2)g (T – 1)2 = (1/ 4)gT2  (T-1)/T =  1/2   T = 2T - 2  T(2  1) = 2  T=

v1 = v 0

(b)

(a) + (b)  S1 + S2 = (V1 + V2)t h  t  VV (S1 + S2 = h & V1 = V2 = V)  t 

 s1 p

2 2 1

A t =0

B

C

t =T-1

t =T

s

Since T < 1, we accept the other value. 2

T=

2 1

= 2(2+1)= (2+2) s.

(b) The height of fall = h = (1/2) gT2 = (1/ 2)(9.8)(2 + 2)2  57 m. 8.

     S1  S2 v1t1  v 2 t 2 Vav   t1  t 2 t1  t 2

= 9.

5 ˆ ˆ 5  10 ˆi  10  5 ˆj [ i  j ] .  5 2 due north of east. = 3 3 10  5

Let the particle move from P to Q in a time interval t. The change in position vector 

 r



  r2  r1



r12  r22  2r1r2 cos 

Putting r1 = r2 = r we obtain

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Part-I-KM(S) -PH-4

 r 

2r 2 (1  cos ) 

    2

2r 2 2 sin 2 

 . 2 Now the magnitude of average velocity



r  2 r sin

 2 r sin 2 v t

r = v = ; t

(i) we know that  v   &   (when w t r constant) r  t (ii) v Putting t from (ii) in (i) we obtain  2 v sin 2 r sin ( / 2) 2 v   r  v 10.

v

Qt=t  r

 r2 O

 v



P t=0

 r1

=

v 2 sin (/2) .  v θ

   We know that velocity of rain w.r.t. man v rm is given as v rm  v r  v m .   Let v m1 & v m 2 are the initial & final velocities of the man, then       v rm1  v r - v m1 & v rm2  v r - v m2 v rm2 v rm1        v r  v rm1  v m1  v rm2  v m2 v m1

Referring the vector diagram we obtain v r  v m 2  v rm 2 1

v m2

1

v r  v 2  v rm2 1

. . .(a)

Putting BC = v = nv we obtain v rm1  nv cot  (b) Using (a) & (b) we find vr 

v 2  (v cot ) 2

vr  v 1  η2 cot 2 θ

A

v v m1

B v rm1

  D

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nv v m2

C v rm2

Part-I-KM(S) -PH-5

LEVEL – II 1.

Let after a time t the particles attain the ground level. The particles acquire equal vertical velocity after time t, that is equal to gt. The horizontal components of the velocities remain uncharged. Therefore the total velocities of the particle after time t are given as   v 1   v 1 ˆi  gt ˆj & v 2  v 2 ˆi  gt ˆj Since they move perpendicularly just before touching the ground, therefore   v1 . v 2  0 (  v1ˆi  gt ˆj ) . (v 2 ˆi  gt ˆj )  0  

 t

g2t2 = v1v2

v 1v 2 g

The vertical displacement of each particle = h 

1 2 gt where h = height of the 2

pole. &

t

V1

v 1v 2

t=0 t=0

g

1   h  g  2 

v1v 2   g 

2



v1v 2 2g

V1

t=t

V2

gt v1

2.

V2

t=t gt

v2

At the point of collision, x1 = x2 and y1 = y2

 cos  

1 t2 =   t1.  cos 2 

 vo cos1t1 = vo cos2t2 or

Where 1 and 2 are the angles of projection and t1, t2 are the time of flight. Similarly for equal vertical displacement (y1 = y2) 1 1 2 vosin1t1  g t12 = vosin1t1  gt 2 2 12 cos 1 g t1  t12  t 22 2 vosin1t1  vosin2 cos 2



vot1



sin 1  2  g 2  cos2 2  cos 2 1   t1   cos 2 2  cos2 2 

 t1 =

2v o sin 1  2  cos 2 cos2 2  cos2 1

t2 = t1 =

cos 1 cos 2

 Time interval t = t1  t2 =

2v o sin 1  2  g cos 1  cos 2

Substituting the given values, Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM(S) -PH-6

3.

t = 10.7 s Forward Journey Vector CB represents wind velocity. AB represents velocity plane relative to ground. AC represents velocity of plane in still air. 400cos ˆi + (400 sin + 50) ˆj = AB tan45 =

N B

45 W

C 

E

A

400 sin   50 400 cos 

S

sec   tan  + =1 8 tan2   1 64 2  63 tan   128 tan + 63 = 0

 tan2  2tan + 1 =

 tan =

105.5 128  508 140.5 or  126 126 126

  = 4811 or 3993 Hence;

| AB

|=

267.65ˆi  347.25ˆj

= 438.42 km/hr.

