NEET Exam  >  NEET Notes  >  Physics Class 11  >  DPP for NEET: Daily Practice Problems, Ch 6: Motion in a Plane- 1 (Solutions)

DPP for NEET: Daily Practice Problems, Ch 6: Motion in a Plane- 1 (Solutions) | Physics Class 11 PDF Download

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 Page 1


1. (b) Given, that y = 
3
 x – (1/2) x
2
....(1)
The above equation is similar to equation of trajectory
of the projectiles
y = tan q  x – 1/2 
q
2 2
cos u
g
 x
2
....(2)
Comparing (1) & (2) we get
tan q = 
3
Þ q = 60º
and 1/2 = (1/2) 
q
2 2
cos u
g
Þ  u
2
 cos
2
 q = g  Þ u
2
 cos
2
 60° = 10
Þ u
2
 (1/4)  = 10  Þ u
2 
= 40  Þ u = 2
10
 m/s
2. (a) For angle of elevation of 60º, we have maximum height
h
1
 = 
22
u sin 60º
2g
 = 
g 8
u 3
2
For angle of elevation of 30º, we have maximum height
h
2 
= 
g 2
º 30 sin u
2 2
 = 
g 8
u
2
; 
2
1
h
h
 = 
1
3
3. (b) T otal time of flight = 2 × time taken to reach max. height
Þ t
2
 = 2t
1
 Þ t
2
/t
1
 = 2/1
4. (d) v
x
 = dx/dt  = 2ct, v
y
 = dy/dt = 2bt
\ v = 
2
y
2
x
v v + = 2t 
22
cb +
5. (c) v
y
 = dy/dt = 8 – 10 t = 8, when t = 0 (at the time of
projection.)
v
x
 = dx/dt = 6, v = 
2
y
2
x
v v + = 
2 2
6 8 +
 = 10 m/s
6. (a) Horizontal component of velocity
v
x
 = u
x
 = u cos q = 30 × cos 30º = 15 3 m/s
V ertical component of the  velocity
v
y
 = u sin q – gt = 30 sin 30º – 10 × 1 = 5 m/s
v
2
 = v
x
2
 + v
y
2
 = 700 Þ u = 10 7 m/s
7. (a) Let u
x
 and u
y
 be the components of the velocity of the
particle along the  x- and y-directions. Then
u
x
 = dx/dt = u
0
 and u
y
  = dy/dt = wa cos wt
Integration : x = u
0
 t and y = a sin wt
Eliminating t : y = a sin (wx/u
0
)
This is the equation of the trajectory
At t = 3p/2w, we have,
x = u
0
 3p/2w and  y = a sin 3p/2 = – a
\ The distance of the particle from the origin is
2 2
y x + = 
ú
ú
û
ù
ê
ê
ë
é
+ ÷
ø
ö
ç
è
æ
w
p
2
2
0
a
2
u 3
8. (b) T = 
g
sin u 2 q
 Þ 2 = 
g
sin u 2 q
Þ u sin q = g
H = 
g 2
sin u
2 2
q
 = 
g 2
g
2
 = 
2
g
 = 5 m
9. (b) Let u
1
 and u
2
 be the initial velocities respectively . If h
1
and h
2
 are the heights attained by them, then
h
1
 = 
g 2
u
2
1
 and  h
2
 = 
g 2
sin u
2 2
2
q
...(1)
The times of ascent of balls are equal,
we have t = u
1
/g = u
2
 sin q/g
\ u
1
 = u
2
 sin q ... (2)
From eq. (1) 
2
1
h
h
 = 
q
2 2
2
2
1
sin u
u
... (3)
From (2) & (3), 
2
1
h
h
 = 
1
1
10. (a) h
1
 = 
g 2
sin u
2 2
q
 and h
2
 = 
g 2
) 90 ( sin u
2 2
q -
\ h
1
 + h
2
 = u
2
/2g (sin
2
q + cos
2
q)
   = u
2
/2g = 
10 2
98
2
´
 = 490
h
1
 – h
2
 = 50,   \ h
1
 = 270 m and h
2
 = 220 m
11. (c) Greatest height attained
h = 
g 2
sin u
2 2
q
... (1)
Horizontal range
R = 
g
2 sin u
2
q
 = 
g
cos sin u 2
2
q q
... (2)
Given that R=2h
Þ 
g
cos sin u 2
2
q q
 = 
g 2
sin u 2
2 2
q
Þ tan q = 2 ... (3)
Hence sin q = 2/ 5 , cos q = 1/ 5 ,
\ From (2) R = 4u
2
/5g
12. (d) R = 
g
2 sin u
2
q
 = (19.6)
2
 sin 90º /10 = 39.2 m
Man must run (67.4m – 39.2 m) = 28.2 m in the time
taken by the ball to come to ground. Time taken by the
ball.
