Respuesta :
Answer:
Speed of the car 1 =[tex]V_1=8.98m/s[/tex]
Speed of the car 2 =[tex]V_2=17.96m/s[/tex]
Explanation:
Given:
Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)
mathematically,
M₁ = 2M₂
Kinetic Energy of the car 1 = Half the kinetic energy of the car 2
KE₁ = 0.5 KE₂
Now, the kinetic energy for a body is given as
[tex]KE =\frac{1}{2}mv^2[/tex]
where,
m = mass of the body
v = velocity of the body
thus,
[tex]\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2[/tex]
or
[tex]\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2[/tex]
or
[tex]2M_2V_1^2=0.5\times M_2V_2^2[/tex]
or
[tex]2V_1^2=0.5\times V_2^2[/tex]
or
[tex]4V_1^2= V_2^2[/tex]
or
[tex]2V_1= V_2[/tex] .................(1)
also,
[tex]\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2[/tex]
or
[tex]\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2[/tex]
or
[tex]2(V_1+9.0)^2=(2V_1+9.0)^2[/tex]
or
[tex]\sqrt{2}(V_1+9.0)=(2V_1+9.0)[/tex]
or
[tex](\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)[/tex]
or
[tex](\sqrt{2}V_1+ 12.72)=(2V_1+9.0)[/tex]
or
[tex](2V_1-\sqrt{2}V_1)=(12.72-9.0)[/tex]
or
[tex](0.404V_1)=(3.72)[/tex]
or
[tex]V_1=8.98m/s[/tex]
and, from equation (1)
[tex]V_2=2\times 8.98m/s = 17.96m/s[/tex]
Hence,
Speed of car 1 =[tex]V_1=8.98m/s[/tex]
Speed of car 2 =[tex]V_2=17.96m/s[/tex]
Answer:
The original speeds of the two cars were :
[tex]6.36\frac{m}{s},12.72\frac{m}{s}[/tex]
Explanation:
Let's start reading the question and making our equations in order to find the speeds.
The first equation is :
[tex]m_{1}=2m_{2}[/tex] (I)
The kinectic energy can be calculated using the following equation :
[tex]K=(\frac{1}{2}).m.v^{2}[/tex] (II)
Where ''K'' is the kinetic energy
Where ''m'' is the mass and where ''v'' is the speed.
By reading the exercise we find that :
[tex]K_{1}=\frac{K_{2}}{2}[/tex] (III)
If we use (II) in (III) :
[tex](\frac{1}{2}).m_{1}.v_{1}^{2}=(\frac{1}{2}).m_{2}.v_{2}^{2}.(\frac{1}{2})[/tex]
[tex]m_{1}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex] (IV)
If we replace (I) in (IV) ⇒
[tex]2.m_{2}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex]
[tex]4.v_{1}^{2}=v_{2}^{2}[/tex]
[tex]2.v_{1}=v_{2}[/tex] (V)
'' When both cars increase their speed by [tex]9.0\frac{m}{s}[/tex], they then have the same kinetic energy ''
The last equation is :
[tex](\frac{1}{2}).m_{1}.(v_{1}+9)^{2}=(\frac{1}{2}).m_{2}.(v_{2}+9)^{2}[/tex] (VI)
If we use (I) in (VI) ⇒
[tex]2.m_{2}.(v_{1}+9)^{2}=m_{2}.(v_{2}+9)^{2}[/tex]
[tex]2.(v_{1}+9)^{2}=(v_{2}+9)^{2}[/tex]
If we use (V) in this last expression ⇒
[tex]2.(v_{1}+9)^{2}=(2.v_{1}+9)^{2}[/tex]
[tex]2.(v_{1}^{2}+18v_{1}+81)=4v_{1}^{2}+36v_{1}+81[/tex]
[tex]2v_{1}^{2}+36v_{1}+162=4v_{1}^{2}+36v_{1}+81[/tex]
[tex]2v_{1}^{2}=81[/tex]
[tex]v_{1}^{2}=40.5[/tex]
[tex]v_{1}=\sqrt{40.5}=6.36[/tex]
We find that the original speed [tex]v_{1}[/tex] is [tex]6.36\frac{m}{s}[/tex]
If we replace this value in the equation (V) ⇒
[tex]2.(6.36\frac{m}{s})=v_{2}[/tex]
[tex]v_{2}=12.72\frac{m}{s}[/tex]
