Page 1
(1) (a).
Form figure
T cos q = mg ...... (1)
T sin q =
2
mv
r
=
2
mv
sin q l
...... (2)
Form eq. (1), T =
mg
cosq
When the string is horizontal, q must be 90º i.e.,cos 90º = 0
\ T =
mg
0
= ¥
Thus the tension must be infinite which is impossible, so
the string can not be in horizontal plane.
The maximum angle q is given by the breaking tension of
the string in the equation T cos q = m.g
Here T (Maximum) = 8 N and m = 0.4 kg
\ 8 cos q = 0.4 × g = 0.4 × 10 = 4
cos q = (4/8) =
1
2
, q = 60º
The angle with horizontal = 90º – 60º = 30º
From equation (2), 8 sin 60º =
2
o
0.4v
4 sin 60
´
v
2
=
2
32sin 60º
0.4
= 80 sin
2
60º
Þ v =
80
sin 60º = 7.7 m/sec
(2) (a). Let m be the mass of the ball. When the ball comes
down to B, its potential energy mgh which is converted
into kinetic energy . Let v
B
, be the velocity of the ball at B.
Then, mgh =
1
2
m v
B
2
The ball now rises to a point D, where its potential energy
is mg(h – 2r). If v
D
be the velocity of the ball at D, then,
m g (h – 2 r) =
1
2
m
2
D
v ......(2)
Now to complete the circular path, it is necessary that the
centrifugal force acting upward at point D, should be equal
or greater than the force mg acting downward at point D
should be equal or greater than the force mg acting
downward. Therefore
2
D
mv
r
³ mg or
2
D
v ³ r g
From equation (2)
2
D
v = 2g (h – 2r),
\ 2g (h – 2r) ³ r g Þ h ³
5
2
r
(3) (d). See fig, Here v = 360 km/hr = 100 m/sec
At lower point, N – mg =
2
mv
R
,
N = weight of the flyer = mg +
2
mv
R
N = 70 × 10 +
70 (10000)
500
´
= 2100 N
At upper point, N + mg =
2
mv
R
,
N =
2
mv
R
– mg = 1400 – 700 = 700 N
At middle point, N =
2
mv
R
= 1400 N
(4) (a). Given that U(r) = 10r
3
So the force F acting on the particle is given by ,
F = –
U
r
¶
¶
= –
r
¶
¶
(10 r
3
) = –10 × 3 r
2
= –30 r
2
For circular motion of the particle,
F =
2
mv
r
= 30 r
2
Substituting the given values, we have,
2
3v
10
´
= 30 × (10)
2
or v = 100 m/s
N
N
N
mg
mg
·
·
·
O
l
Tsinq
T
Tcosq
mg
·A
T
mg
q
q
Page 2
(1) (a).
Form figure
T cos q = mg ...... (1)
T sin q =
2
mv
r
=
2
mv
sin q l
...... (2)
Form eq. (1), T =
mg
cosq
When the string is horizontal, q must be 90º i.e.,cos 90º = 0
\ T =
mg
0
= ¥
Thus the tension must be infinite which is impossible, so
the string can not be in horizontal plane.
The maximum angle q is given by the breaking tension of
the string in the equation T cos q = m.g
Here T (Maximum) = 8 N and m = 0.4 kg
\ 8 cos q = 0.4 × g = 0.4 × 10 = 4
cos q = (4/8) =
1
2
, q = 60º
The angle with horizontal = 90º – 60º = 30º
From equation (2), 8 sin 60º =
2
o
0.4v
4 sin 60
´
v
2
=
2
32sin 60º
0.4
= 80 sin
2
60º
Þ v =
80
sin 60º = 7.7 m/sec
(2) (a). Let m be the mass of the ball. When the ball comes
down to B, its potential energy mgh which is converted
into kinetic energy . Let v
B
, be the velocity of the ball at B.
