Question

A bullet when fired at a target with a velocity of
100 m/sec penetrates one metre into it. If the
bullet is fired at a similar target with a thickness
0.5 metre, then it will emerge from it with a
velocity of:​

Answer 1

Explanation:

The following equation holds for constant acceleration:

v1^2=v2^2+2as where:

v1 = initial velocity

v2 final velocity ( after acceleration or deceleration period)\

s = distance traveled

So in our case v1 = 100 m/s ( in the firs case, the bullet comes to a stop).

s= 1m

So v1^2 = 2as

Therefore, a = v1^2/2s = 100^2/(2*1) = 5000 m/s^2 v1 = 100 m/s , and a = 5000 m/s^2 ( assuming the same material as in the firs case so the decceleration is the same)

then :

v1= 100 , v2 =?

s= 0.5m

and 100^2= v2^2+2*a*s=v2^2+2*5000*0.5 = v2^2+5000

Hence: v2^2 = 10000-5000 = 5000

and v2= sqrt(5000) = 71 m/s.

The emerging velocity for 0.5 m is 71 m/s.