Question

Q4. A bullet of mass 6g travelling at speed of 120 m/s penetrates deeply into [3]

a fixed target and is brought to rest in 0.1 s. Calculate the

i) distance of penetration in the target, and

ii) average retarding force exerted on the bullet​

Answer 1

Answer :

➥ The Distance of penetration in the target = 6 m

➥ The force exerted on the bullet is -7200 N

Given :

➤ Mass of a bullet (m) = 6 g

➤ Intial velocity of a bullet (u) = 120 m/s

➤ Final velocity of a bullet (v) = 0 m/s

To Find :

➤ Distance covered by a bullet (s) = ?

➤ Force exerted on a bullet (F) = ?

Required Solution :

✒ To find the Distance penetration in the target and Force exerted on a bullet, first we need to find the acceleration of the a bullet, then after we will find the distance penetration in the target and Force exerted on a bullet. $\:$

We can find Acceleration of a bullet by using the first equation of motion which says v = u + at.

Here,

• v is the Final velocity in m/s.
• u is the Intial velocity in m/s.
• a is the Acceleration in m/s².
• t is the time taken in second.

✎ So let's calculate Acceleration (a) !

From first equation of motion

→ v = u + at

→ 0 = 120 + a × 0.1

→ 0 = 120 + 0.1a

→ 0 - 120 = 0.1a

→ -120 = 0.1a

→ -120/0.1 = a

→ -1200 = a

→ a = -1200 m/s²

Now, we have Intial velocity, Final velocity, Time taken and it's Acceleration of a bullet.

• Intial velocity = 120 m/s
• Final velocity = 0 m/s
• Time taken = 0.1 sec
• Acceleration = -1200 m/s²

We can find Distance of a bullet by using the second equation of motion which says s = ut + ½ at².

Here,

• s is the distance travelled in m.
• u is the Intial velocity in m/s.
• a is the Acceleration in m/s².
• t is the time taken in second.

✎ So let's find the Distance of penetration in the target (s) !

From second equation of motion

→ s = ut + ½ at²

→ s = 120 × 0.1 + ½ × (-1200) × 0.1²

→ s = 12 + ½ × (-1200) × 0.01

→ s = 12 + 1 × (-600) × 0.01

→ s = 12 + (-600) × 0.01

→ s = 12 + (-6)

→ s = 12 - 6

→ s = 6 m

║Hence, the Distance of penetration in the target is 6 m.║

We can find the force by using the force formula which says F = ma.

Here,

• F is the force in N.
• m is the mass in g.
• a is the Acceleration in m/s².

✎ So let's find the force exerted on the bullet (F) !

As we know that

→ F = ma

→ F = 6 × (-1200)

→ F = -7200 N

║Hence, the force exerted on the bullet is -7200 N.║