Page 2


1. (b) Given, that y = 
3
 x – (1/2) x
2
....(1)
The above equation is similar to equation of trajectory
of the projectiles
y = tan q  x – 1/2 
q
2 2
cos u
g
 x
2
....(2)
Comparing (1) & (2) we get
tan q = 
3
Þ q = 60º
and 1/2 = (1/2) 
q
2 2
cos u
g
Þ  u
2
 cos
2
 q = g  Þ u
2
 cos
2
 60° = 10
Þ u
2
 (1/4)  = 10  Þ u
2 
= 40  Þ u = 2
10
 m/s
2. (a) For angle of elevation of 60º, we have maximum height
h
1
 = 
22
u sin 60º
2g
 = 
g 8
u 3
2
For angle of elevation of 30º, we have maximum height
h
2 
= 
g 2
º 30 sin u
2 2
 = 
g 8
u
2
; 
2
1
h
h
 = 
1
3
3. (b) T otal time of flight = 2 × time taken to reach max. height
Þ t
2
 = 2t
1
 Þ t
2
/t
1
 = 2/1
4. (d) v
x
 = dx/dt  = 2ct, v
y
 = dy/dt = 2bt
\ v = 
2
y
2
x
v v + = 2t 
22
cb +
5. (c) v
y
 = dy/dt = 8 – 10 t = 8, when t = 0 (at the time of
projection.)
v
x
 = dx/dt = 6, v = 
2
y
2
x
v v + = 
2 2
6 8 +
 = 10 m/s
6. (a) Horizontal component of velocity
v
x
 = u
x
 = u cos q = 30 × cos 30º = 15 3 m/s
V ertical component of the  velocity
v
y
 = u sin q – gt = 30 sin 30º – 10 × 1 = 5 m/s
v
2
 = v
x
2
 + v
y
2
 = 700 Þ u = 10 7 m/s
7. (a) Let u
x
 and u
y
 be the components of the velocity of the
particle along the  x- and y-directions. Then
u
x
 = dx/dt = u
0
 and u
y
  = dy/dt = wa cos wt
Integration : x = u
0
 t and y = a sin wt
Eliminating t : y = a sin (wx/u
0
)
This is the equation of the trajectory
At t = 3p/2w, we have,
x = u
0
 3p/2w and  y = a sin 3p/2 = – a
\ The distance of the particle from the origin is
2 2
y x + = 
ú
ú
û
ù
ê
ê
ë
é
+ ÷
ø
ö
ç
è
æ
w
p
2
2
0
a
2
u 3
8. (b) T = 
g
sin u 2 q
 Þ 2 = 
g
sin u 2 q
Þ u sin q = g
H = 
g 2
sin u
2 2
q
 = 
g 2
g
2
 = 
2
g
 = 5 m
9. (b) Let u
1
 and u
2
 be the initial velocities respectively . If h
1
and h
2
 are the heights attained by them, then
h
1
 = 
g 2
u
2
1
 and  h
2
 = 
g 2
sin u
2 2
2
q
...(1)
The times of ascent of balls are equal,
we have t = u
1
/g = u
2
 sin q/g
\ u
1
 = u
2
 sin q ... (2)
From eq. (1) 
2
1
h
h
 = 
q
2 2
2
2
1
sin u
u
... (3)
From (2) & (3), 
2
1
h
h
 = 
1
1
10. (a) h
1
 = 
g 2
sin u
2 2
q
 and h
2
 = 
g 2
) 90 ( sin u
2 2
q -
\ h
1
 + h
2
 = u
2
/2g (sin
2
q + cos
2
q)
   = u
2
/2g = 
10 2
98
2
´
 = 490
h
1
 – h
2
 = 50,   \ h
1
 = 270 m and h
2
 = 220 m
11. (c) Greatest height attained
h = 
g 2
sin u
2 2
q
... (1)
Horizontal range
R = 
g
2 sin u
2
q
 = 
g
cos sin u 2
2
q q
... (2)
Given that R=2h
Þ 
g
cos sin u 2
2
q q
 = 
g 2
sin u 2
2 2
q
Þ tan q = 2 ... (3)
Hence sin q = 2/ 5 , cos q = 1/ 5 ,
\ From (2) R = 4u
2
/5g
12. (d) R = 
g
2 sin u
2
q
 = (19.6)
2
 sin 90º /10 = 39.2 m
Man must run (67.4m – 39.2 m) = 28.2 m in the time
taken by the ball to come to ground. Time taken by the
ball.
t = 
g
sin u 2 q
= 
8 . 9
º 45 sin 6 . 19 2´
 =
2
4
= 2.82 sec
V elocity of man = 
sec 82 . 2
m 2 . 28
 = 10 m/s
13. (a) Referring to (fig.) let P be a point on the trajectory
whose co-ordinates are (4, 4). As the ball strikes the
ground at a distance 14 metre from the wall, the range is
4 + 14 = 18 metre. The equation of trajectory is
 y = x tan q -(1/2) g 
q
2 2
2
cos u
x
or   y = x tan q 
ú
û
ù
ê
ë
é
q q
-
tan . cos u 2
gx
1
2 2
or    y = x tan q 
ú
ú
û
ù
ê
ê
ë
é
q q
-
cos sin
x
g
u 2
1
2
          = x tan q 
ú
û
ù
ê
ë
é
-
R
x
1
... (1)
Here x = 4, y = 4 and R = 18
\ 4 = 4 tan q 
ú
û
ù
ê
ë
é
-
18
4
1
 = 4 tan q
÷
ø
ö
ç
è
æ
9
7
or   tan q = 9/7, sin q = 9/
130
 and
  cos q = 7/ 130
Again R = (2/g) u
2
 sin q cos q
    = (2/9.8) × u
2
 × (9/ 130 ) × (7/ 130 )
u
2
 = 
7 9 2
130 130 8 . 9 18
´ ´
´ ´ ´
 = 
7
13 98 ´
= 182,
u = 
182
 metre per second.
