Question

A machine gun of 5kg fires 50g bullet at 30 bullet per minute a speed 400m/s. What is the velocity of gun

Answer 1

Given :

▪ Mass of machine gun = 5kg

▪ Mass of bullet = 50g = 0.05kg

▪ Velocity of bullet = 400m/s

▪ No. of bullets fired per minute = 30

To Find :

▪ Recoil velocity of gun.

Solution :

→ This question can be solved by concept of momentum conservation.

→ We can apply this concept only when net external force acts on system be zero.

☞ Initial momentum = Final momentum

____________________________

✴ M1u1 + M2u2 = M1v1 + M2v2

• M1 = mass of gun
• M2 = mass of bullet
• u1 = initial velocity of gun
• u2 = initial velocity of bullet
• v1 = final velocity of gun
• v2 = final velocity of bullet

✏ No. of bullets fired per second = 1/2

✒ 0 + 0 = M1v1 + nM2v2

✒ M1v1 = -nM2v2

✒ v2 = -nM2v2/M1

✒ v2 = -(0.5×0.05×400)/5

✒ v2 = -10/5

✒ v2 = -2mps

✏ Negative sign shows opposite direction.

Answer 2

Given :

• Mass of gun (M) = 5kg
• Mass of bullet (m) = 50g=0.05 kg
• Number of bullets in 1 min = 30
• Velocity of bullet (v) = 400 ms¯¹

_____________________

To Find :

• Velocity of the gun (V)

_____________________

Formula used :

According to law of conservation of momentum,

Initial momentum = Final momentum

MU + mu = MV +mv

Here,

• M is mass of gun
• m is mass of bullet
• U is initial velocity of gun
• u is initially velocity of bullet
• V is final velocity of gun
• v is final velocity of bullet.

_____________________

Solution :

According to the question,

Before firing, the gun and bullet was at rest. So, U= u = 0 ms¯¹.

Number of bullets per min or 60 sec = 30

Number of bullets in 1 sec = 0.5

And,

Mass of 1 bullet = 0.05 kg

Mass of 0.5 bullets(m). = 0.05 ×0.5kg = 0.025kg

Now, on solving further

→MU + mu = MV +mv

→5×0 + 0.025×0 = 5× V + 0.025×400

→0 + 0 = 5V + 10

→-5V = 10

→V = - 2 ms¯¹

Therefore, the velocity of the gun is 2 ms¯¹ in the opposite direction.