Page 1
5.1
SOLUTIONS TO CONCEPTS
CHAPTER – 5
1. m = 2kg
S = 10m
Let, acceleration = a, Initial velocity u = 0.
S= ut + 1/2 at
2
? 10 = ½ a (2
2
) ? 10 = 2a ? a = 5 m/s
2
Force: F = ma = 2 × 5 = 10N (Ans)
2. u = 40 km/hr =
3600
40000
= 11.11 m/s.
m = 2000 kg ; v = 0 ; s = 4m
acceleration ‘a’ =
s 2
u v
2 2
?
=
? ?
4 2
11 . 11 0
2 2
?
?
=
8
43 . 123
? = –15.42 m/s
2
(deceleration)
So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10
4
N (Ans)
3. Initial velocity u = 0 (negligible)
v = 5 × 10
6
m/s.
s = 1cm = 1 × 10
–2
m.
acceleration a =
s 2
u v
2 2
?
=
? ?
2
2
6
10 1 2
0 10 5
?
? ?
? ?
=
2
12
10 2
10 25
?
?
?
= 12.5 × 10
14
ms
–2
F = ma = 9.1 × 10
–31
× 12.5 × 10
14
= 113.75 × 10
–17
= 1.1 × 10
–15
N.
4.
g = 10m/s
2
T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N
T
1
– (0.2g + T) =0 ? T
1
= 0.2g + T = 0.2 × 10 + 3 = 5N
?Tension in the two strings are 5N & 3N respectively.
5.
T + ma – F = 0 T – ma = 0 ? T = ma …………(i)
? F= T + ma ? F= T + T from (i)
? 2T = F ? T = F / 2
6. m = 50g = 5 × 10
–2
kg
As shown in the figure,
Slope of OA = Tan?
OD
AD
=
3
15
= 5 m/s
2
So, at t = 2sec acceleration is 5m/s
2
Force = ma = 5 × 10
–2
× 5 = 0.25N along the motion
fig 1
0.2kg
0.3kg
0.2kg
fig 3
0.2g
T
1
T
fig 2
0.3g
0.3kg
A
Fig 2
mg
T
R
F
B
Fig 3
mg
ma
R
T
B
Fig 1
S
A
2
v(m/s)
180°– ? ?
? ?
A B
C D E
? ?
4 6
5
15
10
Page 2
5.1
SOLUTIONS TO CONCEPTS
CHAPTER – 5
1. m = 2kg
S = 10m
Let, acceleration = a, Initial velocity u = 0.
S= ut + 1/2 at
2
? 10 = ½ a (2
2
) ? 10 = 2a ? a = 5 m/s
2
Force: F = ma = 2 × 5 = 10N (Ans)
2. u = 40 km/hr =
3600
40000
= 11.11 m/s.
m = 2000 kg ; v = 0 ; s = 4m
acceleration ‘a’ =
s 2
u v
2 2
?
=
? ?
4 2
11 . 11 0
2 2
?
?
=
8
43 . 123
? = –15.42 m/s
2
(deceleration)
So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10
4
N (Ans)
3. Initial velocity u = 0 (negligible)
v = 5 × 10
6
m/s.
s = 1cm = 1 × 10
–2
m.
acceleration a =
s 2
u v
2 2
?
=
? ?
2
2
6
10 1 2
0 10 5
?
? ?
? ?
=
2
12
10 2
10 25
?
?
?
= 12.5 × 10
14
ms
–2
F = ma = 9.1 × 10
–31
× 12.5 × 10
14
= 113.75 × 10
–17
= 1.1 × 10
–15
N.
4.
g = 10m/s
2
T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N
T
1
– (0.2g + T) =0 ? T
1
= 0.2g + T = 0.2 × 10 + 3 = 5N
?Tension in the two strings are 5N & 3N respectively.
5.
T + ma – F = 0 T – ma = 0 ? T = ma …………(i)
? F= T + ma ? F= T + T from (i)
? 2T = F ? T = F / 2
6. m = 50g = 5 × 10
–2
kg
As shown in the figure,
Slope of OA = Tan?
OD
AD
=
3
15
= 5 m/s
2
So, at t = 2sec acceleration is 5m/s
2
Force = ma = 5 × 10
–2
× 5 = 0.25N along the motion
fig 1
0.2kg
0.3kg
0.2kg
fig 3
0.2g
T
1
T
fig 2
0.3g
0.3kg
A
Fig 2
mg
T
R
F
B
Fig 3
mg
ma
R
T
B
Fig 1
S
A
2
v(m/s)
180°– ? ?
? ?
A B
C D E
? ?
