Question

A bullet of mass 0.02 kg is fired by a gun of mass 20 kg. If he speed of the bullet is 150 m/s, calculate the recoil speed of the gun?

Answer 1

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According to the law of conservation of momentum:

Mass of bullet x Velocity of bullet=Mass of gun x Recoil velocity of gun

0.02 x 200 = 20 x Recoil velocity of gun

Recoil velocity of gun = 0.2 m/sec.

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,

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

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

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

$\rm\blue{Final\:velocity\:of\:the\:bullet(v_1)=150m/s^{-1}}$

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

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

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

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

$\implies\bf\red{(0.02)(0)+(20)(0)=(0.02)(150)+(20)(v_2)}$

$\implies\bf\red{0=3+20v_2}$

$\implies\bf\red{-3=20v_2}$

$\implies\bf\red{v_2=\dfrac{\cancel{-3}}{\cancel{20}}}$

$\implies\bf\red{v_2=0.15m/s^{-1}}$

Thus,The recoil Velocity of the gun will be 0.15m/s^-1