Question

a bullet has a mass 0.02 kg and is moving with a speed of 10m/s . It can penetrate 10cm of a given target before coming to rest. If the same target is only 6cm thick, what will be the speed and kinetic energy of the bullet, when it comes out the target?

Answer 1

mass of the bullet, m = 0.02 kg 
Initial speed of the bullet, u = 10 m/s 
Final speed of the bullet, v = 0 m/s 
Penetration, x = 10 cm = 0.10 m 
Using the Newton’s law of motion, we get 
v^2 = u^2 – 2ax
or, 0 = 100 – 2a*0.1 
or, a = 100/0.2 
or, a = 500 m/s^2 
Thus, retardation a = 500 m/s^2 
Now, when thickness of the wooden block x = 6 cm = 0.06 m 
v^2 = u^2 – 2ax 
or, v^2 = 100 – 2*500*0.06 
or, v^2 = 40 
or, v = 6.325 m/s 
Kinetic energy of the bullet, KE = ½ mv^2 
= ½*0.02*40 = 0.4 J 

Answer 2

mass of the bullet, m = 0.02 kg  

Initial speed of the bullet, u = 10 m/s  

Final speed of the bullet, v = 0 m/s  

Penetration, s = 10 cm = 0.10 m  

Using the Newton’s law of motion, we get  

v^2-u^2 =  2as  

-10*10 =  2a*0.1  

a = 100/0.2  

a = 500 m/s^2  

when thickness of the wooden block S = 6 cm = 0.06 m  

v^2 = u^2 – 2aS  

v^2 = 100 – 2*500*0.06  

v^2 = 40  

v = 6.325 m/s  

Kinetic energy of the bullet, KE = ½ mv^2  

= ½*0.02*40 = 0.4 J