Respuesta :
Answer:
Explanation:
Mass of station wagon M = 1200 kg , velocity V = 12 m/s
Mass of car m = 1800 kg , velocity v = 20 m/s
a ) Let centre of mass of car and station wagon be at a distance d from wagon
Taking moment of weight about it
1200 x d = 1800 x ( 40 - d )
3000 x d = 1800 x 40
d = 24 m
b ) Total momentum = MV +mv
= 1200 x 12 + 1800 x 20
= 50400 kg ms⁻¹
c ) Speed of centre of mass
v₀ = (MV + mv )/(m+M)
= 50400/3000
= 16.8 m/s
d) System Total momentum = velocity of cm x total mass
= 16.8 x 3000 = 50400 m/s .
It is equal to what was calculated in part b)
Here we want to find different measures of a system and the system's center of mass.
We will get:
a) [tex]CM(t) = 24m + (16.8m/s)*t[/tex]
b) [tex]P = 50,400 kg\cdot m/s[/tex]
c) [tex]v_{cm}(t) = 16.8 m/s[/tex]
d) [tex]P_{cm} = 50,400 kg\cdot m/s[/tex]
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First, let's state what we know:
We have a 1200kg wagon moving with a velocity of 12 m/s.
We also have a car with a mass of 1800kg moving with a velocity of 20m/s, and its center of mass is 40m ahead the center of mass of the wagon.
So if we define the initial center of mass of the wagon as our zero in the position, the initial position of the car will be 40m.
a) We want to find the position of the center of mass of the whole system.
First, we need to write the positions of both cars, remember that the position is the initial position plus t times the speed, where t is time in seconds.
[tex]P_w(t) = (12m/s)*t[/tex]
[tex]P_c(t) = 40m + (20m/s)*t[/tex]
The center of mass is computed as:
[tex]CM = \frac{m_1*x_1 + ... + m_n*x_n}{M}[/tex]
Where the "m" terms are the mass of each component of the system, and the "x" are the positions. M is the total mass of the system.
So here, we will have:
[tex]CM(t) = \frac{1200kg*p_w(t) + 1800kg*P_c(t) }{1200kg + 1800kg}[/tex]
Notice that because the position of the vehicles depend on the time, the position of the center of mass also does.
Then we have:
[tex]CM(t) = \frac{1200kg*p_w(t) + 1800kg*P_c(t) }{1200kg + 1800kg}\\\\CM(t) = \frac{1200kg*(12m/s)*t + 1800kg*(40m + (20m/s)*t) }{3000kg}\\\\CM(t) = 24m + (16.8m/s)*t[/tex]
This is the equation of the center of mass.
b) Momentum is defined as the amount of motion, and its given by the mass times the velocity.
Here the total momentum of the system is the momentum of the wagon plus the momentum of the car, we will have:
[tex]P = (1200kg)*(12 m/s) + (1800kg)*(20m/s) = 50,400 kg\cdot m/s[/tex]
c) The speed of the center of mass comes from the term that is being multiplied by the time in the center of mass equation. The velocity is:
[tex]v_{cm}(t) = 16.8 m/s[/tex]
d) The system total momentum by using the velocity of the center of mass is computed as the total mass of the system times the velocity of the center of mass, here we get:
[tex]P_{cm} = (1200kg + 1800kg)*16.8m/s = 5400 kg\cdot m/s[/tex]
So as expected, we got the exact same result than in point b, the momentum is independent of the frame of reference (for not moving frames of reference).
If you want to learn more, you can read:
https://brainly.com/question/16698727