Then, mgh =
1
2
m v
B
2
The ball now rises to a point D, where its potential energy
is mg(h – 2r). If v
D
be the velocity of the ball at D, then,
m g (h – 2 r) =
1
2
m
2
D
v ......(2)
Now to complete the circular path, it is necessary that the
centrifugal force acting upward at point D, should be equal
or greater than the force mg acting downward at point D
should be equal or greater than the force mg acting
downward. Therefore
2
D
mv
r
³ mg or
2
D
v ³ r g
From equation (2)
2
D
v = 2g (h – 2r),
\ 2g (h – 2r) ³ r g Þ h ³
5
2
r
(3) (d). See fig, Here v = 360 km/hr = 100 m/sec
At lower point, N – mg =
2
mv
R
,
N = weight of the flyer = mg +
2
mv
R
N = 70 × 10 +
70 (10000)
500
´
= 2100 N
At upper point, N + mg =
2
mv
R
,
N =
2
mv
R
– mg = 1400 – 700 = 700 N
At middle point, N =
2
mv
R
= 1400 N
(4) (a). Given that U(r) = 10r
3
So the force F acting on the particle is given by ,
F = –
U
r
¶
¶
= –
r
¶
¶
(10 r
3
) = –10 × 3 r
2
= –30 r
2
For circular motion of the particle,
F =
2
mv
r
= 30 r
2
Substituting the given values, we have,
2
3v
10
´
= 30 × (10)
2
or v = 100 m/s
N
N
N
mg
mg
·
·
·
O
l
Tsinq
T
Tcosq
mg
·A
T
mg
q
q
22
DPP/ P 08
The total energy in circular motion
E = K.E. + P .E. =
1
2
mv
2
+ U(r)
=
1
2
× 3 × (100)
2
+ 10 + (10)
3
= 2.5 × 10
4
joule
Angular momentum
= mvr = 3 × 100 × 10 = 3000 kg–m
2
/sec
Also time period T =
2r
v
p
=
2 10
100
´p´
=
5
p
sec
(5) (a). Let T be the tension, q the angle made by the string
with the vertical through the point of suspension.
The time period t = 2p
h
g
=
1
frequency
= p/2
Therefore w =
g
h
= 4 Þ
h
g
=
1
16
cos q =
h
l
=
g
16
= 0.6125 Þ q = 52º 14'
Linear velocity
= (l sin q)w =1 × sin 52º 14' × 4
= 3.16 m/s
(6) (d). Centripetal acceleration, a
c
=
2
v
r
= k
2
rt
2
\ V ariable velocity v =
2 22
k rt
= k r t
The force causing the velocity to varies
F = m
dv
dt
= m k r
The power delivered by the force is,
P = Fv = mkr × krt = mk
2
r
2
t
(7) (a). We know centripetal acceleration
a
c
=
2
(tangential velocity)
radius
=
2
(200)
100
= 400 m/sec
2
a
c
a
t
O
Tangential acceleration
a
t
= 100 m/sec
2
(given)
\ a
net
=
22o
c t ct
a a 2a a cos90 ++
=
22
ct
aa +
=
22
(400) (100) + = 100
17
m/s
2
(8) (b). Suppose v be the velocity of particle at the lowest
position B.
According to conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + mgl =
1
2
mv
2
+ 0 Þ v = 2gl
(9) (a). Maximum tension T =
2
mv
r
+ mg
\
2
mv
r
= T – mg
or
2
mv
r
= 163.6 – 4 × 9.8 Þ v = 6 m/s
(10) (c). The situation is shown in fig. Let v be the velocity of
the bob at the lowest position. In this position the P .E. of
bob is converted into K.E. hence 
mgl =
1
2
mv
2
Þ v
2
= 2gl ....(1)
If T be the tension in the string,
then T – mg =
l
2
mv
....(2)
From (1) & (2), T = 3 mg
(11) (b). The velocity of the swimmer w .r.t. water
SR
v
r
= 4.0 km/
h in the direction perpendicular to the river. The velocity of
river w.r.t. the ground is
RG
v
r
= 3.0 km/h along the length
of river.
