Motion in a Straight Line- 2 Practice Questions - DPP for NEET

 Page 1


1. (c) If a stone is dropped from height h
then  
2
1
2
h gt = ......... (i)
If a stone is thrown upward with velocity u then
h = 
2
11
1
2
ut gt -+ ......... (ii)
If a stone is thrown downward with velocity u then
2
22
1
2
h ut gt =+ ......... (iii)
From (i), (ii) and (iii) we get
22
11
11
22
ut gt gt - += ......... (iv)
22
22
11
22
ut gt gt += ......... (v)
Dividing (iv) and (v) we get
22
1
1
22
2
2
1
()
2
1
()
2
gtt
ut
ut
gtt
-
-
\=
-
or 
22
11
22
2
2
t tt
t
tt
-
-=
-
By solving 
12
t tt =
2. (c) Since direction of v is opposite to the direction of g
and h so from equation of motion
2
1
2
=-n+ h t gt
Þ
2
2 20 -n-= gt th
Þ
2
2 48
2
n± n+
=
gh
t
g
Þ
2
2
11
éù
n
= ++
êú
n
ëû
gh
t
g
3. (c)
22
1 11
11
1 0 2/
22
h ut gt t gt t g = + Þ = ´ + Þ=
V elocity after travelling 1m distance
2 2 22
2 (0)212 = + Þ n = + ´ Þ n= vugh gg
For second 1 meter distance
22
2 2 22
1
1 2 2 2 20
2
g t gt gt gt = ´ + Þ + -=
2
22 88 22
2
g gg
t
g g
- ±+ -±
==
Taking + ve sign 
2
(2 2)/ tg =-
1
2
2/ 1
(2 2)/ 21
g t
t g
\==
--
 and so on.
4. (d) Interval of ball throw = 2 sec.
If we want that minimum three (more than two) ball
remain in air then time of flight of first ball must be
greater than 4 sec.
T > 4 sec.
2
4sec 19.6/
u
u ms
g
> Þ>
for u = 19.6 First ball will just strike the ground(in sky)
Second ball will be at highest point (in sky)
Third ball will be at point of projection or at ground
(not in sky)
5. (a) The distance covered by the ball during the last t
seconds of its upward motion = Distance covered by it
in first t seconds of its downward motion
From 
2
1
2
h ut gt =+
2
1
2
h gt =    [As u = 0 for it downward motion]
6. (d) In the positive region the velocity decreases linearly
(during rise) and in the negative region velocity
increases linearly (during fall) and the direction is
opposite to each other during rise and fall, hence fall is
shown in the negative region.
7. (a) For the given condition initial height h = d and velocity
of the ball is zero. When the ball moves downward its
velocity increases and it will be maximum when the ball
hits the ground & just after the collision it becomes
half and in opposite direction. As the ball moves
upward its velocity again decreases and becomes zero
at height d/2. This explanation match with graph (a).
8. (c) Acceleration of body along AB is g cos q
Distance travelled in time 
2
1
sec ( cos )
2
t AB gt = =q
From DABC, AB = 2R cosq; 
2
1
2 cos cos
2
R gt q=q
2
2
4
or ==
RR
tt
gg
9. (b) It has lesser initial upward velocity .
10. (b) At maximum height velocity n = 0
Page 2


