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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?

1 Answer

6 votes

Answer:

(a)
v_f=2.414\ m.s^(-1)

(b)
v_s=4.828\ m.s^(-1)

Step-by-step explanation:

Let:

mass of father be,
m_f

mass of son be,
m_s=(m_f)/(2)

speed of son be,
v_s

initial speed of father be,
v_f

After speeding up, speed of father is
v_f+1

We know Kinetic Energy is given as


KE=(1)/(2) m.v^2 .....................................(1)

where:

m = mass

v = velocity

Hence, according to the initial condition the father is having kinetic energy half the kinetic energy of the son.


KE_s=2.KE_f


(1)/(2) m_s.v_s^2=2* (1)/(2) m_f.v_f^2


((m_f)/(2))* v_s^2=2* m_f* v_f^2


v_s=2v_f .................................................(2)

According to the final condition:


(1)/(2) m_s.v_s^2= (1)/(2) m_f.(v_f+1)^2


((m_f)/(2))* v_s^2=m_f.(v_f+1)^2


v_s^2=2(v_f+1)^2


v_s=√(2)(v_f+1).....................................................(3)

(a)

From eq. 2 & 3


2v_f=√(2)(v_f+1)


√(2)\ v_f=(v_f+1)


v_f(√(2)-1)=1


v_f=(1)/((√(2)-1))


v_f=2.414\ m.s^(-1)

(b)

putting the above value in eq. (2)


v_s=4.828\ m.s^(-1)

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