Jee 2014 Booklet2 Hwt Solutions Motion in 2d

Vidyamandir Classes

Solutions to Home Work Test/Physics Motion in Two Dimensions 1.(C)

Range = Hmax

2.(B)

R = 100 m

u 2 sin 2 u 2 sin 2   g 2g



HWT - 1   tan 1  4 



u 2 sin 2  100 g 

C 

sin 2  1 / 250

A

2  19.6  2s 9.8

2h  g

4.(B)

T 

6.(A)

Horizontal displacement is equal for both.

7.(B)

Maximum height that can be reached =

8.(B)

H max  5m

9.(A)



  30

Range 

u 2 sin 2  20 3 m g

2g



 300 3  2  10



2

 13.5 km

2u sin  2  20  1   2s g 2  10 10t 



Let total time = t

10.(C)

2

2h g

u 2 sin 2   5 ; u  20 m / s . 2g



Time of flight =

uy

2km

  60



B

Horizontal displacement of bomb = v 

5.(C)

300  600 cos 

h



100  20 cm 500

h  R tan  



  1 / 500



t 2  2t  8  0



t = 4s

1 10 t 2   40 2

Ratio = 4 : 2 = 2 : 1.

R  20 cos 30  4 = 40 3 m  70 m

Motion in Two Dimensions 1.(C)

P should satisfy equation of trajectory

P

V0

h



HWT - 2

h  R tan  

O

gR 2 2v02 cos 2 

R u 3.(D)

45

car

v 10 2 m / s

150 m u 2 sin 90 2u sin 45  150  10 2  g g

VMC/Motion in Two Dimensions



u2 = 1500 + 200 u

19

U = 50 m/s.

HWT-Solutions/Physics

Vidyamandir Classes u 2 sin 90  250m g

4.(A)

R

6.(B)

Rmax 

u2  200m g

H max 

u2  100 m 2g

8.(A)

x = at 



y = bt2 + ct 

vx = a

v 1 sec  

9.(D) 10.(B)

u  20 5 m / s

v x2  v 2y

vy = 2bt + c a 2   2b  c 



2

 uy   c   tan 1    tan 1    a  ux 

ay   g



2b   g



g   2b

| g |  | 2 b |  2 b

Motion in Two Dimensions 1.(B)

R

u 2 sin 2 g

3.(B)

h1 

u 2 sin 2 30 2g

4.(C)

h2  1.1  h1

u22

6.(B)

R  u2

if R become double , u'  4 2

 35 h2 



T'  T 2



x = 2t vx = 2

v at t  0.2 s  9.(A)

u 2 sin 2 60 2g

 1.1 

u12



8.(B)

HWT - 3

v x  10 3

v y  10  2t

10 3 

Initial velocity =

2

5.(D)

H 

1 g



6

y = 5t2 vy = 10t

22  22  2 2

y  10t  t 2

R1 = R2

h2 = 3h1

R2  1.1  10% R1



x  10 3 t



2.(B)

(Please note that time = 0.2 sec )

 102

 20 m / s

 v  1    tan 1    tan 1   30 w / horizontal .  3   vx 

Motion in Two Dimensions 1.(C)

HWT - 4

x  at  bt 2  ct 3 v  a  2bt  3ct 2 a  2b  6ct

a 0



 b  b  b v    a  2b    3c    3c   3c   3c 

= a 

t  b / 3c 2

b2 3c

VMC/Motion in Two Dimensions

20

HWT-Solutions/Physics

Vidyamandir Classes 2.(D)

B could be any vector in x  y plane.

4.(C)

10 sin   5

  30

3.(D)



10 30 west of North.



5.(A)

5   2 g  AB 

C

vB2

B A

vC2  VB2   2 g  BC 



v 2A

2

 1  1       3 2

t1

t1 

t2

2  H  h

H h 

9.(A)

2

AB 20   2 2 BC 7  1  1      4 3



8.(C)

North

2 g

T  t1  t2 



AB 20  AC 27

2h g

t2 

g



H h 

h



Time is minimum when h = H/2.

Height already fallen =

u 2 202   20 m 2g 20

 20  Total storeys = 15     20  4



Motion in Two Dimensions 1.(A)

v   3i  4 j  m / s

2.(A)

T 

4.(A)

y  ax  bx 2

2u y g





H max 

u 2y 2g



HWT - 5

u2  0.8 m . 20

24  0.8 s 10

3.(D)

t 

u cosec  g

Range is solution of y = 0 

5.(D)

ax  bx 2  0





x = a/b

Range, a/b

y  ax  bx 2 Maximum height is attained at solution of

dy 0 dx 2



a  2bx  0

2H g

2



x = a/2b



2a 2 2 a 4bg bg

6.(A)

T 2

7.(A)

