Question

A bullet of mass 10 gram travelling horizontally with a velocity of 150 metre per second strikes a stationary wooden block and come to rest in 0.03 seconds calculate the distance of an iteration into the block also calculate the magnitude of the force exerted by the wooden block on the bullet​

Answer 1

Answer:

The distance penetrated by the bullet is 2.25 m

The force exerted by the wooden block on the bullet  is 50 N

Explanation:

according to the question, we should calculate the distance penetrated by the bullet into the block 

given that :

⇒ u = 150 m/s

⇒ v = 0 

⇒ t = 0.03 sec 

⇒ m = 10 g = 0,01 Kg

we know that , S = ( v² - u²) / 2a  ------------( 1 )

we should calculate acceleration ,  v = u + at 

a =  ( v - u ) / t

a = ( 0 - 150 ) / 0.03

a = -5000 m/s² ---------------( 2 ) 

substituting equation ( 2 ) in ( 1 ) gives ,

S = { 0 - (150)² } / - ( 2 x 5000 )

S= 22500 / 10000

S = 2.25 m 

therefore, the distance penetrated by the bullet is 2.25 m

to calculate force ,we know

F = m x a 

⇒ F = 0.01 x 5000

⇒ F = 50 N 

the force exerted by the wooden block on the bullet  is 50 NN

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Answer 2

Answer:

S=2.25 m and F=50 N

Solution:-

Time=0.03 sec

mass=10 g

$\frac{10}{1000} = \frac{1}{100} \: kg$

U=150 m/s

V=0

For finding Distance by third equation we have to find a

So

$a = \frac{v - u}{t}$

$= \frac{0 - 150}{0.03}$

$= \frac{ - 150}{3} \times 100 = - 5000$

Now by using 3rd equation

${v}^{2} = {u}^{2} + 2as$

${v}^{2} = {150}^{2} + 2 \times (- 5000)s$

$0 = 22500 - 10000s$

$s = \frac{22500}{10000}$

$= 2.25 \: m$

Now Force=ma

$\frac{1}{100} \times - 5000$

-50 N

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The distance of penetration=2.25 m and Force exerter by the wooden block=50 N