14. (b) The situation is shown in fig.
(a) Let the ball collide after t sec
From fig. OC = OB cos 37° = 500 cos 37°
= 500 × 0.8 = 400 cm ...(1)
Horizontal velocity = 700 × cos 37º
\ OC = 700 × cos 37° × t
= 700 × 0.8 × t = 560 t ... (2)
From eqs. (1) and (2) 560 t = 400
or t = (5/7) sec.
Now h = (1/2) g t
2
 = (1/2) × 1000 × (5/7)
2
 = 255.1 cm
(b) Let at the time of impact, v
x
 and v
y
 be the  horizontal
and vertical velocities respectively , then
v
x
 = 700 × cos 37° = 700 × 0.8  = 560 cm/s
and v
y
 = – 700 × sin 37
°
 + 1000 × (5/7)
           = – 700 × 0.6 + (5000/7) = – 420 + 714.3
         = + 294.3 cm/sec (downward)
V elocity of the ball at the time of collision
   v = ) v v (
2
y
2
x
+
\ v = ] ) 3 . 294 ( ) 560 [(
2 2
+ = 632.6 cm/sec
Again tan q = 
x
y
v
v
 = 
560
3 . 294
or q = tan
–1
 
÷
ø
ö
ç
è
æ
560
3 . 294
 = 27
°
 43'
15. (d) Initial velocity is constant let the ball touches the
ground at an angle q and velocity 3u
Hence 3u cos q = u or cos q  = 1/3 or sin q = 8 /3
The vertical component of velocity at the ground
= 3u sin q = 
3
8 3
 = 8 u
For a freely falling body it covers 20 m to acquire veloc-
ity 
8
u
\ (
8
u)
2
 – 0 = 2 × 9.8 × 20 or u = 7 m/s
16. (b) The horizontal range of the projectile on the ground
R = u
g
h 2
 Þ R = 
2
10
10 2 ´
 =
2
.
2
= 2m
17. (a) R = ut  Þ t = R/u = 12/8
Now h = (1/2) gt
2
   = (1/2) × 9.8 × (12/8)
2
  = 11 m
18. (b) The situation is shown in the adjoining figure.
The time taken by the body is equal to the time taken
by the freely falling body from the height 29.4 m. Initial
velocity of  body
500cm
700cm/s
h
C
O
37º
400cm
B
v
x
v
y
v
A
u
q
2 0 m
u
v sina
v cosa
ucosq
Y
usinq
u
h
X
B
v
P
·
q
A
Page 3


1. (b) Given, that y = 
3
 x – (1/2) x
2
....(1)
The above equation is similar to equation of trajectory
of the projectiles
y = tan q  x – 1/2 
q
2 2
cos u
g
 x
2
....(2)
Comparing (1) & (2) we get
tan q = 
3
Þ q = 60º
and 1/2 = (1/2) 
q
2 2
cos u
g
Þ  u
2
 cos
2
 q = g  Þ u
2
 cos
2
 60° = 10
Þ u
2
 (1/4)  = 10  Þ u
2 
= 40  Þ u = 2
10
 m/s
2. (a) For angle of elevation of 60º, we have maximum height
h
1
 = 
22
u sin 60º
2g
 = 
g 8
u 3
2
For angle of elevation of 30º, we have maximum height
h
2 
= 
g 2
º 30 sin u
2 2
 = 
g 8
u
2
; 
2
1
h
h
 = 
1
3
3. (b) T otal time of flight = 2 × time taken to reach max. height
Þ t
2
 = 2t
1
 Þ t
2
/t
1
 = 2/1
4. (d) v
x
 = dx/dt  = 2ct, v
y
 = dy/dt = 2bt
\ v = 
2
y
2
x
v v + = 2t 
22
cb +
5. (c) v
y
 = dy/dt = 8 – 10 t = 8, when t = 0 (at the time of
projection.)