Total distance = D  D   1 Total time =      t0 and (4.5 + 7.5) (t0) = D/2  2   3

t0  Total time =

a2  a  a y  a    b  =  2b   2b  4b

D D  2  12 24

D D 5D   6 24 24

VMC/Motion in Two Dimensions



vav 

D  4.8 m / s 5D / 24

21

HWT-Solutions/Physics

Vidyamandir Classes Displacement 2 r   2 ms 1 time 1s

9.(B)

vav 

10.(D)

u   4i  3 j  m / s a  0.4 ˆi  0.3 ˆj





v  u  at   4i  3 j   0.4 ˆj  0.3 j 10 = 8ˆi  6 ˆj | v |  10 m / s

Motion in Two Dimensions 2.(A)

t 

x 3

x   t  3

2

v  2  t  3





v=0 3.(A)

HWT - 6



t = 3s

x=0

a  k v 6

dv v  k dt t



 u

dv v

t



  k dt 0

2 u   kt

t  5.(D)



2 u k

v v0

v = mx + c

Where m < 0, c > 0 vdv a   v  m   m 2 x  mc dx

a  m 2 x  mc

x0

x



Straight line with positive slope (m2) and y. intercept negative (mc)



Hits nth step Height descended = nh



T 



 2nh  n 2 b2  u 2    g 

u 7.(B)

h b

Horizontal range = nb = Ut 8.(A)

2nh g n

2u 2 h gb2

vx = 500 m/s. v y  gt   10 10  100 m / s  vy   1 Angle with horizontal = tan 1   = tan 1    5 v  x

10.(B)

  a   3ˆi  5 ˆj  m / s 2

u  2ˆi  4 ˆj m / s

s  s0  u t 







 



1 a t 2 = 0  2 2ˆi  4 ˆj  2 3ˆi  5 ˆj = 10 ˆi  18 ˆj m 2

VMC/Motion in Two Dimensions

22

HWT-Solutions/Physics

Vidyamandir Classes

Motion in Two Dimensions 4.(D)

y   x2 ;

d2y dt 2



dy dx  2 x dt dt



HWT - 7

d2y



dt 2

2

 dx  = 2      dt 

u

T 

h

d2x dt 2

 dx   2    dt 

2

 2

dx  vx  dt



6.(B)

 2 x

2h g

x = uT x

8.(B)

g

 20m



u x

g 2h

u  14 m / s .

s x  20  5  100 m ; s y 

s

10.(A)

2h g

Rmax  20 m

4u 2y

9.(C)

xu

v

1 2 g  5  125 m 2

s x2  s 2y  160 m

2h  250 m g

x



v 2

8h  250 m g

Motion in Two Dimensions 1.(B)

HWT - 8

s y   122.5 m 



u

1 2 gt   122.5 2

145m

B

t  5s u  5  40



22.5 m

u  8 m/s

40m 4.(D)

5.(B)



ux = vx v cos   V sin 



V  v cot 

y 

3x 





gx 2 2

tan   3



  60

2h  6s  g

7.(C)

Time of flight =

8.(A)

d   6s   10 m / s   60 m

9.(C)

v x  u x  10 m / s .

vy 

V

v

2 gh 

h

62  10  180 m 2

2  10  180  60 m / s

VMC/Motion in Two Dimensions



v

23

v x2  v 2y 

3600  100  60.8 m / s

HWT-Solutions/Physics

Vidyamandir Classes 10.(C)

v  2v0  2  10 = 20 m/s.

v x2  v 2y

 20



v y  10 3 m / s  .

gt  10 3



t 

3s

Motion in Two Dimensions

HWT - 9

v 5.(D)

t 6.(D)

T = 10 s

H = 100 m. 1 Must satisfy H  gT 2 8 100 

7.(C)

1 2 gt   30 2

Data is incorrect.



t2  t  6  0



t = 3s

Horizontal range = 3  10 sin 60  15 3 m .

30

tan   9.(B)



u y  10 cos 60  5 m / s 5t 

8.(D)

1  10  102 8

15 3

 2/ 3





  tan 1 2 / 3



w   t  2t

a r  w 2 r  4t 2 r



at  r   2 r

 4t 2  a  1 2 Angle between velocity and acceleration = tan 1  r  = tan 1    tan 2t  at   2r 

 

Motion in Two Dimensions 2.(B)

If bomb’s time of fall = t  53

3.(C)

1 2 gt  73 2 Solve to get: v = 202 m/s.  v cos 53 t 

v

Range 

 

102 sin 120 g

2h  g



3

r 

t = 5 s.

3s.

h  15 m .

v  20 / 3 m / s .

g cos 30  v 2 / r .

10.(A)

time of flight of B 

v cos 30  u x  20 cos 60 



 5 3 m.

5t  5 3 m

5.(C)

HWT - 10

g

400 2   80 / 3 3 3 3

 30  30

v



r = 15.4 m.

v  u x  v cos  .

VMC/Motion in Two Dimensions

24

HWT-Solutions/Physics