Question

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

Answer 1

By the third equation of motion
 v^2 = u^2+ 2as

Here let initial velocity u=x then after traveling 3 cm v=x/2 and s=3cm=0.03 m.
So
$a = \frac{(v {}^{2} - u {}^{2} )}{2s} = \frac{ \frac{x {}^{2} }{4} - x {}^{2} }{2 \times 0.03}$
$= \frac{ - 3x {}^{2} }{4 \times 0.06} = - \frac{ - 3x {}^{2} }{0.24}$
when the. bullet comes to rest final vel v=0
$a = \frac{0 - x {}^{2} }{2s}$
now equating the two values of a we get
$\frac{ - 3x {}^{2} }{0.24} = \frac{ - x {}^{2} }{2s}$
so s=0.04 m=4cm
Now distance moved when velocity changes from v/2 to 0 is
4cm-3cm=1cm Ans
PS Note that the acceleration expression is negative which means that the bullet is decelerating i.e. its speed is decreasing.

Answer 2

Answer:

hope it helps you friend