Physics, 10.03.2020 02:04, Briii582

A father racing his son has 1/3 the kinetic energy of the son, who has 1/2 the mass of the father. The father speeds up by 1.4 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son

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answered: Danyboy10

The initial speed of the father is 1.9 m/s and the initial speed of the son is 4.65 m/s.

The given parameters;

kinetic of the son = K.Ekinetic energy of the father = $\frac{K.E}{3}$ mass of the father = mmass of the son = $\frac{m}{2}$ father's final speed = v + 1.4let the initial speed of father = v

The kinetic energy of the son is written as follows;

$K.E_s = \frac{1}{2} m v^2$

The kinetic energy of the father is written as follows;

$K.E_f =\frac{1}{3} K.E_s$

When the fathers speed increases by 1.4 m/s;

$\frac{1}{2} mv^2 = \frac{1}{3} [\frac{1}{2} m(v + 1.4)^2]\\\\ v^2 = \frac{1}{3} (v^2 + 2.8v + 1.96)\\\\ v^2 = \frac{v^2}{3} + 0.93v + 0.653\\\\\frac{2}{3} v^2 - 0.93v - 0.653 = 0\\\\solve \ quadratic \ equation \ using \ formula \ method;\\\\a = \frac{2}{3} , \ b = -0.93 , \ \ c = -0.653\\\\v = \frac{-b \ \ +/- \sqrt{b^2 - 4ac} }{2a} \\\\v = \frac{-(-0.93) \ \ +/- \sqrt{(-0.93)^2 - 4(\frac{2}{3} \times -0.653)} }{2(\frac{2}{3})}\\\\v = 1.9 \ m/s$

The velocity of the son is calculated as follows;

$K.E_f = \frac{1}{3} K.E_s\\\\\frac{1}{2} mv_f^2 = \frac{1}{3} [\frac{1}{2} . \frac{m}{2} v_s^2]\\\\v_f^2 = \frac{1}{6} v_s^2\\\\v_s^2 = 6v_f^2\\\\v_s^2 = 6(1.9)^2\\\\v_s^2 = 21.66 \\\\v_s = \sqrt{21.66} \\\\v_s = 4.65\ m/s$

Thus, the initial speed of the father is 1.9 m/s and the initial speed of the son is 4.65 m/s.

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answered: aubreystechschu11331

The initial velocity of the father is 1.91m/s

and the one of the son is 4.68 m/s

Explanation:

First, the kinetic energy is written as:

K = (m/2)*v^2

where m is mass and v is velocity.

The information that we have is that:

If we use tabs (M, K, V) as the data for the father and (m, k, v) as the data for the son:

K = k/3

m = M/2

now, let's proced with these two equations:

the first one says:

M*V^2 = (m*v^2)/3

(where i cancelled the 1/2 in both sides)

now, we can replace the second equation into this and get:

M*V^2 = ((M/2)*v^2)/3 = (M*v^2)/6

Here we can divide by M in both sides, and we get:

V^2 = (v^2)/6

We apply the square root to both sides, and we get:

V = v/(√6)

Afther that, the father increases his velcity by 1.4m/s and then the kinetic energys are equal, so now we have:

M*(V + 1.4)^2 = m*v^2

again, using that m = M/2

M*(V + 1.4m/s)^2 = (M/2)*v^2

We divide by M in both sides:

(V + 1.4m/s)^2 = (v^2)/2

Now we apply the square root to both sides:

V + 1.4m/s = v/√2

Now we have two equations and two variables:

V + 1.4m/s = v/√2

V = v/(√6)

We can replace the second equation in the first one and obtain the velocity of the son:

v/(√6) + 1.4m/s = v/√2

v( 1/√6 - 1/√2) = -1.4m/s

v = 1.4m/s/( -1/√6 + 1/√2) = 1.4m/s*0.299 = 4.68m/s

Now, knowing the velocity of the son, we can find the velocity of the father using:

V = v/(√6) = (4.68m/s)/√6 = 1.91m/s

answered: Guest

Full change in the electric plus system

answered: Guest

spare some change and give it to the poor

explanation:

plz : ((

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A father racing his son has 1/3 the kinetic energy of the son, who has 1/2 the mass of the father. T...