4 6
5
15
10
Chapter-5
5.2
At t = 4 sec slope of AB = 0, acceleration = 0 [ tan 0° = 0]
?Force = 0
At t = 6 sec, acceleration = slope of BC.
In ?BEC = tan ? =
EC
BE
=
3
15
= 5.
Slope of BC = tan (180° – ?) = – tan ? = –5 m/s
2
(deceleration)
Force = ma = 5 × 10
–2
5 = 0.25 N. Opposite to the motion.
7. Let, F ? contact force between m
A
& m
B
.
And, f ? force exerted by experimenter.
F + m
A
a –f = 0 m
B
a –f =0
? F = f – m
A
a ……….(i) ? F= m
B
a ……...(ii)
From eqn (i) and eqn (ii)
? f – m
A
a = m
B
a ? f = m
B
a + m
A
a ? f = a (m
A
+ m
B
).
? f =
B
m
F
(m
B
+ m
A
) =
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F [because a = F/m
B
]
? The force exerted by the experimenter is
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F
8. r = 1mm = 10
–3
‘m’ = 4mg = 4 × 10
–6
kg
s = 10
–3
m.
v = 0
u = 30 m/s.
So, a =
s 2
u v
2 2
?
=
3
10 2
30 30
?
?
? ?
= –4.5 × 10
5
m/s
2
(decelerating)
Taking magnitude only deceleration is 4.5 × 10
5
m/s
2
So, force F = 4 × 10
–6
× 4.5 × 10
5
= 1.8 N
9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg.
Acceleration a =
m
F
=
x
kx ?
=
? ?
3 . 0
2 . 0 15 ?
=
3 . 0
3
? = –10m/s
2
(deceleration)
So, the acceleration is 10 m/s
2
opposite to the direction of motion
10. Let, the block m towards left through displacement x.
F
1
= k
1
x (compressed)
F
2
= k
2
x (expanded)
They are in same direction.
Resultant F = F
1
+ F
2
? F = k
1
x + k
2
x ? F = x(k
1
+ k
2
)
So, a = acceleration =
m
F
=
m
) k x(k
2 1
?
opposite to the displacement.
11. m = 5 kg of block A.
ma = 10 N
? a 10/5 = 2 m/s
2
.
As there is no friction between A & B, when the block A moves, Block B remains at rest in its position.
Fig 1
s
m 1 m 2 f
F
R ?
m Bg
m Ba
Fig 3
F
R
m Ag
m Ba
Fig 2
F 1
m
x
K 1
F 2
K 2
10N
A
B
0.2m
Page 3
5.1
SOLUTIONS TO CONCEPTS
CHAPTER – 5
1. m = 2kg
S = 10m
Let, acceleration = a, Initial velocity u = 0.
S= ut + 1/2 at
2
? 10 = ½ a (2
2
) ? 10 = 2a ? a = 5 m/s
2
Force: F = ma = 2 × 5 = 10N (Ans)
2. u = 40 km/hr =
3600
40000
= 11.11 m/s.
m = 2000 kg ; v = 0 ; s = 4m
acceleration ‘a’ =
s 2
u v
2 2
?
=
? ?
4 2
11 . 11 0
2 2
?
?
=
8
43 . 123
? = –15.42 m/s
2
(deceleration)
So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10
4
N (Ans)
3. Initial velocity u = 0 (negligible)
v = 5 × 10
6
m/s.
s = 1cm = 1 × 10
–2
m.
acceleration a =
s 2
u v
2 2
?
=
? ?
2
2
6
10 1 2
0 10 5
?
? ?
? ?
=
2
12
10 2
10 25
?
?
?
= 12.5 × 10
14
ms
–2
F = ma = 9.1 × 10
–31
× 12.5 × 10
14
= 113.75 × 10
–17
= 1.1 × 10
–15
N.
4.
g = 10m/s
2
T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N
T
1
– (0.2g + T) =0 ? T
1
= 0.2g + T = 0.2 × 10 + 3 = 5N
?Tension in the two strings are 5N & 3N respectively.
5.
T + ma – F = 0 T – ma = 0 ? T = ma …………(i)
? F= T + ma ? F= T + T from (i)
? 2T = F ? T = F / 2
6. m = 50g = 5 × 10
–2
kg
As shown in the figure,
Slope of OA = Tan?
OD
AD
=
3
15
= 5 m/s
2
So, at t = 2sec acceleration is 5m/s
2
Force = ma = 5 × 10
–2
× 5 = 0.25N along the motion
fig 1
0.2kg
0.3kg
0.2kg
fig 3
0.2g
T
1
T
fig 2
0.3g
0.3kg
A
Fig 2
mg
T
R
F
B
Fig 3
mg
ma
R
T
B
Fig 1
S
A
2
v(m/s)
180°– ? ?