X
Y
V
SR
V
SG
V
RG
·
h
T
mg
q
l
l
O
A
B
mg
Page 3
(1) (a).
Form figure
T cos q = mg ...... (1)
T sin q =
2
mv
r
=
2
mv
sin q l
...... (2)
Form eq. (1), T =
mg
cosq
When the string is horizontal, q must be 90º i.e.,cos 90º = 0
\ T =
mg
0
= ¥
Thus the tension must be infinite which is impossible, so
the string can not be in horizontal plane.
The maximum angle q is given by the breaking tension of
the string in the equation T cos q = m.g
Here T (Maximum) = 8 N and m = 0.4 kg
\ 8 cos q = 0.4 × g = 0.4 × 10 = 4
cos q = (4/8) =
1
2
, q = 60º
The angle with horizontal = 90º – 60º = 30º
From equation (2), 8 sin 60º =
2
o
0.4v
4 sin 60
´
v
2
=
2
32sin 60º
0.4
= 80 sin
2
60º
Þ v =
80
sin 60º = 7.7 m/sec
(2) (a). Let m be the mass of the ball. When the ball comes
down to B, its potential energy mgh which is converted
into kinetic energy . Let v
B
, be the velocity of the ball at B.
Then, mgh =
1
2
m v
B
2
The ball now rises to a point D, where its potential energy
is mg(h – 2r). If v
D
be the velocity of the ball at D, then,
m g (h – 2 r) =
1
2
m
2
D
v ......(2)
Now to complete the circular path, it is necessary that the
centrifugal force acting upward at point D, should be equal
or greater than the force mg acting downward at point D
should be equal or greater than the force mg acting
downward. Therefore
2
D
mv
r
³ mg or
2
D
v ³ r g
From equation (2)
2
D
v = 2g (h – 2r),
\ 2g (h – 2r) ³ r g Þ h ³
5
2
r
(3) (d). See fig, Here v = 360 km/hr = 100 m/sec
At lower point, N – mg =
2
mv
R
,
N = weight of the flyer = mg +
2
mv
R
N = 70 × 10 +
70 (10000)
500
´
= 2100 N
At upper point, N + mg =
2
mv
R
,
N =
2
mv
R
– mg = 1400 – 700 = 700 N
At middle point, N =
2
mv
R
= 1400 N
(4) (a). Given that U(r) = 10r
3
So the force F acting on the particle is given by ,
F = –
U
r
¶
¶
= –
r
¶
¶
(10 r
3
) = –10 × 3 r
2
= –30 r
2
For circular motion of the particle,
F =
2
mv
r
= 30 r
2
Substituting the given values, we have,
2
3v
10
´
= 30 × (10)
2
or v = 100 m/s
N
N
N
mg
mg
·
·
·
O
l
Tsinq
T
Tcosq
mg
·A
T
mg
q
q
22
DPP/ P 08
The total energy in circular motion
E = K.E. + P .E. =
1
2
mv
2
+ U(r)
=
1
2
× 3 × (100)
2
+ 10 + (10)
3
= 2.5 × 10
4
joule
Angular momentum
= mvr = 3 × 100 × 10 = 3000 kg–m
2
/sec
Also time period T =
2r
v
p
=
2 10
100
´p´
=
5
p
sec
(5) (a). Let T be the tension, q the angle made by the string
with the vertical through the point of suspension.
