Question

A bullet of mass 10 g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and come to rest in 0.03s.Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of force exerted by the wooden block of the bullet ?​

Answer 1

Initial velocity, u= 150 m/s

Final velocity, v= 0 (since the bullet finally comes to rest)

Time taken to come to rest, t= 0.03 s

According to the first equation of motion, v= u + at

Acceleration of the bullet, a

0 = 150 + (a — 0.03 s)a = -150 / 0.03 = -5000 m/s2

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:

v2= u2+ 2as

0 = (150)2+ 2 (-5000)

= 22500 / 10000

= 2.25 m

Hence, the distance of penetration of the bullet into the block is 2.25 m.

From Newton’s second law of motion:

Force, F = Mass * Acceleration

Mass of the bullet, m = 10 g = 0.01 kg

Acceleration of the bullet, a = 5000 m/s2

F = ma = 0.01 * 5000 = 50 N

Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N....

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Answer 2

Solution :-

As per the given data ,

• Mass of the bullet (m) = 10 g = 0.01 kg

[ In order to convert g into kg simply divide by 1000 ]

• Initial velocity of the bullet (u)= 150 m/s
• Final velocity of the bullet (v) = 0 m/s [ rest]

• Time taken by the bullet to come to rest (t) = 0.03 s

First , we need to find the acceleration of the bullet as the bullet is moving with uniform acceleration throughout it's motion we can apply the first equation of motion in order to find acceleration (a) .

(1)

As per first equation of motion ,

➜ v = u + at

On rearranging ,

➜ a = v - u / t

➜ a = 0 - 150 / 0.03

➜ a = - 15000/ 3

➜ a = - 5000 m/s ²

Note : Here negative sign denotes retardation .

Now , let us find the force by applying Newton's second law of motion ,

➜ F = ma

➜ F = 0.01 x - 5000

➜ F = - 50 N

The magnitude of force exerted by the block on the bullet is 50 N .

(2)

By applying the third equation of motion ,

➜ v² = u ²  + 2as

➜ s = v² - u² / 2a

➜ s = 0 - 22500 / 2 x - 5000

➜ s = 22500 / 10000

➜ s = 2.25 m

The distance of penetration of the bullet into the block is 2.25 m