Kinemat-Hints and Solution

LEVEL – I

1. Comparing the general expression v0



= (v0 cos 0) iˆ + (v0 sin 0) ˆj with the

given equation v0 = 20 iˆ + 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, vx = 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 iˆ + vy ˆj = 20 iˆ

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



 t0 0 V dt; t 1 V Putting V = (V0 – gt) we obtain



  _  _   0 0 t 0 0 0 0 t 2 t g V dt ) gt V ( V

Where t0 = total time of ascent. Since V = 0 at t = t0 t0 = _g

V₀ . Putting g V t 0 0  , we obtain 2 V V  0 .

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 1 gy v 2 2 0 _ By putting v0 = 100 m/sec, g = 10, y = 1000 m, we obtain r = 1732 m (approximately) And  = tan-1         0 v 2 gy = tan-1         _ 100 2 1000 10   = tan-1 _0.707

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.  t = 0 0cos v 50   v0 cos 0 = 50/t = 50/5 = 10 m/s.

The kinematical equation for vertical motion is – y = (v0 sin 0)t- gt2 2 1 50 m t = 5 s (50, - 1.5) x 0 v0 y t = 0 y = 1.5 m (0,0)  (v0sin 0) (t) = gt2 2 1 - y  (v0sin 0) (5) = (10)(5)2 2 1 - 1.5  v0sin 0 = 24.7 5 5 . 123  m/s. jˆ ) sin v ( iˆ ) cos v ( v₀  ₀ ₀  ₀ ₀ ) jˆ 7 . 24 iˆ 10 ( v₀   m/s.  v0 = 26.64 m/s & 0 = tan1 10 7 . 24 = 67.960_.

5. The distance travelled by the rocket in the first 1 minute in which resultant acceleration is vertically upwards and 10 m/s2 _{will be}

h1 = 0  60 +

2 1

 10  602 _{= 18000 m}

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  h} 2  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.

6. Let they meet after time t from the instant of release at point P as shown in the figure. Since S1 is downwards,

S1 = Sg + Si = ½gt2 + V1t (a) & S2 is upwards  S2 = Si - Sg  S2 = V2t - ½ gt2 (b) (a) + (b)  S1 + S2 = (V1 + V2)t  V V h t   (



S1 + S2 = h & V1 = V2 = V)  2V h t  t = t p t = 0 v1 = v 0 t = 0 v2 = v 1 s  p 2 s  p

7. (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)gT}2  (T-1)/T =  1/2   T = 2T - 2  T(2



1) = 2 t =0 A t =T-1 t =T C B  T = 1 2 2  s

Since T < 1, we accept the other value.

 T =

1 2

2

 = 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. 2 1 2 2 1 1 2 1 2 1 av t t t v t v t t S S V            = 5 10 jˆ 5 10 iˆ 10 5     _{= [iˆ jˆ]} 3 5 _ .  3 2

5 _{due north of east.}

9. Let the particle move from P to Q in a time interval t. The change in position vector r r r 2 1         r  r  2r1r2 cos 2 2 2 1 Putting r1 = r2 = r we obtain

            2 sin 2 2r ) cos 1 ( 2r r 2 2 2   2 sin r 2 r    .

Now the magnitude of average velocity = v = t r   _; t 2 sin r 2 v   (i) we know that r v & t      (when w = constant)  v r t   (ii) r   v Qt=t t=0 P  v 1 r  2 r  O

Putting t from (ii) in (i) we obtain

     2vsin2 v r ) 2 / ( sin r 2 v _ θ /2) ( sin 2 v v _  _.

10. We know that velocity of rain w.r.t. man vrm is given as vrm vr vm

     . Let vm₁ & vm₂  

are the initial & final velocities of the man, then

2 m r 2 rm 1 m r 1 rm v - v & v v - v v         vr vrm1 vm1 vrm2 vm2         

Referring the vector diagram we obtain

2 1 2 1 rm m r v v v   1 m v 1 rm v vrm2 _ 2 m v 2 1 rm 2 r v v v   . . .(a) Putting BC = v = nv we obtain  nv cot vrm₁ (b)

Using (a) & (b) we find

2 2 r v ( v cot ) v     θ cot η 1 v v_r   2 2   1 rm v vrm2 1 m v 2 m v D C A v nv B

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

jˆ gt iˆ v

v₁  ₁  & v₂ v₂ˆ igt jˆ

Since they move perpendicularly just before touching the ground, therefore 0 v . v₁ ₂   (v1iˆ gt jˆ) . (v2iˆ  gt jˆ)  0  g2_t2 _{= v} 1v2  g v v t_ 1 2

The vertical displacement of each particle = 2 1

h gt2 _{where h = height of the}

pole. & g v v t_ 1 2  g 2 v v_{1 2}           2 2 1 g v v g 2 1 h V1 V2 t=0 t=0 t=t t=t V2 V1 2 v gt 1 v gt

2. At the point of collision, x1 = x2 and y1 = y2  vo cos1t1 = vo cos2t2 or t2 =       2 1 cos cos t1.

Where 1 and 2 are the angles of projection and t1, t2 are the time of flight.

