Question

A 5g bullet acquires a speed of 120ms^-1 in a gun with a barrel length 2.0m the average force exerted on the bullet is
A)3.6N
B)18N
C)36N
D)72N
note:dont put any silly questions ok then i will follow you ok​

Answer 1

Answer:

B) 18N

Explanation:

Here,

u = 0 m/a

v = 120 m/a

s = 2 m

m = 5 g = 0.005 Kg

Applying equation of motion

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

${(120)}^{2} = {0}^{2} + 2 \times a \times 2$

$4a = 120 \times 120$

$a = \frac{120 \times 120}{4}$

$a = 3600 \: m{s}^{ - 2}$

Therefore,

Force exerted on bullet is F = ma

= 0.005 Kg × 3600 m/s^2

= 5× 3.6 N

= 18 N

Answer 2

Given :

• Initial velocity, u = 0 m/s

• Final velocity, v = 120 m/s

• Distance, s = 2 m

• Mass, m = 5 g

To find :

• Force exerted on the bullet

According to the question,

By using Newtons third equation of motion,

➞ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

➞ (120)² = (0)² + 2 × a × 2

➞ 14400 = 0 + 4a

➞ 14400 - 0 = 4a

➞ 14400 = 4a

➞ 14400 ÷ 4 = a

➞ 3600 = a

So,the acceleration is 3600 m/s².

Now,

Note : We have to change 5 g into kg. So, it will be 0.005 kg

➞ Force = Mass × Acceleration

Or,

➞ F = ma

➞ F = 0.005 × 3600

➞ F = 18 N

• Option B) 18 N is correct .

So,the force exerted on the bullet is 18 Newton.

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