Question

A bullet of mass 20g is formed a gun of mass 2kg moves with a velocity of 150m/s. What is the recoil velocity of the gun.


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Answer 1

$\huge \mathfrak \purple{answer}$

Let the direction in which the bullet is fired be positive.

Therefore,

Mass of Bullet(m)=20 grams=0.02 kg

Mass of Pistol(M)=2 kg

Initial Velocity of Bullet(u)=0 m/s

Initial Velocity of Pistol(U)=0 m/s

Final Velocity of Bullet(v)=150 m/s

Final Velocity of Pistol(V)= Recoil Velocity of Bullet = ?

According to the Law of Conservation of Momentum,

mu + MU = mv + MV

(0.02)(0) + (2)(0) = (0.02)(150) + (2)(V)

0 + 0 = 3 + 2V

-3/2 = V

V = -1.5 m/s

Therefore, the recoil velocity of the pistol will be 1.5 m/s in the opposite direction.(As the sign is negative)

Answer 2

$\huge \mathfrak \purple{answer}$

Let the direction in which the bullet is fired be positive.

Therefore,

Mass of Bullet(m)=20 grams=0.02 kg

Mass of Pistol(M)=2 kg

Initial Velocity of Bullet(u)=0 m/s

Initial Velocity of Pistol(U)=0 m/s

Final Velocity of Bullet(v)=150 m/s

Final Velocity of Pistol(V)= Recoil Velocity of Bullet = ?

According to the Law of Conservation of Momentum,

mu + MU = mv + MV

(0.02)(0) + (2)(0) = (0.02)(150) + (2)(V)

0 + 0 = 3 + 2V

-3/2 = V

V = -1.5 m/s

Therefore, the recoil velocity of the pistol will be 1.5 m/s in the opposite direction.(As the sign is negative)