Question

A machine gun of mass 24 kg fires 20 g bullets at the rate of 8 bullets per second, with a speed of 250 ms The force
required to hold the gun in position is
​

Answer 1

Answer:

In the given problem,

mg = 24 kg ; vb = 250 m/s

mass of the bullet = 20g = 20 x 10^-3 kg

no. of bullets fired/sec = 8

F = Chnage in momentum/time = mb*vb - mb*ub/t

For t = 1 sec

Total mass of bullets in 1 second mb = (8 x 20)/1000 = 0.16 kg

F = [(0.16 x 250) - 0]/1 = 40 N

Explanation:

That's Right method with Answer

Answer 2

Answer:

• Force required to hold the Gun in position = 40 N

Explanation:

Given that,

• Mass of gun = 24 kg
• Mass of each bullet = 20 g
• Number of bullets fired per second = 8 bullets
• Bullets are fired with a velocity, v = 250 m/s

To find,

• Force required to hold the Gun in position, F =?

Knowledge required,

• Force is known as the rate of change of momentum.

      Force = ( Final momentum - initial momentum ) / time

• Momentum is given by the the product of mass and velocity.

    momentum = mass × velocity

Solution,

Since, bullet will be at rest initially

therefore, Initial momentum of bullets will be zero

• Calculating the mass of bullets fired per second (m=?)

→ total mass of bullets = 20 g × 8

→ total mass of bullets = 160 g = 160/1000 kg

→ total mass of bullets = 0.16 kg

So,

• Now calculating final momentum of bullets

→ Final momentum of bullets = 0.16 kg × 250 m/s

→ Final momentum of bullets = 40 kg m/s

Further, Calculating force

→ Force = ( final momentum - initial momentum ) / time

→ Force = ( 40 - 0 ) / ( 1 )

→ Force = 40 N

Therefore,

• A force of 40 N will be needed to hold the gun in position.