1000  2.28hr 438.42 During backward journey BC = Velocity of plane in shift air. CA = Velocity of wind. BA = velocity of plane in air. ACB   ; BAE  45 Now, BA  400 sin ˆi  (400cos  50) ˆj

Time taken =

Now;

N

B W

45 A



C

400 cos   50 400 sin 

tan45 = 1 =

  = 42 or 50. Hence, | BA | = |268 ˆi  247 ˆj | = 364 km/hr.

S

Hence, time required 1000 = =2.74 hr. 364 4.

  v 1   u1ˆi  gtˆj , v 2   u2 ˆi  gtˆj

  v1.v 2  0  t =

u1u2 g

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

E

Part-I-KM(S) -PH-7

 s = s1 + s2 = u1t + u2t = (u1 + u2)

u1u2 g

Substituting the values, the distance between the particles =

5.

Deceleration 

7  12 = 2.47 m. 9 .8

dv a v dt

dv =  a v  v dv =  ads ds Integrating under specified limits,

or, v



0

v dv   a

vo



3

s

ds

0

0

Again,



v0

 2  2  vo  3a 



s= 



t=

t

dv



  adt

v

0

2 v0 a

3

 2  2  2  vo ,    3a   a

Ans.:  6.

7.

vo

From the vector diagram ar 1  tan 30  at 3 ar 1 Required ratio is which is as above. at 3

 a

 ar

600

900  15 revolution/second. 60  0 = 2f = 2 (15) = 30  rad/s = initial angular velocity. Let after a time t the fan stops Angular displacement before it stops = 75 revolutions = 75  2 = 150 radian. Angular retardation;

The frequency of rotation f =

o = 3 rad/sec2. 2  Time taken to stops = o = 10 sec.  =

8.

Let us take the X-axis along west –east direction and Y- axis along south-north  direction. Let the velocity of the motor cyclist = v m,  the velocity of wind = v  = vx ˆi  v y ˆj    v m  10 ˆi and v m  v   v m Case I : 

 v m  ( v x  10)ˆi  v y ˆj

 Since the wind appears to blow from north, v m must have no x component  vx – 10 = 0  vx = 10 m/s     Case II : v m = 20 ˆi and v m = v   v m Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

300

 at  v

Part-I-KM(S) -PH-8 

 v   ( v x  20 )ˆi + vy ˆj

 Since the wind appears to blow from north east, the x & y –components of v m must be equal.  (vm)y = (vm)x and both are negative hence vx - 20 = 10 - 20 = -10 m/s  wind speed = 102 m/s and the wind is blowing from north – west.

9.

a

VdV  x dx



VdV 



V 2  V02 2 3 / 2  x 2 3

 VdV



x dx

 3( V 2  V02 )   4  



x dx 2/3

x

 9 2 V0   4 

Putting V = 2 V0 we obtain , x = 

10.



2/3

 3 V0  2

 

4/3

  

Let us consider that velocity of train with respect to ground =vT and velocity of car in the first part of journey. = ve and velocity of car in the last part of journey. =ve. Also distance of the point where the object fell and where the car turns back =xkm. 1 Hence: ve  vT = km/min . . . (1) 6 1 ve + vT = km/min . . . (2) 6 x 1 . . . (3) ve X =2 . . . (4) v e Solving these (1), (2), (3) and (4). ve = 13.34 km/hr, ve = 6.67 km/hr; vT = 3.34 km/hr; x = 13.34 km.

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Part-I-KM(S) -PH-9

Solutions to Objective Assignments LEVEL – I 1.

The stone is subjected to earth's gravitational field after losing contact with the elevator. Therefore its acceleration will be equal to g = 9.8 m/s 2 pointed vertically downwards.