? ?
A B
C D E
? ?
4 6
5
15
10
Chapter-5
5.2
At t = 4 sec slope of AB = 0, acceleration = 0 [ tan 0° = 0]
?Force = 0
At t = 6 sec, acceleration = slope of BC.
In ?BEC = tan ? =
EC
BE
=
3
15
= 5.
Slope of BC = tan (180° – ?) = – tan ? = –5 m/s
2
(deceleration)
Force = ma = 5 × 10
–2
5 = 0.25 N. Opposite to the motion.
7. Let, F ? contact force between m
A
& m
B
.
And, f ? force exerted by experimenter.
F + m
A
a –f = 0 m
B
a –f =0
? F = f – m
A
a ……….(i) ? F= m
B
a ……...(ii)
From eqn (i) and eqn (ii)
? f – m
A
a = m
B
a ? f = m
B
a + m
A
a ? f = a (m
A
+ m
B
).
? f =
B
m
F
(m
B
+ m
A
) =
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F [because a = F/m
B
]
? The force exerted by the experimenter is
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F
8. r = 1mm = 10
–3
‘m’ = 4mg = 4 × 10
–6
kg
s = 10
–3
m.
v = 0
u = 30 m/s.
So, a =
s 2
u v
2 2
?
=
3
10 2
30 30
?
?
? ?
= –4.5 × 10
5
m/s
2
(decelerating)
Taking magnitude only deceleration is 4.5 × 10
5
m/s
2
So, force F = 4 × 10
–6
× 4.5 × 10
5
= 1.8 N
9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg.
Acceleration a =
m
F
=
x
kx ?
=
? ?
3 . 0
2 . 0 15 ?
=
3 . 0
3
? = –10m/s
2
(deceleration)
So, the acceleration is 10 m/s
2
opposite to the direction of motion
10. Let, the block m towards left through displacement x.
F
1
= k
1
x (compressed)
F
2
= k
2
x (expanded)
They are in same direction.
Resultant F = F
1
+ F
2
? F = k
1
x + k
2
x ? F = x(k
1
+ k
2
)
So, a = acceleration =
m
F
=
m
) k x(k
2 1
?
opposite to the displacement.
11. m = 5 kg of block A.
ma = 10 N
? a 10/5 = 2 m/s
2
.
As there is no friction between A & B, when the block A moves, Block B remains at rest in its position.
Fig 1
s
m 1 m 2 f
F
R ?
m Bg
m Ba
Fig 3
F
R
m Ag
m Ba
Fig 2
F 1
m
x
K 1
F 2
K 2
10N
A
B
0.2m
Chapter-5
5.3
Initial velocity of A = u = 0.
Distance to cover so that B separate out s = 0.2 m.
Acceleration a = 2 m/s
2
? s= ut + ½ at
2
? 0.2 = 0 + (½) ×2 × t
2
? t
2
= 0.2 ? t = 0.44 sec ? t= 0.45 sec.
12. a) at any depth let the ropes make angle ? with the vertical
From the free body diagram
F cos ? + F cos ? – mg = 0
? 2F cos ? = mg ? F =
? cos 2
mg
As the man moves up. ? increases i.e. cos ??decreases. Thus F
increases.
b) When the man is at depth h
cos ? =
2 2
h ) 2 / d (
h
?
Force =
2 2
2
2
h 4 d
h 4
mg
h
4
d
h
mg
? ?
?
13. From the free body diagram
? R + 0.5 × 2 – w = 0
? R = w – 0.5 × 2
= 0.5 (10 – 2) = 4N.
So, the force exerted by the block A on the block B, is 4N.
14. a) The tension in the string is found out for the different conditions from the free body diagram as
shown below.
T – (W + 0.06 × 1.2) = 0
? T = 0.05 × 9.8 + 0.05 × 1.2
= 0.55 N.
b) ? T + 0.05 × 1.2 – 0.05 × 9.8 = 0
? T = 0.05 × 9.8 – 0.05 × 1.2
= 0.43 N.
c) When the elevator makes uniform motion
T – W = 0
? T = W = 0.05 × 9.8
= 0.49 N
d) T + 0.05 × 1.2 – W = 0
? T = W – 0.05 × 1.2
= 0.43 N.
e) T – (W + 0.05 × 1.2) = 0
? T = W + 0.05 × 1.2
= 0.55 N
s
10N
R
w
ma
F
?
d
Fig-1
?? ?
d/2
F
?