1. (c) If a stone is dropped from height h
then  
2
1
2
h gt = ......... (i)
If a stone is thrown upward with velocity u then
h = 
2
11
1
2
ut gt -+ ......... (ii)
If a stone is thrown downward with velocity u then
2
22
1
2
h ut gt =+ ......... (iii)
From (i), (ii) and (iii) we get
22
11
11
22
ut gt gt - += ......... (iv)
22
22
11
22
ut gt gt += ......... (v)
Dividing (iv) and (v) we get
22
1
1
22
2
2
1
()
2
1
()
2
gtt
ut
ut
gtt
-
-
\=
-
or 
22
11
22
2
2
t tt
t
tt
-
-=
-
By solving 
12
t tt =
2. (c) Since direction of v is opposite to the direction of g
and h so from equation of motion
2
1
2
=-n+ h t gt
Þ
2
2 20 -n-= gt th
Þ
2
2 48
2
n± n+
=
gh
t
g
Þ
2
2
11
éù
n
= ++
êú
n
ëû
gh
t
g
3. (c)
22
1 11
11
1 0 2/
22
h ut gt t gt t g = + Þ = ´ + Þ=
V elocity after travelling 1m distance
2 2 22
2 (0)212 = + Þ n = + ´ Þ n= vugh gg
For second 1 meter distance
22
2 2 22
1
1 2 2 2 20
2
g t gt gt gt = ´ + Þ + -=
2
22 88 22
2
g gg
t
g g
- ±+ -±
==
Taking + ve sign 
2
(2 2)/ tg =-
1
2
2/ 1
(2 2)/ 21
g t
t g
\==
--
 and so on.
4. (d) Interval of ball throw = 2 sec.
If we want that minimum three (more than two) ball
remain in air then time of flight of first ball must be
greater than 4 sec.
T > 4 sec.
2
4sec 19.6/
u
u ms
g
> Þ>
for u = 19.6 First ball will just strike the ground(in sky)
Second ball will be at highest point (in sky)
Third ball will be at point of projection or at ground
(not in sky)
5. (a) The distance covered by the ball during the last t
seconds of its upward motion = Distance covered by it
in first t seconds of its downward motion
From 
2
1
2
h ut gt =+
2
1
2
h gt =    [As u = 0 for it downward motion]
6. (d) In the positive region the velocity decreases linearly
(during rise) and in the negative region velocity
increases linearly (during fall) and the direction is
opposite to each other during rise and fall, hence fall is
shown in the negative region.
7. (a) For the given condition initial height h = d and velocity
of the ball is zero. When the ball moves downward its
velocity increases and it will be maximum when the ball
hits the ground & just after the collision it becomes
half and in opposite direction. As the ball moves
upward its velocity again decreases and becomes zero
at height d/2. This explanation match with graph (a).
8. (c) Acceleration of body along AB is g cos q
Distance travelled in time 
2
1
sec ( cos )
2
t AB gt = =q
From DABC, AB = 2R cosq; 
2
1
2 cos cos
2
R gt q=q
2
2
4
or ==
RR
tt
gg
9. (b) It has lesser initial upward velocity .
10. (b) At maximum height velocity n = 0
DPP/ P 04
9
We know that n = u + at, hence
0 u gT u gT = - Þ=
When ,
2
n=
u
 then
2 2 22
u u gTT
u gt gt gt t = - Þ = Þ = Þ=
Hence at ,
2
T
t = it acquires velocity 
2
u
.
11. (b)
11
22
2 th h
t
g th
= Þ=
12. (c) Speed of the object at reaching the ground 2 n= gh
If heights are equal then velocity will also be equal.
13. (b)
3
10
10 (2 3 1) 35
2
rd
Sm = + ´ -=
3
2
2
107
10 (2 2 1) 25
25
rd
nd
nd
S
S
S
= + ´-= Þ=
14. (c) (2 1)
n
Sn µ- . In equal time interval of 2 seconds
Ratio of distance = 1 : 3 : 5
15. (c) Net acceleration of a body when thrown upward
= acceleration of body – acceleration due to gravity
= a – g
16. (d) The initial velocity of aeroplane is horizontal, then the
vertical component of velocity of packet will be zero.
So 
2h
t
g
=
17. (b) Time = 
Totallength 50 50 100
4sec
Relative velocity 10 15 25
+
= ==
+
18. (d) Relative velocity
= 10 + 5 = 15 m/sec
150
10 sec
15
\== t
19. (a) When two particles moves towards each other then
v
1
 + v
2
 = 4 ....... (i)
When these particles moves in the same direction then
v
1
 – v
2
 = 4 ....... (ii)
By solving v
1
 = 5 and v
2
 = 1 m/s
20. (b) n = n -n
uuu r uu r u u r
ct ct
() n = n + -n
uuu r uu r uu r
ct ct
45°
ct
v
c
v
t
v
-
t
v
Velocity of car w.r.t. train () n
uuur
ct
 is towards
West – North
21. (a) As the trains are moving in the same direction. So the
initial relative speed (v
1
 – v
2
) and by applying
retardation final relative speed becomes zero.
From 
12
12
0()
-
= - Þ = - - Þ=
vv
v u at v v at t
a
22. (d) Let 
A
v
r
 and 
B
v
r
 be the respective velocities of the
particles at A and B. The relative velocity of particle at
A.  w.r.t. to that at B is given by
) v ( v v v
B A B A
r r r r
- + = -
300m
A B
v =25m/s
A
(see figure). From triangle law of velocities if 
OP
 and
PQ
 represent 
A
v
r
 and 
B
v
r
- , then the required relative
velocity 
R
v
r
 is given by 
OQ
.
22
R
| v | 25 20 625 400 32.02 m / s = + = +=
r
If , PQO q = Ð then 
1
255
tan tan
204
-
æö
q= Þ q=
ç÷
èø
O
Q
P
25
20
v
A
v =20m/s
B
Thus, the particle at A, appears to approach B, in a direction
making an angle of tan
–1
 (5/4) with its direction of motion.
Let us draw a line from A, as AC, such that BCA Ð is equal
to 
q
.
A
C
B
M
300m
Thus, to B, A appears to move along AC. From B, draw a
perpendicular to AC as BM.
 BM is the shortest distance between them.
\
  
