Question

A gun weighing 10 kg fires a bullet of 10 g with
a velocity of 330 m/s. With what velocity does
the gun recoil?​

Answer 1

Given:

• Mass of gun, $\sf{m_g}$ = 10 kg
• Mass of bullet, $\sf{m_b}$ = 10 g = 10/1000 kg = 0.01 kg
• Initial velocity of gun, $\sf{u_g}$ = 0
• Initial velocity of bullet, $\sf{u_b}$ = 0
• Final velocity of bullet, $\sf{v_b}$ = 330 m/s

To find:

• Recoil velocity of gun = final velocity of gun, $\sf{v_g}$ =?

Knowledge required:

• Law of conservation of momentum

During a collision between two bodies the total momentum of the system before collision is equal to the total momentum of bodies after collision.

$\purple{\bigstar}\boxed{\sf{m_1u_1+m_2u_2=m_1v_1+m_2v_2}}$

[ Where, $\sf{m_1\;and\;m_2}$ are masses of two bodies with initial velocities $\sf{u_1\;and\;u_2}$ and final velocities after collision being $\sf{v_1\;and\;v_2}$. ]

Calculation:

Using law of conservation of momentum

$\implies\sf{m_gu_g+m_bu_b=m_gv_g+m_bv_b}$

$\implies\sf{(10)(0)+(0.01)(0)=(10)(v_g)+(0.01)(330)}$

$\implies\sf{0+0=10\;\;v_g+3.3}$

$\implies\sf{0=10\;\;v_g+3.3}$

$\implies\sf{10\;\;v_g=-3.3}$

$\implies\boxed{\boxed{\sf{v_g=-0.33\;\;m/s}}}\purple{\bigstar}$

Therefore,

• Recoil velocity of gun would be 0.33 m/s in the direction opposite to the motion of bullet.

Answer 2

$\huge\bf{\underbrace{\gray{GIVEN:-}}}$

• Weight of a gun is 10 kg .

• The gun fires a bullet of 10g with a velocity 330m/s .

$\huge\bf{\underbrace{\gray{TO\:FIND:-}}}$

• The velocity, with which the gun recoils .

$\huge\bf{\underbrace{\gray{SOLUTION:-}}}$

Let,

• $\bf\red{Velocity\:of\:gun}$ = V

According to the question,

• $\bf\red{Mass\:of\:gun}$ (M) = 10kg

• $\bf\red{Mass\:of\:bullet}$ (m) = 10g = 0.01 kg

• $\bf\red{Velocity\:of\:bullet}$ (v) = 330 m/s .

BEFORE FIRING :-

☯︎ Momentum of the system = 0

AFTER FIRING :-

☯︎ Momentum of the system = mv + MV

✯ According to Conservation of Linear Momentum,

$\orange\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{Momentum\:before\:firing\:=\:Momentum\:\atop{after\:firing}\:}}}}}$

➪ 0 = mv + MV

➪ (0.01 × 330) + (10 × V) = 0

➪ 3.3 + 10V = 0

➪ 10V = -3.3

➪ V = -3.3/10

➪ V = - 0.33 m/s

[NOTE → The negative sign indicates direction of velocity of gun is opposite to that of the bullet .]

$\huge\red\therefore$ The velocity with which the gun recoils is "0.33 m/s" .