Question

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?

Answer 1

(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)`