Fig-2
?
mg
F
F
?
h
d/2
A
mg
2 m/s
2
B
A
0.5×2
R
W=mg=0.5×10
Fig-1
2m/s
2
W
0.05×1.2
T
Fig-2
W
0.05×1.2
T
Fig-4
–a
Fig-3
1.2m/s
2
W
T
Fig-6
a=0
Fig-5
Uniform
velocity
Fig-7
a=1.2m/s
2
W
0.05×1.2
T
Fig-8
W
0.05×1.2
T
Fig-10
–a
Fig-9
1.2m/s
2
Page 4
5.1
SOLUTIONS TO CONCEPTS
CHAPTER – 5
1. m = 2kg
S = 10m
Let, acceleration = a, Initial velocity u = 0.
S= ut + 1/2 at
2
? 10 = ½ a (2
2
) ? 10 = 2a ? a = 5 m/s
2
Force: F = ma = 2 × 5 = 10N (Ans)
2. u = 40 km/hr =
3600
40000
= 11.11 m/s.
m = 2000 kg ; v = 0 ; s = 4m
acceleration ‘a’ =
s 2
u v
2 2
?
=
? ?
4 2
11 . 11 0
2 2
?
?
=
8
43 . 123
? = –15.42 m/s
2
(deceleration)
So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10
4
N (Ans)
3. Initial velocity u = 0 (negligible)
v = 5 × 10
6
m/s.
s = 1cm = 1 × 10
–2
m.
acceleration a =
s 2
u v
2 2
?
=
? ?
2
2
6
10 1 2
0 10 5
?
? ?
? ?
=
2
12
10 2
10 25
?
?
?
= 12.5 × 10
14
ms
–2
F = ma = 9.1 × 10
–31
× 12.5 × 10
14
= 113.75 × 10
–17
= 1.1 × 10
–15
N.
4.
g = 10m/s
2
T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N
T
1
– (0.2g + T) =0 ? T
1
= 0.2g + T = 0.2 × 10 + 3 = 5N
?Tension in the two strings are 5N & 3N respectively.
5.
T + ma – F = 0 T – ma = 0 ? T = ma …………(i)
? F= T + ma ? F= T + T from (i)
? 2T = F ? T = F / 2
6. m = 50g = 5 × 10
–2
kg
As shown in the figure,
Slope of OA = Tan?
OD
AD
=
3
15
= 5 m/s
2
So, at t = 2sec acceleration is 5m/s
2
Force = ma = 5 × 10
–2
× 5 = 0.25N along the motion
fig 1
0.2kg
0.3kg
0.2kg
fig 3
0.2g
T
1
T
fig 2
0.3g
0.3kg
A
Fig 2
mg
T
R
F
B
Fig 3
mg
ma
R
T
B
Fig 1
S
A
2
v(m/s)
180°– ? ?
? ?
A B
C D E
? ?
4 6
5
15
10
Chapter-5
5.2
At t = 4 sec slope of AB = 0, acceleration = 0 [ tan 0° = 0]
?Force = 0
At t = 6 sec, acceleration = slope of BC.
In ?BEC = tan ? =
EC
BE
=
3
15
= 5.
Slope of BC = tan (180° – ?) = – tan ? = –5 m/s
2
(deceleration)
Force = ma = 5 × 10
–2
5 = 0.25 N. Opposite to the motion.
7. Let, F ? contact force between m
A
& m
B
.
And, f ? force exerted by experimenter.
F + m
A
a –f = 0 m
B
a –f =0
? F = f – m
A
a ……….(i) ? F= m
B
a ……...(ii)
From eqn (i) and eqn (ii)
? f – m
A
a = m
B
a ? f = m
B
a + m
A
a ? f = a (m
A
+ m
B
).
? f =
B
m
F
(m
B
+ m
A
) =
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F [because a = F/m
B
]
? The force exerted by the experimenter is
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F
8. r = 1mm = 10
–3
‘m’ = 4mg = 4 × 10
–6
kg
s = 10
–3
m.
v = 0
u = 30 m/s.
So, a =
s 2
u v
2 2
?
=
3
10 2
30 30
?
?
? ?
= –4.5 × 10
5
m/s
2
(decelerating)
Taking magnitude only deceleration is 4.5 × 10
5
m/s
2
So, force F = 4 × 10
–6
× 4.5 × 10
5
= 1.8 N
9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg.
Acceleration a =
m
F
=
x
kx ?
=
? ?
3 . 0
2 . 0 15 ?