v
x
 = dx/dt = 6, v = 
2
y
2
x
v v + = 
2 2
6 8 +
 = 10 m/s
6. (a) Horizontal component of velocity
v
x
 = u
x
 = u cos q = 30 × cos 30º = 15 3 m/s
V ertical component of the  velocity
v
y
 = u sin q – gt = 30 sin 30º – 10 × 1 = 5 m/s
v
2
 = v
x
2
 + v
y
2
 = 700 Þ u = 10 7 m/s
7. (a) Let u
x
 and u
y
 be the components of the velocity of the
particle along the  x- and y-directions. Then
u
x
 = dx/dt = u
0
 and u
y
  = dy/dt = wa cos wt
Integration : x = u
0
 t and y = a sin wt
Eliminating t : y = a sin (wx/u
0
)
This is the equation of the trajectory
At t = 3p/2w, we have,
x = u
0
 3p/2w and  y = a sin 3p/2 = – a
\ The distance of the particle from the origin is
2 2
y x + = 
ú
ú
û
ù
ê
ê
ë
é
+ ÷
ø
ö
ç
è
æ
w
p
2
2
0
a
2
u 3
8. (b) T = 
g
sin u 2 q
 Þ 2 = 
g
sin u 2 q
Þ u sin q = g
H = 
g 2
sin u
2 2
q
 = 
g 2
g
2
 = 
2
g
 = 5 m
9. (b) Let u
1
 and u
2
 be the initial velocities respectively . If h
1
and h
2
 are the heights attained by them, then
h
1
 = 
g 2
u
2
1
 and  h
2
 = 
g 2
sin u
2 2
2
q
...(1)
The times of ascent of balls are equal,
we have t = u
1
/g = u
2
 sin q/g
\ u
1
 = u
2
 sin q ... (2)
From eq. (1) 
2
1
h
h
 = 
q
2 2
2
2
1
sin u
u
... (3)
From (2) & (3), 
2
1
h
h
 = 
1
1
10. (a) h
1
 = 
g 2
sin u
2 2
q
 and h
2
 = 
g 2
) 90 ( sin u
2 2
q -
\ h
1
 + h
2
 = u
2
/2g (sin
2
q + cos
2
q)
   = u
2
/2g = 
10 2
98
2
´
 = 490
h
1
 – h
2
 = 50,   \ h
1
 = 270 m and h
2
 = 220 m
11. (c) Greatest height attained
h = 
g 2
sin u
2 2
q
... (1)
Horizontal range
R = 
g
2 sin u
2
q
 = 
g
cos sin u 2
2
q q
... (2)
Given that R=2h
Þ 
g
cos sin u 2
2
q q
 = 
g 2
sin u 2
2 2
q
Þ tan q = 2 ... (3)
Hence sin q = 2/ 5 , cos q = 1/ 5 ,
\ From (2) R = 4u
2
/5g
12. (d) R = 
g
2 sin u
2
q
 = (19.6)
2
 sin 90º /10 = 39.2 m
Man must run (67.4m – 39.2 m) = 28.2 m in the time
taken by the ball to come to ground. Time taken by the
ball.
t = 
g
sin u 2 q
= 
8 . 9
º 45 sin 6 . 19 2´
 =
2
4
= 2.82 sec
V elocity of man = 
sec 82 . 2
m 2 . 28
 = 10 m/s
13. (a) Referring to (fig.) let P be a point on the trajectory
whose co-ordinates are (4, 4). As the ball strikes the
ground at a distance 14 metre from the wall, the range is
4 + 14 = 18 metre. The equation of trajectory is
 y = x tan q -(1/2) g 
q
2 2
2
cos u
x
or   y = x tan q 
ú
û
ù
ê
ë
é
q q
-
tan . cos u 2
gx
1
2 2
or    y = x tan q 
ú
ú
û
ù
ê
ê
ë
é
q q
-
cos sin
x
g
u 2
1
2
          = x tan q 
ú
û
ù
ê
ë
é
-
R
x
1
... (1)
Here x = 4, y = 4 and R = 18
\ 4 = 4 tan q 
ú
û
ù
ê
ë
é
-
18
4
1
 = 4 tan q
÷
ø
ö
ç
è
æ
9
7
or   tan q = 9/7, sin q = 9/
130
 and
  cos q = 7/ 130
Again R = (2/g) u
2
 sin q cos q
    = (2/9.8) × u
2
 × (9/ 130 ) × (7/ 130 )
u
2
 = 
7 9 2
130 130 8 . 9 18
´ ´
´ ´ ´
 = 
7
13 98 ´
= 182,
u = 
182
 metre per second.
14. (b) The situation is shown in fig.
(a) Let the ball collide after t sec
From fig. OC = OB cos 37° = 500 cos 37°
= 500 × 0.8 = 400 cm ...(1)
Horizontal velocity = 700 × cos 37º
\ OC = 700 × cos 37° × t
= 700 × 0.8 × t = 560 t ... (2)
From eqs. (1) and (2) 560 t = 400
or t = (5/7) sec.
Now h = (1/2) g t
2
 = (1/2) × 1000 × (5/7)
2
 = 255.1 cm
(b) Let at the time of impact, v
x
 and v
y
 be the  horizontal
and vertical velocities respectively , then
v
x
 = 700 × cos 37° = 700 × 0.8  = 560 cm/s
and v
y
 = – 700 × sin 37
°
 + 1000 × (5/7)
           = – 700 × 0.6 + (5000/7) = – 420 + 714.3
         = + 294.3 cm/sec (downward)
V elocity of the ball at the time of collision
   v = ) v v (
2
y
2
x
+
\ v = ] ) 3 . 294 ( ) 560 [(
2 2
+ = 632.6 cm/sec
Again tan q = 
x
y
v
v
 = 
560
3 . 294
or q = tan
–1
 
÷
ø
ö
ç
è
æ
560
3 . 294
 = 27
°
 43'
15. (d) Initial velocity is constant let the ball touches the
ground at an angle q and velocity 3u
Hence 3u cos q = u or cos q  = 1/3 or sin q = 8 /3
The vertical component of velocity at the ground
= 3u sin q = 
3
8 3
 = 8 u
For a freely falling body it covers 20 m to acquire veloc-
ity 
8
u
\ (
8
u)
2
 – 0 = 2 × 9.8 × 20 or u = 7 m/s
16. (b) The horizontal range of the projectile on the ground
R = u
g
h 2
 Þ R = 
2
10
10 2 ´
 =
2
.
