Question

The muzzle velocity of a 50.0 g shell leaving a 3.00 kg rifle is 400 m/s. What is the recoil velocity of the rifle?
pls give full explanation...​

Answer 1

recoil velocity of rifle will be 6.67 m/s.

concept used- conservation of momentum (m₁v₁=m₂v₂)

 given, m₁=mass of shell=50 x 10⁻³ kg

            v₁ (muzzle velocity)=400 m/s

             m₂(mass of rifle)=3 kg

               v₂=?

using the given formula,

    (50 x 10⁻³ x 400)/3 =v2

   v₂=6.67 m/s

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hope it helped !!

Answer 2

m₁v₁ + m₂v₂ (before collision) = m₁v₁ + m₂v₂ (after collision)

Here,

The bullet and the gun are of one mass before the collision, and are sitting still, which cancels out the first half of the equation.

⇒ m₁v₁ = -m₂v₂

⇒ 0.05 × 400 = -3 ×  V₂

⇒ 20 = -3 × V₂

⇒ V₂ = 20/-3

⇒ V₂ = -6.67 m/s

$\boxed{\bold{Recoil \:Velocity \:of \:the \:rifle=-6.67\:m/s}}$