2.

Let this projectiles be projected with velocities  v 1  v x 1 ˆi  y y 1 ˆj  ( v 1 cos 1 )ˆi  ( v1 sin 1 )ˆj  and v 2  v x 2 ˆi  y y 2 ˆj  ( v 2 cos 2 )ˆi  ( v 2 sin 2 )ˆj The velocities of the projectiles after a time t are given as  v1  ( v 1 cos 1 )ˆi  ( v 1 sin 1  gt )ˆj  v 2  ( v 2 cos 2 )ˆi  ( v 2 sin 2  gt )ˆj   v1  v 2  ( v 1 cos 1  v 2 cos 2 )ˆi  ( v 1 sin 1  v 2 sin 2 )ˆj   Since the relative velocity v1  v2 is a constant and so does not vary with time, the locus of one w.r.t the other is a straight line.

3.

v = a1t1 = a2t2  t1 = (a2/a1) t2 = (4/2)t2 = 2t2. t1 + t2 = 3 or 2t2 + t2 = 3t2 t2 = 1 sec. and t1  v = 2  2 = 4 m/s.

4.

X =ct2 & dx  2ct  dt Since v =  v=

dy  2bt dt

&

ds  dt

 dx     dt 

2

 dy    dt 

2



( 2ct )2  ( 2bt )2

 v = 2t 5.

y = bt2

b2  c 2

.

Let OP = r. Angular speed about the origin =  =

 v p0  r

t

 , where v p0

component of velocity of P w.r.t. O perpendicular to OP. v sin   where r = b cosec  r v sin2 b







(C) is correct.

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

t

= The

Part-I-KM(S) -PH-10

The angular speed of any point P w.r.t. another point Q = pq = [(The component of relative velocity between them is perpendicular to pq) / (The distance of separation pq)] Therefore you can neither write

v v v v  nor  op b cosec op sin b

The last choice (D) is dimensionally wrong. 6.

Let the particle be projected vertically up-ward with velocity u. further, let t be the time when projectile be at height h. Then 1 2 1 2 h = ut gt or gt – ut + h = 0 2 2 2u 2h t 0 g g

 t2 -

. . . . (1)

from equation (1) b 2u  a g c 2h  product of roots, t1t2 = a g

sum of roots, t1 + t2 = -

. . .. (2) . .. (3)

t1 1  or t2 = 3t1 . .. . (4) t2 3 from equation (2) and (4), we get Given that

4t1 =

2u u ort1  g 2g

. .. . (5)

from equation (3) and (4) we get  u  2h  or 3 g  2g 

3 t12 = or

2



2h g

u2 4h  2g 3

. .. . (6)

u2 Also, maximum height, H = . . . (7) 2g

from eq. (6) and (7) we obtain, 4h 3H orh  H= . 3 4 7.

By geometry of the figure the length of the string is  = 4xA + xB

dx dx d 4 A  B dt dt dt



Since the length of the string is constant, Putting

vA  VAB =

d =0 dt

dx A dx B  vA &  vB dt dt

vB numerically. 4

  v A - vB

= vA + vB =

vB + vB = 1.25vB = 1.25v 4

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Part-I-KM(S) -PH-11

Since the displacement of B is four times that of A, their speeds can not be equal. 8.

= =

9.

d

=

2 v mw  v 2w

480 102  82

 vw

  The angle between v mw and and v w is  say.  The required time of crossing = d t= v mw sin  1 .5 5 sin 600

=

3 5 3

hr 

3 5

d

vmw  vw

hr

The time period of revolution of the disc is equal to the time of motion of water drop in air.  2 / =

2H g



 = 2

g 2g .  2H H

5 t 2 Now dv = adt, integrating we have, 5 2 10t  t = 0  t = 0 or 8 sec., Accepted value 8 sec. 4  (C)

11.

a = 10 

12.

v1 + v2 = 4 m/s v1  v2 = 0.4 m/s  v1 = 2.2 m/s, v2 = 1.8 m/s.  (D)

13.

h = vo(5) 

14.

 vm

 vmw

480  80 s. 6

 t= 10.

 vw

d vm

Time taken =

1  10  (5)2 2 1 h =  vo(1)   10  (1)2 2  v0 = 20 m/s, t = 5 sec.  (C)

. . . (1) . . . (2)

   v  u  at  7ˆi  ˆj 

 v  (C)

5

2m/s

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Part-I-KM(S) -PH-12

15.