2
= 2m
17. (a) R = ut  Þ t = R/u = 12/8
Now h = (1/2) gt
2
   = (1/2) × 9.8 × (12/8)
2
  = 11 m
18. (b) The situation is shown in the adjoining figure.
The time taken by the body is equal to the time taken
by the freely falling body from the height 29.4 m. Initial
velocity of  body
500cm
700cm/s
h
C
O
37º
400cm
B
v
x
v
y
v
A
u
q
2 0 m
u
v sina
v cosa
ucosq
Y
usinq
u
h
X
B
v
P
·
q
A
u sin q = 9.8 sin 30
º
  = 4.9 m/s
From the relation, h = u sin q t + (1/2) gt
2
,
we get 29.4 = 4.9 t + (1/2) × 9.8 t
2
  Þ t = 2 sec
19. (b) The horizontal and vertical velocities of the bomb are
independent to each other. The time taken by the bomb
to hit the target can be calculated by its vertical mo-
tion. Let this time be t. Putting h = 490 m and g = 9.8 m/
s
2
 in the formula h = 1/2 gt
2
,
we have  490 = (1/2) × 9.8 × t
2
,
\ t = 
8 . 9
490 2 ´
 = 10 sec
The bomb will hit the target after 10 sec of its dropping.
The horizontal velocity of the bomb is 60 km/hr which
is constant. Hence the horizontal distance travelled by
the bomb in 10 sec (horizontal velocity × time)
= 60 km/hr × 10 sec
= 60 km/hr × 10/(60 × 60) hr = 1/6 km
Hence the distance of aeroplane from the enemy post
is  1/6 km = 1000/6 m = 500/3 meter.
The trajectory of the bomb as seen by an observer on
the ground is parabola. Since the horizontal velocity of
the bomb is the same as that of the aeroplane, the fall-
ing bomb will always remain below the aeroplane. Hence
the person sitting inside the aeroplane will observe the
bomb falling vertically downward.
20. (a) The angle of projection of the ball is q
0
 ( = 30º) and the
velocity of projection is u ( = 10 m/s). Resolving u in
horizontal and vertical components,
we have horizontal component,
u
x
 = u cos q
0
 = 10 cos 30º = 8.65 m/s
and vertical component (upward),
u
y
 = u sin 30º = 5.0 m/s
If the ball hit the ground after t sec of projection, then
the horizontal range is  R = u
x
 × t = 8.65 t meter
\ t = 
65 . 8
R
 = 
s / m 65 . 8
m 3 . 17
 = 2.0 s
If h be the height of the tower,
then h = u'
y
 t + (1/2) g t
2
,
where u
y
' is the vertical component
(downward) of the velocity of the ball.
Here u
y
 = – u
y
' = – 5.0 m/s and t = 2.0 s
\ h = (–5.0) × 2.0 + 1/2 × 10 × (2.0)
2
= – 10 + 20 = 10 meter
21. (b) Let the ball B hits the ball A after t sec
The X-component of velocity of A is
v
0
 cos 37º = 700 cos 37º
The X-compoment of position of B is 300 cos 37º
The collision will take place when the X-coordinate of
A is the same as that of B.
As the collision takes place at a time t, hence
700 cos 37º × t = 300 cos 37º
or t = (300/700) = (3/7) sec
In this time the ball B has fallen through a distance
y = – 1/2  gt
2
 (Free fall of body B)
 = – 1/2 × 980 × (3 / 7)
2
 = – 90 cm
Hence the ball B falls a distance 90 cm
22. (b)
(1) Because force is constant hence acceleration will be
constant. When force is in oblique direction with initial
velocity , the resultant path is parabolic path.
(2) Total time of flight = T = 
g
sin u 2 q
,
Maximum height attained H = 
g 2
sin u
2 2
q
Now  
T
H
 = 
4
sin u q
(3) Initially the height of the monkey = MB = y = x tan q
Let the monkey drop to along line MA and the bullet
reach along the parabolic path OA. If both reach at A
simultaneously , the monkey is hit by the bullet.
AB  = x tan q – 
q
2 2
2
cos u 2
gx
,
\ MA = MB – AB
MA = x tan q – x tan q + 
q
2 2
2
cos u 2
gx
= 
q
2 2
2
cos u 2
gx
....(i)
Time taken by the bullet to reach point A,
t = 
q cos u
x
....(ii)
29.4 m
30º
B A
v
Tower
u
y
u
x
X
Y
u
h
R=17.3m
X
v
M
A
O B
q
Page 4


1. (b) Given, that y = 
3
 x – (1/2) x
2
....(1)
The above equation is similar to equation of trajectory
of the projectiles
y = tan q  x – 1/2 
q
2 2
cos u
g
 x
2
....(2)
Comparing (1) & (2) we get
tan q = 
3
Þ q = 60º
and 1/2 = (1/2) 
q
2 2
cos u
g
Þ  u
2
 cos
2
 q = g  Þ u
2
 cos
2
 60° = 10
Þ u
2
 (1/4)  = 10  Þ u
2 
= 40  Þ u = 2
10
 m/s
2. (a) For angle of elevation of 60º, we have maximum height
h
1
 = 
22
u sin 60º
2g
 = 
g 8
u 3
2
For angle of elevation of 30º, we have maximum height
h
2 
= 
g 2
º 30 sin u
2 2
 = 
g 8
u
2
; 
2
1
h
h
 = 
1
3
3. (b) T otal time of flight = 2 × time taken to reach max. height
Þ t
2
 = 2t
1
 Þ t
2
/t
1
 = 2/1
4. (d) v
x
 = dx/dt  = 2ct, v
y
 = dy/dt = 2bt
\ v = 
2
y
2
x
v v + = 2t 
22
cb +
5. (c) v
y
 = dy/dt = 8 – 10 t = 8, when t = 0 (at the time of
projection.)
