Question

A bullet of mass 20g is horizontally fired
with a velocity 150 m/s from a pistol of
mass 3kg. What is the recoil velocity of the
pistol?

Answer 1

$\frak{Given} \:\begin{cases}\:\quad \sf Mass \:of\:bullet \:,\:m_1 \:=\:\pmb{\frak{ 20 \:g \:or \:0.02\:kg\:}} \:\\\:\quad \sf Mass \:of\:pistol \:,\:m_2 \:=\:\pmb{\frak{ \:3\:kg\:}}\\ \sf\:\quad Initial \:Velocity \:of \: bullet \:,\:u_1\:and \:pistol\:,\:u_2\:=\:\pmb{\frak{ 0 \:m/s \:}}\:\\ \sf \quad \: Final \:Velocity \:of \: bullet \: ,\:v_1\:=\:\pmb{\frak{ 150\:m/s\:}}.\:\end{cases}$

Need To Find : The Recoil Velocity of Pistol ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider v m/s be the Recoil Velocity of pistol.

⠀⠀⠀▪︎⠀⠀According to Principle of CONSERVATION of MOMENTUM :

⠀

$\qquad \twoheadrightarrow \pmb{\pink{\sf Initial \:Momentum \:=\:Final\:Momentum \:}}\:$

$\qquad \twoheadrightarrow \sf Initial \:Momentum \:=\:Final\:Momentum \:\:\\\\\\\qquad \twoheadrightarrow \sf m_1 u_1\:+\:m_2u_2 \:=\:m_1 v_1 \:+\:m_2v_2\:\:\\\\\\\qquad \twoheadrightarrow \sf \big( 0.02 \times 0 \big) \:+\:\big( 2 \times 0 \big) \:=\:\big( 0.02 \times 150 \big) \:+\:\big( 2 \times v \big) \:\:\\\\\\\qquad \twoheadrightarrow \sf \big( 0 \big) \:+\:\big( 0 \big) \:=\:\big( 3 \big) \:+\:\big( 2 \times v \big) \:\:\\\\\\\qquad \twoheadrightarrow \sf 0 \:=\:\big( 3 \big) \:+\:\big( 2v \big) \:\:\\\\\\\qquad \twoheadrightarrow \sf 0 \:=\: 3 \:+\:2v \:\:\\\\\\\qquad \twoheadrightarrow \sf v \:=\:\cancel {\dfrac{-3}{2}} \:\:\\\\\\\qquad \twoheadrightarrow \pmb{\underline {\boxed {\frak{ \purple { v \:=\: 1.5\:m/s\:}}}}}\:\:\bigstar \:\:$

∴ Hence Recoil Velocity of Pistol is – 1.5 m/s .

Answer 2

❥❥$\large\underline\color{pink}{Question:}$❥❥

✒ A bullet of mass 20g is horizontally fired with a velocity 150 m/s from a pistol of mass 3kg. What is the recoil velocity of the pistol?

$\large\underline\color{orange}{Answer:}$࿐

✒Recoil Velocity of Pistol is – 1.5 m/s .