Question

From a rifle of mass 4 kg, a bullet of mass 50 gm is fired with an initial velocity of 35 m/s calculate the initial recoil velocity of rifle

Do it fast please.......................................​

Answer 1

Answer:-

$\red{\bigstar}$The Initial recoil velocity of the rifle is $\\ \huge{\leadsto{\boxed{-0.44 \: m/s}}}$

• Given:-

Mass of rifle = 4 kg

Mass of bullet = 50 gm → 50 × 10⁻³ kg

Initial velocity = 35 m/s

• To Find:-

Initial recoil velocity = ?

• Solution:-

Let the recoil velocity of the Pistol be "v".

From Law of Conservation of momentum;

We know,

Initial Momentum = Final Momentum.

$\boxed{\sf{ P_i = P_f}}$

Before firing the gun and the bullet are at Rest.

Therefore, Initial momentum will be zero. Then the equation will be:-

$\boxed{\sf{P_f = 0}}$

Now,

$\boxed{\sf{mu + Mv = 0}}$

here,

M = Mass of Rifle.

m = Mass of Bullet.

u = Velocity of bullet.

v = Velocity of Rifle.

$\implies\sf{50 \times 10^{-3} \times 35 + 4 \times v = 0}$

→ $\sf{1750 \times 10^{-3} + 4v = 0}$

→ $\sf{1.750 + 4v = 0}$

→ $\sf{4v = - 1.750}$

→ $\sf{v = \dfrac{- 1.750}{4}}$

→ $\sf{v = - 0.4375}$

→ $\boxed{\sf{v = - 0.44 \: m/s}}$

Therefore, the Initial Recoil velocity of Rifle is -0.44 m/s.

$\red{\bigstar}$ As it is moving opposite to the motion of the bullet hence, the negative sign of recoil velocity.

Answer 2

m1 = 4kg

u1 = 0m/s

v1 = ?

m2 = 0.05kg

u2 = 0m/s

V2 = 35 m/s

Conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2

=> 4v1 + 1.75 = 0 + 0

=> v1 = -1.75/4

=> v1 = -0.4375 m/s

-ve sign shows gun goes in backward direction

Ans. -0.4375 m/s

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