Similarly for equal vertical displacement (y1 = y2)

vosin1t1 2 1 g 2 1 t = vosin1t1 2 1 g 2 12 t vosin1t1 vosin2



2



2 2 1 1 2 1 _{t t} 2 g t cos cos     vot1



































2 2 1 2 2 2 2 1 2 2 1

cos

cos

cos

t

2

g

cos

sin

 t1 =





1 2 2 2 2 2 1 o

cos

cos

cos

sin

v

2















t2 = t1 = 2 1 cos cos    Time interval t = t1 t2 =





2 1 2 1 o

cos

cos

g

sin

v

2













t = 10.7 s

3. Forward Journey

Vector CB represents wind velocity. AB represents velocity plane relative to

ground.

AC represents velocity of plane in still air.

400cos iˆ + (400 sin + 50)ˆj = AB

tan45 = 400₄₀₀sin_cos_50

45 N W E S  B C A  tan  + 8 sec = 1  tan2_{ 2tan + 1 =} 64 1 tan2_  63 tan2_{ 128 tan + 63 = 0}  tan = 126 5 . 140 126 508 128   _or 126 5 . 105   = 4811 or 3993 Hence; |AB| =

267

.

65

iˆ



347

.

25

jˆ

= 438.42 km/hr. Time taken = 2.28hr 42 . 438 1000 

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iˆ (400cos 50)ˆj

Now; tan45 = 1 = 400₄₀₀cos_sin_50

  = 42 or 50. 45 N W E S  B C A Hence, |BA| = |268 iˆ  247ˆj| = 364 km/hr.

Hence, time required = 364 1000 =2.74 hr. 4. v1  u1iˆgtjˆ  , v2  u2iˆ gtjˆ  0 v . v₁ ₂  t = g u u_{1 2}

 s = s1 + s2 = u1t + u2t = (u1 + u2)

g u u_{1 2}

Substituting the values, the distance between the particles =

8 . 9 12 7 _{= 2.47 m.} 5. Deceleration  a v dt dv  or, v ds dv =  a v  v dv =  ads

Integrating under specified limits,





 s 0 0 o v ds a dv v _{ s =} 2 3 o v a 3 2       Again,







t 0 0 v adt v dv 0  t = a v 2 0 Ans.: 2 _o 3 o v a 2 , v a 3 2            

6. From the vector diagram

3 1 30 tan a a t r _{ } Required ratio is t r a a which is 3 1 as above.

7. The frequency of rotation f = 15

60 900  revolution/second. 300 600 ar  a  at  v 

 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;  =   2 o _{= 3 rad/sec}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 iˆvyjˆ Case I : v_m 10iˆ andv_m  v v_m  v_m (v_x 10)iˆv_yjˆ



Since the wind appears to blow from north, vm



must have no x component

 vx – 10 = 0  vx = 10 m/s Case II : vm  = 20 iˆ and vm  = v vm    

 v (vx 20)iˆ 

+ vyˆj

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. x dx VdV a   VdV  x dx 



VdV  x dx  2 02 _x3/2 3 2 2 V V   _ 2 2/3 0 2 4 ) V V ( 3 x         _  Putting V = 2 V0 we obtain , x = 3 / 4 0 3 / 2 2 0 2 V 3 V 4 9           

10. 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. Hence: ve vT = 6 1 km/min . . . (1) ve + vT = 6 1 km/min . . . (2) 1 v x e  . . . (3) e v X  = 2 . . . (4)

Solving these (1), (2), (3) and (4).

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/s2 _{pointed vertically}

downwards.

2. Let this projectiles be projected with velocities jˆ ) sin v ( iˆ ) cos v ( jˆ y iˆ v v_{1 x y 1 1 1 1} 1 1       

and v₂ v_x iˆ y_y jˆ (v₂cos ₂)iˆ (v₂sin ₂)jˆ

2

2     





The velocities of the projectiles after a time t are given as jˆ ) gt sin v ( iˆ ) cos v ( v₁  ₁ ₁  ₁ ₁ jˆ ) gt sin v ( iˆ ) cos v ( v2  2 2  2 2   jˆ ) sin v sin v ( iˆ ) cos v cos v ( v v₁ ₂  ₁ ₁ ₂ ₂  ₁ ₁ ₂ ₂

Since the relative velocity v₁ v₂is 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 _{& y = bt}2  2ct dt dx  & 2bt dt dy  Since v = 2 2 dt dy dt dx dt ds                v = _(2ct)2 _(2bt)2  v = 2t _b2_c2 _.

5. Let OP = r. Angular speed about the origin =  = t p0 _t

0 p v where , r v    = The component of velocity of P w.r.t. O perpendicular to OP.

 r sin v    where r = b cosec   b sin v 2_    (C) is correct.

