Afather racing his son has 1/2 the kinetic energy of the son, who has 1/3 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?

a) 0.78 m/s b) 1.57 m/s

Explanation:

M = father's mass


m = son's mass = M/3


V = father's initial speed


v = son's initial speed

(1/2)MV^2 = (1/2)*(1/2)*m v^2


M*V^2 = (1/2)(M/3)v^2


V^2/v^2 = 1/4


V = v/2

Second equation:


(1/2)M*(V + 1.4)^2 = (1/2)m*v^2


= (1/2)*(M/3)*(3V)^2


cancel out the M's and (1/2)'s


(V + 1.4)^2 = 3V^2


V^2 + 2.8V + 1.96 = 3V^2


V^2 -1.4V -0.98 = 0

V^2 = 0.98/0.4 = 2.45

V = 1.57

a) The speed of the father is 3.38 m/s

b) The speed of the son is 8.28 m/s

Explanation:

Hi there!

The equation of kinetic energy is as follows:

KE = 1/2 · m · v²

Where:

KE = kinetic energy

m = mass

v = speed

The kinetic energy of the father will be:

KEf = 1/2 · mf · vf²

Where the letter "f" indicates "father"

The kinetic energy of the son will be:

KEs = 1/2 · ms · vs²

KEs = 1/2 · (1/3 · mf) · vs²

Where the letter "s" indicates "son".

We know that the kinetic energy of the father is half the kinetic energy of the son. Then:

KEf = 1/2 KEs

1/2 · mf · vf² = 1/2 · (1/2 · 1/3 · mf · vs²)

Let´s solve this equation for vf:

1/2 · mf · vf² = 1/12 · mf · vs²

Multiply by 2 and divide by mf both sides of the equation:

vf² = 1/6 · vs²

6 · vf² = vs²

We also know that when the father speeds up by 1.4 m/s the kinetic energy of the son and father is equal. Then:

KEf = 1/2 · mf · (vf + 1.4 m/s)²

KEs = 1/2 · 1/3 · mf · vs²

KEf = KEs

1/2 · mf · (vf + 1.4 m/s)² =  1/2 · 1/3 · mf · vs²

divide both sides of the equation by mf and 1/2

(vf + 1.4 m/s)² = 1/3 · vs²

(vf + 1.4 m/s) · (vf + 1.4 m/s) = 1/3 · vs²

vf² +2.8 vf + 1.96 = 1/3 · vs²

Replace vs² by 6 vf²

vf² +2.8 vf + 1.96 = 1/3 · 6 vf²

vf² +2.8 vf + 1.96 = 2 vf²

subtract 2 vf to both sides of the equation:

-2 vf² + vf² +2.8 vf + 1.96 = 0

-1 vf² + 2.8 vf + 1.96 = 0

Let´s solve the quadratic equation using the quadratic formula:

a = -1

b = 2.8

c = 1.96

vf = [-b ± √(b² - 4ac)]/2a

vf = 3.38 m/s

(The other solution of the quadratic equation is negative and therefore discarded).

The speed of the son will be:

6 · vf² = vs²

6 · (3.38 m/s)² = vs²

vs = 8.28 m/s

Then, the velocity of the son is 8.28 m/s and the velocity of the father is 3.38 m/s

explanation:

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this