Question

a bullet of mass 50g is fired from a gun with Initial velocity of 35 m/s. If the mass of the gun is 4 kg, then calculate the recoil velocity of the gun

Answer 1

before recoil =after recoil

m × v = m1 × v1

0.005 × 35 = 4 × v2

(0.005 × 35 ) ÷ 4 = v2

0.04375 m/ s = v2

Answer 2

$\text{\underline{\underline{We\:have:}}}$

$\sf{Mass \: of \: the \: bullet \: m_{1} = 50 \: g = 50 \times {10}^{ - 3} kg}$

$\sf{Speed \: of \: the \: bullet \: v_{1} = 35 \: m/s}$

$\sf{Mass \: of \: the \: gun \: m_{2} = 4 \: kg}$

$\sf{Velocity \: of \: the \: gun \: v_{2} = ? }$


$\sf{\underline{According\:to\:the\:conservation\:of\:momentum:}}$

Total momentum:

$\sf\boxed{Before \: firing = After \: firing}$


But,

$\sf{\underline{Total \: momentum \: before \: firing}}$ = $\boxed{0}$


Because,

$\text{\underline{The\:bullet\:gun\:is\:initially\:at\:rest.}}$


Therefore,

Momentum of the system will be 0 after collision.

$\sf{m_{1}v_{1} + m_{2}v_{2} = 0}$

$\sf{v_{2} = \frac{ - m_{1}v_{1} }{m_{2} }}$

$\sf{v_{2} = \frac{ - 0.05 \times 35}{4}}$

$\sf{v_{2} = - 0.4375 \: m/s}$

$\sf{v_{2} =\boxed{ - 0.44 \: m/s}}$


$\text{\underline{NOTE:}}$

The negative sign shows the gun moves opposite to the direction of motion of the bullet.