Question

A bullet of mass 10g travelling horizontally with speed 150m/s strikes a stationary wooden block and comes to rest in0.03sec.calculate the distance of penetration of bullet into the block. ​

Answer 1

Explanation:

Given,

mass m= 10g = 0.01 kg

initial velocity u= 150 m/sec

Bullet finally come in rest, so final velocity, v= 0 m/sec

Time t= 0.03 sec

Now, from equation

v= u+at

0= 150+a(0.03)

acceleration a= -150/0.03 = -5000 m/sec2

From equation v2 -u2 = 2as

02 - (150)2 = 2×(-5000)×s

So, s= 150×150/(2×5000)

s= 2.25 m

The penetration distance of the bullet in the wooden block =2.25 m

and magnitude of force, F = ma

=0.01×5000 = 50 N

Answer 2

GIVEN :

• Mass of a bullet = 10 g = 10/1000 = 0.01 kg.
• Initial velocity of a bullet, u = 150 m/s.
• Final velocity of a bullet, v = 0 m/s.
• Time, t = 0.03 sec.

FORMULAS USED :

• $\tt V \: = \: u \: + \: at$
• $\tt V^{2} \: = \: u^{2} \: + \: 2as$

Here,

V = Final velocity.

u = Initial velocity.

a = acceleration.

t = time taken.

S = distance.

TO CALCULATE :

• The distance of penetration of bullet into the block.

SOLUTION :

To find the distance, firstly we have to find the acceleration, a = ?

We know that,

$\leadsto \sf V \: = \: u \: + \: at$

Put the given values in the above formula,

$\leadsto \sf 0 \: = \: 150 \: + \: a \times 0.03$

$\leadsto \sf a \: = \: - 5000 \: m/s^{2}$

$\therefore {\boxed {\boxed {\sf a \: = \: -5000 \: m/s^{2}}}}$

Here, The acceleration is in the negative. So the negative acceleration shown the retardation.

Now, To find the distance, S = ?

$\leadsto \sf V^{2} \: = \: u^{2} \: + \: 2as$

Put the given values in the above formula,

we get

$\leadsto \sf 0 \: = \: 150^{2} \: + \: 2 \: \times \: (-5000) \: \times \: S$

$\leadsto \sf S \: = \: 2.35 \: m$

$\therefore {\boxed {\boxed {\sf S \: = \: 2.35 \: m}}}$