Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find
the general expression for the velocities of the two rods after impact. b) If mA = 2 kg, m3 = 1 kg, Vo= 3 m/s, and e = 0.65, find the value of the initial velocity v4 of rod A for it to be at rest after the impact and the final velocity vB of rod B. c) Find the magnitude of the impulse for the condition in part b. (3 marks) d) Find the percent decrease in kinetic energy which corresponds to the impact in part b.
1 answer:
Answer:
A.) Find the answer in the explanation
B.) Ua = 7.33 m/s , Vb = 7.73 m/s
C.) Impulse = 17.6 Ns
D.) 49%
Explanation:
Let Ua = initial velocity of the rod A
Ub = initial velocity of the rod B
Va = final velocity of the rod A
Vb = final velocity of the rod B
Ma = mass of rod A
Mb = mass of rod B
Given that
Ma = 2kg
Mb = 1kg
Ub = 3 m/s
Va = 0
e = restitution coefficient = 0.65
The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.
Please find the attached files for the solution
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Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v =
in this expression the densities are constant
v = A √h
A =
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A =
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h =
Δh = Δv
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
= 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
= 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
Answer:
See it in the pic.
Explanation:
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