4
BM ABcos 300 187.41 m
41
= q= ´=
Also, m 26 . 234 sin AB AM = q =
\ time taken to cover a distance
Page 3


1. (c) If a stone is dropped from height h
then  
2
1
2
h gt = ......... (i)
If a stone is thrown upward with velocity u then
h = 
2
11
1
2
ut gt -+ ......... (ii)
If a stone is thrown downward with velocity u then
2
22
1
2
h ut gt =+ ......... (iii)
From (i), (ii) and (iii) we get
22
11
11
22
ut gt gt - += ......... (iv)
22
22
11
22
ut gt gt += ......... (v)
Dividing (iv) and (v) we get
22
1
1
22
2
2
1
()
2
1
()
2
gtt
ut
ut
gtt
-
-
\=
-
or 
22
11
22
2
2
t tt
t
tt
-
-=
-
By solving 
12
t tt =
2. (c) Since direction of v is opposite to the direction of g
and h so from equation of motion
2
1
2
=-n+ h t gt
Þ
2
2 20 -n-= gt th
Þ
2
2 48
2
n± n+
=
gh
t
g
Þ
2
2
11
éù
n
= ++
êú
n
ëû
gh
t
g
3. (c)
22
1 11
11
1 0 2/
22
h ut gt t gt t g = + Þ = ´ + Þ=
V elocity after travelling 1m distance
2 2 22
2 (0)212 = + Þ n = + ´ Þ n= vugh gg
For second 1 meter distance
22
2 2 22
1
1 2 2 2 20
2
g t gt gt gt = ´ + Þ + -=
2
22 88 22
2
g gg
t
g g
- ±+ -±
==
Taking + ve sign 
2
(2 2)/ tg =-
1
2
2/ 1
(2 2)/ 21
g t
t g
\==
--
 and so on.
4. (d) Interval of ball throw = 2 sec.
If we want that minimum three (more than two) ball
remain in air then time of flight of first ball must be
greater than 4 sec.
T > 4 sec.
2
4sec 19.6/
u
u ms
g
> Þ>
for u = 19.6 First ball will just strike the ground(in sky)
Second ball will be at highest point (in sky)
Third ball will be at point of projection or at ground
(not in sky)
5. (a) The distance covered by the ball during the last t
seconds of its upward motion = Distance covered by it
in first t seconds of its downward motion
From 
2
1
2
h ut gt =+
2
1
2
h gt =    [As u = 0 for it downward motion]
6. (d) In the positive region the velocity decreases linearly
(during rise) and in the negative region velocity
increases linearly (during fall) and the direction is
opposite to each other during rise and fall, hence fall is
shown in the negative region.
7. (a) For the given condition initial height h = d and velocity
of the ball is zero. When the ball moves downward its
velocity increases and it will be maximum when the ball
hits the ground & just after the collision it becomes
half and in opposite direction. As the ball moves
upward its velocity again decreases and becomes zero
at height d/2. This explanation match with graph (a).
8. (c) Acceleration of body along AB is g cos q
Distance travelled in time 
2
1
sec ( cos )
2
t AB gt = =q
From DABC, AB = 2R cosq; 
2
1
2 cos cos
2
R gt q=q
2
2
4
or ==
RR
tt
gg
9. (b) It has lesser initial upward velocity .
10. (b) At maximum height velocity n = 0
DPP/ P 04
9
We know that n = u + at, hence
0 u gT u gT = - Þ=
When ,
2
n=
u
 then
2 2 22
u u gTT
u gt gt gt t = - Þ = Þ = Þ=
Hence at ,
2
T
t = it acquires velocity 
2
u
.
11. (b)
11
22
2 th h
t
g th
= Þ=
12. (c) Speed of the object at reaching the ground 2 n= gh
If heights are equal then velocity will also be equal.
13. (b)
3
10
10 (2 3 1) 35
2
rd
Sm = + ´ -=
3
2
2
107
10 (2 2 1) 25
25
rd
nd
nd
S
S
S
= + ´-= Þ=
14. (c) (2 1)
n
Sn µ- . In equal time interval of 2 seconds
Ratio of distance = 1 : 3 : 5
15. (c) Net acceleration of a body when thrown upward
= acceleration of body – acceleration due to gravity
= a – g
16. (d) The initial velocity of aeroplane is horizontal, then the
vertical component of velocity of packet will be zero.
So 
2h
t
g
=
17. (b) Time = 
Totallength 50 50 100
4sec
Relative velocity 10 15 25
+
= ==
+
18. (d) Relative velocity
= 10 + 5 = 15 m/sec
150
10 sec
15
\== t
19. (a) When two particles moves towards each other then
v
1
 + v
2
 = 4 ....... (i)
When these particles moves in the same direction then
v
1
 – v
2
 = 4 ....... (ii)
By solving v
1
 = 5 and v
2
 = 1 m/s
20. (b) n = n -n
uuu r uu r u u r
ct ct
() n = n + -n
uuu r uu r uu r
ct ct
45°
ct
v
c
v
t
v
-
t
v
Velocity of car w.r.t. train () n
uuur
ct
 is towards
West – North
21. (a) As the trains are moving in the same direction. So the
initial relative speed (v
1
 – v
2
) and by applying
retardation final relative speed becomes zero.
From 
12
12
0()
-
= - Þ = - - Þ=
vv
v u at v v at t
a
22. (d) Let 
A
v
r
 and 
B
v
r
 be the respective velocities of the
particles at A and B. The relative velocity of particle at
A.  w.r.t. to that at B is given by
) v ( v v v
B A B A
r r r r
- + = -
300m
A B
v =25m/s
A
(see figure). From triangle law of velocities if 
OP
 and
PQ
 represent 
A
v
r
 and 
B
v
r
- , then the required relative
velocity 
R
v
r
 is given by 
OQ
.
22
R
| v | 25 20 625 400 32.02 m / s = + = +=
r
If , PQO q = Ð then 
1
255
tan tan
204
-
æö
q= Þ q=
ç÷
èø
O
Q
P
25
20
v
A
v =20m/s
B
Thus, the particle at A, appears to approach B, in a direction
making an angle of tan
–1
 (5/4) with its direction of motion.
Let us draw a line from A, as AC, such that BCA Ð is equal
to 
q
.
A
C
B
M
300m
Thus, to B, A appears to move along AC. From B, draw a
perpendicular to AC as BM.
 BM is the shortest distance between them.
\
  
