Question

A Bullet of mass 10g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03s Calculate the distance of penetration of the bullet into the block. Give reason.
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Answer 1

Given:

• Mass of the bullet = 10 g
• Initial velocity = 150 m/s
• Final velocity = 0 m/s
• time = 0.03 s

To find:

• Distance of penetration

Method:

First, we need to find the acceleration of the bullet.

Assuming that the bullet is moving with uniform acceleration we can apply the first equation of motion in order to find a.

$\clubsuit$ v = u + at

⇒ a = v - u / t

Now, let's substitute the values,

⇒ a = 0 - 150 / 0.03

⇒ a = 15000/3

⇒ a = 5000 m/s²

Now,

By applying the second equation of motion,

$\clubsuit$ s = ut + ½ at²

Now, let's substitute the values,

⇒ s = 150 x 0.03 + ½ x 5000 x 0.0009

⇒ s = 4.5 + 2.25

⇒ s = 6.75 m

The distance of penetration of bullet into the wooden block is 6.75 m.

Answer 2

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Answer:

• 6.75 m

Explanation:

Given:

• Mass of the bullet = 10 g
• Initial velocity = 150 m/s
• Final velocity = 0 m/s
• Time = 0.03 s

Tofind:

• distance of penetration of the bullet into the block.

Calculation:

• v = u + at

>> a = v-u/t

>> a = 0-150/0.03

>> a = 15000/3

>> a = 5000 m/s²

• s = ut + 1/2 at²

>> s = 150*0.03 + 1/2 *5000*0.0009

>> s = 4.5 + 2.25

>> s = 6.75 m

hence, distance of penetration of bullet into the wooden block is 6.75 m.

Extra Information:

equations of motion

• s = ut+½at²
• s = vt- ½at²
• v = u+ at
• u = v-at
• v²-u² = 2as
• v²-u²/2s = a
• v²-u²/2a = s

where,

• s = distance covered
• v = final velocity
• u = initial velocity
• t = time taken
• a = accelartion