The height at which stone was released = 40m. Velocity of balloon in the upward direction at the time of release of stone = 1.25  8=10m/s Hence, when stone reaches ground, 1 40 = 10t   10  t2  t = 4 sec 2  (C)

16.

At time t0 velocity of the ball be equal to gt0 vertically downward. Relative velocity between the two balls remains constant at gt 0 as relative acceleration between them is zero, as both are moving under gravity. At the time 1 2 gt 0 . of dropping the second ball, the distance between the two balls = 2 Separation between the two balls increases linearly with time. Graph between separation and time will be a straight line upto the instant the first ball strikes the ground. The ground is perfectly inelastic means that first ball comes to rest just after striking the ground. After this happens separation between them continuously decreases parabolically.  (D) The tangential component of force = dK ds 2K the centripetal component = . R

17.

19.

Given ac = at  R 2 = R or 2 =  = d/dt d  dt Integrating we get   = 0/(1- 0 t)  d/dt = 0/(1- 0 t) Further integrating we get 1 - 0T = e-2 or t = (R/ v0)(1 - e-2)

20.

a = v 



dv v  dx m

v

dv 

v0

 x=  (A)



x

0

 dx m

mv o 

LEVEL – II

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Part-I-KM(S) -PH-13

1.

dx 2 t dt 1 t3 y 23 dy t 2  dt 2

…(i)

…(ii)

t = 1, vx = 1, vy =

1 2

r 1 v  ˆi  ˆj 2 d2 x  2t …(iii) dt 2 d2 y t …(iv) dt 2 at t = 1 s ax = 2 and ay = 1 r  a  2iˆ  ˆj .

3.

Using constrained equation, v2 cos   v1 . On differentiation a2 cos   a1 .

(y  h)  x 2  h2  l dy x dx  0 dt x 2  h2 dt

dy x dx  dt x 2  h2 dt dy 3   vA dt 5 h= 4cm

2.

y x=3cm

| uB |

3 vA 5

…(i)

d2 y h2  v A dt 2 (x 2  h2 ) 3

2

16 aB  v A (5)3 16 aB  v A …(ii) 125

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM(S) -PH-14

4.

if

if

u2 cos2  u2 sin2   g 2g

  tan 1 2 u2 cos2  u2 sin2   g 2g   tan1 2 .

5.

Final velocity of each ball

v  (2gh) This is independent of path Further, t 

 2h 1   sin   g 

1 for same h sin  t sin   1  . t 2 sin  or

7.

t

As velocity of the particle increases from zero and finally it decreases to zero, hence acceleration can not remain positive for all the time of motion. Minimum magnitude of acceleration can be found out by the velocity- time curve. Displacement = Area bounded by the graph and time axis. Magnitude of minimum acceleration = 4 m/s2.  (A) & (C) COMPREHENSIONS

dv y

1.

 2t dt dy  v y  t2  dt t3 and y = 3

2.

Drift = V0  time of crossing.

3.

x  v0t y

…(i)

3

t …(ii) 3

from (i) and (ii)

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM(S) -PH-15

y

x3 . 3v 30

1 0 m /s 5

3 m /s

v

60º 5 m /s

4.

y

vx  v 30º

Time after which velocity vector becomes perpendicular to initial velocity vector is u 10 2 t   seconds gsin  10 sin 60 3 Let vy be the vertical component of velocity at that instant then vy = u + at 10  2 vy = 5 3  3 v x= 5



v x=

5

vy  

3

 v  5 

 5

2



v

5. 6.

v= 10 3

5 3

10

2

 3

m/s; gcos  

v2 R

3 v2 20 R = R m. gcos  3 3 Acceleration of particle is equal to acceleration due to gravity. at = g sin  5 3  10  = 5 m/s2. 10 3

MATCH THE FOLLOWING 1.



u  60 cos 30o iˆ + 60 sin 30o ˆj a  g  ˆj y

 

ax  0

***

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