Physics, 13.07.2019 22:20 sanafarghal

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?

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Answer from: mainray1792

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,

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

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

$2M_2V_1^2=0.5\times M_2V_2^2$

$2V_1^2=0.5\times V_2^2$

4V_1^2= V_2^2

2V_1= V_2 .................(1)

also,

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

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

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

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

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

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

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

(0.404V_1)=(3.72)

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

Answer from: bellajuliet13

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) ⇒

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

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

Answer from: Quest

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