2.3 Solutions
93
p − mv cos θ = 2mv cos θ
∴ p = 3mv cos θ
∴ velocity =
p
m
= 3v cos θ
2.54 Volume of air moving down per second = Av, where v is the air velocity
moving down through an area A.
Mass of air moving down per second = ρ Av
F =
p
t
=
mass
sec
((v) = ρ Av
2
Reaction force upward = Helicopter’s weight
ρ Av
2
= Mg
v =
Mg
ρ A
=
500 × 9.8
1.3 × 45
= 9.15 m/s
2.55 If v is the velocity of each bullet of mass m and n the number of bullets that
can be fired per second then rate of change of momentum will be
p
t
= mnv
(1)
∴
p
t
= F = mnv
(2)
n =
F
mv
=
150
(0.1)(1000)
= 1.5/s
Thus the number of bullets that can be fired per minute will be 60 × 1.5 = 90.
2.56 If v is the velocity with which a particle of mass m falls on the balance pan,
momentum before impact is mv and after impact −mv so that
p = −mv − mv = −2mv
(1)
If height of fall is h then
v =
2gh =
√
2 × 9.8 × 1.6 = 5.6 m/s
( 2 )
93
p − mv cos θ = 2mv cos θ
∴ p = 3mv cos θ
∴ velocity =
p
m
= 3v cos θ
2.54 Volume of air moving down per second = Av, where v is the air velocity
moving down through an area A.
Mass of air moving down per second = ρ Av
F =
p
t
=
mass
sec
((v) = ρ Av
2
Reaction force upward = Helicopter’s weight
ρ Av
2
= Mg
v =
Mg
ρ A
=
500 × 9.8
1.3 × 45
= 9.15 m/s
2.55 If v is the velocity of each bullet of mass m and n the number of bullets that
can be fired per second then rate of change of momentum will be
p
t
= mnv
(1)
∴
p
t
= F = mnv
(2)
n =
F
mv
=
150
(0.1)(1000)
= 1.5/s
Thus the number of bullets that can be fired per minute will be 60 × 1.5 = 90.
2.56 If v is the velocity with which a particle of mass m falls on the balance pan,
momentum before impact is mv and after impact −mv so that
p = −mv − mv = −2mv
(1)
If height of fall is h then
v =
2gh =
√
2 × 9.8 × 1.6 = 5.6 m/s
( 2 )