The time period t = 2p
h
g
=
1
frequency
= p/2
Therefore w =
g
h
= 4 Þ
h
g
=
1
16
cos q =
h
l
=
g
16
= 0.6125 Þ q = 52º 14'
Linear velocity
= (l sin q)w =1 × sin 52º 14' × 4
= 3.16 m/s
(6) (d). Centripetal acceleration, a
c
=
2
v
r
= k
2
rt
2
\ V ariable velocity v =
2 22
k rt
= k r t
The force causing the velocity to varies
F = m
dv
dt
= m k r
The power delivered by the force is,
P = Fv = mkr × krt = mk
2
r
2
t
(7) (a). We know centripetal acceleration
a
c
=
2
(tangential velocity)
radius
=
2
(200)
100
= 400 m/sec
2
a
c
a
t
O
Tangential acceleration
a
t
= 100 m/sec
2
(given)
\ a
net
=
22o
c t ct
a a 2a a cos90 ++
=
22
ct
aa +
=
22
(400) (100) + = 100
17
m/s
2
(8) (b). Suppose v be the velocity of particle at the lowest
position B.
According to conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + mgl =
1
2
mv
2
+ 0 Þ v = 2gl
(9) (a). Maximum tension T =
2
mv
r
+ mg
\
2
mv
r
= T – mg
or
2
mv
r
= 163.6 – 4 × 9.8 Þ v = 6 m/s
(10) (c). The situation is shown in fig. Let v be the velocity of
the bob at the lowest position. In this position the P .E. of
bob is converted into K.E. hence 
mgl =
1
2
mv
2
Þ v
2
= 2gl ....(1)
If T be the tension in the string,
then T – mg =
l
2
mv
....(2)
From (1) & (2), T = 3 mg
(11) (b). The velocity of the swimmer w .r.t. water
SR
v
r
= 4.0 km/
h in the direction perpendicular to the river. The velocity of
river w.r.t. the ground is
RG
v
r
= 3.0 km/h along the length
of river.
X
Y
V
SR
V
SG
V
RG
·
h
T
mg
q
l
l
O
A
B
mg
DPP/ P 08
23
The velocity of the swimmer w .r.t. the ground is
SG
v
r
where
SG SR RG
V VV
® ®®
=+
2 2 22
SG SR RG
V V V 43 = + =+
16 9 25 5 km / hr = +==
(12) (b). The minimum speed at highest point of a vertical circle
is given by v
c
= rg = 20 9.8 ´ = 14 m/s
(13) (a). The speed at highest point must be
v > gr , v = rw = r
2
T
p
\ r
2
T
p
> rg
T <
2r
rg
p
< 2p
r
g
< 2p
0.5
9.8
< 1.4 sec
Maximum period of revolution = 1.4 sec
(14) (a). Let the particles leaves the sphere at height h,
2
mv
R
= mg cos q – N
When the particle leaves the sphere i.e. N = 0
2
mv
R
= mg cos q
Þ v
2
= gR cos q ....(1)
According to law of conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + mgR =
1
2
mv
2
+ mgh
Þ v
2
= 2g (R – h) ....(2)
From (1) & (2), h =
2
3
R
Also cos q=
2
3
(15) (a). Let the body will have the circular path at height h
above the bottom of circle from figure
h
O
A
mg
v
B
2
mv
l
= T + mg cos a
On leaving the circular path
T = 0
\
2
mv
l
= mg cos a
Þ v
2
= g l cosa ....(1)
According to law of conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + 2mgl =
1
2
mv
2
+ mgh
Þ v
2
= 2g (2l – h) ....(2)
From (1) & (2) h =
5
3
l
Also, cos a =
h  l
l
(16) (d)
2
Tsin MR q=w
2
T sin M L sin q=wq
From (i) and (ii)
2
T ML =w
22
M4 nL =p
L
R
S
T
mR w
2
q
q
2
2
2
M4L
æö
=p
ç÷
èø p
16 ML =
(17) (a). v = 60 km/hr =
50
3
m/s
r = 0.1 km = 100m
\ tan q =
2
v
rg
= 0.283
\ q = tan
–1
(0.283)
(18) (c). We know that tan q =
2
v
rg
..... (1)
Let h be the relative raising of outer rail with respect to
inner rail. Then
tan q =
h
l
...... (2)
(l = separation between rails)
From (1) & (2) , h =
2
v
rg
× l
Hence v = 48 km/hr =
120
9
m/s, (r = 400 m, l = 1m),
\ h =
2
(120 / 9) 1
400 9.8
´
´
= 0.045 m = 4.5 cm
B
N
q
A
mg
q
v
h
Page 4
(1) (a).
