A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 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?

Guest Mar 15, 2017
edited by Guest  Mar 15, 2017

12+0 Answers


Upper case = father, lower case = son


(1/2)MV2 = (1/2)*(1/2)mv2


m = M/2.  So. V2 = v2/4.   


(1/2)M(V+1)2 = (1/2)mv2  So (V+1)2 = v2/2, or  V2 + 2V + 1 = v2/2


So. V2 + 2V + 1 = 4V2/2. 


V2 - 2V - 1 = 0


V2 - 2V + 1 = 2


(V-1)2 = 2


V = 1 + √2.   (Since only the positive V is needed) This is the father's original speed (in m/s)


Use the fact that v2 = 4V2 to find the son's speed.

Alan  Mar 15, 2017

This assumes the father and son have the same start velocity.

MathDude  Mar 15, 2017

No, it assumes the father's starting speed is V and the son's is v.  It also assumes the son's speed doesn't change when the father's does (indeed, it assumes the son's speed doesn't change at any time).



Alan  Mar 15, 2017

(1/2)M(V+1)2 = (1/2)mv2  So (V+1)2 = v2/2

Instead it has to cancel out the M and m by (1/2)M(V+1)2 = (1/2)mv2  So (V+1)2 = v2/4

that should account for the discrepancies not the velocity thing I said before

MathDude  Mar 15, 2017

K = 1/2*mv2

K10  = 2 * K20

K10 = K2F

m1 = 1/2 m2

K10 = 1/4*m2v2

K20 = 1/2*m2*v2

1/4*m2v102 = 2 * 1/2*m2v202

1/4*v102 = v202

K2F = 1/2*m2*(v20 + 1)2

1/4*m2*v102 = 1/2*m2*(v20 + 1)2 

1/4*v102 = 1/2*(v20 + 1)2 

1/4*v102 = v202 

v202 = 1/2*(v20 + 1)2 

v20 = 1/2*(v20 + 1)

v20 = 1/2*v20 + 1/2

v20 = 1 m/s

1/4*v102 = v202

1/4*v102 = 1

v102 = 4

v10 = 2 m/s

Son's starting speed = 2 m/s

Father's starting speed = 1 m/s

MathDude  Mar 15, 2017

v20 = 1/4*(v20 + 1)

v20 = 1/4*v20 + 1/4

v20 = 1/3 m/s

1/4*v102 = v202

1/4*v102 = 1/9


v102 = 4/9

v10 = 2/3 m/s

Son's starting speed = 1/3 m/s

Father's starting speed = 2/3 m/s

I have no idea why the answers are different because I can't follow your logic well, but atleast this part of mine was wrong

MathDude  Mar 15, 2017

Solution for original velocities of father & son, via kinetic energy


\(\text {(a)}\\ \small v_f \small \text { velocity of father. } v_s \text { velocity of son. } \\ \frac{1}{2} m_f(v_f + 1.0)^2 = 2 × ½ m_fv^2_f \\ (vf + 1.0)^2 = 2v^2_f \\ v_f +1 = \sqrt{2}v_f \hspace{3em} \text { | square root of both sides}\\ v_f + 1 \approx 1.4142v_f \\ v_f \approx 2.4143v_f \\ 0.4142v_f \approx 1 \\ v_f \approx 1.4143 \\ \text { } \\ \text {(b)}\\ \frac{1}{2}v_s^2 = 2 * \frac{1}{2} (m_fv^2_f) \\ \frac {1}{4}v_s^2 = v^2_f \\ v_s^2 = 4v_f^2\\ v_s = 2v_f \approx 4.8286 m/s \)


\(\small \text{Theory & Formulas: Complements of Gottfried Wilhelm Leibniz, with Christiaan Huygens & René Descartes. }\\ \small \text{ }\hspace{17em}\scriptsize \text {(Amazingly, there is very little of Sir Isaac Newton in this.) }\\ \small \text{Produced by Lancelot Link and company. }\\ \small \text{Directed by GingerAle. }\\ \small \text{Sponsored by Naus Corp. Tortoise and Hair Dye: }\\ \small \text{ }\hspace{9em} \small \text {Artificial intelligence formula for slow blondes and some mathematicians. }\\ \small \text{ }\hspace{9em} \small \text {New formula now works for red-heads. } \scriptsize \text{(Helps to prevent communist sympathies, too.) } \)

GingerAle  Mar 15, 2017
edited by GingerAle  Mar 15, 2017

Your own equation:

(v(f) + 1)^2 =2v^2(f). Substitute your own "answers", where v(f) =Father's speed.

(1.4142135 + 1)^2 =2 x (1.4142135)^2

5.828427...............= 4  ???!!!!.


