2.3 Solutions
89
By measuring h and knowing m and M, the original velocity of the bullet can
be calculated.
2.45 Let the area of the nozzle through where the jet comes be A m 2 . The mass of
water in the jet per second is ρ A v, where ρ is the density of water and v the
get velocity.
The momentum associated with this volume of water is
p = (ρ Av)v = ρ Av
2
(1)
The momentum after hitting the wall will also be equal to ρ Av 2 since the
collision is assumed to be elastic. Resolve the momentum along the x-axis
and y-axis, Fig. 2.35.
Fig. 2.35
The change of the x-component of momentum is
p x = p sin θ − (− p sin θ) = 2 p sin θ
(2)
The change in the y-component of momentum is
p y = p cos θ − p cos θ = 0
( 3 )
Then p = p x = 2 p sin θ = 2ρ Av
2 sin θ
(4)
Pressure exerted on the wall will be
P =
p
A
= 2ρv
2 sin θ
(5)
For normal incidence, θ = 90 ◦ and
P = 2ρv
2
(6)
89
By measuring h and knowing m and M, the original velocity of the bullet can
be calculated.
2.45 Let the area of the nozzle through where the jet comes be A m 2 . The mass of
water in the jet per second is ρ A v, where ρ is the density of water and v the
get velocity.
The momentum associated with this volume of water is
p = (ρ Av)v = ρ Av
2
(1)
The momentum after hitting the wall will also be equal to ρ Av 2 since the
collision is assumed to be elastic. Resolve the momentum along the x-axis
and y-axis, Fig. 2.35.
Fig. 2.35
The change of the x-component of momentum is
p x = p sin θ − (− p sin θ) = 2 p sin θ
(2)
The change in the y-component of momentum is
p y = p cos θ − p cos θ = 0
( 3 )
Then p = p x = 2 p sin θ = 2ρ Av
2 sin θ
(4)
Pressure exerted on the wall will be
P =
p
A
= 2ρv
2 sin θ
(5)
For normal incidence, θ = 90 ◦ and
P = 2ρv
2
(6)