Form figure
T cos q = mg ...... (1)
T sin q =
2
mv
r
=
2
mv
sin q l
...... (2)
Form eq. (1), T =
mg
cosq
When the string is horizontal, q must be 90º i.e.,cos 90º = 0
\ T =
mg
0
= ¥
Thus the tension must be infinite which is impossible, so
the string can not be in horizontal plane.
The maximum angle q is given by the breaking tension of
the string in the equation T cos q = m.g
Here T (Maximum) = 8 N and m = 0.4 kg
\ 8 cos q = 0.4 × g = 0.4 × 10 = 4
cos q = (4/8) =
1
2
, q = 60º
The angle with horizontal = 90º – 60º = 30º
From equation (2), 8 sin 60º =
2
o
0.4v
4 sin 60
´
v
2
=
2
32sin 60º
0.4
= 80 sin
2
60º
Þ v =
80
sin 60º = 7.7 m/sec
(2) (a). Let m be the mass of the ball. When the ball comes
down to B, its potential energy mgh which is converted
into kinetic energy . Let v
B
, be the velocity of the ball at B.
Then, mgh =
1
2
m v
B
2
The ball now rises to a point D, where its potential energy
is mg(h – 2r). If v
D
be the velocity of the ball at D, then,
m g (h – 2 r) =
1
2
m
2
D
v ......(2)
Now to complete the circular path, it is necessary that the
centrifugal force acting upward at point D, should be equal
or greater than the force mg acting downward at point D
should be equal or greater than the force mg acting
downward. Therefore
2
D
mv
r
³ mg or
2
D
v ³ r g
From equation (2)
2
D
v = 2g (h – 2r),
\ 2g (h – 2r) ³ r g Þ h ³
5
2
r
(3) (d). See fig, Here v = 360 km/hr = 100 m/sec
At lower point, N – mg =
2
mv
R
,
N = weight of the flyer = mg +
2
mv
R
N = 70 × 10 +
70 (10000)
500
´
= 2100 N
At upper point, N + mg =
2
mv
R
,
N =
2
mv
R
– mg = 1400 – 700 = 700 N
At middle point, N =
2
mv
R
= 1400 N
(4) (a). Given that U(r) = 10r
3
So the force F acting on the particle is given by ,
F = –
U
r
¶
¶
= –
r
¶
¶
(10 r
3
) = –10 × 3 r
2
= –30 r
2
For circular motion of the particle,
F =
2
mv
r
= 30 r
2
Substituting the given values, we have,
2
3v
10
´
= 30 × (10)
2
or v = 100 m/s
N
N
N
mg
mg
·
·
·
O
l
Tsinq
T
Tcosq
mg
·A
T
mg
q
q
22
DPP/ P 08
The total energy in circular motion
E = K.E. + P .E. =
1
2
mv
2
+ U(r)
=
1
2
× 3 × (100)
2
+ 10 + (10)
3
= 2.5 × 10
4
joule
Angular momentum
= mvr = 3 × 100 × 10 = 3000 kg–m
2
/sec
Also time period T =
2r
v
p
=
2 10
100
´p´
=
5
p
sec
(5) (a). Let T be the tension, q the angle made by the string
with the vertical through the point of suspension.