v
x
 = dx/dt = 6, v = 
2
y
2
x
v v + = 
2 2
6 8 +
 = 10 m/s
6. (a) Horizontal component of velocity
v
x
 = u
x
 = u cos q = 30 × cos 30º = 15 3 m/s
V ertical component of the  velocity
v
y
 = u sin q – gt = 30 sin 30º – 10 × 1 = 5 m/s
v
2
 = v
x
2
 + v
y
2
 = 700 Þ u = 10 7 m/s
7. (a) Let u
x
 and u
y
 be the components of the velocity of the
particle along the  x- and y-directions. Then
u
x
 = dx/dt = u
0
 and u
y
  = dy/dt = wa cos wt
Integration : x = u
0
 t and y = a sin wt
Eliminating t : y = a sin (wx/u
0
)
This is the equation of the trajectory
At t = 3p/2w, we have,
x = u
0
 3p/2w and  y = a sin 3p/2 = – a
\ The distance of the particle from the origin is
2 2
y x + = 
ú
ú
û
ù
ê
ê
ë
é
+ ÷
ø
ö
ç
è
æ
w
p
2
2
0
a
2
u 3
8. (b) T = 
g
sin u 2 q
 Þ 2 = 
g
sin u 2 q
Þ u sin q = g
H = 
g 2
sin u
2 2
q
 = 
g 2
g
2
 = 
2
g
 = 5 m
9. (b) Let u
1
 and u
2
 be the initial velocities respectively . If h
1
and h
2
 are the heights attained by them, then
h
1
 = 
g 2
u
2
1
 and  h
2
 = 
g 2
sin u
2 2
2
q
...(1)
The times of ascent of balls are equal,
we have t = u
1
/g = u
2
 sin q/g
\ u
1
 = u
2
 sin q ... (2)
From eq. (1) 
2
1
h
h
 = 
q
2 2
2
2
1
sin u
u
... (3)
From (2) & (3), 
2
1
h
h
 = 
1
1
10. (a) h
1
 = 
g 2
sin u
2 2
q
 and h
2
 = 
g 2
) 90 ( sin u
2 2
q -
\ h
1
 + h
2
 = u
2
/2g (sin
2
q + cos
2
q)
   = u
2
/2g = 
10 2
98
2
´
 = 490
h
1
 – h
2
 = 50,   \ h
1
 = 270 m and h
2
 = 220 m
11. (c) Greatest height attained
h = 
g 2
sin u
2 2
q
... (1)
Horizontal range
R = 
g
2 sin u
2
q
 = 
g
cos sin u 2
2
q q
... (2)
Given that R=2h
Þ 
g
cos sin u 2
2
q q
 = 
g 2
sin u 2
2 2
q
Þ tan q = 2 ... (3)
Hence sin q = 2/ 5 , cos q = 1/ 5 ,
\ From (2) R = 4u
2
/5g
12. (d) R = 
g
2 sin u
2
q
 = (19.6)
2
 sin 90º /10 = 39.2 m
Man must run (67.4m – 39.2 m) = 28.2 m in the time
taken by the ball to come to ground. Time taken by the
ball.
t = 
g
sin u 2 q
= 
8 . 9
º 45 sin 6 . 19 2´
 =
2
4
= 2.82 sec
V elocity of man = 
sec 82 . 2
m 2 . 28
 = 10 m/s
13. (a) Referring to (fig.) let P be a point on the trajectory
whose co-ordinates are (4, 4). As the ball strikes the
ground at a distance 14 metre from the wall, the range is
4 + 14 = 18 metre. The equation of trajectory is
 y = x tan q -(1/2) g 
q
2 2
2
cos u
x
or   y = x tan q 
ú
û
ù
ê
ë
é
q q
-
tan . cos u 2
gx
1
2 2
or    y = x tan q 
ú
ú
û
ù
ê
ê
ë
é
q q
-
cos sin
x
g
u 2
1
2
          = x tan q 
ú
û
ù
ê
ë
é
-
R
x
1
... (1)
Here x = 4, y = 4 and R = 18
\ 4 = 4 tan q 
ú
û
ù
ê
ë
é
-
18
4
1
 = 4 tan q
÷
ø
ö
ç
è
æ
9
7
or   tan q = 9/7, sin q = 9/
130
 and
  cos q = 7/ 130
Again R = (2/g) u
2
 sin q cos q
    = (2/9.8) × u
2
 × (9/ 130 ) × (7/ 130 )
u
2
 = 
7 9 2
130 130 8 . 9 18
´ ´
´ ´ ´
 = 
7
13 98 ´
= 182,
u = 
182
 metre per second.
