Question

A bullet of mass 20gram is fired by a gun of mass 20 kg .If the muzzle speed of the bullet is 100m/sec ,what is the recoil speed of gun?​

Answer 1

ACC to law of conservation of momentum

mass of gun×velocity of gun= mass of bullet×velocity of bullet

2×10^4×X= 20×100

X=0.1ms^-1

Answer 2

$\bigstar\bf\blue{ANSWER}\bigstar$

$\bigstar\: \: \bf\red{GIVEN}\bigstar$

$\bigstar\rm\red{BEFORE\:FIRING}$

$\rm\blue{Mass\:of\:Bullet(m_1)=0.02kg}$

$\rm\blue{Mass\:of\:gun(m_2)=20kg}$

$\rm\blue{initial\:velocity\:of\:the\:bullet(u_1)=0}$

$\rm\blue{initial\:velocity\:of\:the\:gun(u_2)=0}$

Now,

$\bigstar\rm\red{AFTER\:FIRING}$

$\rm\blue{Mass\:of\:Bullet(m_1)=0.02kg}$

$\rm\blue{Mass\:of\:gun(m_2)=20kg}$

$\rm\blue{Final\:velocity\:of\:the\:bullet(v_1)=100m/s^{-1}}$

$\rm\blue{Final\:velocity\:of\:the\:gun(v_2)=?}$

Now, According to the law of conservation of momentum,Momentum of the system is Conserved.

$\textbf{Total Momentum before firing=Total Momentum after firing}$

$\implies\bf\red{m_1u_1+m_2u_2=m_1v_1+m_2v_2}$

$\implies\bf\red{(0.02)(0)+(20)(0)=(0.02)(100)+(20)(v_2)}$

$\implies\bf\red{0=2+20v_2}$

$\implies\bf\red{-2=20v_2}$

$\implies\bf\red{v_2=\dfrac{\cancel{-2}}{\cancel{20}}}$

$\implies\bf\red{v_2=-0.5m/s^{-1}}$

Thus,The recoil Velocity of the gun will be -0.5m/s^-1