The time period t = 2p
h
g
=
1
frequency
= p/2
Therefore w =
g
h
= 4 Þ
h
g
=
1
16
cos q =
h
l
=
g
16
= 0.6125 Þ q = 52º 14'
Linear velocity
= (l sin q)w =1 × sin 52º 14' × 4
= 3.16 m/s
(6) (d). Centripetal acceleration, a
c
=
2
v
r
= k
2
rt
2
\ V ariable velocity v =
2 22
k rt
= k r t
The force causing the velocity to varies
F = m
dv
dt
= m k r
The power delivered by the force is,
P = Fv = mkr × krt = mk
2
r
2
t
(7) (a). We know centripetal acceleration
a
c
=
2
(tangential velocity)
radius
=
2
(200)
100
= 400 m/sec
2
a
c
a
t
O
Tangential acceleration
a
t
= 100 m/sec
2
(given)
\ a
net
=
22o
c t ct
a a 2a a cos90 ++
=
22
ct
aa +
=
22
(400) (100) + = 100
17
m/s
2
(8) (b). Suppose v be the velocity of particle at the lowest
position B.
According to conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + mgl =
1
2
mv
2
+ 0 Þ v = 2gl
(9) (a). Maximum tension T =
2
mv
r
+ mg
\
2
mv
r
= T – mg
or
2
mv
r
= 163.6 – 4 × 9.8 Þ v = 6 m/s
(10) (c). The situation is shown in fig. Let v be the velocity of
the bob at the lowest position. In this position the P .E. of
bob is converted into K.E. hence 
mgl =
1
2
mv
2
Þ v
2
= 2gl ....(1)
If T be the tension in the string,
then T – mg =
l
2
mv
....(2)
From (1) & (2), T = 3 mg
(11) (b). The velocity of the swimmer w .r.t. water
SR
v
r
= 4.0 km/
h in the direction perpendicular to the river. The velocity of
river w.r.t. the ground is
RG
v
r
= 3.0 km/h along the length
of river.
X
Y
V
SR
V
SG
V
RG
·
h
T
mg
q
l
l
O
A
B
mg
DPP/ P 08
23
The velocity of the swimmer w .r.t. the ground is
SG
v
r
where
SG SR RG
V VV
® ®®
=+
2 2 22
SG SR RG
V V V 43 = + =+
16 9 25 5 km / hr = +==
(12) (b). The minimum speed at highest point of a vertical circle
is given by v
c
= rg = 20 9.8 ´ = 14 m/s
(13) (a). The speed at highest point must be
v > gr , v = rw = r
2
T
p
\ r
2
T
p
> rg
T <
2r
rg
p
< 2p
r
g
< 2p
0.5
9.8
< 1.4 sec
Maximum period of revolution = 1.4 sec
(14) (a). Let the particles leaves the sphere at height h,
2
mv
R
= mg cos q – N
When the particle leaves the sphere i.e. N = 0
2
mv
R
= mg cos q
Þ v
2
= gR cos q ....(1)
According to law of conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + mgR =
1
2
mv
2
+ mgh
Þ v
2
= 2g (R – h) ....(2)
From (1) & (2), h =
2
3
R
Also cos q=
2
3
(15) (a). Let the body will have the circular path at height h
above the bottom of circle from figure
h
O
A
mg
v
B
2
mv
l
= T + mg cos a
On leaving the circular path
T = 0
\
2
mv
l
= mg cos a
Þ v
2
= g l cosa ....(1)
According to law of conservation of energy
(K.E. + P .E.) at A = (K.E. + P .E.) at B
Þ 0 + 2mgl =
1
2
mv
2
+ mgh
Þ v
2
= 2g (2l – h) ....(2)
From (1) & (2) h =
5
3
l
Also, cos a =
h  l
l
(16) (d)
2
Tsin MR q=w
2
T sin M L sin q=wq
From (i) and (ii)
2
T ML =w
22
M4 nL =p
L
R
S
T
mR w
2
q
q
2
2
2
M4L
æö
=p
ç÷
èø p
16 ML =
(17) (a). v = 60 km/hr =
50
3
m/s
r = 0.1 km = 100m
\ tan q =
2
v
rg
= 0.283
\ q = tan
–1
(0.283)
(18) (c). We know that tan q =
2
v
rg
..... (1)
Let h be the relative raising of outer rail with respect to
inner rail. Then
tan q =
h
l
...... (2)
(l = separation between rails)
From (1) & (2) , h =
2
v
rg
× l
Hence v = 48 km/hr =
120
9
m/s, (r = 400 m, l = 1m),
\ h =
2
(120 / 9) 1
400 9.8
´
´
= 0.045 m = 4.5 cm
B
N
q
A
mg
q
v
h
24
DPP/ P 08
(19) (a). The woman has two velocities simultaneously while
running on the deck, one velocity is equal to the velocity
of ship i.e. 12 m/s due east and other velocity is 5 m/s due
north.
