Question

A bullet of mass 40gm is fired from a gun of mass 2 kg with a velocity of 300m/s.Calculate the recoil velocity of gun.

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Answer 1

Given

• Mass of the Bullet = 40 g
• Mass of the gun = 2 kg
• Velocity of the bullet = 300 m/s

To Find

• Recoil Velocity

Solution

● m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ [Conservation of Momentum]

● Here we shall first change the mass of the bullet from g to kg. And then substitute thr values in the formula to get the recoil velocity!!

● Note that here u₁ & u₂ are 0 as the gun is initially at rest

✭ Mass of the Bullet :

→ Mass (Bullet) = 40/1000

→ Mass (Bullet) = 0.004 kg

━━━━━━━━━━━━━━━━━━━━━

✭ Recoil Velocity :

→ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

• m₁ = Mass of bullet = 0.004 kg
• m₂ = Mass of the gun = 2 kg
• v₁ = Velocity of the bullet = 300 m/s
• v₂ = Recoil Velocity = ?

→ 0 + 0 = (0.04 × 300) + (2 × v₂)

→ 0 = 12 + 2v₂

→ -12 = 2v₂

→ -12/2 = v₂

• Here it is negative as it is the recoil Velocity (opposite direction) and so we may ignore the -ve sign

→ v₂ = 6 m/s

∴ The recoil velocity is 6 m/s

Answer 2

Explanation:

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$\text{\Large\underline{\red{Question:-}}}$

:$\Longrightarrow$ A bullet of mass 40gm is fired from a gun of mass 2 kg with a velocity of 300m/s.Calculate the recoil velocity of gun.

$\text{\Large\underline{\orange{Given:-}}}$

:$\Longrightarrow$ Bullet mass = 40g = 0.04kg.

:$\Longrightarrow$ Velocity of bullet, u = 300m/s.

:$\Longrightarrow$ Mass of gun, m = 2kg.

$\text{\Large\underline{\green{Theory:-}}}$

:$\Longrightarrow$ Conservation of linear momentum:-

Law of conservation of linear momentum: If external force acting on a system of bodies is zero , the net momentum of system remains constant.

$\text{\Large\underline{\purple{Solution:-}}}$

:$\Longrightarrow$ Let recoil velocity be v.

:$\Longrightarrow$ Initial both bullet and gun are at rest . Thus, initial momentum before firing = 0. As, no external force acts on the system.

:$\Longrightarrow$ According to the law of conservation of momentum:

:$\Longrightarrow$ Total momentum before firing = Total momentum after the firing.

:$\Longrightarrow$ 0 = mu + Mv

:$\Longrightarrow$ Mv = -mu

Now we put the given values :-

:$\Longrightarrow$ 2 × v = 0.04 × 300

:$\Longrightarrow$ v = $\mathrm{\dfrac{-300×0.04}{2}}$

:$\Longrightarrow$ v = -6ms‐¹

Here, negative sign indicates that the direction in which the rifle would recoil is opposite to that of bullet .

$\text{\Large\underline{\blue{Hence,\ Verified}}}$

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