Kinetic energy of runner race question

AI Thread Summary
In the discussion, a father with double the mass of his son has half the kinetic energy initially. When the father increases his speed by one m/s, their kinetic energies become equal. The calculations reveal that the father's velocity should be approximately 2.41 m/s, leading to the son's velocity being around 4.82 m/s. The confusion arises from incorrect assumptions in the kinetic energy equations, particularly in relating the father's and son's velocities and masses. Correcting these calculations aligns with the book's values of 2.4 m/s for the father and 4.8 m/s for the son.
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Say a father who has a mass that is two times that of his son is racing against him, and his kinetic energy is half of his son. When the father increases his speed by one m/s, the kinetic energies are equal.

I do this:
M = mass of father
V = velocity of father
0.5MV^2 = 0.5(0.5mv^2)

2MV^2 = mv^2
(4m)V^2 = mv^2

4V^2 = v^2

2V = v

0.5M(V+1)^2 = 0.5m(2V)^2
(V+1)^2 = 4V^2
V^2 + 2V + 1 = 4V^2

-3V^2 +2V + 1 = 0

(3V +1) (-V + 1)

V = 1, -1/3

so V = 1 m/s

However, my book says the speeds are 2.4 m/s and 4/8 m/s for father and son, respectively. What am I doing wrong?
 
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M & V =mass and velocity of father right?
m & v =mass and velocity of son?
since the father initial KE is half his son, why did u multiply .5 on the KE of the son instead of the father?

btw,my ans for velocity of son is 3.414m/s
 
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No, that is how it should be. KE of F = 0.5(KE of son)

Does anyone know why the answers are 2.4 and 4.8?
 
OVB said:
Say a father who has a mass that is two times that of his son is racing against him, and his kinetic energy is half of his son. When the father increases his speed by one m/s, the kinetic energies are equal.

I do this:
M = mass of father
V = velocity of father
0.5MV^2 = 0.5(0.5mv^2)

2MV^2 = mv^2
(4m)V^2 = mv^2

4V^2 = v^2

2V = v

0.5M(V+1)^2 = 0.5m(2V)^2
2m(V+1)^2 = m(2V)^2 <== added
(V+1)^2 = 2V^2
V^2 + 2V + 1 = 2V^2

-1V^2 +2V + 1 = 0

etc.

(3V +1) (-V + 1)

V = 1, -1/3

so V = 1 m/s
However, my book says the speeds are 2.4 m/s and 4/8 m/s for father and son, respectively. What am I doing wrong?
See the colors
 
The first thing we do is relate the father's kinetic energy to the son's according to the question. I will keep the father on the LHS and son on the RHS. I will used lowercase v for the father's velocity and uppercase V for the son's velocity.

1) (0.5)(m)(v*v) = (0.5)(0.5)(0.5m)(V*V)
// Now multiply by 8 to remove fraction...
4m(v*v) = m(V*V)
// Now divide by m to simplify...
4(v*v) = (V*V)
// Now take square root of both sides.
2v = V
// This gives us the son's velovity V in terms of the
// father's velocity v. ie: V = 2v.

Now in order to have the father's K equal the son's K we do two things.
- Add 1 to the father's velocity on the LHS.
- Multiply the RHS by 2 since we are not relating the father's K to
half the son's K anymore. ie: Instead of K = 0.5K we now have
K = K since that's what happens when we add 1 to the father's
velocity.

2) (0.5)(m)(v+1)(v+1) = (0.5)(0.5m)(2v)(2v)
// Remember V = 2v
// Multiply by 2 and divide by m to simplify...
(v+1)(v+1) = (2)(v*v)
// Take the square root of both sides...
v+1 = sqrt(2)*v
v = 1 / (sqrt(2) - 1)
v = 2.41 m/s.

Now all we have to do is substitute into V = 2v to get the son's original
velocity...
V = 2 * 2.41 = 4.82 m/s.
 
you went wrong on this part:

0.5M(V+1)^2 = 0.5m(2V)^2
(V+1)^2 = 4V^2
V^2 + 2V + 1 = 4V^2

-3V^2 +2V + 1 = 0

(3V +1) (-V + 1)

V = 1, -1/3

so V = 1 m/s

0.5M(V+1)^2 = 0.5m(2V)^2 ... subtitute M=2m here. this will give you ..
2(V+1)^2 = 4V^2
V^2 + 2V + 1 = 2V^2

-V^2 +2V + 1 = 0
0r V^2-2V-1=0
solving I get V= 2.4142 m/sec.

use this to calculate v=2V=2*2.4142=4.8284 meters per second.
 
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