Question

A bullet of mass 200 gram is fired from a gun of mass 10kg with a velocity of 100 metre per second. Calculate the velocity of recoil.

Answer 1

Answer :

➥ The velocity of recoil = -2 m/s

Given :

➤ Mass of bullet (m₁) = 200 g

➤ Mass of gun (m₂) = 10 kg

➤ Velocity of bullet (v₁) = 100 m/s

To Find :

➤ Velocity of gun (v₂) = ?

Solution :

To find the velocity of gun, first we need to convert the mass of bullet gram into kg, then after we find the velocity of gun$.$

◈ Mass of bullet (m₁) = 200 g

→ 200/1000

→ 0.2 kg

Intial velocity u₁ = u₂ = 0 [both are at rest]

We can find Velocity of gun by using principal of conversation of momentum which says m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂, So let's calculate v₂ !

From principal of conversation of momentum

→ m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

→ 0 + 0 = 20 + 10 × v₂

→ 0 = 20 + 10 × v₂

→ 0 = 20 + 10v₂

→ 0 - 20 = 10v₂

→ -20 = 10v₂

→ -20/10 = v₂

→ -2 = v₂

→ v₂ = -2 m/s

║Hence, the velocity of the gun is -2 m/s.║

【NOTE :- (-) minus sign indicates that velocity of gun in opposite direction.】

Answer 2

$\mathcal{\pink{\underline{\underline{\blue{GIVEN:-}}}}}$

• A bullet of mass 200g is fired from a gun of mass 10kg with a velocity of 100m/s .

• Mass of bullet (m) = 200g = 0.2kg

• velocity of bullet (v) = 100m/s

• Mass of gun (M) = 10kg

$\mathcal{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}$

• The recoil velocity of gun .

$\mathcal{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}$

We have know about

• “ Law of conservation of linear momentum ” .

$\green\bigstar\:\rm{\gray{\overbrace{\underbrace{\red{Linear\:momentum\:of\:conservation\::-\atop{initial\:momentum\:(p_i)\:=\:final\:momentum\:(p_f)\:}\:}}}}}$

✍️ Before firing, the system (gun+bullet) is at rest, therefore, initial momentum of the system = 0 .

$\rm{\implies\:Initial\:momentum\:(p_i)\:=\:(M\:+\:m)\times{v}\:}$

Where,

• $\rm{\pink{v}}$ = 0m/s

$\rm{\implies\:Initial\:momentum\:(p_i)\:=\:(10\:+\:0.2)\times{0}\:}$

$\rm{\green{\implies\:Initial\:momentum\:(p_i)\:=\:0\:}}$

✍️ After firing, The bullet is separated from gun and moving with velocity “ 100m/s ” and the gun is recoiled .

$\rm{\implies\:Final\:momentum\:(p_f)\:=\:m\times{v}\:+\:M\times{v_r}\:}$

Where,

• v = 100m/s

• $\rm{\pink{v_r}}$ = resolve velocity = ?

$\rm{\implies\:Final\:momentum\:(p_f)\:=\:0.2\times{100}\:+\:10\times{v_r}\:}$

$\rm{\green{\implies\:Final\:momentum\:(p_f)\:=\:20\:+\:10\times{v_r}\:}}$

According to Linear conservation of momentum,

$\rm{\pink{p_i\:=\:p_f\:}}$

$\rm{\implies\:0\:=\:20\:+\:10\times{v_r}\:}$

$\rm{\implies\:v_r\:=\:\dfrac{-20}{10}\:}$

$\rm{\purple{\implies\:v_r\:=\:-2\:m/s\:}}$

[NOTE :- -ve sign indicates the velocity of gun in the opposite direction .]

$\rm{\red{\therefore}}$ The recoil velocity is “ 2m/s ” .