Question

A bullet moving at a speed of 400m/s just pierces a wooden block 4 cm thick. What speed will be required to just pierce a wooden block 9 cm thick?

Answer 1

A bullet moving at a speed of 400m/s just pierces a wooden block 4 cm thick. To find : The speed that will be required to just pierce a wooden block 9 cm thick.

We know that, u^2 = v^2 - 2as

where u = Initial velocity

           v = Final velocity

           a = Acceleration

           s = Distance

• Let us consider that the moment the bullet hits the brick be initial velocity.
• And the moment the bullet just pierces the brick be final velocity which will be equal to zero as the bullet stops in that portion.
• And the distance covered be 's' due to deceleration 'a'.
• And convert all the values to a same unit. So first distance covered, s = 0.04m and second distance covered, s= 0.09m.

So, putting the values we get,

(400)^2 = (0)^2 - 2 x (a) x (0.04)  

Calculating the above equation we get, a = - 2 x 10^6 m/s^2.

Here we see that we got the value of a as negative which means that there is deceleration of bullet inside the wooden block.

So, speed required to just pierce a wooden block of 0.09m will be :

u^2 = (0)^2 - 2 x (-2 x 10^6) x (0.09)

Calculating the equation we get, u = 600 m/s.

Hence the Initial Velocity required to just pierce through a wooden block of 9cm will be 600m/s.

 

Answer 2

Speed of bullet $\mathbf{600 \ (\frac{m}{s})}$ will be required to just pierce in a wooden block 9 cm thick.

Explanation:

1. First case

Given

Speed of bullet $\mathbf{(u)=400 \ (\frac{m}{s})}$

Distance travelled by bullet in wooden block (s)= 4 cm =0.04 m

Final speed of bullet after travelling 4 cm in wooden block $\mathbf{(v)=0 \ (\frac{m}{s})}$

2. From equation of motion

$\mathbf{v^{2}=u^{2}-2as}$

$\mathbf{0^{2}=400^{2}-2a\times 0.04}$

So retardation of bullet in wooden block will be obtain by solving above equation

$\mathbf{a=2000000 \ \left ( \frac{m}{s^{2}} \right )}$

3. Second case

Distance of bullet travel in wooden block (s)= 9 cm =0.09 m

Retardation of bullet same as previous  $\mathbf{a=2000000 \ \left ( \frac{m}{s^{2}} \right )}$

Speed of bullet after travelling 9 cm distance $\mathbf{(v)=0 \ (\frac{m}{s})}$

Again from equation of motion

$\mathbf{v^{2}=u^{2}-2as}$

$\mathbf{0^{2}=u^{2}-2\times 2000000\times 0.09}$

After solving above equation, we get

$\mathbf{(u)=600 \ (\frac{m}{s})}$ = This is required speed of bullet to pierces in bullet through 9 cm.