A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm . How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

Options

A. 1.0cm
B. 1.5cm
C. 2.0cm
D. 3.0cm

Answer

A

Explanation

Retardation `(-a)prop` Displacement (x)
`therefore a prop (-x)` or `a=-Kx`
`therefore (dv)/(dt)=-Kx`
`therefore (dv)/(dx).(dx)/(dt)=-Kx " But" (dx)/(dt)=v`
`therefore v(dv)/(dx)=-Kx` or `vdv=-Kxdx`
Suppose that for a displacement from 0 to x the velocity changes from u to v, them.
`int_(u)^(v)v dv=-K int_(0)^(x)x dx`
`therefore (v^(2))/(2)-(u^(2))/(2)=-K(x^(2))/(2)`
`therefore (u^(2))/(2)-(v^(2))/(2)=(Kx^(2))/(2)`
`therefore (1)/(2)mu^(2)-(1)/(2)mv^(2)=(mKx^(2))/(2)=K'x^(2)`
Where `K'=(mK)/(2)` is constant
`therefore` Loss in K.E. `= K' x^(2) " " therefore` Loss in K.E. `prop x^(2)`