=
3 . 0
3
? = –10m/s
2
(deceleration)
So, the acceleration is 10 m/s
2
opposite to the direction of motion
10. Let, the block m towards left through displacement x.
F
1
= k
1
x (compressed)
F
2
= k
2
x (expanded)
They are in same direction.
Resultant F = F
1
+ F
2
? F = k
1
x + k
2
x ? F = x(k
1
+ k
2
)
So, a = acceleration =
m
F
=
m
) k x(k
2 1
?
opposite to the displacement.
11. m = 5 kg of block A.
ma = 10 N
? a 10/5 = 2 m/s
2
.
As there is no friction between A & B, when the block A moves, Block B remains at rest in its position.
Fig 1
s
m 1 m 2 f
F
R ?
m Bg
m Ba
Fig 3
F
R
m Ag
m Ba
Fig 2
F 1
m
x
K 1
F 2
K 2
10N
A
B
0.2m
Chapter-5
5.3
Initial velocity of A = u = 0.
Distance to cover so that B separate out s = 0.2 m.
Acceleration a = 2 m/s
2
? s= ut + ½ at
2
? 0.2 = 0 + (½) ×2 × t
2
? t
2
= 0.2 ? t = 0.44 sec ? t= 0.45 sec.
12. a) at any depth let the ropes make angle ? with the vertical
From the free body diagram
F cos ? + F cos ? – mg = 0
? 2F cos ? = mg ? F =
? cos 2
mg
As the man moves up. ? increases i.e. cos ??decreases. Thus F
increases.
b) When the man is at depth h
cos ? =
2 2
h ) 2 / d (
h
?
Force =
2 2
2
2
h 4 d
h 4
mg
h
4
d
h
mg
? ?
?
13. From the free body diagram
? R + 0.5 × 2 – w = 0
? R = w – 0.5 × 2
= 0.5 (10 – 2) = 4N.
So, the force exerted by the block A on the block B, is 4N.
14. a) The tension in the string is found out for the different conditions from the free body diagram as
shown below.
T – (W + 0.06 × 1.2) = 0
? T = 0.05 × 9.8 + 0.05 × 1.2
= 0.55 N.
b) ? T + 0.05 × 1.2 – 0.05 × 9.8 = 0
? T = 0.05 × 9.8 – 0.05 × 1.2
= 0.43 N.
c) When the elevator makes uniform motion
T – W = 0
? T = W = 0.05 × 9.8
= 0.49 N
d) T + 0.05 × 1.2 – W = 0
? T = W – 0.05 × 1.2
= 0.43 N.
e) T – (W + 0.05 × 1.2) = 0
? T = W + 0.05 × 1.2
= 0.55 N
s
10N
R
w
ma
F
?
d
Fig-1
?? ?
d/2
F
?
Fig-2
?
mg
F
F
?
h
d/2
A
mg
2 m/s
2
B
A
0.5×2
R
W=mg=0.5×10
Fig-1
2m/s
2
W
0.05×1.2
T
Fig-2
W
0.05×1.2
T
Fig-4
–a
Fig-3
1.2m/s
2
W
T
Fig-6
a=0
Fig-5
Uniform
velocity
Fig-7
a=1.2m/s
2
W
0.05×1.2
T
Fig-8
W
0.05×1.2
T
Fig-10
–a
Fig-9
1.2m/s
2
Chapter-5
5.4
f) When the elevator goes down with uniform velocity acceleration = 0
T – W = 0
? T = W = 0.05 × 9.8
= 0.49 N.
15. When the elevator is accelerating upwards, maximum weight will be recorded.
R – (W + ma ) = 0
? R = W + ma = m(g + a) max.wt.
When decelerating upwards, maximum weight will be recorded.
R + ma – W = 0
?R = W – ma = m(g – a)
So, m(g + a) = 72 × 9.9 …(1)
m(g – a) = 60 × 9.9 …(2)
Now, mg + ma = 72 × 9.9 ? mg – ma = 60 × 9.9
? 2mg = 1306.8
? m =
9 . 9 2
8 . 1306
?
= 66 Kg
So, the true weight of the man is 66 kg.
Again, to find the acceleration, mg + ma = 72 × 9.9
? a = 9 . 0
11
9 . 9
66
9 . 9 66 9 . 9 72
? ?
? ? ?
m/s
2
.
16. Let the acceleration of the 3 kg mass relative to the elevator is ‘a’ in the downward direction.
As, shown in the free body diagram
T – 1.5 g – 1.5(g/10) – 1.5 a = 0 from figure (1)
and, T – 3g – 3(g/10) + 3a = 0 from figure (2)
? T = 1.5 g + 1.5(g/10) + 1.5a … (i)
And T = 3g + 3(g/10) – 3a … (ii)
Equation (i) × 2 ? 3g + 3(g/10) + 3a = 2T
Equation (ii) × 1 ? 3g + 3(g/10) – 3a = T
Subtracting the above two equations we get, T = 6a
Subtracting T = 6a in equation (ii)
6a = 3g + 3(g/10) – 3a.
