Answer: 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
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;
The kinetic energy of the son is written as follows;
[tex]K.E_s = \frac{1}{2} m v^2[/tex]
The kinetic energy of the father is written as follows;
[tex]K.E_f =\frac{1}{3} K.E_s[/tex]
When the fathers speed increases by 1.4 m/s;
[tex]\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[/tex]
The velocity of the son is calculated as follows;
[tex]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[/tex]
Thus, the initial speed of the father is 1.9 m/s and the initial speed of the son is 4.65 m/s.
Learn more here:https://brainly.com/question/23503524