Question

A bullet of mass 10gm is fired from a gun of mass 6Kg with a velocity of 300gm/s. Calculate the recoil velocity of the gun. with solution and explanation​

Answer 1

Answer:

- 0.5 m/s

Explanation:

Final momentum of the system = momentum of the gun + momentum of the bullet = Mv/ + mv

(v/ is the recoil velocity of the gun)

Applying conservation of linear momentum,

0 = Mv/ + mv

Implies, v/ = -(mv)/M

The negative sign indicates that the direction of velocity of the gun is opposite to the direction of velocity of the bullet.

m =10g =0.01kg

mass of gun M= 6kg

velocity of bullet = 300m/s

v/ = -(mv)/M

= -(0.01 * 300)/6

= 3/6= -0.5 m/s

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Answer 2

Answer:

Final momentum of the system = momentum of the gun + momentum of the bullet = Mv/ + mv

(v/ is the recoil velocity of the gun)

Applying conservation of linear momentum,

0 = Mv/ + mv

Implies, v/ = -(mv)/M

The negative sign indicates that the direction of velocity of the gun is opposite to the direction of velocity of the bullet.

m =10g =0.01kg

mass of gun M= 6kg

velocity of bullet = 300m/s

v/ = -(mv)/M

= -(0.01 * 300)/6

= 3/6= -0.5 m/s

Explanation:

Final momentum of the system = momentum of the gun + momentum of the bullet = Mv/ + mv

(v/ is the recoil velocity of the gun)

Applying conservation of linear momentum,

0 = Mv/ + mv

Implies, v/ = -(mv)/M

The negative sign indicates that the direction of velocity of the gun is opposite to the direction of velocity of the bullet.

m =10g =0.01kg

mass of gun M= 6kg

velocity of bullet = 300m/s

v/ = -(mv)/M

= -(0.01 * 300)/6

= 3/6= -0.5 m/s