? 9a =
10
g 33
? a = 34 . 32
10
33 ) 8 . 9 (
?
?a = 3.59 ? T = 6a = 6 × 3.59 = 21.55
T
1
= 2T = 2 × 21.55 = 43.1 N cut is T
1
shown in spring.
Mass =
8 . 9
1 . 43
g
wt
? = 4.39 = 4.4 kg
17. Given, m = 2 kg, k = 100 N/m
From the free body diagram, kl – 2g = 0 ? kl = 2g
? l =
100
6 . 19
100
8 . 9 2
k
g 2
?
?
? = 0.196 = 0.2 m
Suppose further elongation when 1 kg block is added be x,
Then k(1 + x) = 3g
? kx = 3g – 2g = g = 9.8 N
? x =
100
8 . 9
= 0.098 = 0.1 m
W
T
Fig-12 Fig-11
Uniform
velocity
m
W
R
a
ma
W
R
a
ma
–a
Fig-1
1.5g
T
1.5(g/10)
1.5a
Fig-2
3g
T
3(g/10)
3a
kl
2g
Page 5
5.1
SOLUTIONS TO CONCEPTS
CHAPTER – 5
1. m = 2kg
S = 10m
Let, acceleration = a, Initial velocity u = 0.
S= ut + 1/2 at
2
? 10 = ½ a (2
2
) ? 10 = 2a ? a = 5 m/s
2
Force: F = ma = 2 × 5 = 10N (Ans)
2. u = 40 km/hr =
3600
40000
= 11.11 m/s.
m = 2000 kg ; v = 0 ; s = 4m
acceleration ‘a’ =
s 2
u v
2 2
?
=
? ?
4 2
11 . 11 0
2 2
?
?
=
8
43 . 123
? = –15.42 m/s
2
(deceleration)
So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10
4
N (Ans)
3. Initial velocity u = 0 (negligible)
v = 5 × 10
6
m/s.
s = 1cm = 1 × 10
–2
m.
acceleration a =
s 2
u v
2 2
?
=
? ?
2
2
6
10 1 2
0 10 5
?
? ?
? ?
=
2
12
10 2
10 25
?
?
?
= 12.5 × 10
14
ms
–2
F = ma = 9.1 × 10
–31
× 12.5 × 10
14
= 113.75 × 10
–17
= 1.1 × 10
–15
N.
4.
g = 10m/s
2
T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N
T
1
– (0.2g + T) =0 ? T
1
= 0.2g + T = 0.2 × 10 + 3 = 5N
?Tension in the two strings are 5N & 3N respectively.
5.
T + ma – F = 0 T – ma = 0 ? T = ma …………(i)
? F= T + ma ? F= T + T from (i)
? 2T = F ? T = F / 2
6. m = 50g = 5 × 10
–2
kg
As shown in the figure,
Slope of OA = Tan?
OD
AD
=
3
15
= 5 m/s
2
So, at t = 2sec acceleration is 5m/s
2
Force = ma = 5 × 10
–2
× 5 = 0.25N along the motion
fig 1
0.2kg
0.3kg
0.2kg
fig 3
0.2g
T
1
T
fig 2
0.3g
0.3kg
A
Fig 2
mg
T
R
F
B
Fig 3
mg
ma
R
T
B
Fig 1
S
A
2
v(m/s)
180°– ? ?
? ?
A B
C D E
? ?
4 6
5
15
10
Chapter-5
5.2
At t = 4 sec slope of AB = 0, acceleration = 0 [ tan 0° = 0]
?Force = 0
At t = 6 sec, acceleration = slope of BC.
In ?BEC = tan ? =
EC
BE
=
3
15
= 5.
Slope of BC = tan (180° – ?) = – tan ? = –5 m/s
2
(deceleration)
Force = ma = 5 × 10
–2
5 = 0.25 N. Opposite to the motion.
7. Let, F ? contact force between m
A
& m
B
.
And, f ? force exerted by experimenter.
F + m
A
a –f = 0 m
B
a –f =0
? F = f – m
A
a ……….(i) ? F= m
B
a ……...(ii)
From eqn (i) and eqn (ii)
? f – m
A
a = m
B
a ? f = m
B
a + m
A
a ? f = a (m
A
+ m
B
).