Alan's answer =Sqrt(2) + 1. Substitute in your own equation:

(2.4142135 + 1)^2........= 2 x (2.4142135)^2

11.656853....= 11.656853.....


Guest Mar 15, 2017

Corrected error (and clarified math for the hoary hair tortuous).


\(\text {(a)}\\ \small v_f \small \text { velocity of father. } v_s \text { velocity of son. } \\ \frac{1}{2} m_f(v_f + 1.0)^2 = (2)* \frac{1}{2} m_fv^2_f \\ (v_f + 1.0)^2 = 2v^2_f \hspace{2em} \small \text {| Divide by } m_f \text { and reduce}\\ v_f +1 = \sqrt{2}v_f \hspace{3em} \text { | square root of both sides}\\ v_f + 1 \approx 1.4142v_f \\ 0.4142v_f \approx 1 \hspace{4em} \text{ | subtract and divide}\\ v_f \approx 2.4142m/s \\\)




\(\text {(b)}\\ \text { } \\ \frac{1}{2}(m_f/2)v_s^2 = (2) * \frac{1}{2} (m_fv^2_f) \\ \frac {1}{4}v_s^2 = v^2_f \hspace{3em} \small \text {|Divide by } m_f \text { and reduce}\\ v_s^2 = 4v_f^2 \hspace{3em} \small \text { |Multiply through by 4)}\\ v_s = 2v_f \hspace{3em} \small \text { |Take square root }\\ v_s = 2*(2.4142) \approx 4.8284 m/s\)



Sometimes I need a second application of the Tortoise & Hair dye AI formula – especially when the blarney banker is nearby.  If you had used it, then this wouldn’t have been a "W T F" meltdown moment –it would be just a mistake.

GingerAle  Mar 16, 2017

Even after your "correction", it turns out that it is exactly the same as Alan's answer!!. Did you plagiarize from him? Or, did you even check his answer first, before you offered your "brilliant" solution? It appears to me that you are as phony as a $3 bill !!. Cheers and have some Canada dry!!.

Guest Mar 16, 2017

Wow! We are having a meltdown, aren’t we? What happened? Did you lose your canteen privileges?


My answer is the same as Sir Alan’s?  Did you use a computer to analyze this or did you do it the old-fashioned way?  Never mind, I already know the answer.  Of course, my answer is the same as Sir Alan’s, because brilliant minds think alike. Our answers are both correct, so they have to be the same.


 Why do you think this is plagiarism?  In this post:  http://web2.0calc.com/questions/big-cheese#r4  I used an analogy of embezzlement to explain plagiarism to you.  This was reasonable because you were a banker, and you’re probably in the “Big House” for this very reason, so, it seemed like you would understand this.  In that post, I said, “It’s always a good idea to make sure the answer you are copying is the correct answer, because you are less likely to get caught (most of us learn that in grade school). . .”


See, Sir Alan’s answer was correct and mine was not, so that kind of, sort of contradicts your accusation, unless I purposely did that to throw the bloodhounds off the scent.   Yep, we brilliant, genetically enhanced chimps will do things like that when we have a mind to.  We know how to use our noodles! We also post parallel solutions when a student has a problem understanding a presentation, or just when we blŏŏdy well feel like practicing. 


Anyway, I can understand why you think having the same answer as someone else is plagiarism.  You post computer generated answers, usually preceded by blarney of the most useless kind, believing them to be solutions.  The only time you ever post a true solution is when you’ve plagiarized it. It’s the “Thick as Thieves” theory. Thieves (embezzlers) usually believe everyone else is a thief, too.  The same applies to plagiarists.   


I am surprised you use the “phony as a $3 bill” cliché.  I would think you being a banker would know $3 notes were issued by several banks in the mid-nineteenth century –there is nothing phony about them.  Of course, you were a little boy then and probably not too bright or aware.


Anyway, Mr. Banker, It is a sad thought to be a “has been.” However, I want you to know I do not think of you as a “has been,” nope, not at all.  I think of you as a “never were.”


Well, I think I’ll have a snack: a few gingersnaps and some Canada Dry Ginger Ale. You should try it, after your canteen privileges are restored. 


Until next time, Mr. Banker, don’t take any wooden nickels. Cheers.

GingerAle  Mar 17, 2017
edited by GingerAle  Mar 17, 2017

GingerAle, I’ve never seen a math post where the names of the discoverers and theorists for the formulas were included. It’s really a nice touch.  For this one, I would have thought Newton for sure.


Guest, there is nothing sick about this. This is very funny.  Maybe you don’t think so because you are not too bright or aware.  Hahaha

Guest Mar 17, 2017

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