Question

A bullet has a mass of 0.02 kg and is moving with a speed of 10 ms-1. It can penetrate 10cm of
a given target before coming to rest. If the same target were only 6 cm thick, what will be the
speed and kinetic energy of the bullet, when it comes out?​

Answer 1

Given:

Mass of the bullet (m) = 0.02 kg  

Initial speed of the bullet (u) = 10 m/s  

And:

Final speed of the bullet (v) = 0 m/s  

Penetration (s) = 10 cm = 0.10 m  

Using Newton’s law of motion,

We get:

$\boxed{\sf{v^{2}-u^{2}= 2as}}$

$\implies$ $\sf{-10\times10=2a\times0.1}$

$\implies$ $\sf{a= 100/0.2 }$

$\implies$ $\sf{a = 500\:m/s^{2}}$

Note:

Thickness of the wooden block (S) = 0.06 m

$\boxed{\sf{v^{2} = u^{2} - 2aS }}$

$\implies$ $\sf{v^{2} = 100 - 2\times500\times0.06 }$

$\implies$ $\sf{v^{2} = 40 }$

$\implies$ $\sf{v = 6.325\:m/s }$

Kinetic energy of the bullet,

$\boxed{\sf{Kinetic\:every = \frac{1}{2} mv^{2}}}$

$\implies$ $\sf{\frac{1}{2} \times0.02\times40}$

$\implies$ $\sf{0.4\:J}$

So:

The bullet comes out with 0.4 J of kinetic energy.

Answer 2

Answer:-

$v = 6.32 \\ K_e = 0.4 J$

Given :-

mass of bullet = 0.02 kg

u = 10 m/s

v = 0 m/s

s = 0.1 m

To find :-

The speed and kinetic energy of bullet.

Solution:-

Since, bullet can penetrate 10 cm before v = 0.

By using 3rd equation of motion:-

The Retardation of the bullet will be :-

→$2as = v^2 - u^2$

→$2 \times a \times 0.1 = (0)^2 - (10)^2$

→$0.2 a = - 100$

→$a = \dfrac{-100}{0.2}$

→$a = - 500 m/s^2$

hence,

Retardation of bullet will be -500 m/s².

• When target is 0.06 m thick.

The velocity of bullet is given by :-

→$2as = v^2 -u^2$

→$2\times -500 \times 0.06 = (v)^2 - (10) ^2$

→$-1000 \times 0.06 = -100$

→$-60 = v^2 -100$

→$v^2 = 40$

→$v = \sqrt{40}$

→$v= 6.32m/s$

The kinetic energy of the bullet will be :-

→$K_e = \dfrac{1}{2}mv^2$

→$K_e = \dfrac{0.02 \times (6.32)^2 }{2}$

→$K_e = \dfrac{0.02 \times 39.94}{2}$

→$K_e = \dfrac{0.80}{2}$

→$K_e = 0.4J$

hence,

The kinetic energy of bullet will be 0.4 J.