? f =
B
m
F
(m
B
+ m
A
) =
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F [because a = F/m
B
]
? The force exerted by the experimenter is
?
?
?
?
?
?
?
?
?
B
A
m
m
1 F
8. r = 1mm = 10
–3
‘m’ = 4mg = 4 × 10
–6
kg
s = 10
–3
m.
v = 0
u = 30 m/s.
So, a =
s 2
u v
2 2
?
=
3
10 2
30 30
?
?
? ?
= –4.5 × 10
5
m/s
2
(decelerating)
Taking magnitude only deceleration is 4.5 × 10
5
m/s
2
So, force F = 4 × 10
–6
× 4.5 × 10
5
= 1.8 N
9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg.
Acceleration a =
m
F
=
x
kx ?
=
? ?
3 . 0
2 . 0 15 ?
=
3 . 0
3
? = –10m/s
2
(deceleration)
So, the acceleration is 10 m/s
2
opposite to the direction of motion
10. Let, the block m towards left through displacement x.
F
1
= k
1
x (compressed)
F
2
= k
2
x (expanded)
They are in same direction.
Resultant F = F
1
+ F
2
? F = k
1
x + k
2
x ? F = x(k
1
+ k
2
)
So, a = acceleration =
m
F
=
m
) k x(k
2 1
?
opposite to the displacement.
11. m = 5 kg of block A.
ma = 10 N
? a 10/5 = 2 m/s
2
.
As there is no friction between A & B, when the block A moves, Block B remains at rest in its position.
Fig 1
s
m 1 m 2 f
F
R ?
m Bg
m Ba
Fig 3
F
R
m Ag
m Ba
Fig 2
F 1
m
x
K 1
F 2
K 2
10N
A
B
0.2m
Chapter-5
5.3
Initial velocity of A = u = 0.
Distance to cover so that B separate out s = 0.2 m.
Acceleration a = 2 m/s
2
? s= ut + ½ at
2
? 0.2 = 0 + (½) ×2 × t
2
? t
2
= 0.2 ? t = 0.44 sec ? t= 0.45 sec.
12. a) at any depth let the ropes make angle ? with the vertical
From the free body diagram
F cos ? + F cos ? – mg = 0
? 2F cos ? = mg ? F =
? cos 2
mg
As the man moves up. ? increases i.e. cos ??decreases. Thus F
increases.
b) When the man is at depth h
cos ? =
2 2
h ) 2 / d (
h
?
Force =
2 2
2
2
h 4 d
h 4
mg
h
4
d
h
mg
? ?
?
13. From the free body diagram
? R + 0.5 × 2 – w = 0
? R = w – 0.5 × 2
= 0.5 (10 – 2) = 4N.
So, the force exerted by the block A on the block B, is 4N.
14. a) The tension in the string is found out for the different conditions from the free body diagram as
shown below.
T – (W + 0.06 × 1.2) = 0
? T = 0.05 × 9.8 + 0.05 × 1.2
= 0.55 N.
b) ? T + 0.05 × 1.2 – 0.05 × 9.8 = 0
? T = 0.05 × 9.8 – 0.05 × 1.2
= 0.43 N.
c) When the elevator makes uniform motion
T – W = 0
? T = W = 0.05 × 9.8
= 0.49 N
d) T + 0.05 × 1.2 – W = 0
? T = W – 0.05 × 1.2
= 0.43 N.
e) T – (W + 0.05 × 1.2) = 0
? T = W + 0.05 × 1.2
= 0.55 N
s
10N
R
w
ma
F
?
d
Fig-1
?? ?
d/2
F
?
Fig-2
?
mg
F
F
?
h
d/2
A
mg
2 m/s
2
B
A
0.5×2
R
W=mg=0.5×10
Fig-1
2m/s
2
W
0.05×1.2
T
Fig-2
W
0.05×1.2
T
Fig-4
–a
Fig-3
1.2m/s
2
W
T
Fig-6
a=0
Fig-5
Uniform
velocity
Fig-7
a=1.2m/s
2
W
0.05×1.2
T
Fig-8
W
0.05×1.2
T
Fig-10
–a
Fig-9
1.2m/s
2
Chapter-5
5.4
f) When the elevator goes down with uniform velocity acceleration = 0
T – W = 0
? T = W = 0.05 × 9.8
= 0.49 N.
15. When the elevator is accelerating upwards, maximum weight will be recorded.
R – (W + ma ) = 0
? R = W + ma = m(g + a) max.wt.
When decelerating upwards, maximum weight will be recorded.
