Question

a machina gun fired 10 bullets per sec with speed 10m/s if mass of bullet is 300gm then the force required to keep the gun stationary is​

Answer 1

Explanation:

force required to hold gun

f=nmv

=10×300x10^-3×10

=30 Newton

Answer 2

The force required to keep the gun stationary is equal to 30 Newton.

Given:

Number of bullets fired per second = 10

Speed of bullets (v) = 10 ms⁻¹

Mass of 1 bullet (m) = 300 gram = 0.3 kg

To Find:

The force required to keep the gun stationary.

Solution:

→ When the bullets are getting ejected from the gun, they will experience a force that will be equal to their change in momentum.

→ Since the gun exerts a force on the bullets, therefore by Newtons' third law of motion the gun will also experience a recoiling force due to the bullets. This reactionary force will also be equal to the rate of change of momentum of the bullets.

.→ Hence to keep the gun stationary we must apply a force equal to the recoiling force (F):

→ Force (F) = Rate of change of momentum of bullets

The initial momentum of bullets (P₁) = 0 (Bullets are at rest)

∵ The final momentum of 1 bullet = mv

∴ The final momentum of 10 bullets (P₂) = 10(mv) = 10(0.3 × 10) = 30 kg ms⁻¹

The time taken to fire 10 bullets (ΔT) = 1 second

Hence the recoiling force (F) is given by:

                                        $\therefore F = \frac{\triangle P}{\triangle T}$

                                               $=\frac{P_{2} -P_{1} }{\triangle T}\\\\=\frac{P_2}{\triangle T}\\\\=\frac{30}{1}$

                                        $\therefore F = 30N$

Hence, the force required to keep the gun stationary is equal to 30 Newton.

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