14. (b) The situation is shown in fig.
(a) Let the ball collide after t sec
From fig. OC = OB cos 37° = 500 cos 37°
= 500 × 0.8 = 400 cm ...(1)
Horizontal velocity = 700 × cos 37º
\ OC = 700 × cos 37° × t
= 700 × 0.8 × t = 560 t ... (2)
From eqs. (1) and (2) 560 t = 400
or t = (5/7) sec.
Now h = (1/2) g t
2
 = (1/2) × 1000 × (5/7)
2
 = 255.1 cm
(b) Let at the time of impact, v
x
 and v
y
 be the  horizontal
and vertical velocities respectively , then
v
x
 = 700 × cos 37° = 700 × 0.8  = 560 cm/s
and v
y
 = – 700 × sin 37
°
 + 1000 × (5/7)
           = – 700 × 0.6 + (5000/7) = – 420 + 714.3
         = + 294.3 cm/sec (downward)
V elocity of the ball at the time of collision
   v = ) v v (
2
y
2
x
+
\ v = ] ) 3 . 294 ( ) 560 [(
2 2
+ = 632.6 cm/sec
Again tan q = 
x
y
v
v
 = 
560
3 . 294
or q = tan
–1
 
÷
ø
ö
ç
è
æ
560
3 . 294
 = 27
°
 43'
15. (d) Initial velocity is constant let the ball touches the
ground at an angle q and velocity 3u
Hence 3u cos q = u or cos q  = 1/3 or sin q = 8 /3
The vertical component of velocity at the ground
= 3u sin q = 
3
8 3
 = 8 u
For a freely falling body it covers 20 m to acquire veloc-
ity 
8
u
\ (
8
u)
2
 – 0 = 2 × 9.8 × 20 or u = 7 m/s
16. (b) The horizontal range of the projectile on the ground
R = u
g
h 2
 Þ R = 
2
10
10 2 ´
 =
2
.
2
= 2m
17. (a) R = ut  Þ t = R/u = 12/8
Now h = (1/2) gt
2
   = (1/2) × 9.8 × (12/8)
2
  = 11 m
18. (b) The situation is shown in the adjoining figure.
The time taken by the body is equal to the time taken
by the freely falling body from the height 29.4 m. Initial
velocity of  body
500cm
700cm/s
h
C
O
37º
400cm
B
v
x
v
y
v
A
u
q
2 0 m
u
v sina
v cosa
ucosq
Y
usinq
u
h
X
B
v
P
·
q
A
u sin q = 9.8 sin 30
º
  = 4.9 m/s
From the relation, h = u sin q t + (1/2) gt
2
,
we get 29.4 = 4.9 t + (1/2) × 9.8 t
2
  Þ t = 2 sec
19. (b) The horizontal and vertical velocities of the bomb are
independent to each other. The time taken by the bomb
to hit the target can be calculated by its vertical mo-
tion. Let this time be t. Putting h = 490 m and g = 9.8 m/
s
2
 in the formula h = 1/2 gt
2
,
we have  490 = (1/2) × 9.8 × t
2
,
\ t = 
8 . 9
490 2 ´
 = 10 sec
The bomb will hit the target after 10 sec of its dropping.
The horizontal velocity of the bomb is 60 km/hr which
is constant. Hence the horizontal distance travelled by
the bomb in 10 sec (horizontal velocity × time)
= 60 km/hr × 10 sec
= 60 km/hr × 10/(60 × 60) hr = 1/6 km
Hence the distance of aeroplane from the enemy post
is  1/6 km = 1000/6 m = 500/3 meter.
The trajectory of the bomb as seen by an observer on
the ground is parabola. Since the horizontal velocity of
the bomb is the same as that of the aeroplane, the fall-
ing bomb will always remain below the aeroplane. Hence
the person sitting inside the aeroplane will observe the
bomb falling vertically downward.
20. (a) The angle of projection of the ball is q
0
 ( = 30º) and the
velocity of projection is u ( = 10 m/s). Resolving u in
horizontal and vertical components,
we have horizontal component,
u
x
 = u cos q
0
 = 10 cos 30º = 8.65 m/s
and vertical component (upward),
u
y
 = u sin 30º = 5.0 m/s
If the ball hit the ground after t sec of projection, then
the horizontal range is  R = u
x
 × t = 8.65 t meter
\ t = 
65 . 8
R
 = 
s / m 65 . 8
m 3 . 17
 = 2.0 s
If h be the height of the tower,
then h = u'
y
 t + (1/2) g t
2
,
where u
y
' is the vertical component
(downward) of the velocity of the ball.
Here u
y
 = – u
y
' = – 5.0 m/s and t = 2.0 s
\ h = (–5.0) × 2.0 + 1/2 × 10 × (2.0)
2
= – 10 + 20 = 10 meter
21. (b) Let the ball B hits the ball A after t sec
The X-component of velocity of A is
v
0
 cos 37º = 700 cos 37º
The X-compoment of position of B is 300 cos 37º
The collision will take place when the X-coordinate of
A is the same as that of B.