R + ma – W = 0
?R = W – ma = m(g – a)
So, m(g + a) = 72 × 9.9 …(1)
m(g – a) = 60 × 9.9 …(2)
Now, mg + ma = 72 × 9.9 ? mg – ma = 60 × 9.9
? 2mg = 1306.8
? m =
9 . 9 2
8 . 1306
?
= 66 Kg
So, the true weight of the man is 66 kg.
Again, to find the acceleration, mg + ma = 72 × 9.9
? a = 9 . 0
11
9 . 9
66
9 . 9 66 9 . 9 72
? ?
? ? ?
m/s
2
.
16. Let the acceleration of the 3 kg mass relative to the elevator is ‘a’ in the downward direction.
As, shown in the free body diagram
T – 1.5 g – 1.5(g/10) – 1.5 a = 0 from figure (1)
and, T – 3g – 3(g/10) + 3a = 0 from figure (2)
? T = 1.5 g + 1.5(g/10) + 1.5a … (i)
And T = 3g + 3(g/10) – 3a … (ii)
Equation (i) × 2 ? 3g + 3(g/10) + 3a = 2T
Equation (ii) × 1 ? 3g + 3(g/10) – 3a = T
Subtracting the above two equations we get, T = 6a
Subtracting T = 6a in equation (ii)
6a = 3g + 3(g/10) – 3a.
? 9a =
10
g 33
? a = 34 . 32
10
33 ) 8 . 9 (
?
?a = 3.59 ? T = 6a = 6 × 3.59 = 21.55
T
1
= 2T = 2 × 21.55 = 43.1 N cut is T
1
shown in spring.
Mass =
8 . 9
1 . 43
g
wt
? = 4.39 = 4.4 kg
17. Given, m = 2 kg, k = 100 N/m
From the free body diagram, kl – 2g = 0 ? kl = 2g
? l =
100
6 . 19
100
8 . 9 2
k
g 2
?
?
? = 0.196 = 0.2 m
Suppose further elongation when 1 kg block is added be x,
Then k(1 + x) = 3g
? kx = 3g – 2g = g = 9.8 N
? x =
100
8 . 9
= 0.098 = 0.1 m
W
T
Fig-12 Fig-11
Uniform
velocity
m
W
R
a
ma
W
R
a
ma
–a
Fig-1
1.5g
T
1.5(g/10)
1.5a
Fig-2
3g
T
3(g/10)
3a
kl
2g
Chapter-5
5.5
18. a = 2 m/s
2
kl – (2g + 2a) = 0
?kl = 2g + 2a
= 2 × 9.8 + 2 × 2 = 19.6 + 4
? l =
100
6 . 23
= 0.236 m = 0.24 m
When 1 kg body is added total mass (2 + 1)kg = 3kg.
elongation be l
1
kl
1
= 3g + 3a = 3 × 9.8 + 6
? l
1
=
100
4 . 33
= 0.0334 = 0.36
Further elongation = l
1
– l = 0.36 – 0.12 m.
19. Let, the air resistance force is F and Buoyant force is B.
Given that
F
a
? v, where v ? velocity
? F
a
= kv, where k ? proportionality constant.
When the balloon is moving downward,
B + kv = mg …(i)
? M =
g
kv B ?
For the balloon to rise with a constant velocity v, (upward)
let the mass be m
Here, B – (mg + kv) = 0 …(ii)
? B = mg + kv
? m =
g
kw B ?
So, amount of mass that should be removed = M – m.
=
g
kv 2
g
kv B kv B
g
kv B
g
kv B
?
? ? ?
?
?
?
?
=
G
) B Mg ( 2 ?
= 2{M – (B/g)}
20. When the box is accelerating upward,
U – mg – m(g/6) = 0
? U = mg + mg/6 = m{g + (g/6)} = 7 mg/7 …(i)
? m = 6U/7g.
When it is accelerating downward, let the required mass be M.
U – Mg + Mg/6 = 0
? U =
6
Mg 5
6
Mg Mg 6
?
?
? M =
g 5
U 6
Mass to be added = M – m = ?
?
?
?
?
?
? ? ?
7
1
5
1
g
U 6
g 7
U 6
g 5
U 6
=
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
g
U
35
12
35
2
g
U 6
=
?
?
?
?
?
?
?
?
?
g
1
6
mg 7
35
12
from (i)
= 2/5 m.
? The mass to be added is 2m/5.
2a
a
kl
2g
a
2×2
kl
3g
2m/s
2
M
Fig-1
v
kV
mg
B
Fig-2
v
kV
mg
B
Fig-1
g/6T
mg/6
mg
V
Fig-2
g/6T
mg/6
mg
V
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