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Lena [83]
3 years ago
6

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars? Express your answers using two significant figures separated by a comma?
Physics
2 answers:
kirill [66]3 years ago
8 0

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

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

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

NeTakaya3 years ago
5 0

Answer:

The original speeds of the two cars were :

6.36\frac{m}{s},12.72\frac{m}{s}

Explanation:

Let's start reading the question and making our equations in order to find the speeds.

The first equation is :

m_{1}=2m_{2} (I)

The kinectic energy can be calculated using the following equation :

K=(\frac{1}{2}).m.v^{2} (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 :  

K_{1}=\frac{K_{2}}{2} (III)

If we use (II) in (III) :

(\frac{1}{2}).m_{1}.v_{1}^{2}=(\frac{1}{2}).m_{2}.v_{2}^{2}.(\frac{1}{2})

m_{1}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2} (IV)

If we replace (I) in (IV) ⇒

2.m_{2}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}

4.v_{1}^{2}=v_{2}^{2}

2.v_{1}=v_{2} (V)

'' When both cars increase their speed by 9.0\frac{m}{s}, they then have the same kinetic energy ''

The last equation is :

(\frac{1}{2}).m_{1}.(v_{1}+9)^{2}=(\frac{1}{2}).m_{2}.(v_{2}+9)^{2} (VI)

If we use (I) in (VI) ⇒

2.m_{2}.(v_{1}+9)^{2}=m_{2}.(v_{2}+9)^{2}

2.(v_{1}+9)^{2}=(v_{2}+9)^{2}

If we use (V) in this last expression ⇒

2.(v_{1}+9)^{2}=(2.v_{1}+9)^{2}

2.(v_{1}^{2}+18v_{1}+81)=4v_{1}^{2}+36v_{1}+81

2v_{1}^{2}+36v_{1}+162=4v_{1}^{2}+36v_{1}+81

2v_{1}^{2}=81

v_{1}^{2}=40.5

v_{1}=\sqrt{40.5}=6.36

We find that the original speed v_{1} is 6.36\frac{m}{s}

If we replace this value in the equation (V) ⇒

2.(6.36\frac{m}{s})=v_{2}

v_{2}=12.72\frac{m}{s}

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