13 m/s
E
N
5 m/s
12m/s
The resultant velocity of woman
=
22
(12) (5) + = 13 m/s
(20) (c). If we consider velocity of rain with respect to the man
is V km/h.
v = 3km/h
mg
Rain v = 4km/h
rg
Road
Relative velocity of man w.r.t. ground
mgg
m
v v v ........(1)
® ®®
=
V elocity of rain w .r.t. ground
rgg
r
v v v ........(2)
® ®®
=
v
rm
v = 4 km/hr
rm
v = 3 km/hr
mg
v
mg
V elocity of rain w .r.t. man rmm
r
v vv
® ®®
=
On subtracting eq
n
. 1 from eq
n
. 2
rm mg
rg
v vv
® ®®
=
2 2 22
rm rg mg
 v  v v 4 3 5 km / hr = + = +=
(21) (d) Since the maximum tension T
B
in the string moving in
the vertical circle is at the bottom and minimum tension T
T
is at the top.
\
22
BT
BT
mv mv
T mg and T mg
LL
=+ =
\
2
B
2
BB
22
T
TT
mv
T v gL 44
L
or
T 11
mv v gL
mg
L
+
= ==


or
2 2 22
B T BT
v gL 4v 4gL but v v 4gL + =  =+
\
2 22
T TT
v 4gL gL 4v 4gL 3v 9gL + + =  Þ=
\
2
TT
10
v 3 g L 3 10 or v 10m / sec
3
= ´´ =´ ´ =
(22) (d). Use definition of relative velocity
PQ PQ
V VV =
r rr
PQ
V const. ; V const. ==
rr
\
PQ QP
 V   V  const. ==
rr
;
PQ
V  V >
rr
\
PQ QP
V ve ; V ve ®+ =
rr
i.e. towards origin.
(23) (c). He can only reach the opposite point if he can cancel
up the velocity of river by his component of velocity.
(24) (a). v = Rw
v
1
> v
2
(25) (b), (26) (b), (27) (c).
The path of a projectile as observed by other projectile is a
straight line.
q
q
u cos
0
q
u cos
0
q
u cos
0
q
u
0
sinq
u
0
sinq
u
0
sinq
v
ba
B
A AB
ˆ ˆˆ
v u cos i (u sin gt) j.v (2u cos ) i = q+ q =q
B
ˆˆ
v u cos i (u sin gt) j =  q + q ;
BA
a g g0 ==
The vertical component u
0
sin q will get cancelled.
The relative velocity will only be horizontal which is equal
to 2u
0
cos q
Hence B will travel horiozontally towards left w .r.t A with
constant speed 2u
0
cos q and minimum distance will be h.
rel
rel0
S
V 2u cos
=
q
l
(28) (a) When two bodies are moving in opposite direction,
relative velocity between them is equal to sum of the
velocity of bodies. But if the bodies are moving in same
direction their relative velocity is equal to difference in
velocity of the bodies.
(29) (b) Time taken is shortest when one aims perpendicular to
the flow .
(30) (d)
22
r/m rm
v vv =+
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