Respuesta :
Answer:
1) [tex]P_c[/tex] = 8721 Kg*m/s
2) [tex]P_t[/tex] = 12450 Kg*m/s
3) θ = 55.03°
4) P = 15182 Kg*m/s
5) V = 8.91 m/s
6) [tex]E_i>E_f[/tex]
145099.5 J > 67638 J
Explanation:
1) The momentum is calculated by the next equation:
P = MV
where M is the mass and V is the velocity
so, the linear momentum of the car is:
[tex]P_c[/tex] = (459)(19)
[tex]P_c[/tex] = 8721 Kg*m/s
2) the linear momentum of the truck is:
[tex]P_t[/tex] = (1245)(10)
[tex]P_t[/tex] = 12450 Kg*m/s
3) For answer this we will use the law of the conservation of the linear momentum where:
[tex]P_i = P_f[/tex]:
So, we will do this for each axis:
First on axis x:
[tex]M_cV_c = V_{sx}M_s[/tex]
Where [tex]M_c[/tex] is the mass of the blue car, [tex]V_c[/tex] is the velocity of the car, [tex]V_{sx}[/tex] is the velocity of the system in x after the collition and [tex]M_s[/tex] is the mass of both cars. Replacing, we get:
[tex](459)(19) = V_{sx}(1245+459)[/tex]
Solving for [tex]V_{sx}[/tex]:
[tex]V_{sx} = 5.11 m/s[/tex]
Second, on axis y:
[tex]M_tV_t = V_{sy}M_s[/tex]
Where [tex]M_t[/tex] is the mass of the truck, [tex]V_t[/tex] is the velocity of the truck, [tex]V_{sy}[/tex] is the velocity of the system in y after the collition and [tex]M_s[/tex] is the mass of both cars.
[tex](1245)(10) = V_{sy}(1245+459)[/tex]
Solving for [tex]V_{sy}[/tex]:
[tex]V_{sy} = 7.306 m/s[/tex]
Now using the definition of tangent:
θ = [tex]tan^{-1}(\frac{7.306}{5.11})[/tex]
θ = 55.03°
4) First, we have to find the magnitude of the velocity using the pythagorean theorem as:
V = [tex]\sqrt{7.306^2+5.11^2}[/tex]
V = 8.91 m/s
With the velocity, we find the momentum as:
P = MV
P =(1245+459)(8.91)
P = 15182 Kg*m/s
5) This was calculated before, so:
V = 8.91 m/s
6) The energy of the total system before the collision is calculated as:
[tex]E_i = \frac{1}{2}M_cV_c+\frac{1}{2}M_tV_t[/tex]
[tex]E_i = \frac{1}{2}(459)(19)^2+\frac{1}{2}(1245)(10)^2[/tex]
[tex]E_i = 145099.5 J [/tex]
The energy of the total system after the collision is calculated as:
[tex]E_f = \frac{1}{2}M_sV^2[/tex]
[tex]E_f = \frac{1}{2}(1245+459)(8.91)^2[/tex]
[tex]E_f = 67638 J[/tex]
So, the energy of the system before the collision is bigger than the energy of the system after the collision.