4
BM ABcos 300 187.41 m
41
= q= ´=
Also, m 26 . 234 sin AB AM = q =
\ time taken to cover a distance
10
DPP/ P 04
AB = 234.26 m with a velocity of 32.02 m/s
  = 
234.26
7.32 sec.
32.02
=
23. (d) Since the wind is blowing toward the east, the plane
must head west of north as shown in figure. The velocity
of the plane relative to the ground 
r
v
pg
 will be the sum of
the velocity of the plane relative to the air 
r
v
pa
 and the
velocity of the air relative to the ground 
r
v
ag
.
(i) 1. The velocity of the plane relative to the ground is
given by equation :
r
v
pg 
= 
r
v
pa
 + 
r
v
ag
2. The sine of the angle 
q
 between the velocity of
the plane and north equals the ratio of    v
ag
 and
v
pa
.
sin 
q
= 
v
v
ag
pa
 = 
90 km/h
200 km / h
 = 0.45   
\ q
= 26.74
(ii) Since v
ag
 and v
pg
 are perpendicular, we can use the
Pythagorean theorem to find the magnitude of 
r
v
pg
.
v
2
pg
 = v
2
ag
 + v
2
pg
v
pg 
= v v
pa ag
2 2
- = ( /)( /) 200 90
2 2
kmh kmh -
        = 179 km/h.
24. (a) Using,     
2
1
S uT aT
2
=+
r
r r
(i)
2
1
40 10T gT
2
- =-
or  
2
40 10T 5T - =-
or   
2
5T 10T 40 0 - -= h
v
u
u=10m/s
40m
or 
2
10 10 4 5( 40) 10 100 800
T
2 5 10
+ - ´- ++
==
´
   
10 30
4 sec.
10
+
==
(ii)
2 10
t 2 sec.
g
´
==
(iii) v = 10 + g × 2 = 30 m/s
25. (b) V
plane
 = 100 m/s
v
plane
v
total
v
wind
22
total
V (20) (100) 2 20 100 cos135 = + +´´ ´°
= 
1
400 10000 2 20 100
2
æö
+ + ´ ´ ´-
ç÷
èø
= 87 m/s
26. (c)
wind
plane
v
tan
v
f=
= 
20
100
v
plane
v
total
v
wind
f
1
20
tan
100
-
æö
\ f=
ç÷
èø
27. (d)
plane
total
v
cos
v
f=
plane
total
v
100
v m/s
cos cos
\ ==
ff
28. (a) If components of velocities of boat relative to river is
same normal to river flow (as shown in figure) are same,
both boats reach other bank simultaneously.
v
v
q q
Boat 1
Boat 2
River
29. (a) Both statement - 1 & statement - 2 are correct and
statement - 2 is correct explanation of statement - 2
30. (d) Statement - 1 is true but statement - 2 is false.