As the collision takes place at a time t, hence
700 cos 37º × t = 300 cos 37º
or t = (300/700) = (3/7) sec
In this time the ball B has fallen through a distance
y = – 1/2  gt
2
 (Free fall of body B)
 = – 1/2 × 980 × (3 / 7)
2
 = – 90 cm
Hence the ball B falls a distance 90 cm
22. (b)
(1) Because force is constant hence acceleration will be
constant. When force is in oblique direction with initial
velocity , the resultant path is parabolic path.
(2) Total time of flight = T = 
g
sin u 2 q
,
Maximum height attained H = 
g 2
sin u
2 2
q
Now  
T
H
 = 
4
sin u q
(3) Initially the height of the monkey = MB = y = x tan q
Let the monkey drop to along line MA and the bullet
reach along the parabolic path OA. If both reach at A
simultaneously , the monkey is hit by the bullet.
AB  = x tan q – 
q
2 2
2
cos u 2
gx
,
\ MA = MB – AB
MA = x tan q – x tan q + 
q
2 2
2
cos u 2
gx
= 
q
2 2
2
cos u 2
gx
....(i)
Time taken by the bullet to reach point A,
t = 
q cos u
x
....(ii)
29.4 m
30º
B A
v
Tower
u
y
u
x
X
Y
u
h
R=17.3m
X
v
M
A
O B
q
Hence from (1), MA = (1/2) gt
2
The monkey drops through distance (1/2) gt
2
 in the
same time. So the monkey is hit by the bullet.
(4) The range R = 
g
2 sin u
2
q
\ Maximum range R
max
 = d = 
g
u
2
....(iii)
Height H = 
g 2
sin u
2 2
q
\ Maximum height
H
max
= 
g 2
u
2
....(iv)
From (iii) & (iv), H
max 
= d/2
23. (a) Range of projectile, R = 
g
2 sin u
2
q
The range is same for two angle q
1
 and q
2
 provided
q
2
 = 90º – q
1
At an angle q
1
, range R
1
 = 
g
2 sin u
1
2
q
At an angle of projection q
2
,
Range R
2
 =
g
2 sin u
2
2
q
= 
g
) º 90 ( 2 sin u
1
2
q -
= 
g
2 sin u
1
2
q
Þ  R
1
 = R
2
\ other angle = 90º - q
1
 = 90º – 15º = 75º
24. (a) t
1
 = 
g
sin u 2 q
t
2 
 = 
2u sin(90º )
g
-q
= 
g
cos u 2 q
t
1
t
2
  = 
g
2
 
2
u sin2
g
q
 = 
2
g
. R
where R is the range, Hence t
1
t
2
 µ R
25. (c) If the ball hits the n
th
 step, the horizontal
and vertical distances traversed are nb and nh
respectively.
n step
th
h
1
2
u
b
R
Let t be the time taken by the ball for these horizontal
and vertical displacement. Then velocity along
horizontal direction remains constant = u ; initial vertical
velocity is zero
\ nb = ut ....(1)
nh = 0 + (1/2) gt
2
....(2)
From (1) & (2) we get
nh = (1/2) g (nb/u)
2
Þ  n = 
2
2
gb
hu 2
 (eleminating t)
26. (a) y = (1/2) gt
2
 (downward)
Þ 1000 = (1/2) × 10 × t
2
 Þ  t = 14.15 sec
x = ut  = 
÷
÷
ø
ö
ç
ç
è
æ
´
´
60 60
10 144
3
 × 14.15= 571.43 m
27. (b) Horizontal component of velocity
= 720 × 5/8 = 200 m/s
Let t be the time taken for a freely falling body from 490.
Then y  = (1/2) gt
2
Þ  490 = (1/2) × 9.8 × t
2
  Þ  t = 10 second
Now horizontal distance = Velocity × time
= 200 × 10 =2000m
Hence the bomb missed the target by 2000 m
28. (a) Since W = D K implies that the final speed will be same.
29. (a) The time of flight depends only on the vertical
component of velocity which remains unchanged in
collision with a vertical wall.
30. (a) In statement-2, if speed of both projectiles are same,
horizontal ranges will be same. Hence statement-2 is
correct explanation of statement-1.
Read More
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FAQs on DPP for NEET: Daily Practice Problems, Ch 6: Motion in a Plane- 1 (Solutions) - Physics Class 11

1. What is motion in a plane?
Ans. Motion in a plane refers to the movement of an object in two dimensions, typically described by the coordinates of its position on a Cartesian plane.
2. What are the types of motion in a plane?
Ans. There are two types of motion in a plane: rectilinear motion, which occurs along a straight line, and curvilinear motion, which occurs along a curved path.
3. How is motion in a plane different from motion in a straight line?
Ans. Motion in a plane involves movement in two dimensions, while motion in a straight line occurs only along a single dimension. In motion in a plane, the object can move both horizontally and vertically, while in motion in a straight line, it can only move along one direction.
4. How is velocity calculated in motion in a plane?
Ans. Velocity in motion in a plane is calculated using the same formula as velocity in motion in a straight line. It is the rate of change of displacement over time and can be determined by dividing the change in position by the change in time.
5. What is projectile motion in a plane?
Ans. Projectile motion in a plane refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. It is a combination of horizontal and vertical motion and is commonly seen in the trajectory of a thrown ball or a projectile fired from a cannon.
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