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 nor _opsinv _bv

cosec b v op v     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

h = ut - 2 1 gt2 _or 2 1 gt2 _{– ut + h = 0}  t2 _{- 0} g h 2 t g u 2   . . . . (1) from equation (1) sum of roots, t1 + t2 = - _g u 2 a b  . . .. (2) product of roots, t1t2 = _g h 2 a c  . .. (3) Given that 3 1 t t 2 1 _ or t2 = 3t1 . .. . (4)

from equation (2) and (4), we get

4t1 = _2g u ort g u 2 1  . .. . (5)

from equation (3) and (4) we get

3 2 1 t = g h 2 g 2 u 3 or g h 2 2      or 3 h 4 g 2 u2  . .. . (6)

Also, maximum height, H =

g 2 u2

. . . (7) from eq. (6) and (7) we obtain,

H = 4 H 3 orh 3 h 4  .

7. By geometry of the figure the length of the string is  = 4xA + xB  dt dx dt dx 4 dt d_ A _ B

Since the length of the string is constant, dt d = 0 Putting B _B A A _v dt dx & v dt dx   4 v v B A  numerically.

v

-

v





v

Since the displacement of B is four times that of A, their speeds can not be equal. 8. Time taken = m v d = ₂ w 2 mw v v d  = ₁₀2 ₈2 480  =  6 480 80 s. vw  vm  vmw  vw 

9. The angle between vmw and and vw

 

is 

say.

 The required time of crossing = t =  sin v d mw  t = _5sin600 5 . 1 = hr 5 3 hr 3 5 3   vmw d vw

10. The time period of revolution of the disc is equal to the time of motion of water drop in air.  2 / = g H 2   = 2 H g 2 H 2 g   . 11. a = 10  2 5 t

Now dv = adt, integrating we have, 10t 

4 5

t2 _{= 0  t = 0 or 8 sec., Accepted value 8 sec.}  (C) 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)  2 1  10  (5)2 _{. . . (1)} h =  vo(1)  2 1  10  (1)2 _{. . . (2)}  v0 = 20 m/s, t = 5sec.  (C) 14. vuat7iˆ jˆ  v 5 2m/s  (C)

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,

40 = 10t 

2 1

 10  t2 _{ t = 4 sec}

 (C)

16. At time t0 velocity of the ball be equal to gt0 vertically downward.

Relative velocity between the two balls remains constant at gt0 as relative

acceleration between them is zero, as both are moving under gravity. At the time of dropping the second ball, the distance between the two balls = gt20

2 1

. 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)

17. The tangential component of force = dK ds the centripetal component = 2K

R . 19. Given ac = at  R 2 _{= R} or 2 _{=  = d/dt} dt d     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 m v dx dv _{ } 







x  0 v 0 v dx m dv  x = o mv  (A) LEVEL – II

1. dx_t2 dt …(i) 3 1 t y 2 3  2 dy t dt  2 …(ii) t = 1, vx = 1, vy = 1 2 1 ˆ ˆ v i j 2   r 2 2 d x 2t dt  …(iii) 2 2 d y t dt  …(iv) at t = 1 s ax = 2 and ay = 1 ˆ ˆ a 2i j  r  .

2. Using constrained equation, v₂cos v₁. On differentiation a₂cos a₁. 3.

_{(y h)}

_{ }

_x

2

_

_h

2

_

_l

2 2

dy

x

dx

0

dt



_x

_

_h

dt



2 2

dy

x

dx

dt

 

_x

_

_h

dt

A dy 3 v dt  5 x = 3 c m h= 4c m y B A 3 | u | v 5  …(i) 2 2 A 2 2 2

d y

h

v

3

dt



(x



_{h ) 2}

B A 3 16 a v (5)  B A 16 a v 125  …(ii)

4. if 2 2 2 2

u cos

u sin

g

2g



_



1

tan



2

 

if 2 2 2 2

u cos

u sin

g

2g



_



1

tan



2

 

.

5. Final velocity of each ball

v



(2gh)

This is independent of path

Further, t 1 2h sin g    _{ }    or t 1 sin   for same h 1 2 t sin t sin    .

7. 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 1.

dv

y

2t

dt



 2 y dy v t dt   and y = 3

t

3

2. Drift = V0 time of crossing.

3.

x v t



₀ …(i) 3

t

y

3



…(ii)

3 3 0

x

y

3v



. 4. 6 0 º 3 0 º 1 0 m / s 5 m / s  vx vy 5 3 m / s _v

Time after which velocity vector becomes perpendicular to initial velocity vector is

u 10 2

t

gsin 10 sin 60 3

  

 seconds

Let vy be the vertical component of velocity at that instant then

vy = u + at vy = 10 2 5 3 3   v = v = 5x  v =x 5 3 1 0 3 y 5 v 3   2 2 5 v 5 3     _{  }   10 v 3  m/s; gcos v2 R   2 v R gcos  = 20 R 3 3  m.

5. Acceleration of particle is equal to acceleration due to gravity. 6. at = g sin  5 3 10 10 3   = 5 m/s2_.

MATCH THE FOLLOWING

1. _{u 60 cos30}_ o_{i +ˆ 60 sin30}oˆ_j

 

y ˆ a  g j ax 0

* * *